mm-1072 === Subject: Re: Root Finder ix.,final > stuff. Why hasn't anyone found a way to find the root to an > infinite polynomial series? Submit your results to http://www.crank.net/submit.html They are guaranteed to publish. gtoomey === Subject: Re: Root Finder ix.,final > This closes the book on a problem that I've worked on for about > 5 years, which began as a surveying problem posed by a British > engineer. What is the angle that subtends a known arc and known > chord, on a circle of unknown radius? It was the solution to a > polynomial series. That is how I got started on all this root > stuff. Why hasn't anyone found a way to find the root to an > infinite polynomial series? (carried out to any number of terms) Because in general it is impossible. Did you ever take any time at all to look at published work? How about looking up Abel's Impossibility Theorem and Gaolis Theory. (I hope my spelling is correct.) > I finally found out how to do it, to my own satisfaction, > anyway. There is some clerical work needed, having to do with > the right signs (+/-) of the vectors involved. Otherwise, thank > you all for your feedback and comments, sometimes cynical. If > anyone can find a better way, the sky's the limit. Unfortunately, it is still sloppy and inaccurate at its best. It is pretty sad that in previous posts, you went through all that work and still said 'use Newton's Method' in the end. > Jon Giffen > Root Finder ix. > by Jon Giffen > The polynomial > a[0]+a[1]t+a[2]t^2+a[3]t^3+ ... + a[n] = 0 > may be expressed as the dot product of the two vectors, > T = (t,t^2,t^3,...,t^n) and > N = (a[1],a[2],a[3],...,a[n]) or, > (t,t^2,t^3,..,t^n)*(a[1],a[2],a[3],..,a[n])+a[0]=0 or > T*N+a[0]=0 is the nth degree polynomial > N may be considered the normal to the more general case > of the plane, > a[0] + a[1]x[1] + a[2]x[2] + ... + a[n]x[n] = 0 > where x[1]=t x[2]=t^2 x[3]=t^3 ... x[n]=t^n > Following T up from the origin, R is the variable vector > parallel to N and orthogonal to T-R, as T varies along > its curve. then > R*(T-R)=0 > taking the derivative with respect to t, > R'*(T-R)+R*(T'-R')=0 > or > 2R'*R = R'*T + R*T' > The projection of T' onto N/|N| is R' > T'*N N > ---- ---- = R' > |N| |N| > define > Q = (T*N)N/|N|^2 = -a[0]N/|N|^2 > Q = (-a[0]/|N|^2)(a[1],a[2].a[3],..a[n]) > or > Q = (q[1],q[2],q[3],...,q[4]) > Since R' is in the same direction as Q, it holds that > since Q and T'-(T'*N)N/|N|^2 are orthogonal, > Q*(T'-(T'*N)N/|N|^2) = 0 > where Q is the shortest vector from the origin to the plane. > T = (t,t^2,t^3,...,t^n) > T'= (1,2t,3t^2,..,nt^(n-1) ) > T'*Q =q[1]+2q[2]t+3q[3]t^2+....+ nq[n]t^(n-1) > =(1/t)T*(q[1],2q[2],3q[3],...,nq[n]) > T'*N = a[1]+2a[2]t+3a[3]t^2+....+ na[n]t^(n-1) > =(1/t)T*(a[1],2a[2],3a[3],...,na[n]) > Suppose the vectors D,S,U are defined, > D=( 1 , 2 , 3 ,...,n ) > S=(a[1],2a[2],3a[3],...,na[n]) > U=( 1 , 1 , 1 ,...,1 ) > Then > T'* U =(1/t)T*D > T'* N =(1/t)T*S > for instance, > (1,2t,3t^2,..,nt^(n-1) )*(1,1,1,..) > =(1/t)(t,t^2,t^3,...,t^n)*(1,2,3,..) > and > (1,2t,3t^2,..,nt^(n-1) )*(a[1],a[2],a[3],..) > =(1/t)(t,t^2,t^3,...,t^n)*(a[1],2a[2],3a[3],..) > dividing the below equations with each other, > T'* U =(1/t)T*D > T'* N =(1/t)T*S > (T'*U)(T*S) - (T'*N)(T*D) = 0 > factoring out T, > T*((T'*U)S - (T'*N)D) = 0 > so T is orthogonal with (T'*U)S - (T'*N)D > or T is orthogonal with (1/t)((T*D)S - (T*S)D) > or T is orthogonal with (T*D)S - (T*S)D > this is true, since > T*((T*D)S - (T*S)D) = 0 > Let, > G = (T*D)S - (T*S)D > If m is some ratio of G such that, > (mG-Q)*N = 0 > mG*N - Q*N = 0 since Q*N=T*N, > m[(T*D)(S*N)-(T*S)(D*N)]-(T*N) = 0 > or > T*[m((S*N)D-(D*N)S )-N] = 0 > Notice that this can just as easily be, > T*[m((S*N)D+(D*N)S )-N] = 0 > One solution for m, among others, can be found from > m((S*N)D-(D*N)S - N = 0 or > m((S*N)D-(D*N)S = N taking the dot product of both > sides with D and solving for m, > N*D > m = --------------------- > (S*N)(D*D)-(D*N)(S*D) > Otherwise, T and m((S*N)D-(D*N)S )-N are orthogonal > Notice that since N*((S*N)D-(D*N)S )= 0, > ((S*N)D-(D*N)S ) is orthogonal to N. > T is orthogonal with m((S*N)D-(D*N)S)-N > N is orthogonal with (S*N)D-(D*N)S > OY=m((S*N)D-(D*N)S)-N > *------------ N T > | / > | / > | / > | / > | / > | / > Z _________|/ > |O > OZ= | > (S*N)D-(D*N)S | > | > | > | > -N > T is orthogonal with OY > N is orthogonal with OZ > The plane formed by OY and OT is not necessarily > the plane formed by OY and OT, but > OT*OY=0 > Suppose OP is orthogonal to ON and OZ then > OP*ON=0 and > OP*OZ=0 > OP*OY=0 > OP*OT ?=0 > OT*OY=0 > or > OT*(mOZ-ON)=0 > N = (Q/|Q|)|N| > T*(m(|Q|/|N|)OZ - Q)=0 > Q T > *----------* > | / > | / > | / > | / > | / > angle(OQT)=pi/2 > QT=m(|Q|/|N|)OZ > Hence, OP*OT=0 > since angle YOZ = angle TON, > OY*OZ T*N OY*N > -------- = ------ = {1 - (-------)^2 }^(1/2) > |OY||OZ| |T||N| |OY||N| > (OY*OZ)^2 (OY*N)^2 > ------------ = 1 - ----------- > |OY|^2|OZ|^2 |OY|^2|N|^2 > Multiplying both sides of this equation by > |OY|^2|OZ|^2|N|^2, > (OY*OZ)^2|N|^2 = |OY|^2|OZ|^2|N|^2 - (OY*N)^2|OZ|^2 eqn i. > Letting (S*N)D+/-(D*N)S = C, OZ=C and OY=mC-N and > (OY*OZ)^2=((mC-N)*C)^2|N|^2 = m^2|C|^2|N|^2 > |OY|^2=(mC-N)*(mC-N) = m^2|C|^2+|N|^2 > |OZ|^2=|C|^2 > (OY*N)^2=((mC-N)*N)^2=-|N|^4 > Substituting these in eqn i, > (m^2|C|^2|N|^2)|N|^2 > (m^2|C|^2+|N|^2)|C|^2|N|^2 > |N|^4|C|^2 or > m^2|N|^2 = m^2(|C|^2+|N|^2) + N^2 > N^2 = m^2(|N|^2 - |C|^2 - |N|^2) > |N| > m = - ----- > |C| > So > OY=m((S*N)D-(D*N)S)-N or > |N| > OY= - ---((S*N)D-(D*N)S)-N > |C| > or > |N| > OY= - ---------------((S*N)D-(D*N)S) - N > |(S*N)D-(D*N)S| > or > |N| > OY= - --- C - N > |C| > (qOY-Q)*N=0 > qOY*N-Q*N=0 > q(0-|N|^2) = Q*N > q(-|N|^2) = -a[0] > q = a[0]/|N|^2 > angle ZOY > OY*OZ T*N > cos(ZOY)=-------- = ------ = cos(NOT) > |OY||OZ| |T||N| > T*N= -a[0] and > a[0]|OY||OZ| > |T| = - ------------- > |N|(OY*OZ) > (a[0])^2|OY|^2|OZ|^2 > |T|^2 = --------------------- > |N|^2(OY*OZ)^2 > |OY|^2 = 2|N|^2 > (OY*OZ)^2 = (|N||C|)^2 > (a[0])^2 2|N|^2 |C|^2 > |T|^2 = --------------------- = 2|Q|^2 > |N|^2 |N|^2 |C|^2 > T = Q + {|T|^2-|Q|^2}^(1/2)(-C/|C|) > |Q| > T = Q +/- ----- C > |C| > Where > C = (S*N)D-(D*N)S > D=( 1 , 2 , 3 ,...,n ) > S=(a[1],2a[2],3a[3],...,na[n]) > |Q| > T = Q +/- --- C > |C| > |N| > OY= - --- C - N > |C| > T*OY = -Q*N+/-|Q||N|=0 when > |Q| > T[0] = Q - --- C this is the shortest possible T > |C| > Since T must be perpendicular to OY, OY is the axis > that T rotates on. It sweeps out a line on the plane > that has the equation, (X-T[0])*T[0]=0 A vector L that > is parallel to the resulting line, is the solution to, > L*N=0 > L*C=0 > L becomes the new C and T[0] becomes the new Q in, > |Q| > T[0] = Q - --- C this is the shortest possible T > |C| > or > T[0] > T[1] = T[0] - ----- L > |L| > This process is repeated, until in general, if T[i] is > the ith solution, and L[i] is the ith vector > perpendicular to L[i-1] and T[i-2] , then > |Q| > T[0] = Q - --- C > |C| > L[1]*C=0 > L[1]*Q=0 > |T[0]| > T[1] = T[0] - -------- L[1] > |L[1]| > L[2]*L[1]=0 > L[2]*T[0]=0 > |T[1]| > T[2] = T[1] - ------ L[2] > |L[2]| > L[3]*L[2]=0 > L[3]*T[1]=0 > . |T[i-1]| > T[i] = T[i-1] - -------- L[i] > |L[i]| > t^6+t-10=0 > N=(1,0,0,0,0,1) > |N|^2=2 > Q=(10/2)(1,0,0,0,0,1)=5(1,0,0,0,0,1) > |Q|=(5)2^(1/2) > D=( 1 , 2 , 3 ,...,n )=(1,2,3,4,5,6) > S=(a[1],2a[2],3a[3],...,na[n])=(1,0,0,0,0,6) > S*N=(1,0,0,0,0,6)*(1,0,0,0,0,1)=1+6=7 > D*N=(1,2,3,4,5,6)*(1,0,0,0,0,1)=1+6=7 > C=(S*N)D+(D*N)S > =7[(1,2,3,4,5,6)+(1,0,0,0,0,6) > =7(2,2,3,4,5,12) > |C|=(7)202^(1/2) > T=+/-Q+/-C|Q|/|C| > =5(1,0,0,0,0,1)+/-7(2,2,3,4,5,12)[(5)2^(1/2)]/[(7)202^(1/2)] > =5(1,0,0,0,0,1)+/-5(2,2,3,4,5,12)/101^(1/2) > t^6 = 5 + 5(12)/101^(1/2) = 10.9702 t=1.4906 > 10.9702+1.49-10=2.46 > t=1.4906-2.46/(6(1.4906^5)+1)=1.4361 N.M. > t^6=8.7721 > t^5=6.1083 > t^4=4.2534 > t^3=2.9618 > t^2=2.0624 > T[0]=(1.4361,2.0624,2.9618,4.2534,6.1083,8.7721) > |T[0]|=12.1425 > L*Q=(a,b,0,0,0,c)*(1,0,0,0,0,1)=0 a=-c > L*T[0]=(-c,b,0,0,0,c)*T[0] > = -1.43617c + 8.7721c + 2.0624b=0 c=1 b=-3.5570 > L=(-1,-3.5570,0,0,0,1) |L|=3.8278 > T[1]=+/-T[0]+/-L|T[0]|/|L| > =+/-(1.4361, 2.0624,2.9618,4.2534,6.1083,8.7721) > +/-( -1,-3.5570, 0, 0, 0, > 1)(12.1425/3.8278) > ----------------------------------------------------------------- > =+/-(1.4361, 2.0624,2.9618,4.2534,6.1083,8.7721) > +/-( -1,-3.5570, 0, 0, 0, 1)(3.1728) > ----------------------------------------------------------- > =+/-( 1.4361, 2.0624,2.9618,4.2534,6.1083,8.7721) > +/-(-3.1728,-11.2858, 0, 0, 0,3.1728) > t^6=8.7721+3.1728=11.9449 t=-1.5119 > 11.9449-1.5119-10=0.433 > t=-1.5119-0.433/(6(-1.5119^5)+1)=-1.5025 > (-1.5025)^6-1.5025-10=0.0056~0 > end === Subject: Re: Root Finder ix.,final > This closes the book on a problem that I've worked on for about > 5 years, which began as a surveying problem posed by a British > engineer. What is the angle that subtends a known arc and known > chord, on a circle of unknown radius? It was the solution to a > polynomial series. That is how I got started on all this root > stuff. Why hasn't anyone found a way to find the root to an > infinite polynomial series? (carried out to any number of terms) Because there are no exact formulas for roots, if the degree is at least 5. You've been told that for a couple years now. None of your examples have proven that your method works, either; when you provide specific examples, to claim that your method works, at least make sure that the examples work ahead of time. If they don't, then there's something wrong with your method. -- Christopher Heckman === Subject: Re: Root Finder ix.,final I've outlined the solution on my web site, http://mypeoplepc.com/members/jon8338/polynomial/ along with a 6th degree polynomial and its (correct) solution... Which is more than any of you have done too lazy to see it's right, too presumptuous with ignorance in your vanity. Victory, Jon Giffen >>This closes the book on a problem that I've worked on for about >>5 years, which began as a surveying problem posed by a British >>engineer. What is the angle that subtends a known arc and known >>chord, on a circle of unknown radius? It was the solution to a >>polynomial series. That is how I got started on all this root >>stuff. Why hasn't anyone found a way to find the root to an >>infinite polynomial series? (carried out to any number of terms) > Because there are no exact formulas for roots, if the degree is > at least 5. You've been told that for a couple years now. > None of your examples have proven that your method works, either; > when you provide specific examples, to claim that your method works, > at least make sure that the examples work ahead of time. If they don't, > then there's something wrong with your method. > -- Christopher Heckman === Subject: Re: Root Finder ix.,final > I've outlined the solution on my web site, > http://mypeoplepc.com/members/jon8338/polynomial/ > along with a 6th degree polynomial and its (correct) > solution... Wow, you finally found a polynomial where the result is decently close. But all the other examples you posted are still very, very wrong. One example does not make a case for a method which is supposed to be true for all possibilities. > Which is more than any of you have done too lazy to see it's > right, too presumptuous with ignorance in your vanity. And you are presumptuous, ignorant, and vain for suggesting that one example alone proves you right. > Victory, You won one battle out of an infinite number of wars. > Jon Giffen > >>This closes the book on a problem that I've worked on for about >>5 years, which began as a surveying problem posed by a British >>engineer. What is the angle that subtends a known arc and known >>chord, on a circle of unknown radius? It was the solution to a >>polynomial series. That is how I got started on all this root >>stuff. Why hasn't anyone found a way to find the root to an >>infinite polynomial series? (carried out to any number of terms) > > > Because there are no exact formulas for roots, if the degree is > at least 5. You've been told that for a couple years now. > None of your examples have proven that your method works, either; > when you provide specific examples, to claim that your method works, > at least make sure that the examples work ahead of time. If they don't, > then there's something wrong with your method. > -- Christopher Heckman === Subject: Root finder XI Root Finder xi. by Jon Giffen. It is found that the roots to the polynomial, a[0]+a[1]t+a[2]t^2+...+a[n]t^n where T=(t,t^2,t^3,..,t^n) and N=(a[1],a[2],a[3],...,a[n]) are given by, (D*N)|S|^2 D + (S*N)|D|^2 S T = (-a[0]){---------------------------} (D*N)^2|S|^2 + (S*N)^2|D|^2 where D=(1,2,3,...,n) S=(a[1],2a[2],3a[3],...,na[n]) Solve the 6th degree polynomial, t^6 + t - 10=0 a[0] = -10 N=(1,0,0,0,0,1) D=(1,2,3,4,5,6) S=(1,0,0,0,0,6) D*N=7 S*N=7 |D|^2 = 91 |S|^2 = 37 37(1,2,3,4,5,6)+91(1,0,0,0,0,6) T = (10)------------------------------- 7(37+91) (1280,740,1110,1480,1850,7680) = ------------------------------ = (t,t^2,t^3,t^4,t^5,t^6) 896 since t and t^6 need only be considered, t = 1280/896 = 10/7 t^6 = 7680/896 = 60/7 60/7 + (60/7)^(1/6) - 10 = 0.002 which is almost zero. (10/7)^6 + 10/7 - 10 = -0.07 which is also close Solve the 6th degree polynomial, t^6 - t - 10=0 t = -(60/7)^(1/6) from the prior example f(t) = t^7 + t^3 + t - 20 = 0 a[0] = -20 N=(1,0,1,0,0,0,1) D=(1,2,3,4,5,6,7) S=(1,0,3,0,0,0,7) D*N = 11 S*N = 11 |S|^2 = 59 |D|^2 = 138 59(1,2,3,4,5,6,7)+138(1,0,3,0,0,0,7) T = (20)------------------------------------ 11(59+138) t^7 = 27580/2167 t=1.4382 f(1.4382) = -2.8597 applying this to Newton's Method, 1.4382 - (-2.857)/[7(1.4382^6)+3(1.4382^2) + 1] = 1.4795 f(1.4795)=0.2361 1.4795 - (0.2361)/[7(1.4795^6)+3(1.4795^2) + 1] = 1.4766 f(1.4766)=0.000092 ~ 0 f(t) = t^7 + t^3 + t + 20 = 0 t = -1.4766 from last example f(t) = 2t^4 + 3t^3 + 2t^2 + t - 13 = 0 T*(1,2,3,2)-13 = 0 a[0]= -13 N=(1,2,3,2) D=(1,2,3,4) S=(1,4,9,8) |D|^2=30 |S|^2 = 162 D*N=22 S*N=52 13[22(162)(1,2,3,4)+52(30)(1,4,9,8)] T = -------------------------------------- 162(22^2) + 30(52^2) t^4 = 347568/159520 = 2.1787 t = 1.2149 f(1.2149) = 0.9038 applying Newton's Method, 1.2149 - 0.9038/[8(1.2149^3)+9(1.2149^2)+4(1.2149)+1] = 1.1879 f(1.1879) = 0.0213 ~ 0 f(t)=t^6 - t^5 + 4t^4 - 5t^3 + t^2 - t - 100 = 0 (-1,1,-5,4,-1,1)*T - 100 = 0 dividing the negatives from the positives, (0,1,0,4,0,1)*T - (1,0,5,0,1,0)*T = 100 (0,1,0,4,0,1)*T - 100p = (1,0,5,0,1,0)*T + 100(1-p) = 0 p is some ratio Applying the formula to each partition of N, (D*N)|S|^2 D + (S*N)|D|^2 S T = (-a[0]){---------------------------} (D*N)^2|S|^2 + (S*N)^2|D|^2 where D=(1,2,3,...,n) S=(a[1],2a[2],3a[3],...,na[n]) for (0,1,0,4,0,1)*T - 100p , a[0]=-100p D*N=2+16+6=24 |S|^2=4+16^2+36=296 S*N=2+16+6=24 |D|^2=1+4+9+16+25+36=91 for (1,0,5,0,1,0)*T + 100(1-p) , a[0]=100(1-p) D*N=1+15+5=21 |S|^2=1+15^2+25=251 S*N=1+15+5=21 |D|^2=91 296(1,2,3,4,5,6)+91(0,2,0,16,0,6) (100p)---------------------------------- 24(296+91) 251(1,2,3,4,5,6)+91(1,0,15,0,5,0) =100(p-1)--------------------------------- 21(251+91) Taking the magnitude of both sides |296(1,2,3,4,5,6)+91(0,2,0,16,0,6)| = |(296,774,888,2640,1480,2322)|=4003.36 |251(1,2,3,4,5,6)+91(1,0,15,0,5,0)| = |(342,502,2118,1004,1710,1506|=3324.91 4003.36 3324.91 p--------- = (p-1)------- 9288 7182 (0.9310)p = p-1 1 = (1-0.9310)p p = 14.5 then f(t)=t^6 - t^5 + 4t^4 - 5t^3 + t^2 - t - 100 = 0 t^6 = 1450/4 =362.5 t=2.67 f(2.67 )=239.38 t^5 = 858400/9288= 92.420 t=2.1263 f(2.1263)=-15.2200 Selecting -15.2200 for Newton's Method, f(t)=t^6 - t^5 + 4t^4 - 5t^3 + t^2 - t - 100 = 0 2.1263+15.22/ [6(2.1263^5)-5(2.1263^4)+16(2.1263^3)-15(2.1263^2)+2(2.1263)-1] t=2.1877 f(2.1877)= 1.3927 The development of this is given at, http://mypeoplepc.com/members/jon8338/polynomial/id7.html Jon Giffen === Subject: Re: Root finder XI > Root Finder xi. > by Jon Giffen. > It is found that the roots to the polynomial, > a[0]+a[1]t+a[2]t^2+...+a[n]t^n where > [...] > f(t) = t^7 + t^3 + t - 20 = 0 > [...] > applying this to Newton's Method, You've been told about this before. There are already good places to start off Newton's Method, so you haven't provided anything new. Strike Eleven. -- Christopher Heckman === Subject: Re: Root finder XI > Root Finder xi. You told us that ix was the final one. === Subject: Re: Root finder XI > > Root Finder xi. > > You told us that ix was the final one. Based on his posts, did you really think you could believe him? -- Christopher Heckman === Subject: Root Finder 12 Root Finder 12 by Jon Giffen. This solution was so simple that I couldn't believe it. But I tried it, and it works. It is found that the roots to the polynomial, a[0]+a[1]t+a[2]t^2+...+a[n]t^n=0 where T=(t,t^2,t^3,..,t^n) and N=(a[1],a[2],a[3],...,a[n]) are given by, -a[0] T= -----D D*N D=(1,2,3,4,...,n) Solve the 6th degree polynomial, f(t)=t^6 + t - 10=0 a[0] = -10 N=(1,0,0,0,0,1) D=(1,2,3,4,5,6) a[0]=-10 D*N=7 -a[0] 10 T = -----D = ---(1,2,3,4,5,6) D*N 7 t =10/7 t=1.42835 f(1.42835)=-0.071 t^6=60/7 t=1.43056 f(1.43056)= 0.002 Solve the 49th degree polynomial, f(t)=t^49 + t^16 + 40t^5 - 6000 = 0 a[0]=-6000 N=(1,0,0,0,40,0,..,1,0,...,1) D=(1,2,3,..,49) D*N=1+200+16+49=266 -a[0] 6000 T = -----D = ------(1,2,3,..,49) D*N 266 t^49=294000/266=1105.26 t=1.15375 f(1.15375)=-3575.99 Applying Newton's Method, t=1.15375-(-3575)/[49(1.15375^48)+16(1.15375^15)+200(1.15375)^4] =1.15375 again, so the answer must depend on distant decimal places. mixed signs, no pattern f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t - 100 = 0 a[0]=-100 N=(-1,1,5,4,-1,1) D=(1,2,3,4,5,6) D*N=-1+2+15+16-5+6=33 -a[0] 100 T = -----D = ---(1,2,3,4,5,6) D*N 33 t^6=600/33=18.1818 t=1.62158 f(1.62158)=-43.0453 too low t^5=500/33=15.1515 t=1.72223 f(1.72223)=-27.0815 too low f^4=400/33=12.1212 t=1.86589 f(1.86589)= 2.16509 pretty close Applying Newton's Method, 1.86589- 2.16509 ------------------------------------------------------------------ 6(1.86589^5)-5(1.86589^4)+16(1.86589^3)+15(1.86589^2)+2(1.86589)-1 t=1.856637 f(1.856637)=0.0192 t^2 + 2t - 3 = 0 a[0]=-3 N=(2,1) D=(1,2) D*N=4 -a[0] 3 T = -----D =---(1,2) t^2=3/4 t=0.866 ~ 1 D*N 4 The lengthy development of this is given at, http://mypeoplepc.com/members/jon8338/polynomial/id7.html Jon Giffen === Subject: Root Finder 13 Root Finder 13 Jon Giffen Another approach is considered, along with a possibility for finding the root to an Infinite Series. It is discovered that the property of the nth degree polynomial, a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 where N=(a[1],a[2],a[3],...,a[n]) T=(t.t^2.t^3,t^4,..., t^n ) Is, |C|^2 |T|^2=(-a[0])^2{-------------------} |N|^2|C|^2-(N*C)^2 where C=(a[1],2a[2],3a[3],4a[4],...,na[n]) f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t - 100 = 0 a[0]=-100 N=(-1,1,5,4,-1,1) C=(-1,2,15,16,-5,6) |C|^2=547 |N|^2=48 N*C=153 |N|^2 |C|^2 - (N*C)^2 = 48(547)-153^2=2847 |C|^2 D T =(-a[0]){----------------------}^(1/2) --- |N|^2 |C|^2 - (N*C)^2 |D| where D=(1,2,3,4,5,6) and 100 547 T = -------- {-----}^(1/2) (1,2,3,4,5,6) = 4.594933(1,2,3,4,5,6) 91^(1/2) 2847 t^6=27.5696 t=1.738097 t^5=22.9746 t=1.871758 ---------- t=1.856637 is the correct root Notice that the root to a polynomial that is so long, that it is virtually an infinite Power Series; is found by using the solution to the Geometric Series, |C|^2 |T|^2=t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{------------------} |N|^2|C|^2-(N*C)^2 adding 1 to both sides, |C|^2 1+t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{------------------}+1 |N|^2|C|^2-(N*C)^2 then the sum S=1/(1-t^2) but |C|^2 S=(-a[0])^2{------------------}+1 (1-t^2)=1/S t^2=1-1/S and |N|^2|C|^2-(N*C)^2 t ={1 - 1/S}^(1/2) where N=(a[1],a[2],a[3],...,a[n]) T=(t,t^2,t^3,t^4,..., t^n ) C=(a[1],2a[2],3a[3],4a[4],...,na[n]) to the nth degree power series, a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 Development Suppose the polynomial, a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 Is expressed as, a[0]+a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n=0 Where c is almost 1 then dividing the two, a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n= -a[0] --------------------------------------------------- a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n = -a[0] Suppose K=(a[1]c,a[2]c^2,a[3]c^3,...a[n]c^n) then simply, T*(K-N)=0 so (K-N) is orthogonal to T. Consequently, N*(K-N) T is parallel to N - ------- K and |K-N|^2 N*(K-N) [m(N - -------- K) - Q]*N=0 solve this for m. Then |K-N|^2 N*(K-N) T= m(N - -------- K) |K-N|^2 Substitute m in the above and take the square of the magnitude of both sides. Then (K-N)*(K-N) 0 |T|^2=Lim (-a[0])^2 ---------------------------- = --- c->1 |N|^2(K-N)*(K-N)-[N*(K-N)]^2 0 applying L'Hopital two times with respct to c, |C|^2 |T|^2=(-a[0])^2{-------------------} |N|^2 |C|^2-(N*C)^2 where d C = Lim ---- K = =(a[1],2a[2],3a[3],4a[4],...,na[n]) c->1 dc E.O.P. Jon Giffen http://mypeoplepc.com/members/jon8338/polynomial/id7.html === Subject: Re: Root Finder 13 > Root Finder 13 Strike Thirteen. > Jon Giffen > Another approach is considered, along with a possibility for > finding the root to an Infinite Series. > It is discovered that the property of the nth degree polynomial, > a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 where > N=(a[1],a[2],a[3],...,a[n]) > T=(t.t^2.t^3,t^4,..., t^n ) The reason that Root Finder N will be wrong for all values of N is that you have a bunch of equations of the form t^n = A_n t^(n-1) = A_(n-1) t^(n-2) = A_(n-2) etc. Now, if you can find a value of t that satisfies ALL equations SIMULTANEOUSLY, then you indeed have found a root of the original polynomial. However, if there is no solution, your claim that you have found a root is not automatic; it may be a root, or it may not. In fact, you may also be close to a root. But this is like firing a bullet at a wall, then drawing the target around it. -- Christopher Heckman > [non sequitor part of the post has been cut] === Subject: Re: Root Finder 13 > Root Finder 13 > Jon Giffen > Another approach is considered, along with a possibility for > finding the root to an Infinite Series. > It is discovered that the property of the nth degree polynomial, Your discovery is wrong, I will explain a few more lines down. > a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 where > N=(a[1],a[2],a[3],...,a[n]) > T=(t.t^2.t^3,t^4,..., t^n ) > Is, > |C|^2 > |T|^2=(-a[0])^2{-------------------} > |N|^2|C|^2-(N*C)^2 > where > C=(a[1],2a[2],3a[3],4a[4],...,na[n]) > f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t - 100 = 0 Try changing the constant term of the polynomial to 1000. You will see that your method fails miserably. > a[0]=-100 N=(-1,1,5,4,-1,1) C=(-1,2,15,16,-5,6) > |C|^2=547 |N|^2=48 N*C=153 > |N|^2 |C|^2 - (N*C)^2 = 48(547)-153^2=2847 > |C|^2 D > T =(-a[0]){----------------------}^(1/2) --- > |N|^2 |C|^2 - (N*C)^2 |D| where This equation is your method in a nutshell. If it is wrong, your whole method is wrong. Lets say we have two polynomials who differ only in the constant term. Lets call the polynomials W and V and their respective constant terms w and v. Since the equation above has a scalar term of a[0], that equation then implies that if (x*w) is a root of W then that (x*v) is a root of V. That is clearly false, just compare the two polynomials x^5 - 2x - 28 = 0 and x^5 - 2x - 237 = 0. Assume that the root of the first polynomial, which is 2, was found by your method. By the fact that a[0] is scalar in the equation the root was derived from, your method would say the second polynomial would have a root of 237/14=16.93, which is clearly false, since the root is 3. So if you choose any polynomial where your method created an accurate answer, I can find an infinite number polynomials where your method fails miserably by simply changing the constant term. Because of this, your method is fatally flawed and totally wrong. > D=(1,2,3,4,5,6) and > 100 547 > T = -------- {-----}^(1/2) (1,2,3,4,5,6) = 4.594933(1,2,3,4,5,6) > 91^(1/2) 2847 > t^6=27.5696 t=1.738097 > t^5=22.9746 t=1.871758 > ---------- > t=1.856637 is the correct root > Notice that the root to a polynomial that is so long, that > it is virtually an infinite Power Series; is found by using > the solution to the Geometric Series, > |C|^2 > |T|^2=t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{------------------} > |N|^2|C|^2-(N*C)^2 > adding 1 to both sides, > |C|^2 > 1+t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{------------------}+1 > |N|^2|C|^2-(N*C)^2 > then the sum S=1/(1-t^2) but > |C|^2 > S=(-a[0])^2{------------------}+1 (1-t^2)=1/S t^2=1-1/S and > |N|^2|C|^2-(N*C)^2 > t ={1 - 1/S}^(1/2) where > N=(a[1],a[2],a[3],...,a[n]) > T=(t,t^2,t^3,t^4,..., t^n ) > C=(a[1],2a[2],3a[3],4a[4],...,na[n]) > to the nth degree power series, > a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 > Development > Suppose the polynomial, > a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 > Is expressed as, > a[0]+a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n=0 > Where c is almost 1 then dividing the two, > a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n= -a[0] > --------------------------------------------------- > a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n = -a[0] > Suppose K=(a[1]c,a[2]c^2,a[3]c^3,...a[n]c^n) then simply, > T*(K-N)=0 so (K-N) is orthogonal to T. Consequently, > N*(K-N) > T is parallel to N - ------- K and > |K-N|^2 > N*(K-N) > [m(N - -------- K) - Q]*N=0 solve this for m. Then > |K-N|^2 > N*(K-N) > T= m(N - -------- K) > |K-N|^2 > Substitute m in the above and take the square of the > magnitude of both sides. Then > (K-N)*(K-N) 0 > |T|^2=Lim (-a[0])^2 ---------------------------- = --- > c->1 |N|^2(K-N)*(K-N)-[N*(K-N)]^2 0 > applying L'Hopital two times with respct to c, > |C|^2 > |T|^2=(-a[0])^2{-------------------} > |N|^2 |C|^2-(N*C)^2 > where > d > C = Lim ---- K = =(a[1],2a[2],3a[3],4a[4],...,na[n]) > c->1 dc > E.O.P. > Jon Giffen > http://mypeoplepc.com/members/jon8338/polynomial/id7.html === Subject: Re: Root Finder 13 >>f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t - 100 = 0 > Try changing the constant term of the polynomial to 1000. You will > see that your method fails miserably. f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t + 1000 = 0 a[0]=1000 N=(-1,1,5,4,-1,1) D=(1,2,3,4,5,6) N*D=33 t^6 = (-1000/33)(6) = -2000/11 (2k+1)(pi) (2k+1)(pi) t = {2000/11}^(1/6)[cos ---------- + i sin -----------] 6 6 Binomial Equation, where k=0,1,2,3,4,5 > Lets say we have two polynomials who differ only in the constant term. > Lets call the polynomials W and V and their respective constant terms > w and v. Since the equation above has a scalar term of a[0], that > equation then implies that if (x*w) is a root of W then that (x*v) is > a root of V. > That is clearly false, just compare the two polynomials x^5 - 2x - 28 > = 0 and x^5 - 2x - 237 = 0. Assume that the root of the first > polynomial, which is 2, was found by your method. By the fact that > a[0] is scalar in the equation the root was derived from, your method > would say the second polynomial would have a root of 237/14=16.93, > which is clearly false, since the root is 3. > So if you choose any polynomial where your method created an accurate > answer, I can find an infinite number polynomials where your method > fails miserably by simply changing the constant term. Because of > this, your method is fatally flawed and totally wrong. x^5 - 2x - 28 a[0]=-28 N=(-2,0,0,0,1) D=( 1,2,3,4,5) -a[0]/(N*D)=28/3 (t,t^2,t^3,t^4,t^5)=(28/3)(1,2,3,4,5) t^5 = (28/3)(5) t=(140/3)^(1/5)=2.1567 (should be 2) x^5 - 2x - 237 a[0]=-237 N=(-2,0,0,0,1) D=( 1,2,3,4,5) -a[0]/(N*D)=237/3 (t,t^2,t^3,t^4,t^5)=(237/3)(1,2,3,4,5) t^5 = (237/3)(5) t=(1185/3)^(1/5)=3.3061 (should be 3) In general, the root to the nth degree polynomial a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n where T=(t,t^2,t^3,...,t^n) N=(a[1],a[2],a[3],...,a[n]) D=(1,2,3,...,n) is approximated by the identity, -a[0] T = ----- D decoding, N*D -a[0] t = -------------------------- a[1]+2a[2]+3a[3]+...+na[n] -2a[0] t^2= -------------------------- a[1]+2a[2]+3a[3]+...+na[n] -3a[0] t^3= -------------------------- a[1]+2a[2]+3a[3]+...+na[n] . . . -na[0] t^n= -------------------------- a[1]+2a[2]+3a[3]+...+na[n] >>http://mypeoplepc.com/members/jon8338/polynomial/id7.html === Subject: Re: Root Finder 13 Yes I see that squaring all terms leads to a lack of inheritance of the property of negatives in the series. N*C destroys them as well. Root Finder 13 appears to be inferior. Root Finder 12, though, still seem to hold promise. >>Root Finder 13 >>Jon Giffen >>Another approach is considered, along with a possibility for >>finding the root to an Infinite Series. >>It is discovered that the property of the nth degree polynomial, > Your discovery is wrong, I will explain a few more lines down. >>a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 where >>N=(a[1],a[2],a[3],...,a[n]) >>T=(t.t^2.t^3,t^4,..., t^n ) >>Is, >> |C|^2 >>|T|^2=(-a[0])^2{-------------------} >> |N|^2|C|^2-(N*C)^2 >>where >>C=(a[1],2a[2],3a[3],4a[4],...,na[n]) >>f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t - 100 = 0 > Try changing the constant term of the polynomial to 1000. You will > see that your method fails miserably. >>a[0]=-100 N=(-1,1,5,4,-1,1) C=(-1,2,15,16,-5,6) >>|C|^2=547 |N|^2=48 N*C=153 >>|N|^2 |C|^2 - (N*C)^2 = 48(547)-153^2=2847 >> |C|^2 D >>T =(-a[0]){----------------------}^(1/2) --- >> |N|^2 |C|^2 - (N*C)^2 |D| where > This equation is your method in a nutshell. If it is wrong, your > whole method is wrong. > Lets say we have two polynomials who differ only in the constant term. > Lets call the polynomials W and V and their respective constant terms > w and v. Since the equation above has a scalar term of a[0], that > equation then implies that if (x*w) is a root of W then that (x*v) is > a root of V. > That is clearly false, just compare the two polynomials x^5 - 2x - 28 > = 0 and x^5 - 2x - 237 = 0. Assume that the root of the first > polynomial, which is 2, was found by your method. By the fact that > a[0] is scalar in the equation the root was derived from, your method > would say the second polynomial would have a root of 237/14=16.93, > which is clearly false, since the root is 3. > So if you choose any polynomial where your method created an accurate > answer, I can find an infinite number polynomials where your method > fails miserably by simply changing the constant term. Because of > this, your method is fatally flawed and totally wrong. >>D=(1,2,3,4,5,6) and >> 100 547 >>T = -------- {-----}^(1/2) (1,2,3,4,5,6) = 4.594933(1,2,3,4,5,6) >> 91^(1/2) 2847 >>t^6=27.5696 t=1.738097 >>t^5=22.9746 t=1.871758 >> ---------- >>t=1.856637 is the correct root >>Notice that the root to a polynomial that is so long, that >>it is virtually an infinite Power Series; is found by using >>the solution to the Geometric Series, >> |C|^2 >>|T|^2=t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{------------------} >> |N|^2|C|^2-(N*C)^2 >>adding 1 to both sides, >> |C|^2 >>1+t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{------------------}+1 >> |N|^2|C|^2-(N*C)^2 >>then the sum S=1/(1-t^2) but >> |C|^2 >>S=(-a[0])^2{------------------}+1 (1-t^2)=1/S t^2=1-1/S and >> |N|^2|C|^2-(N*C)^2 >>t ={1 - 1/S}^(1/2) where >>N=(a[1],a[2],a[3],...,a[n]) >>T=(t,t^2,t^3,t^4,..., t^n ) >>C=(a[1],2a[2],3a[3],4a[4],...,na[n]) >>to the nth degree power series, >>a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 >>Development >>Suppose the polynomial, >>a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 >>Is expressed as, >>a[0]+a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n=0 >>Where c is almost 1 then dividing the two, >>a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n= -a[0] >>--------------------------------------------------- >> a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n = -a[0] >>Suppose K=(a[1]c,a[2]c^2,a[3]c^3,...a[n]c^n) then simply, >>T*(K-N)=0 so (K-N) is orthogonal to T. Consequently, >> N*(K-N) >>T is parallel to N - ------- K and >> |K-N|^2 >> N*(K-N) >>[m(N - -------- K) - Q]*N=0 solve this for m. Then >> |K-N|^2 >> N*(K-N) >>T= m(N - -------- K) >> |K-N|^2 >>Substitute m in the above and take the square of the >>magnitude of both sides. Then >> (K-N)*(K-N) 0 >>|T|^2=Lim (-a[0])^2 ---------------------------- = --- >> c->1 |N|^2(K-N)*(K-N)-[N*(K-N)]^2 0 >>applying L'Hopital two times with respct to c, >> |C|^2 >>|T|^2=(-a[0])^2{-------------------} >> |N|^2 |C|^2-(N*C)^2 >>where >> d >>C = Lim ---- K = =(a[1],2a[2],3a[3],4a[4],...,na[n]) >> c->1 dc >>E.O.P. >>Jon Giffen >>http://mypeoplepc.com/members/jon8338/polynomial/id7.html === Subject: Root Finder: Angle Subtending Arc and Chord ANGLE SUBTENDING ARC AND CHORD On a circle of unknown radius, arc A and chord B subtend angle x, which is obtained by the formula, x=5.062792913(1-B/A)^(1/2) B/A x from FORMULA x SHOULD BE 0.995892735 0.324463996 0.314159265 0.983631643 0.647728053 0.628318531 0.963397762 0.968598927 0.942477796 0.935489284 1.28589674 1.256637061 0.900316316 1.598461583 1.570796327 0.858393691 1.905160043 1.884955592 0.810331958 2.204891605 2.199114858 0.756826729 2.496594918 2.513274123 0.698646585 2.77925389 2.827433388 0.636619772 3.051903588 3.141592654 1 <= B/A <= 2/pi Development Let x=angle (unknown) A=arc B=chord then from trigonometry, B x - - = sin(x/2) A 2 and letting t=(x/2)^2, 1 1 1 1 (1-B/A)- ---t + ---t^2 - ---t^3 + ---t^4 - .... 3! 5! 7! 9! then N=(-1/3! , 1/5! , -1/7! , 1/9! , .... ) and -a[0] T = ----- Root Approximation Formula D*N where a[0]=(1-B/A) D=(1,2,3,4,5..) N=(-1/3! , 1/5! , -1/7! , 1/9! , .... ) then oo n 1 2 3 4 D*N = Sum(-1)^n ------- = - --- + --- - --- + --- - .... n=1 (2n+1)! 3! 5! 7! 9! Carrying out the calculation up to n=7, D*N= -0.1505843395 1/(D*N)=-6.640796802 (t,t^2,t^3,t^4,t^5,t^6,t^7)=6.640796802(1-B/A)(1,2,3,4,5,6,7) Selecting the first component, t=6.640796802(1-B/A)(1) but since t=(x/2)^2 , x=2(t^1/2) and x=5.062792913(1-B/A)^(1/2) This sheds merit on -a[0] T = ----- Root Approximation Formula D*N proving its utility in arriving within enough accuracy to require only a few iterations of Newton's Method to obtain a very precise (and accurate) result. Jon Giffen === Subject: Re: Root Finder: Angle Subtending Arc and Chord > ANGLE SUBTENDING ARC AND CHORD > On a circle of unknown radius, arc A and chord B > subtend angle x, which is obtained by the formula, > x=5.062792913(1-B/A)^(1/2) > B/A x from FORMULA x SHOULD BE > 0.995892735 0.324463996 0.314159265 > 0.983631643 0.647728053 0.628318531 > 0.963397762 0.968598927 0.942477796 > 0.935489284 1.28589674 1.256637061 > 0.900316316 1.598461583 1.570796327 > 0.858393691 1.905160043 1.884955592 > 0.810331958 2.204891605 2.199114858 > 0.756826729 2.496594918 2.513274123 > 0.698646585 2.77925389 2.827433388 > 0.636619772 3.051903588 3.141592654 > 1 <= B/A <= 2/pi What this table tells me is that FORMULA, whatever it is, is wrong. Check out the difference between pi and (2143/22)^(1/4). pi = 3.14159265353... (2143/22)^(1/4) = 3.14159265258... Just because two things are close doesn't mean they're equal. -- Christopher Heckman === Subject: Re: Root Finder: Angle Subtending Arc and Chord > ANGLE SUBTENDING ARC AND CHORD > On a circle of unknown radius, arc A and chord B > subtend angle x, which is obtained by the formula, > x=5.062792913(1-B/A)^(1/2) > B/A x from FORMULA x SHOULD BE > 0.995892735 0.324463996 0.314159265 > 0.983631643 0.647728053 0.628318531 > 0.963397762 0.968598927 0.942477796 > 0.935489284 1.28589674 1.256637061 > 0.900316316 1.598461583 1.570796327 > 0.858393691 1.905160043 1.884955592 > 0.810331958 2.204891605 2.199114858 > 0.756826729 2.496594918 2.513274123 > 0.698646585 2.77925389 2.827433388 > 0.636619772 3.051903588 3.141592654 Using your approximation formula, the worst |relative error| is roughly 3%. > 1 <= B/A <= 2/pi That's not possible, and so I suppose that you intended to write 2/pi <= B/A <= 1 instead. > Development > Let > x=angle (unknown) > A=arc > B=chord > then from trigonometry, > B x > - - = sin(x/2) > A 2 > and letting t=(x/2)^2, > t=6.640796802(1-B/A)(1) but since > t=(x/2)^2 , x=2(t^1/2) and > x=5.062792913(1-B/A)^(1/2) I fail to see how the coefficient 5.062792913 was obtained. Indeed, it seems clear from the end of your development that we should instead have had x = 2*t^(1/2) = 2*(6.640796802*(1 - B/A))^(1/2) = 5.1539487*(1 - B/A)^(1/2) Using that approximation, relative error exceeds 5% when B/A is near 1. > This sheds merit on > -a[0] > T = ----- Root Approximation Formula > D*N > proving its utility in arriving within enough accuracy to > require only a few iterations of Newton's Method to obtain > a very precise (and accurate) result. I'm not sure that it sheds merit. How did 5.06... arise, rather than 5.15...? It is easy to establish that a simple approximation is x = 2*Sqrt(6*(1 - B/A)). The relative error in that approximation is worst, about -6%, when B/A = 2/pi, and its |relative error| decreases to 0 as B/A approaches 1. Instead of the coefficient 2*Sqrt(6), which is about 4.9, if we use a somewhat larger coefficient, then we can reduce worst |relative error|, assuming of course that B/A is restricted to the interval [2/pi, 1]. Specifically, to minimize worst |relative error|, it can be shown that the coefficient should be 5.0504... In any event, it's not clear how you arrived at 5.062792913 for the coefficient. Please explain. David Cantrell === Subject: Re: Root Finder 12 > Root Finder 12 > by Jon Giffen. > This solution was so simple that I couldn't believe it. > But I tried it, and it works. > It is found that the roots to the polynomial, > a[0]+a[1]t+a[2]t^2+...+a[n]t^n=0 where > T=(t,t^2,t^3,..,t^n) and > N=(a[1],a[2],a[3],...,a[n]) are given by, > -a[0] > T= -----D > D*N > D=(1,2,3,4,...,n) Ok then, attempt to construct the quadratic formula for your method: at^2 + bt + c = 0 T = (t, t^2) N = (b, a) D = (1, 2) T = -c/(b+2a) * (1,2) = (t, t^2) t = -c/(b+2a) t^2 = -2c/(b+2a) Clearly you are dead wrong, since the correct t's are t = (-b + sqrt[b^2-4ac])/(2a) and t = (-b - sqrt[b^2-4ac])/(2a) If you can not reconstruct the quadratic formula, then you are wrong. Any polynomials with a root approximated by your method are just coincidences. === Subject: Re: Root Finder 12 ex. t^2+2t-3=0 N=(2,1) |N|^2=5 Q=(3/5)(2,1) D=(1,2) |D|^2=5 Q*D=12/5 (mD-Q)*D=0 m|D|^2=Q*D m=(Q*D)/|D|^2 = 12/25 T=mD = (12/25)(1,2) t^2=24/25 ~ 1 >>Root Finder 12 >>by Jon Giffen. >>This solution was so simple that I couldn't believe it. >>But I tried it, and it works. >>It is found that the roots to the polynomial, >>a[0]+a[1]t+a[2]t^2+...+a[n]t^n=0 where >>T=(t,t^2,t^3,..,t^n) and >>N=(a[1],a[2],a[3],...,a[n]) are given by, >> -a[0] >>T= -----D >> D*N >>D=(1,2,3,4,...,n) > Ok then, attempt to construct the quadratic formula for your method: > at^2 + bt + c = 0 > T = (t, t^2) > N = (b, a) > D = (1, 2) > T = -c/(b+2a) * (1,2) = (t, t^2) > t = -c/(b+2a) > t^2 = -2c/(b+2a) > Clearly you are dead wrong, since the correct t's are > t = (-b + sqrt[b^2-4ac])/(2a) > and > t = (-b - sqrt[b^2-4ac])/(2a) > If you can not reconstruct the quadratic formula, then you are wrong. > Any polynomials with a root approximated by your method are just > coincidences. === Subject: Re: Root Finder 12 > ex. > t^2+2t-3=0 > N=(2,1) |N|^2=5 Q=(3/5)(2,1) D=(1,2) |D|^2=5 Q*D=12/5 > (mD-Q)*D=0 > m|D|^2=Q*D m=(Q*D)/|D|^2 = 12/25 > T=mD = (12/25)(1,2) > t^2=24/25 ~ 1 So, you're saying that t=sqrt(24/25) is a root of x^2+2t-3=0. Remember, and you have been told this repeatedly: YOU ARE NOT ALLOWED TO USE APPROXIMATIONS. You also say that Newton's method is great. However, there are established results saying where a good place is to start. So what you have done here is not new, and what you claim to do (to FIND roots, not just The reason that Root Finder N will be wrong for all values of N is that you have a bunch of equations of the form t^n = A_n t^(n-1) = A_(n-1) t^(n-2) = A_(n-2) etc. Now, if you can find a value of t that satisfies ALL equations SIMULTANEOUSLY, then you indeed have found a root of the original polynomial. However, if there is no solution, your claim that you have found a root is not automatic; it may be a root, or it may not. In fact, you may also be close to a root. But this is like firing a bullet at a wall, then drawing the target around it. -- Christopher Heckman > >>Root Finder 12 >by Jon Giffen. >This solution was so simple that I couldn't believe it. >>But I tried it, and it works. >It is found that the roots to the polynomial, >a[0]+a[1]t+a[2]t^2+...+a[n]t^n=0 where >T=(t,t^2,t^3,..,t^n) and >>N=(a[1],a[2],a[3],...,a[n]) are given by, > -a[0] >>T= -----D >> D*N >D=(1,2,3,4,...,n) >> > Ok then, attempt to construct the quadratic formula for your method: > > at^2 + bt + c = 0 > T = (t, t^2) > N = (b, a) > D = (1, 2) > > T = -c/(b+2a) * (1,2) = (t, t^2) > t = -c/(b+2a) > t^2 = -2c/(b+2a) > > Clearly you are dead wrong, since the correct t's are > t = (-b + sqrt[b^2-4ac])/(2a) > and > t = (-b - sqrt[b^2-4ac])/(2a) > > If you can not reconstruct the quadratic formula, then you are wrong. > Any polynomials with a root approximated by your method are just > coincidences. === Subject: Re: Root Finder 12 Dodging the issue, heh? I proved your method failed for the general formula, and all you did was post one of the coincidences I mentioned. Why don't you even try to discuss the fact I proved that the general quadratic can not be factored by your method? > ex. > t^2+2t-3=0 > N=(2,1) |N|^2=5 Q=(3/5)(2,1) D=(1,2) |D|^2=5 Q*D=12/5 > (mD-Q)*D=0 > m|D|^2=Q*D m=(Q*D)/|D|^2 = 12/25 > T=mD = (12/25)(1,2) > t^2=24/25 ~ 1 > >>Root Finder 12 >by Jon Giffen. >This solution was so simple that I couldn't believe it. >>But I tried it, and it works. >It is found that the roots to the polynomial, >a[0]+a[1]t+a[2]t^2+...+a[n]t^n=0 where >T=(t,t^2,t^3,..,t^n) and >>N=(a[1],a[2],a[3],...,a[n]) are given by, > -a[0] >>T= -----D >> D*N >D=(1,2,3,4,...,n) >> > Ok then, attempt to construct the quadratic formula for your method: > > at^2 + bt + c = 0 > T = (t, t^2) > N = (b, a) > D = (1, 2) > > T = -c/(b+2a) * (1,2) = (t, t^2) > t = -c/(b+2a) > t^2 = -2c/(b+2a) > > Clearly you are dead wrong, since the correct t's are > t = (-b + sqrt[b^2-4ac])/(2a) > and > t = (-b - sqrt[b^2-4ac])/(2a) > > If you can not reconstruct the quadratic formula, then you are wrong. > Any polynomials with a root approximated by your method are just > coincidences. === Subject: Re: Root Finder 12 > Root Finder 12 > by Jon Giffen. > This solution was so simple that I couldn't believe it. > But I tried it, and it works. > [...] > Solve the 6th degree polynomial, > f(t)=t^6 + t - 10=0 > a[0] = -10 N=(1,0,0,0,0,1) D=(1,2,3,4,5,6) > a[0]=-10 D*N=7 > -a[0] 10 > T = -----D = ---(1,2,3,4,5,6) > D*N 7 > t =10/7 t=1.42835 f(1.42835)=-0.071 10/7 is not a root of t^6 + t - 10 = 0. The Rational Root Test says that any rational solutions to this polynomial are +/-1, +/-2, +/5, or +/10. It can only be an approximation. Strike Twelve. (The Rational Root Theorem -- a.k.a. the Rational Zero Theorem -- can be found at http://mathworld.wolfram.com/RationalZeroTheorem.html . > Solve the 49th degree polynomial, > f(t)=t^49 + t^16 + 40t^5 - 6000 = 0 > a[0]=-6000 N=(1,0,0,0,40,0,..,1,0,...,1) D=(1,2,3,..,49) > D*N=1+200+16+49=266 > -a[0] 6000 > T = -----D = ------(1,2,3,..,49) > D*N 266 > t^49=294000/266=1105.26 t=1.15375 f(1.15375)=-3575.99 Once again, the Rational Root Theorem says that the only possible rational roots are integers. > Applying Newton's Method, Ah, so. You aren't finding roots after all, only approximations to them. You've been told repeatedly that this isn't the same as finding the roots. > t=1.15375-(-3575)/[49(1.15375^48)+16(1.15375^15)+200(1.15375)^4] > =1.15375 again, so the answer must depend on distant decimal > places. The problem here is you don't have enough precision to make Newton's Method work. > [...] > t^2 + 2t - 3 = 0 > a[0]=-3 N=(2,1) D=(1,2) D*N=4 > -a[0] 3 > T = -----D =---(1,2) t^2=3/4 t=0.866 ~ 1 > D*N 4 What? Your method can't even solve a quadratic equation? That's when you know it's really bad. -- Christopher Heckman === Subject: Re: Root Finder 12 ex. t^2+2t-3=0 N=(2,1) |N|^2=5 Q=(3/5)(2,1) D=(1,2) |D|^2=5 Q*D=12/5 (mD-Q)*D=0 m|D|^2=Q*D m=(Q*D)/|D|^2 = 12/25 T=mD = (12/25)(1,2) t^2=24/25 ~ 1 ex. at^2+bt+c=0 N=(b,a) |N|^2=b^2+a^2 Q=(-c/[b^2+a^2])(b,a) D=(1,2) |D|^2=5 Q*D=(-c/[b^2+a^2])(b+2a) (mD-Q)*D=0 m=(Q*D)/|D|^2 = (1/5)(-c/[b^2+a^2])(b+2a) T=mD=(1/5)(-c/[b^2+a^2])(b+2a)(1,2) t^2 =(2/5)(-c/[b^2+a^2])(b+2a) b+/-{b^2-4ac}^(1/2) t ={(2/5)(-c/[b^2+a^2])(b+2a)}^(1/2)=-------------------- 2a solve and find the required correction >>Root Finder 12 >>by Jon Giffen. >>This solution was so simple that I couldn't believe it. >>But I tried it, and it works. >>[...] >>Solve the 6th degree polynomial, >>f(t)=t^6 + t - 10=0 >>a[0] = -10 N=(1,0,0,0,0,1) D=(1,2,3,4,5,6) >>a[0]=-10 D*N=7 >> -a[0] 10 >>T = -----D = ---(1,2,3,4,5,6) >> D*N 7 >>t =10/7 t=1.42835 f(1.42835)=-0.071 > 10/7 is not a root of t^6 + t - 10 = 0. The Rational Root Test says > that any rational solutions to this polynomial are +/-1, +/-2, +/5, > or +/10. It can only be an approximation. > Strike Twelve. > (The Rational Root Theorem -- a.k.a. the Rational Zero Theorem -- > can be found at http://mathworld.wolfram.com/RationalZeroTheorem.html . >>Solve the 49th degree polynomial, >>f(t)=t^49 + t^16 + 40t^5 - 6000 = 0 >>a[0]=-6000 N=(1,0,0,0,40,0,..,1,0,...,1) D=(1,2,3,..,49) >>D*N=1+200+16+49=266 >> -a[0] 6000 >>T = -----D = ------(1,2,3,..,49) >> D*N 266 >>t^49=294000/266=1105.26 t=1.15375 f(1.15375)=-3575.99 > Once again, the Rational Root Theorem says that the only possible rational > roots are integers. >>Applying Newton's Method, > Ah, so. You aren't finding roots after all, only approximations to them. > You've been told repeatedly that this isn't the same as finding the roots. >>t=1.15375-(-3575)/[49(1.15375^48)+16(1.15375^15)+200(1.15375)^4] >> =1.15375 again, so the answer must depend on distant decimal >>places. > The problem here is you don't have enough precision to make Newton's > Method work. >>[...] >>t^2 + 2t - 3 = 0 >>a[0]=-3 N=(2,1) D=(1,2) D*N=4 >> -a[0] 3 >>T = -----D =---(1,2) t^2=3/4 t=0.866 ~ 1 >> D*N 4 > What? Your method can't even solve a quadratic equation? That's when > you know it's really bad. > -- Christopher Heckman === Subject: Re: Root Finder 12 > ex. > t^2+2t-3=0 > N=(2,1) |N|^2=5 Q=(3/5)(2,1) D=(1,2) |D|^2=5 Q*D=12/5 > (mD-Q)*D=0 > m|D|^2=Q*D m=(Q*D)/|D|^2 = 12/25 > T=mD = (12/25)(1,2) > t^2=24/25 ~ 1 No, this says that t = sqrt(24/25), which is not a root of t^2+2t-3. My claim that your method can't solve quadratic equations is thus proven by your example. Remember, and you have been told this repeatedly: YOU ARE NOT ALLOWED TO USE APPROXIMATIONS. You also say that Newton's method is great. However, there are established results saying where a good place is to start. So what you have done here is not new, and what you claim to do (to FIND roots, not just The reason that Root Finder N will be wrong for all values of N is that you have a bunch of equations of the form t^n = A_n t^(n-1) = A_(n-1) t^(n-2) = A_(n-2) etc. Now, if you can find a value of t that satisfies ALL equations SIMULTANEOUSLY, then you indeed have found a root of the original polynomial. However, if there is no solution, your claim that you have found a root is not automatic; it may be a root, or it may not. In fact, you may also be close to a root. But this is like firing a bullet at a wall, then drawing the target around it. > ex. > at^2+bt+c=0 > N=(b,a) |N|^2=b^2+a^2 Q=(-c/[b^2+a^2])(b,a) D=(1,2) |D|^2=5 > Q*D=(-c/[b^2+a^2])(b+2a) > (mD-Q)*D=0 > m=(Q*D)/|D|^2 = (1/5)(-c/[b^2+a^2])(b+2a) > T=mD=(1/5)(-c/[b^2+a^2])(b+2a)(1,2) > t^2 =(2/5)(-c/[b^2+a^2])(b+2a) > b+/-{b^2-4ac}^(1/2) > t ={(2/5)(-c/[b^2+a^2])(b+2a)}^(1/2)=-------------------- > 2a > solve and find the required correction What is this correction? A fudge factor, maybe? -- Christopher Heckman > >>Root Finder 12 >by Jon Giffen. >This solution was so simple that I couldn't believe it. >>But I tried it, and it works. >>[...] >Solve the 6th degree polynomial, >f(t)=t^6 + t - 10=0 >a[0] = -10 N=(1,0,0,0,0,1) D=(1,2,3,4,5,6) >a[0]=-10 D*N=7 > -a[0] 10 >>T = -----D = ---(1,2,3,4,5,6) >> D*N 7 >t =10/7 t=1.42835 f(1.42835)=-0.071 > > > 10/7 is not a root of t^6 + t - 10 = 0. The Rational Root Test says > that any rational solutions to this polynomial are +/-1, +/-2, +/5, > or +/10. It can only be an approximation. > > Strike Twelve. > > (The Rational Root Theorem -- a.k.a. the Rational Zero Theorem -- > can be found at http://mathworld.wolfram.com/RationalZeroTheorem.html . > > >Solve the 49th degree polynomial, >f(t)=t^49 + t^16 + 40t^5 - 6000 = 0 >a[0]=-6000 N=(1,0,0,0,40,0,..,1,0,...,1) D=(1,2,3,..,49) >D*N=1+200+16+49=266 > -a[0] 6000 >>T = -----D = ------(1,2,3,..,49) >> D*N 266 >t^49=294000/266=1105.26 t=1.15375 f(1.15375)=-3575.99 > > > Once again, the Rational Root Theorem says that the only possible rational > roots are integers. > > >>Applying Newton's Method, > > > Ah, so. You aren't finding roots after all, only approximations to them. > You've been told repeatedly that this isn't the same as finding the roots. > > >>t=1.15375-(-3575)/[49(1.15375^48)+16(1.15375^15)+200(1.15375)^4] > =1.15375 again, so the answer must depend on distant decimal >>places. > > > The problem here is you don't have enough precision to make Newton's > Method work. > > >>[...] >t^2 + 2t - 3 = 0 >a[0]=-3 N=(2,1) D=(1,2) D*N=4 > -a[0] 3 >>T = -----D =---(1,2) t^2=3/4 t=0.866 ~ 1 >> D*N 4 > > > What? Your method can't even solve a quadratic equation? That's when > you know it's really bad. > -- Christopher Heckman === Subject: Re: Root Finder 12 I thought you already posted the last word on this subject. Are you suffering from some kind of attention deficit disorder, or are you being deliberately misleading? If there's a third possibility, I'd welcome your explanation. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Root Finder 12 You people are savage, and there's no call for it. I suppose you are one of the anal perfectionist obsessed with keeping on course down to the Angstrom to offset the Gudermanian. Newton's Method is an invention of genius. Why not use it? Newton just didn't come up with the approximations to plug into it... or did he? > I thought you already posted the last word on this subject. Are you > suffering from some kind of attention deficit disorder, or are you > being deliberately misleading? If there's a third possibility, I'd > welcome your explanation. > -- > There are two things you must never attempt to prove: the unprovable > -- and the obvious. > -- > Democracy: The triumph of popularity over principle. > -- > http://www.crbond.com === Subject: Re: Root Finder 12 > You people are savage, and there's no call for it. I suppose > you are one of the anal perfectionist obsessed with keeping on > course down to the Angstrom to offset the Gudermanian. > Newton's Method is an invention of genius. Why not use it? > Newton just didn't come up with the approximations to plug into > it... > or did he? No, but they have been done, and in mathematics, it doesn't matter who does it, just that it is true and can be used. Remember, and you have been told this repeatedly: YOU ARE NOT ALLOWED TO USE APPROXIMATIONS. You also say that Newton's method is great. However, there are established results saying where a good place is to start. So what you have done here is not new, and what you claim to do (to FIND roots, not just The reason that Root Finder N will be wrong for all values of N is that you have a bunch of equations of the form t^n = A_n t^(n-1) = A_(n-1) t^(n-2) = A_(n-2) etc. Now, if you can find a value of t that satisfies ALL equations SIMULTANEOUSLY, then you indeed have found a root of the original polynomial. However, if there is no solution, your claim that you have found a root is not automatic; it may be a root, or it may not. In fact, you may also be close to a root. But this is like firing a bullet at a wall, then drawing the target around it. -- Christopher Heckman P.S. I can use cut and paste, too. > > > I thought you already posted the last word on this subject. Are you > suffering from some kind of attention deficit disorder, or are you > being deliberately misleading? If there's a third possibility, I'd > welcome your explanation. > > -- > There are two things you must never attempt to prove: the unprovable > -- and the obvious. > -- > Democracy: The triumph of popularity over principle. > -- > http://www.crbond.com > === Subject: Re: Root Finder 12 There's always room for improvement > I thought you already posted the last word on this subject. Are you > suffering from some kind of attention deficit disorder, or are you > being deliberately misleading? If there's a third possibility, I'd > welcome your explanation. > -- > There are two things you must never attempt to prove: the unprovable > -- and the obvious. > -- > Democracy: The triumph of popularity over principle. > -- > http://www.crbond.com === Subject: Re: Root Finder ix.,final > I've outlined the solution on my web site, > http://mypeoplepc.com/members/jon8338/polynomial/ > along with a 6th degree polynomial and its (correct) > solution... > Which is more than any of you have done too lazy to see it's > right, too presumptuous with ignorance in your vanity. (1) You just posted it, so no one has had a chance to look at it. (2) The fact that there ARE examples which show your algorithm doesn't work is enough. Otherwise I could say the following: Theorem. If a and b are real numbers, then a + b = a * b. Proof: 2 + 2 = 2 * 2. 0 + 0 = 0 * 0. 3 + 3/2 = 3 * 3/2. QED. > Victory, You evidently have a weird definition of victory. -- Christopher Heckman > >>This closes the book on a problem that I've worked on for about >>5 years, which began as a surveying problem posed by a British >>engineer. What is the angle that subtends a known arc and known >>chord, on a circle of unknown radius? It was the solution to a >>polynomial series. That is how I got started on all this root >>stuff. Why hasn't anyone found a way to find the root to an >>infinite polynomial series? (carried out to any number of terms) > > > Because there are no exact formulas for roots, if the degree is > at least 5. You've been told that for a couple years now. > None of your examples have proven that your method works, either; > when you provide specific examples, to claim that your method works, > at least make sure that the examples work ahead of time. If they don't, > then there's something wrong with your method. > -- Christopher Heckman === Subject: EZ Root This shows the steps on how to find roots to the nth degree polynomial. An example of a 6th degree polynomial is provided, and two of its solutions are calculated. by Jon Giffen The polynomial a[0]+a[1]t+a[2]t^2+a[3]t^3+ ... + a[n] = 0 Has the roots, |Q| (t,t^2,t^3,..t^n)= T = --- C + Q |C| Where C=(D*N)S+(S*N)D and N=(a[1], a[2], a[3],....,a[n]) D=( 1 , 2 , 3 ,...,n ) S=(a[1],2a[2],3a[3],...,na[n]) Q=(-a[0]/|N|^2)N |N|^2=a[1]^2+a[2]^2+a[3]^2+..+a[n]^2 |Q| T[0] = Q + --- C |C| L[1]*C=0 L[1]*Q=0 Solve for L[1] |T[0]| T[1] = T[0] + -------- L[1] |L[1]| L[2]*L[1]=0 L[2]*T[0]=0 Solve for L[2] |T[1]| T[2] = T[1] + ------ L[2] |L[2]| L[3]*L[1]=0 L[3]*L[2]=0 L[3]*T[1]=0 Solve for L[3] . . . |T[i-1]| T[i] = T[i-1] + -------- L[i] |L[i]| t^6+t-10=0 N=(1,0,0,0,0,1) |N|^2=2 Q=(10/2)(1,0,0,0,0,1)=5(1,0,0,0,0,1) |Q|=(5)2^(1/2) D=( 1 , 2 , 3 ,...,n )=(1,2,3,4,5,6) S=(a[1],2a[2],3a[3],...,na[n])=(1,0,0,0,0,6) S*N=(1,0,0,0,0,6)*(1,0,0,0,0,1)=1+6=7 D*N=(1,2,3,4,5,6)*(1,0,0,0,0,1)=1+6=7 C=(S*N)D+(D*N)S =7[(1,2,3,4,5,6)+(1,0,0,0,0,6)] =7(2,2,3,4,5,12) |C|=(7)202^(1/2) T=Q+C|Q|/|C| =5(1,0,0,0,0,1)+7(2,2,3,4,5,12)[(5)2^(1/2)]/[(7)202^(1/2)] =5(1,0,0,0,0,1)+5(2,2,3,4,5,12)/101^(1/2) t^6 = 5 + 5(12)/101^(1/2) = 10.9702 t=1.4906 10.9702+1.49-10=2.46 t=1.4906-2.46/(6(1.4906^5)+1)=1.4361 N.M. t^6=8.7721 t=1.491 t^5=6.1083 t=1.436 t^4=4.2534 t=1.436 t^3=2.9618 t=1.436 t^2=2.0624 t=1.436 T[0]=(1.4361,2.0624,2.9618,4.2534,6.1083,8.7721) |T[0]|=12.1425 L*Q=(a,b,0,0,0,c)*(1,0,0,0,0,1)=0 a=-c L*T[0]=(-c,b,0,0,0,c)*T[0] = -1.43617c + 8.7721c + 2.0624b=0 c=1 b=-3.5570 L=(-1,-3.5570,0,0,0,1) |L|=3.8278 T[1]=T[0]+L|T[0]|/|L| = (1.4361, 2.0624,2.9618,4.2534,6.1083,8.7721) +( -1,-3.5570, 0, 0, 0, 1)(12.1425/3.8278) ----------------------------------------------------------------- = (1.4361, 2.0624,2.9618,4.2534,6.1083,8.7721) +( -1,-3.5570, 0, 0, 0, 1)(3.1728) ----------------------------------------------------------- = ( 1.4361, 2.0624,2.9618,4.2534,6.1083,8.7721) +(-3.1728,-11.2858, 0, 0, 0,3.1728) ----------------------------------------------------------- (-1.7367, -9.2234,2.9618,4.4534,6.1083,11.9449) t= -1.7367, 3.304i ,1.436 ,1.452 ,1.436 ,1.5119 t^6=8.7721+3.1728=11.9449 t=-1.5119 11.9449-1.5119-10=0.433 t=-1.5119-0.433/(6(-1.5119^5)+1)=-1.5025 N.M. (-1.5025)^6-1.5025-10=0.0056~0 === Subject: Re: EZ Root > This shows the steps on how to find roots to the nth degree > polynomial. An example of a 6th degree polynomial is provided, > and two of its solutions are calculated. OK, I have a little test for this. > by Jon Giffen > The polynomial > a[0]+a[1]t+a[2]t^2+a[3]t^3+ ... + a[n] = 0 > Has the roots, > |Q| > (t,t^2,t^3,..t^n)= T = --- C + Q > |C| > Where > C=(D*N)S+(S*N)D and > N=(a[1], a[2], a[3],....,a[n]) > D=( 1 , 2 , 3 ,...,n ) > S=(a[1],2a[2],3a[3],...,na[n]) > Q=(-a[0]/|N|^2)N > |N|^2=a[1]^2+a[2]^2+a[3]^2+..+a[n]^2 > |Q| > T[0] = Q + --- C > |C| > L[1]*C=0 > L[1]*Q=0 Solve for L[1] > |T[0]| > T[1] = T[0] + -------- L[1] > |L[1]| > L[2]*L[1]=0 > L[2]*T[0]=0 Solve for L[2] > |T[1]| > T[2] = T[1] + ------ L[2] > |L[2]| > L[3]*L[1]=0 > L[3]*L[2]=0 > L[3]*T[1]=0 Solve for L[3] > . |T[i-1]| > T[i] = T[i-1] + -------- L[i] > |L[i]| > t^6+t-10=0 Try using the polynomial created when translating the given polynomial 1 unit to the right: (t-1)^6+(t-1)-10 = t^6-6t^5+15t^4-20t^3+15t^2-5t-10 = 0. > N=(1,0,0,0,0,1) > |N|^2=2 > Q=(10/2)(1,0,0,0,0,1)=5(1,0,0,0,0,1) > |Q|=(5)2^(1/2) > D=( 1 , 2 , 3 ,...,n )=(1,2,3,4,5,6) > S=(a[1],2a[2],3a[3],...,na[n])=(1,0,0,0,0,6) > S*N=(1,0,0,0,0,6)*(1,0,0,0,0,1)=1+6=7 > D*N=(1,2,3,4,5,6)*(1,0,0,0,0,1)=1+6=7 > C=(S*N)D+(D*N)S > =7[(1,2,3,4,5,6)+(1,0,0,0,0,6)] > =7(2,2,3,4,5,12) > |C|=(7)202^(1/2) > T=Q+C|Q|/|C| > =5(1,0,0,0,0,1)+7(2,2,3,4,5,12)[(5)2^(1/2)]/[(7)202^(1/2)] > =5(1,0,0,0,0,1)+5(2,2,3,4,5,12)/101^(1/2) > t^6 = 5 + 5(12)/101^(1/2) = 10.9702 t=1.4906 > 10.9702+1.49-10=2.46 > t=1.4906-2.46/(6(1.4906^5)+1)=1.4361 N.M. > t^6=8.7721 t=1.491 > t^5=6.1083 t=1.436 > t^4=4.2534 t=1.436 > t^3=2.9618 t=1.436 > t^2=2.0624 t=1.436 > T[0]=(1.4361,2.0624,2.9618,4.2534,6.1083,8.7721) > |T[0]|=12.1425 > L*Q=(a,b,0,0,0,c)*(1,0,0,0,0,1)=0 a=-c > L*T[0]=(-c,b,0,0,0,c)*T[0] > = -1.43617c + 8.7721c + 2.0624b=0 c=1 b=-3.5570 > L=(-1,-3.5570,0,0,0,1) |L|=3.8278 > T[1]=T[0]+L|T[0]|/|L| > = (1.4361, 2.0624,2.9618,4.2534,6.1083,8.7721) > +( -1,-3.5570, 0, 0, 0, 1)(12.1425/3.8278) > ----------------------------------------------------------------- > = (1.4361, 2.0624,2.9618,4.2534,6.1083,8.7721) > +( -1,-3.5570, 0, 0, 0, 1)(3.1728) > ----------------------------------------------------------- > = ( 1.4361, 2.0624,2.9618,4.2534,6.1083,8.7721) > +(-3.1728,-11.2858, 0, 0, 0,3.1728) > ----------------------------------------------------------- > (-1.7367, -9.2234,2.9618,4.4534,6.1083,11.9449) > t= -1.7367, 3.304i ,1.436 ,1.452 ,1.436 ,1.5119 > t^6=8.7721+3.1728=11.9449 t=-1.5119 > 11.9449-1.5119-10=0.433 > t=-1.5119-0.433/(6(-1.5119^5)+1)=-1.5025 N.M. > (-1.5025)^6-1.5025-10=0.0056~0 === Subject: Finding unique sums. I'm trying to find a series where the sum of a subset of that is unique. Or in other words, I can find the numbers from sum. An example would be f(n)=2^n, you add any number of elements in this set, you'll get a number with all those bits set. But I'm looking for a series that does not grow geometrically.. Pradeep === Subject: Re: Finding unique sums. > I'm trying to find a series where the sum of a subset of that > is unique. Or in other words, I can find the numbers from sum. > An example would be f(n)=2^n, you add any number of elements > in this set, you'll get a number with all those bits set. > But I'm looking for a series that does not grow geometrically.. > Pradeep Something a bit different (not better) from the other answers given... How about the sequence 0.4142135623730950488016887242... 0.8284271247461900976033774484... 0.6568542494923801952067548968... 0.3137084989847603904135097936... 0.6274169979695207808270195873... 0.2548339959390415616540391747... 0.5096679918780831233080783494... .... (Puzzle - is the sequence dense in [0,1]? (The sequence is 2^(n + 1/2) mod 1).) Mike. === Subject: Re: Finding unique sums. > How about the sequence > 0.4142135623730950488016887242... > 0.8284271247461900976033774484... > 0.6568542494923801952067548968... > 0.3137084989847603904135097936... > 0.6274169979695207808270195873... > 0.2548339959390415616540391747... > 0.5096679918780831233080783494... > .... > (Puzzle - is the sequence dense in [0,1]? (The sequence is 2^(n + 1/2) mod > 1).) Puzzle, or open problem? Think of the sequence as the fractional part of 2^n sqrt2, and you'll see it depends on the binary expansion of sqrt2. I don't think enough is known about the binary expansion of sqrt2 to answer the question. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Finding unique sums. >> How about the sequence >> 0.4142135623730950488016887242... >> 0.8284271247461900976033774484... >> 0.6568542494923801952067548968... >> 0.3137084989847603904135097936... >> 0.6274169979695207808270195873... >> 0.2548339959390415616540391747... >> 0.5096679918780831233080783494... >> .... >> (Puzzle - is the sequence dense in [0,1]? (The sequence is 2^(n + 1/2) mod >> 1).) >Puzzle, or open problem? >Think of the sequence as the fractional part of 2^n sqrt2, and you'll >see it depends on the binary expansion of sqrt2. I don't think enough >is known about the binary expansion of sqrt2 to answer the question. It's equivalent to the question of whether every finite string of 0's and 1's occurs in the binary expansion of sqrt(2). This is indeed open (although I would bet that the answer is yes). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Finding unique sums. > How about the sequence > 0.4142135623730950488016887242... > 0.8284271247461900976033774484... > 0.6568542494923801952067548968... > 0.3137084989847603904135097936... > 0.6274169979695207808270195873... > 0.2548339959390415616540391747... > 0.5096679918780831233080783494... > .... > (Puzzle - is the sequence dense in [0,1]? (The sequence is 2^(n + 1/2) mod > 1).) > Puzzle, or open problem? > Think of the sequence as the fractional part of 2^n sqrt2, and you'll > see it depends on the binary expansion of sqrt2. I don't think enough > is known about the binary expansion of sqrt2 to answer the question. Well I don't know the answer myself, so I suppose I should have said open problem... > -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Finding unique sums. > I'm trying to find a series where the sum of a subset of that > is unique. Or in other words, I can find the numbers from sum. > An example would be f(n)=2^n, you add any number of elements > in this set, you'll get a number with all those bits set. > But I'm looking for a series that does not grow geometrically.. I was looking for a series which consists of positive integers which has this property. Is it helpful if the size of the subset is known? For example, the sum of random 10 elements of a series is given, is it possible to get the individual elements(I mean is there such a series of positive integers other that 2^n and doesn't grow exponentially)? === Subject: Re: Finding unique sums. >> I'm trying to find a series where the sum of a subset of that >> is unique. Or in other words, I can find the numbers from sum. >> An example would be f(n)=2^n, you add any number of elements >> in this set, you'll get a number with all those bits set. >> But I'm looking for a series that does not grow geometrically.. >I was looking for a series which consists of positive integers >which has this property. This has already been answered. Since the 2^n series is the 'most efficient' such series, all other series must grow faster. >Is it helpful if the size of the subset is known? >For example, the sum of random 10 elements of a series is given, >is it possible to get the individual elements(I mean is there >such a series of positive integers other that 2^n and doesn't >grow exponentially)? I believe this doesn't really change the outcome. All size-10 subsets must produce a unique sum implies (with handwave) that all subsets must produce a unique sum, so your sequence is going to grow exponentially. You can probably do away with the handwave above, by showing that assuming the existence of two size-N+1 subsets with identical sum but no size-N subsets leasds to a contradiction. -- Patrick Hamlyn posting from Perth, Western Australia Windsurfing capital of the Southern Hemisphere Moderator: polyforms group (polyforms-subscribe@egroups.com) === Subject: Re: Finding unique sums. >> >> I'm trying to find a series where the sum of a subset of that >> is unique. Or in other words, I can find the numbers from sum. >> >> An example would be f(n)=2^n, you add any number of elements >> in this set, you'll get a number with all those bits set. >> But I'm looking for a series that does not grow geometrically.. >I was looking for a series which consists of positive integers >which has this property. > This has already been answered. Since the 2^n series is the 'most efficient' > such series, all other series must grow faster. Not true. See below. >Is it helpful if the size of the subset is known? >For example, the sum of random 10 elements of a series is given, >is it possible to get the individual elements(I mean is there >such a series of positive integers other that 2^n and doesn't >grow exponentially)? > I believe this doesn't really change the outcome. All size-10 subsets must > produce a unique sum implies (with handwave) that all subsets must produce a > unique sum, so your sequence is going to grow exponentially. I don't think I believe this. It's anyway not true that if all size-2 subsets produce a unique sum then all subsets produce a unique sum, and it's not clear to me why size-10 should differ from size-2 in this regard. There's a lot of literature on these problems. A good starting place is Guy's Unsolved Problems in Number Theory. The 3rd edition is out, but I don't have it yet. In the 2nd edition, C9 concerns m, the maximum number of integers between 1 and n inclusive with all sums of pairs distinct. The conjecture is that m - sqrt n is bounded. C11 concerns the requirement that sums of h terms all be distinct for some given h. C8 asks about the maximum number of positive integers not exceeding 2^k with all subset sums distinct. Conway & Guy conjecture it's k + 2 - notice that this is better than the k + 1 you get by taking powers of 2. An example that achieves k + 2 is given. C5 says you can reconstruct a set of N numbers from the set of sums of pairs provided N is not a power of 2. Sums of size 3 determine the set except perhaps when N = 27 or 486. Sums of larger size have also been considered. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Finding unique sums. > C8 asks about the maximum number of positive integers not exceeding > 2^k with all subset sums distinct. Conway & Guy conjecture it's > k + 2 - notice that this is better than the k + 1 you get by taking > powers of 2. An example that achieves k + 2 is given. I too was convinced that you couldn't do better than powers of 2, and I'd be very interested to see that example. For those of us who don't have the book, is the example simple enough for you to reproduce here? (I'm not asking you to type in ten pages of text!) === Subject: Re: Finding unique sums. > C8 asks about the maximum number of positive integers not exceeding > 2^k with all subset sums distinct. Conway & Guy conjecture it's > k + 2 - notice that this is better than the k + 1 you get by taking > powers of 2. An example that achieves k + 2 is given. > > I too was convinced that you couldn't do better than powers of 2, and > I'd be very interested to see that example. For those of us who don't > have the book, is the example simple enough for you to reproduce here? > (I'm not asking you to type in ten pages of text!) Here are some reviews from Math Reviews, you can probably extract the information you want (and more!) from them, especially if TeX doesn't bother you. MR0917837 (89a:11019) Lunnon, W. F.(4-WALC) Integer sets with distinct subset-sums. Math. Comp. 50 (1988), no. 181, 297--320. 11B13 (94A60) The set of integers $p_i=2^{i-1}$, $i=1,2,cdots,n$, has the interesting property that all of its distinct subsets have distinct sums. Let us call this property SSD. The question is whether there are sets $overline{p}={p_0=01)$. It is easy to see that the polynomial $g_m(x)=prod_{k=1}^m(1+z^{J_k})$ has the above-defined properties and that $$deg g_m=frac43.872^m-frac32+frac{(-1)^m}6.$$ Therefore, $$d(m)lefrac43.872^m-frac32+frac{(-1)^m}6.$$ The authors show that the equality holds if and only if $mle5$. In the general case, they prove that $$2^m+c_1mle d(m)le frac{103}{96}.872^m+c_2$$ and they conjecture that for any $epsilon>0$ the inequality $d(m)<(1+epsilon)2^m$ holds for sufficiently large $m$. Also, they consider the related problem of finding a set of $m$ positive integers with distinct subset sums and minimal largest element and show that the well-known Conway--Guy sequence yields the optimal solution for $mle9$. Reviewed by Sergeui V. Konyagin MR1486396 (98k:11014) Bohman, Tom(1-MIT) A construction for sets of integers with distinct subset sums. (English. English summary) Electron. J. Combin. 5 (1998), Research Paper 3, 14 pp. (electronic). 11B75 (05D10) A finite set $S$ of positive integers has distinct subset sums if the $2sp{|S|}$ sums $sumsb{ain A}a$, where $Asubseteq S$, are pairwise distinct. For brevity, call sets with distinct subset sums DSS-sets. The paper investigates the following questions: How small can a positive integer $N$ be such that ${1,2,cdots,N}$ contains an $n$-element DSS-set? For every $n$ let $f(n)$ be the smallest $N$ with this property. In different terms: $f(n) = min max_S N$, where the minimum is taken over all $n$-element DSS-sets. Obviously, $f(n)le 2sp {n-1}$ (take $S={1,2,4,cdots,2sp{n-1}})$. Erd.9as conjectured that $f(n)gg 2sp n$ (the implicit constant is absolute). Together with Moser he proved in 1955 a weaker inequality $f(n)ge 2sp n /(4sqrt n)$, which remains, up to the constant, the best known lower bound for $f(n)$. In the opposite direction, J. H. Conway and R. K. Guy ref[Notices Amer. Math. Soc. 15 (1968), no. 2, 345, Abstract 654-32] constructed short DSS-sets, using a special sequence of integers they discovered (the Conway-Guy sequence). Their result implied an estimate $f(n) le 0.23513.872sp n$ for $nge 40$. W. F. Lunnon ref[Math. Comp. 50 (1988), no. 181, 297--320; MR0917837 (89a:11019)] suggested a similar construction, which implied $f(n) le 0.22096 .872sp n$ for $nge 67$. In his paper Bohman presents two parametric families of infinite sequences, which include, for small values of parameters, the sequences of Conway-Guy and Lunnon. Using his sequences, he finds many new examples of DSS-sets, and, in particular, obtains a new upper estimate $f(n) le 0.22002 .872sp n$ for sufficiently large $n$. Bohman's construction is very subtle and interesting and is likely to find different applications in combinatorics, cryptography, and related fields. Reviewed by Yuri Bilu MR1464377 (98i:11012) Maltby, Roy(3-CALG) Bigger and better subset-sum-distinct sets. (English. English summary) Mathematika 44 (1997), no. 1, 56--60. 11B75 A set of natural numbers is called subset-sum-distinct (SSD) if all pairwise distinct subsets have unequal sums. If $A$ is any SSD set, define $alpha(A)=(max A)/2^{|A|-1}$, where $max A$ is the biggest element of $A$. Given an SSD set, it is shown how to construct a bigger SSD set whose $alpha$-ratio is smaller. This shows that $inf{alpha(A)colon A$ is an SSD set}is not realised by any SSD set. (Erd.9as asked if the inf is positive.) The author also points out that one of the claims of W. F. Lunnon ref[Math. Comp. 50 (1988), no. 181, 297--320; MR0917837 (89a:11019)] concerning SSD sets is false. Reviewed by Ian Anderson MR1363448 (97b:11027) Bohman, Tom(1-RTG) A sum packing problem of Erd.9as and the Conway-Guy sequence. (English. English summary) Proc. Amer. Math. Soc. 124 (1996), no. 12, 3627--3636. 11B75 A set $A$ of positive integers has distinct subset sums if the set ${sum_{xin X}xcolon Xsubseteq A}$ has $2^{|A|}$ distinct elements. J. H. Conway and R. K. Guy ref[Notices Amer. Math. Soc. 15 (1968), 345] defined a sequence, ${A_k}$, of sets of integers as follows: (1) let $u_0=0, u_1=1$, and, for $ngeq1, u_{n+1}=2u_n-u_{n-r}$, where $r$ is the closest integer to $sqrt{2n}$; (2) for each $kgeq1$, define $A_k={u(k+1)-u(i)colon1leq ileq k}$. They conjectured that for every $k, A_k$ has distinct subset sums, and they showed this to be true for $1leq kleq40$. W. F. Lunnon ref[Math. Comp. 50 (1988), no. 181, 297--320; MR0917837 (89a:11019)] extended this to all $kleq80$. In this paper the author establishes the conjecture of Conway and Guy for all $k$. Define $f(n)=min{max_{sin S}scolon|S|=n$ and $S$ has distinct subset sums}. The Conway-Guy sequence mentioned above gives rise to $2^{n-2}$ as an upper bound on $f(n)$. This bound was improved by Lunnon to $0.2246(2^n)$. In this paper, the author presents a modification of the Conway-Guy sequence, and states that this leads to a slight improvement over Lunnon's bound, namely $0.22002(2^n)$. The author does not include a proof of this statement here, but plans to include it in a later paper. Reviewed by Bruce Landman -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Finding unique sums. matt a .8ecrit : >> C8 asks about the maximum number of positive integers not exceeding >> 2^k with all subset sums distinct. Conway & Guy conjecture it's >> k + 2 - notice that this is better than the k + 1 you get by taking >> powers of 2. An example that achieves k + 2 is given. > I too was convinced that you couldn't do better than powers of 2, and > I'd be very interested to see that example. For those of us who don't > have the book, is the example simple enough for you to reproduce here? > (I'm not asking you to type in ten pages of text!) C'est un forum en fran.8dais, respectez la charte ! -- Denis L.8eger === Subject: Re: Finding unique sums. > matt a .8ecrit : >> C8 asks about the maximum number of positive integers not exceeding >> 2^k with all subset sums distinct. Conway & Guy conjecture it's >> k + 2 - notice that this is better than the k + 1 you get by taking >> powers of 2. An example that achieves k + 2 is given. >> > > I too was convinced that you couldn't do better than powers of 2, and > I'd be very interested to see that example. For those of us who don't > have the book, is the example simple enough for you to reproduce here? > (I'm not asking you to type in ten pages of text!) > C'est un forum en fran.8dais, respectez la charte ! Please accept my apologies. I didn't notice that this was being posted to a French-language group. === Subject: Re: Finding unique sums. Vous avez aussi poste ce message a des forums anglais, donc vous n'avez pas respecte leurs chartes. -ilan > matt a .8ecrit : >> C8 asks about the maximum number of positive integers not exceeding >> 2^k with all subset sums distinct. Conway & Guy conjecture it's >> k + 2 - notice that this is better than the k + 1 you get by taking >> powers of 2. An example that achieves k + 2 is given. >> > > I too was convinced that you couldn't do better than powers of 2, and > I'd be very interested to see that example. For those of us who don't > have the book, is the example simple enough for you to reproduce here? > (I'm not asking you to type in ten pages of text!) > C'est un forum en fran.8dais, respectez la charte ! === Subject: Re: Finding unique sums. >> I'm trying to find a series where the sum of a subset of that >> is unique. Or in other words, I can find the numbers from sum. >> An example would be f(n)=2^n, you add any number of elements >> in this set, you'll get a number with all those bits set. >> But I'm looking for a series that does not grow geometrically.. >I was looking for a series which consists of positive integers >which has this property. >Is it helpful if the size of the subset is known? >For example, the sum of random 10 elements of a series is given, >is it possible to get the individual elements(I mean is there >such a series of positive integers other that 2^n and doesn't >grow exponentially)? Knowing the number of elements as well as their sum adds some possibilities. For example, f(n)=1000000+2^(n-1). But the core problem is that for every number you add, the number of possible subsets doubles. In order for each combination of numbers to produce a sum different from all other combinations, there need to be 2^N possible sums, where N is the number of values which might be part of the sum. Is this for a magic trick? --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Finding unique sums. > I'm trying to find a series where the sum of a subset of that > is unique. Or in other words, I can find the numbers from sum. > An example would be f(n)=2^n, you add any number of elements > in this set, you'll get a number with all those bits set. > But I'm looking for a series that does not grow geometrically.. What about f(n) = 1/2^n (or 1/a^n, where a >= 2)? Does that count? -- Christopher Heckman === Subject: Re: Finding unique sums. - Super-increasing sequences and knapsack problems A super-increasing sequence is a sequence of integers >= 1 in which each element is larger than the sum of all the preceding elements. Starting with 1 and aiming at the slowest growing rate you get 1, 2, 4, 8, ... . This is the only super-increasing sequence that allows any integer >= 1 to be represented as the sum of some of its elements. No proof given here; must be not too difficult. Because of the super-increasing property these representations are unique. One could ask: given a sheath (one-dimensional knapsack) of integer length N >= 1; given a set S of rods of integer lengths, their total length being >= N; determine what subsets of S exactly fit into the sheath, if any such subsets exist. This problem is known as the knapsack problem. In general it is an NP-hard problem, reason why it plays an important role in cryptography. If the lengths of the rods in S form a super-increasing sequence, then the knapsack problem can be solved at once; for cryptographic applications you need sequences that are much slower increasing than 2^n. The farther below 2^n, the better. The 2^n sequence is at the borderline between (A) representing all integers but not uniquely, and (B) representing not all integers, but uniquely if possible. Literature: Denning: Cryptography and Data Security. Addison-Wesley, 1982. ISBN 0-201-10150-5 >> I'm trying to find a series where the sum of a subset of that >> is unique. Or in other words, I can find the numbers from sum. >> An example would be f(n)=2^n, you add any number of elements >> in this set, you'll get a number with all those bits set. >> But I'm looking for a series that does not grow geometrically.. >> >What about f(n) = 1/2^n (or 1/a^n, where a >= 2)? Does that count? > -- Christopher Heckman === Subject: Re: Finding unique sums. > I'm trying to find a series where the sum of a subset of that > is unique. Or in other words, I can find the numbers from sum. > An example would be f(n)=2^n, you add any number of elements > in this set, you'll get a number with all those bits set. > But I'm looking for a series that does not grow geometrically.. > Pradeep How about ANY series of positive numbers where each term is greater than the sum of all those preceding it? Would that work? (Still grows fast, but not necessarily geometrically.) === Subject: Re: Finding unique sums. >> I'm trying to find a series where the sum of a subset of that is unique. Or >> in other words, I can find the numbers from sum. >> An example would be f(n)=2^n, you add any number of elements in this set, >> you'll get a number with all those bits set. But I'm looking for a series >> that does not grow geometrically.. > How about ANY series of positive numbers where each term is greater than the > sum of all those preceding it? Would that work? (Still grows fast, but not > necessarily geometrically.) Actually, that is geometric. Say the first term is a_0. Then the second term is a_1 > a_0, the third term is a_2 > a_1 + a_0 > 2a_0, the fourth is a_3 > a_2 + a_1 + a_0 > 2a_0 + a_0 + a_0 = 4a_0, etc. Apply induction if you wish; the answer is a sequence bounded below by a constant multiple of a geometric sequence. -- Ryan Reich ryanr@uchicago.edu === Subject: Re: Finding unique sums. > >> >> I'm trying to find a series where the sum of a subset of that is unique. Or >> in other words, I can find the numbers from sum. >> >> An example would be f(n)=2^n, you add any number of elements in this set, >> you'll get a number with all those bits set. But I'm looking for a series >> that does not grow geometrically.. >> > > > How about ANY series of positive numbers where each term is greater than the > sum of all those preceding it? Would that work? (Still grows fast, but not > necessarily geometrically.) > Actually, that is geometric. Say the first term is a_0. Then the second term > is a_1 > a_0, the third term is a_2 > a_1 + a_0 > 2a_0, the fourth is a_3 > a_2 > + a_1 + a_0 > 2a_0 + a_0 + a_0 = 4a_0, etc. Apply induction if you wish; the > answer is a sequence bounded below by a constant multiple of a geometric sequence. Oh ... I understood grow geometrically to mean than a[i+1]/a[i] was constant. Is that not right? === Subject: Re: Finding unique sums. > How about ANY series of positive numbers where each term is greater than the > sum of all those preceding it? Would that work? (Still grows fast, but not > necessarily geometrically.) > > Actually, that is geometric. Say the first term is a_0. Then the second term > is a_1 > a_0, the third term is a_2 > a_1 + a_0 > 2a_0, the fourth is a_3 > a_2 > + a_1 + a_0 > 2a_0 + a_0 + a_0 = 4a_0, etc. Apply induction if you wish; the > answer is a sequence bounded below by a constant multiple of a geometric sequence. > Oh ... I understood grow geometrically to mean than a[i+1]/a[i] was > constant. Is that not right? You're right. Ryan appears to think that n!, 2^2^n and 2^n! grow geometrically. This is either a gross abuse of terminology, or just plain wrong. Phil -- ... one Marine noticed one of the prisoners was still breathing. A Marine can be heard saying on the pool footage provided to Reuters Television: He's ing faking he's dead. He faking he's ing dead. The Marine then raises his rifle and fires into the man's head. The pictures are too graphic for us to broadcast, Sites said. === Subject: Re: Finding unique sums. > >> How about ANY series of positive numbers where each term is greater >> than the sum of all those preceding it? Would that work? (Still grows >> fast, but not necessarily geometrically.) > > Actually, that is geometric. Say the first term is a_0. Then the second > term is a_1 > a_0, the third term is a_2 > a_1 + a_0 > 2a_0, the fourth > is a_3 > a_2 + a_1 + a_0 > 2a_0 + a_0 + a_0 = 4a_0, etc. Apply induction > if you wish; the answer is a sequence bounded below by a constant > multiple of a geometric sequence. >> Oh ... I understood grow geometrically to mean than a[i+1]/a[i] was >> constant. Is that not right? > You're right. > Ryan appears to think that n!, 2^2^n and 2^n! grow geometrically. This is > either a gross abuse of terminology, or just plain wrong. > Phil Well, if you are trying to avoid geometric growth it is reasonable for me to expect you to be trying to avoid greater-than-geometric growth as well. I was thinking of this from an efficiency perspective. The original question seemed to me to desire such an answer, which is impossible, of course. Totally off-topic for the thread, but pertinent to the terminology: has anyone else read Orson Scott Card's Memory of Earth series (I think it's called) and winced at the Overmind's description of its memory layout? It says something like This made my memory grow geometrically, but that is not enough; it needed to be exponential. -- Ryan Reich ryanr@uchicago.edu === Subject: Re: Finding unique sums. > I'm trying to find a series where the sum of a subset of that > is unique. Or in other words, I can find the numbers from sum. > An example would be f(n)=2^n, you add any number of elements > in this set, you'll get a number with all those bits set. > But I'm looking for a series that does not grow geometrically.. How about sqrt{p_n} where p_n is the n-th prime? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Finding unique sums. > I'm trying to find a series where the sum of a subset of that > is unique. Or in other words, I can find the numbers from sum. > An example would be f(n)=2^n, you add any number of elements > in this set, you'll get a number with all those bits set. > But I'm looking for a series that does not grow geometrically.. The series f(n)=2^n grows exponentially. And it is clearly the most efficient series with your desired property, since all sums can be made with the minimum of terms. Any other such series would therefore be less efficiently 'packed', and would grow at a greater rate. Allowing negative numbers would be one way to 'break' this argument. -- Patrick Hamlyn posting from Perth, Western Australia Windsurfing capital of the Southern Hemisphere Moderator: polyforms group (polyforms-subscribe@egroups.com) === Subject: Re: Finding unique sums. Distribution: aus >> I'm trying to find a series where the sum of a subset of that >> is unique. Or in other words, I can find the numbers from sum. >> An example would be f(n)=2^n, you add any number of elements >> in this set, you'll get a number with all those bits set. >> But I'm looking for a series that does not grow geometrically.. > The series f(n)=2^n grows exponentially. And it is clearly the most > efficient series with your desired property, since all sums can be > made with the minimum of terms. > Any other such series would therefore be less efficiently 'packed', > and would grow at a greater rate. To put it another way, looking at only the first k terms, the number of subsets (and hence sums) is exponential. Since the next term must avoid these values, density arguments indicate that the terms must also grow exponentially. As a side point, if the number of terms is finite then it may be possible to find such sets with maximum less than 2^n. For instance, all subsets of { 11, 17, 20, 22, 23, 24 } have distinct sums. Geoff. ---------------------------------------------------------------------------- - Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@maths.usyd.edu.au | Gameplayer by vocation. ---------------------------------------------------------------------------- - === Subject: Re: Mathematics Books free for Download - 2 kasee_pl@hotmail.com schrieb in Nachricht >Great Webistes. Yep, and pretty long ago I posted Note: For Herbert Wilf's Algorithms and Complexity you should use the URL http://www.cis.upenn.edu/%7Ewilf/AlgComp3.html or http://www.cis.upenn.edu/~wilf/AlgComp3.html Hermann -- === Subject: Re: Mathematics Books free for Download - 2 Die erste Wahl bei der Suche nach Mathe-B.9fchern ist die etwa : Archimedes Suche in: de.sci.mathematik Autor: Hermann Kremer Dank an den flei¤igen Hermann Hero === Subject: Re: logic of the Cantorian followers mind The > mainstream accepts the existence of objects that can't be finitely > described. Exactly what objects are those? === Subject: Re: logic of the Cantorian followers mind >> The >> mainstream accepts the existence of objects that can't be finitely >> described. >Exactly what objects are those? Well, consider the set of uncomputable real numbers. Now that set is an acceptable mainstream object, and it can be finitely described. So it presumably exists in the sense you're intending. Well, in set theory, if a set exists then each of its individual elements also exists, right? Therefore particular uncomputable real numbers exist, right? But no particular uncomputable real number can be finitely described. -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: logic of the Cantorian followers mind <6yPod.89276$T02.37355@twister.rdc-kc.rr.com> Discussion, linux) >> The > mainstream accepts the existence of objects that can't be finitely > described. >>Exactly what objects are those? > Well, consider the set of uncomputable real numbers. Now that set is an > acceptable mainstream object, and it can be finitely described. So it > presumably exists in the sense you're intending. Well, in set theory, if > a set exists then each of its individual elements also exists, right? > Therefore particular uncomputable real numbers exist, right? But no > particular uncomputable real number can be finitely described. Depends on what you mean by finitely described. Chaitin's Omega can be finitely described, can't it? -- Jesse F. Hughes I don't know if you noticed but I had a tremendous drop in confidence concomittant [sic] with a dramatic grip of existential crisis. --- James S. Harris even has better diseases than you === Subject: Re: logic of the Cantorian followers mind >> The >> mainstream accepts the existence of objects that can't be finitely >> described. >Exactly what objects are those? >> Well, consider the set of uncomputable real numbers. Now that set is an >> acceptable mainstream object, and it can be finitely described. So it >> presumably exists in the sense you're intending. Well, in set theory, if >> a set exists then each of its individual elements also exists, right? >> Therefore particular uncomputable real numbers exist, right? But no >> particular uncomputable real number can be finitely described. >Depends on what you mean by finitely described. Chaitin's Omega can >be finitely described, can't it? You're right, that is a flawed argument (as I've already admitted). The cardinalities argument is the appropriate one to use: there are finitely-defined sets (such as R) which have more elements than there are possible labels for. -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: logic of the Cantorian followers mind [snipped] >>Depends on what you mean by finitely described. Chaitin's Omega can >>be finitely described, can't it? > You're right, that is a flawed argument (as I've already admitted). By convention, you must be a troll. === Subject: Re: logic of the Cantorian followers mind >> The > mainstream accepts the existence of objects that can't be finitely > described. >>Exactly what objects are those? > Well, consider the set of uncomputable real numbers. Now that set is an > acceptable mainstream object, and it can be finitely described. So it > presumably exists in the sense you're intending. Well, in set theory, > if > a set exists then each of its individual elements also exists, right? > Therefore particular uncomputable real numbers exist, right? But no > particular uncomputable real number can be finitely described. Of course I pointed out you were wrong. You didn't know what you were talking about. Subsequently you called me a troll. Why do you bother to even post. I didn't call you a troll or any other name for that matter when you posted the bull above. === Subject: Re: logic of the Cantorian followers mind <6yPod.89276$T02.37355@twister.rdc-kc.rr.com> Agreed, this is poorly stated: > But no > particular uncomputable real number can be finitely described. Barb is swapping between label and encapsulate when she uses the word describe. Herc === Subject: Re: logic of the Cantorian followers mind > But no > particular uncomputable real number can be finitely described. Doesn't Chaitin describe a uncomputable number? === Subject: Re: logic of the Cantorian followers mind >> But no >> particular uncomputable real number can be finitely described. >Doesn't Chaitin describe a uncomputable number? That's a good point. My initial reasoning was overly complex to begin with. Here's a simpler version that handles Chaitin's omega too: Consider the set of real numbers. That set is an acceptable mainstream object, and it can be finitely described. So it presumably exists in the sense you're intending. Well, in set theory, if a set exists then each of its particular elements also exists, right? Therefore every particular real number exists. But, the reals are uncountable and the set of possible finite descriptions (over a finite alphabet) is countable, so most particular reals do not have any finite description -- there are not enough finite descriptions to go around. -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: logic of the Cantorian followers mind > But no > particular uncomputable real number can be finitely described. >>Doesn't Chaitin describe a uncomputable number? > That's a good point. My initial reasoning was overly complex to begin > with. > Here's a simpler version that handles Chaitin's omega too: > Consider the set of real numbers. That set is an acceptable mainstream > object, and it can be finitely described. So it presumably exists in > the > sense you're intending. Well, in set theory, if a set exists then each > of > its particular elements also exists, right? Therefore every particular > real > number exists. But, the reals are uncountable and the set of possible > finite descriptions (over a finite alphabet) is countable, so most > particular reals do not have any finite description -- there are not > enough > finite descriptions to go around. Let's start here: Can't I put a point on a number line? Conceptually that point can go anywhere, can't it? Which reals can't be represented by that number line and point? Is that a finite description? Or how about this: One of the reals. is a sentence that applies to each and every real number. What part of that sentence isn't finite? What about this: Well, in set theory, if a set exists then each of its particular elements can be selected, right? Therefore we must be able to describe it. Hmmm.... Aren't all objects that are in existence in this finite world forced into the world of finiteness? OTOH - Objects that don't really exist might not have a finite description. === Subject: Re: logic of the Cantorian followers mind > But no > particular uncomputable real number can be finitely described. >Doesn't Chaitin describe a uncomputable number? > That's a good point. My initial reasoning was overly complex to begin > with. > Here's a simpler version that handles Chaitin's omega too: > Consider the set of real numbers. That set is an acceptable mainstream > object, and it can be finitely described. So it presumably exists in > the > sense you're intending. Well, in set theory, if a set exists then each > of > its particular elements also exists, right? Therefore every particular > real > number exists. But, the reals are uncountable and the set of possible > finite descriptions (over a finite alphabet) is countable, so most > particular reals do not have any finite description -- there are not > enough > finite descriptions to go around. > Let's start here: > Can't I put a point on a number line? Conceptually that point can go > anywhere, can't it? Which reals can't be represented by that number > line and point? Is that a finite description? > Or how about this: > One of the reals. is a sentence that applies to each and every real > number. What part of that sentence isn't finite? > What about this: > Well, in set theory, if a set exists then each of its particular elements > can be selected, right? Therefore we must be able to describe it. Yes, as a set. A real (between 0 an 1) is represented by a function from N, the natural numbers, to the set {0,1}. We interpret that as the binary expansion of a real. A function is a particular type of set. It is not necessary in the definition of a function for the function to be able to be DESCRIBE (which I interpret as, say, generated by a finite algorithm). Take the real number .0101010101010101010101001111011110011101010100111... If you ask me, what is the 47-th place, I might say, 1. And if you ask me the 348784845784-th place, I say 0. For any n, I will give you back a 0 or 1. And if you give the the same n repeatedly, each time you give me a particular n, I'll always give you That is what makes f a FUNCTION. It is NOT REQUIRED (sorry for the shouting, I'm getting agitated) to be able to describe f with an algorithm or a process. It's only necessary that every time you ask what's the 47-th digit, I give you the same answer. === Subject: Re: logic of the Cantorian followers mind <6yPod.89276$T02.37355@twister.rdc-kc.rr.com> <1pQod.89283$T02.68184@twister.rdc-kc.rr.com> >particular elements also exists, right? Therefore every particular real >number exists. But, the reals are uncountable and the set of possible >finite descriptions (over a finite alphabet) is countable, so most >particular reals do not have any finite description -- there are not enough >finite descriptions to go around. marvellous! worthy of men in suits to go doorknocking with. reminds me of the morman bible page I read where a mystical spirit told the mormans to go and tell other people, and to tell them to tell more people! wow, an unlistable infinite universe of undescribable numbers, all because you can flip the digit sequence down a list and extrapolate it (wrongly) to infinity. did you ever notice the bable in undescribable? Herc === Subject: Re: logic of the Cantorian followers mind >> The > mainstream accepts the existence of objects that can't be finitely > described. >>Exactly what objects are those? > Well, consider the set of uncomputable real numbers. Now that set is an > acceptable mainstream object, and it can be finitely described. So it > presumably exists in the sense you're intending. Well, in set theory, > if > a set exists then each of its individual elements also exists, right? > Therefore particular uncomputable real numbers exist, right? But no > particular uncomputable real number can be finitely described. > -- > --------------------------- > | BBB b Barbara at LivingHistory stop co stop uk > | B B aa rrr b | > | BBB a a r bbb | Quidquid latine dictum sit, > | B B a a r b b | altum viditur. > | BBB aa a r bbb | > ----------------------------- Not a bad description of objects you can't finitely describe. === Subject: Re: logic of the Cantorian followers mind : :> The :> mainstream accepts the existence of objects that can't be finitely :> described. : Exactly what objects are those? Most of the real numbers. Most of the sets of integers. Most of the languages over any alphabet. Stephen === Subject: Re: logic of the Cantorian followers mind > : :> The > :> mainstream accepts the existence of objects that can't be finitely > :> described. > : Exactly what objects are those? > Most of the real numbers. Most of the sets of integers. > Most of the languages over any alphabet. > Stephen That's a finite description of those things. A poor one, but finite. === Subject: Re: logic of the Cantorian followers mind Originator: joshp@xoxy.net (joshp) >> :> The mainstream accepts the existence of objects that can't be finitely >> :> described. >> : Exactly what objects are those? >> Most of the real numbers. Most of the sets of integers. >> Most of the languages over any alphabet. >That's a finite description of those things. No, those are finite descriptions of the *class* of each of those things. But by Cantor's argument, those classes each contain elements that are not finitely describable. -- Josh Purinton === Subject: Re: logic of the Cantorian followers mind > :> The mainstream accepts the existence of objects that can't be > finitely > :> described. > : Exactly what objects are those? > Most of the real numbers. Most of the sets of integers. > Most of the languages over any alphabet. >>That's a finite description of those things. > No, those are finite descriptions of the *class* of each of those > things. But by Cantor's argument, those classes each contain elements > that are not finitely describable. > -- > Josh Purinton Which real numbers can't be finitely described? === Subject: Re: logic of the Cantorian followers mind Originator: joshp@xoxy.net (joshp) > Which real numbers can't be finitely described? Let Ident be a set of identifiers (finite strings over a countable alphabet), and let X be a class with uncountably many elements, such as the class of all reals. Let F be a function from Ident into X. Then there are uncountably many elements of X that are not in the range of F. -- Josh Purinton === Subject: Re: logic of the Cantorian followers mind >> Which real numbers can't be finitely described? > Let Ident be a set of identifiers (finite strings over a countable > alphabet), and let X be a class with uncountably many elements, such as > the class of all reals. Let F be a function from Ident into X. > Then there are uncountably many elements of X that are not in the > range of F. > -- > Josh Purinton Does Cantor's diagonal proof show this? What does identifiers have to do with Cantor's diagonal proof? === Subject: Re: logic of the Cantorian followers mind Originator: joshp@xoxy.net (joshp) >> Let Ident be a set of identifiers (finite strings over a countable >> alphabet), and let X be a class with uncountably many elements, such as >> the class of all reals. Let F be a function from Ident into X. >> Then there are uncountably many elements of X that are not in the >> range of F. > Does Cantor's diagonal proof show this? Given a function F from a countable set into an uncountable set X, diagonalization demonstrates the existence of an element of X that is not in the range of F. The absence of uncountably many elements of X from the range of F is an easy corollary. > What does identifiers have to do with Cantor's diagonal proof? Ident is a countable set. -- Josh Purinton === Subject: Re: logic of the Cantorian followers mind > Let Ident be a set of identifiers (finite strings over a countable > alphabet), and let X be a class with uncountably many elements, such as > the class of all reals. Let F be a function from Ident into X. > Then there are uncountably many elements of X that are not in the > range of F. >> Does Cantor's diagonal proof show this? > Given a function F from a countable set into an uncountable set X, > diagonalization demonstrates the existence of an element of X that is > not in the range of F. The absence of uncountably many elements of X > from the range of F is an easy corollary. >> What does identifiers have to do with Cantor's diagonal proof? > Ident is a countable set. Why countable? Why does the alphabet need to be countable? === Subject: Re: logic of the Cantorian followers mind > What does identifiers have to do with Cantor's diagonal proof? >> Ident is a countable set. > Why countable? Why does the alphabet need to be countable? Are you talking about the alphabet, or about the set of strings that can be formed from that alphabet? The English alphabet has only 26 letters, but there's a countable infinity of strings of letters. And no, using an infinite alphabet won't change anything, as long as it's a countable infinity. If you are going to talk about uncountable alphabets, then you may as well let each real number be a character in this alphabet. According to that scheme, each real number has a description that is one character long, but the character is no more describable than the number is. What have you gained? -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: logic of the Cantorian followers mind Originator: joshp@xoxy.net (joshp) > Why does the alphabet need to be countable? Are you familiar with the Church-Turing thesis? -- Josh Purinton === Subject: Re: logic of the Cantorian followers mind >> Why does the alphabet need to be countable? > Are you familiar with the Church-Turing thesis? > -- > Josh Purinton I've been told Cantor's proof has nothing to do with computability. I don't find it strange that I was able to lead you here though. === Subject: Re: logic of the Cantorian followers mind :> :> The mainstream accepts the existence of objects that can't be :> finitely :> :> described. :> : Exactly what objects are those? :> Most of the real numbers. Most of the sets of integers. :> Most of the languages over any alphabet. :>>That's a finite description of those things. :> No, those are finite descriptions of the *class* of each of those :> things. But by Cantor's argument, those classes each contain elements :> that are not finitely describable. :> -- :> Josh Purinton : Which real numbers can't be finitely described? Most of them. Stephen === Subject: Re: logic of the Cantorian followers mind <41Qod.89279$T02.75468@twister.rdc-kc.rr.com> Discussion, linux) > : Which real numbers can't be finitely described? > Most of them. Name one. -- Jesse F. Hughes C is for Cookie. That's good enough for me. Cookie Monster === Subject: Re: logic of the Cantorian followers mind >> : Which real numbers can't be finitely described? >> Most of them. >Name one. I don't see any smiley or other indication of irony there, so I assume you're considering that to be a reasonable request. But in fact it's not a reasonable request to be asking for a description of something that by definition lacks a description, n'est-ce pas? The answer most of them is thoroughly accurate: the ones that do have finite descriptions are a vanishingly small fraction of the total. BTW, this fact is used in Shanon's elegent information-theory proof that most codings are completely random (and hence perfect); it's only (and exactly) the ones that we can actually generate that aren't! -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: logic of the Cantorian followers mind <41Qod.89279$T02.75468@twister.rdc-kc.rr.com> <87act7fyz1.fsf@phiwumbda.org> Discussion, linux) > : Which real numbers can't be finitely described? > Most of them. >>Name one. > I don't see any smiley or other indication of irony there, so I > assume you're considering that to be a reasonable request. I don't do smileys. -- I don't want to wine and dine and date you once or twice. I want to hold you now. I just want to spend the night. You tell me a better plan. Baby, I'm not a patient man. -- Jimmy Lafave, the romantic troubadour. === Subject: Re: logic of the Cantorian followers mind Please translate the latin. Quidquid latine dictum sit, altum viditur. Bob Kolker === Subject: Re: logic of the Cantorian followers mind > Please translate the latin. > Quidquid latine dictum sit, altum viditur. Whatsoever is said in Latin, is seen as high(?). === Subject: Re: logic of the Cantorian followers mind >> Please translate the latin. >> Quidquid latine dictum sit, altum viditur. >Whatsoever is said in Latin, is seen as high(?). High, or deep (go figure). In this case, a good translation is profound. BTW, for the OP, a Google search gets lots of hits for this. -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: logic of the Cantorian followers mind :> : Which real numbers can't be finitely described? :> Most of them. : Name one. 'bob'. :) I am not sure that 'naming' and 'describing' are at all related in this case. I can name some indescribable number 'bob', but that does not help you describe or even identify that number, unless he is wearing his name tag. So I claim that 'bob' is a indescribable number. I do not claim that 'bob' is a description of the indescribable, just the name. Of course there are only countably infinite names, so there are unnameable real numbers as well. :) Stephen === Subject: Re: logic of the Cantorian followers mind <41Qod.89279$T02.75468@twister.rdc-kc.rr.com> <87act7fyz1.fsf@phiwumbda.org> Discussion, linux) > :> :> : Which real numbers can't be finitely described? > :> :> Most of them. > : Name one. > 'bob'. :) > I am not sure that 'naming' and 'describing' > are at all related in this case. Well, probably neither of us should be spending so much time on a throwaway line, but... I'd say that the plain English interpretation of name as in name one is: specify one. Giving a name for which I don't know the referent surely wouldn't count as satisfying my demand that you name one. -- Jesse F. Hughes That's what's annoying about Usenet as some loser will state a case, get their ass kicked, but STILL keep coming back as if nothing happened. -- James Harris explains his strategy. === Subject: Re: logic of the Cantorian followers mind >> : Which real numbers can't be finitely described? >> Most of them. >Name one. x. Lee Rudolph === Subject: Re: logic of the Cantorian followers mind <41Qod.89279$T02.75468@twister.rdc-kc.rr.com> <87act7fyz1.fsf@phiwumbda.org> Discussion, linux) > : Which real numbers can't be finitely described? > Most of them. >>Name one. > x. x has a finite description. It is the least real number greater than or equal to x. Nice try. -- Jesse F. Hughes I have written many words to sci.math, some of them are not even meaningless. --Ross Finlayson === Subject: Re: logic of the Cantorian followers mind > : Which real numbers can't be finitely described? > Most of them. >Name one. >> x. > x has a finite description. It is the least real number greater than > or equal to x. > Nice try. Berry nice try. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: logic of the Cantorian followers mind > :> :> The mainstream accepts the existence of objects that can't be > :> finitely > :> :> described. > :> : Exactly what objects are those? > :> Most of the real numbers. Most of the sets of integers. > :> Most of the languages over any alphabet. > :>>That's a finite description of those things. > :> :> No, those are finite descriptions of the *class* of each of those > :> things. But by Cantor's argument, those classes each contain elements > :> that are not finitely describable. > :> -- > :> Josh Purinton > : Which real numbers can't be finitely described? > Most of them. > Stephen Most of the real numbers might include 3/4 and pi. Are you wrong, incabable of answering, or what? === Subject: Re: logic of the Cantorian followers mind :> :> :> The mainstream accepts the existence of objects that can't be :> :> finitely :> :> :> described. :> :> : Exactly what objects are those? :> :> Most of the real numbers. Most of the sets of integers. :> :> Most of the languages over any alphabet. :> :>>That's a finite description of those things. :> :> :> :> No, those are finite descriptions of the *class* of each of those :> :> things. But by Cantor's argument, those classes each contain elements :> :> that are not finitely describable. :> :> -- :> :> Josh Purinton :> : Which real numbers can't be finitely described? :> Most of them. :> Stephen : Most of the real numbers might include 3/4 and pi. What is that suppose to mean? 3/4 and pi are describable, so they are clearly not in the set of indescribable numbers. There is no 'might' about it. : Are you : wrong, incabable of answering, or what? Most real numbers are indescribable. It is a simple consequence of the definitions. The set of descriptions is countably infinite. The set of real numbers is not countably infinite. Are you capable of understanding that? There are more real numbers than descriptions. There are so many more real numbers than descriptions that by any reasonable defintion of 'most', most real numbers are not describable. Stephen === Subject: Re: logic of the Cantorian followers mind Most real numbers are indescribable. It is a simple > consequence of the definitions. The set of descriptions > is countably infinite. The set of real numbers is > not countably infinite. > Are you capable of understanding that? There are more > real numbers than descriptions. There are so many more > real numbers than descriptions that by any reasonable > defintion of 'most', most real numbers are not describable. > Stephen Which numbers can't be described by a point on a number line? === Subject: Re: logic of the Cantorian followers mind : :> Most real numbers are indescribable. It is a simple :> consequence of the definitions. The set of descriptions :> is countably infinite. The set of real numbers is :> not countably infinite. :> Are you capable of understanding that? There are more :> real numbers than descriptions. There are so many more :> real numbers than descriptions that by any reasonable :> defintion of 'most', most real numbers are not describable. :> Stephen : Which numbers can't be described by a point on a number line? How do you describe pi, or any number for that matter, using a number line? Stephen === Subject: Re: logic of the Cantorian followers mind > : :> Most real numbers are indescribable. It is a simple > :> consequence of the definitions. The set of descriptions > :> is countably infinite. The set of real numbers is > :> not countably infinite. > :> :> Are you capable of understanding that? There are more > :> real numbers than descriptions. There are so many more > :> real numbers than descriptions that by any reasonable > :> defintion of 'most', most real numbers are not describable. > :> :> Stephen > : Which numbers can't be described by a point on a number line? > How do you describe pi, or any number for that matter, using a number > line? > Stephen The same way most people do. === Subject: Re: logic of the Cantorian followers mind :> : :> :> Most real numbers are indescribable. It is a simple :> :> consequence of the definitions. The set of descriptions :> :> is countably infinite. The set of real numbers is :> :> not countably infinite. :> :> :> :> Are you capable of understanding that? There are more :> :> real numbers than descriptions. There are so many more :> :> real numbers than descriptions that by any reasonable :> :> defintion of 'most', most real numbers are not describable. :> :> :> :> Stephen :> : Which numbers can't be described by a point on a number line? :> How do you describe pi, or any number for that matter, using a number :> line? :> Stephen : The same way most people do. I am quite sure most people have never tried to describe pi using a number line. So I guess that means that you do not know how to describe pi using a number line. Stephen === Subject: Re: logic of the Cantorian followers mind > :> :> : :> :> Most real numbers are indescribable. It is a simple > :> :> consequence of the definitions. The set of descriptions > :> :> is countably infinite. The set of real numbers is > :> :> not countably infinite. > :> :> :> :> Are you capable of understanding that? There are more > :> :> real numbers than descriptions. There are so many more > :> :> real numbers than descriptions that by any reasonable > :> :> defintion of 'most', most real numbers are not describable. > :> :> :> :> Stephen > :> :> : Which numbers can't be described by a point on a number line? > :> :> How do you describe pi, or any number for that matter, using a number > :> line? > :> :> Stephen > : The same way most people do. > I am quite sure most people have never tried to describe pi > using a number line. So I guess that means that you do not > know how to describe pi using a number line. > Stephen I guess that means you don't know how to do it. Otherwise you wouldn't need for me to tell you these things. === Subject: Re: logic of the Cantorian followers mind In sci.logic, Poker Joker <7FTod.89499$T02.4671@twister.rdc-kc.rr.com>: >> :>> :> : > :> :> Most real numbers are indescribable. It is a simple >> :> :> consequence of the definitions. The set of descriptions >> :> :> is countably infinite. The set of real numbers is >> :> :> not countably infinite. >> :> :>> :> :> Are you capable of understanding that? There are more >> :> :> real numbers than descriptions. There are so many more >> :> :> real numbers than descriptions that by any reasonable >> :> :> defintion of 'most', most real numbers are not describable. >> :> :>> :> :> Stephen >> :>> :> : Which numbers can't be described by a point on a number line? >> :>> :> How do you describe pi, or any number for that matter, using a number >> :> line? >> :>> :> Stephen >> : The same way most people do. >> I am quite sure most people have never tried to describe pi >> using a number line. So I guess that means that you do not >> know how to describe pi using a number line. >> Stephen > I guess that means you don't know how to do it. Otherwise > you wouldn't need for me to tell you these things. I'm not sure one can describe an arbitrary point on the line anyway, unless it happens to be in one of the following sets. [1] Q. [2] A known transcendental, such as e, pi, log(2). [3] The solution of an equation, usually (but not always!) polynomial in nature. [4] Any arithmetic combination of the above. [5] The limit point of certain sequences or series, where the terms can be computed easily (e.g., sum(i=1,+oo) (1/i!) yields e; sum(i=1,+oo) (1/i^2) yields pi^2/6, IIRC). Note that an infinite decimal expansion fits into this category; the number .012345678910111213141516171819... is one example of many. This territory becomes fairly murky quickly, as you might well imagine. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Elementary probability of the infinite for the overworked mathematicians You are one person tossing a coin. I have infinite people, for every time you toss a coin, my infinite people all toss a coin aswell. A subset of my people have all got the same sequence as you, I never run out of people (infinite), you can toss your coin as long as you want, you NEVER form a new sequence to what an infinite set of people can make. Whatever sequence you form must be a *possible outcome* of tossing coins, infinite trials at the possible always succeed. With finite people, the length of the sequence where every combination is covered is finite, about log(#P). With infinite people, the length of the sequence where every combination is covered is without bound, INFINITE. Infinite length sequences are all covered, the diagonal is a 'cheat' and in standard theory it does not provide any new information to the dataset. Herc === Subject: Re: Confused about DFT and Fourier Series and Fourier Transform? Check http://www.ee.vt.edu/~ee4624ss/week4.pdf > I am confused by the four transforms in Signal & Systems... > The Continuous Time Fourier Transform(CTFT) is most understandable; DFT and > Fourier Series alone are individually recoginizable and understandable... > Not sure about how does DTFT kick in... > Anyway, remembering all of these four transforms' formulas are already very > headache... very easily got confuse one with another... > Even worse, homework and test problems often asks for conversion among these > four transforms... > Given a signal's CTFT, how do you get DFT for N-point? How does the DFT > compare to the Fourier Series(looks to me they are all discrete spectrum, > etc.) so on and so forth, how are they related and how to get one from > another? > Are there any good resources that clearly demonstrate the relationship and > conversion among these 4 transforms? === Subject: Two Interesting Second Order Linear Recurrence Generators Two interesting second order linear recurrence sequences are { 2, 2, 0, 1 } and { 2, 4, 0, 7 }. Can you tell why? They come from FLT, that is why. Each sequence is made of the values of the dual exponential generator most strongly associated with a^n + b^n = c^n. It is: (a^n + b^n) mod c. There are of course, two other generators. The specific values are a=3; b=4; c=5 and a=5; b=12; c=13. Can you solve for a.0 and a.1 in the following equations? x.2 = a.0 * x.0 + a.1 * x.1 (mod c) x.i = a.0 * x.(i-2) + a.1 * x.(i-1) (mod c) where {x} is one of the set of values above and c is the associated 5 or 13? For such generators a.0 and a.1 are, as you might expect, mod c, that is, they can have any value from 0 to c-1. I have tried this and found multiple solutions. Should there be none, one, or more than one for each Pythagorean triple (a,b,c)? Note there are two terms, these are second order generators, two is prime and a Galois field of order two exists. Also, two and the trivial one are the only possible values for n in FLT. And I have two feet. :) Coincidence? Perhaps... You decide. Can you solve this? Basically it's 2,2 * A = 0 2,0 * A = 1 0,1 * A = 2 1,2 * A = 2 and you solve for A, a 2x2 matrix of configuration: 0 a.0 1 a.1 At some point we start leaving out the mod c. I tolerance everything and tolerate everyone. I love: Dona, Jeff, Kim, Kimmie, Mom, Neelix, Tasha, and Teri, alphabetically. I drive: A double-step Thunderbolt with 657% range. I fight terrorism by: Using less gasoline. === Subject: Re: Two Interesting Second Order Linear Recurrence Generators That is, 0 == 2 * a.0 + 2 * a.1 (mod 5) 1 == 2 * a.0 + 0 * a.1 (mod 5) and with that zero in the second equation we have 2 * a.0 mod c == 1 Let's try a.0 = 0. Nope. 1. Nope. 2. Nope. 3. Yep. Any more? 4. Nope. 5. Nope. 6. Nope. a.0 = 3 2*3 + 2*a.1 == 0 mod 5 2 looks good. Any more? 0. Nope. 1. Nope. 2. Well, we have 2. 3. Nope. 4. Nope. 5. Nope. 6. Nope. a.1 = 2; a.0 = 3. >{ 2, 2, 0, 1 } 2 == 3*0 + 2*1 mod 5. 2 == 3*1 + 2*2 mod 5. It works! I've shown that for *a* dual exponential congurential sequence associated with *a* solution to FLT there is *exactly one* linear recurrence generator and associated matrix. And there are mentions of Galois fields in the literature on these sequences. Can you show that for *any* dual exponential congurential generator or sequence there is an associated unique linear recurrence generator? I tolerance everything and tolerate everyone. I love: Dona, Jeff, Kim, Kimmie, Mom, Neelix, Tasha, and Teri, alphabetically. I drive: A double-step Thunderbolt with 657% range. I fight terrorism by: Using less gasoline. === Subject: Re: Two Interesting Second Order Linear Recurrence Generators >a.1 = 2; a.0 = 3. Mathcad verifies this is the right solution, and shows that for the dual subtractive generator, there is more than one equivalent recurrence generator with identical sequence. I tolerance everything and tolerate everyone. I love: Dona, Jeff, Kim, Kimmie, Mom, Neelix, Tasha, and Teri, alphabetically. I drive: A double-step Thunderbolt with 657% range. I fight terrorism by: Using less gasoline. === Subject: Re: Computer language and category theory > Is there any work done one computer languages and category theory? [...] > However, I don't like such animals. So I wonder if there is some use > of category theory or something else that have been used to model > language like constructs. Any formalism for transition systems generalizes to something involving categories. Suppose G = (Q,s,M,P) is a grammar; i.e., s is in Q, Q is a set of variables, M is a monoid (the standard formalism admits only free monoids M = X*, where X is then deemed the 'alphabet', but everything works in the general case), and P is a subset of Q x M[Q], where M[Q] is the free extension of M over the set Q (note then that (X*)[Q] is just (X union Q)*). A transition relation -> can be defined over M[Q] recursively by the following conditions: (a) a -> a, for any a in M[Q] (b) if a -> b, b -> c, then a -> c, where a, b, c are in M[Q] (c) if (q,b) is in P then aqc -> abc, where a, c are in M[Q] This has the following properties: * Let [a] = { m in M: a -> m } then [m] = {m} for all m in M Context freeness: [ab] = [a][b] for all a, b in M[Q] [q] = union {[a]: (q,a) in P} * if a -> b, c -> d then ac -> bd * the set { q = [q]: q in Q } is the least solution to the set-theoretic system defined from the grammar (i.e., where a rule q -> xry becomes the inequality [q] superseet of {x}[r]{y} for q,r in Q, x,y in M similarly for the other rules). The morphisms are defined recursively by: I: a -> a if f: a -> b, g: b -> c then gf: a -> c (q,b): aqc -> abc Alternatively, one can restrict the last family of morphisms only to the following: (q,b): q -> b and then add another family of the form if f: a -> b, g: c -> d then f x g: ac -> bd given the property just cited above. When the grammar is not cyclic then s is an initial object. All the elements m of M are terminal objects. The language L(G) is just [s] which is the set of elements m of M for which morphisms f: s -> m exist. Each morphism corresponds roughly to a derivation sequence. === Subject: General Harmonic analysis question I know there doesn't exist a general theory for harmonic analysis on non-locally compact groups. What useful applications would a theory for harmonic analysis on non-locally compact groups have? Are there real world practical applications? In particular, does anyone need to study continuous functions from non-compact groups to the real line or anything similar? In order to create a theory for non-locally compact groups, would there have to be a Haar integral analog, a Peter-Weyl analog, and Plancherel theorem analog? Isaac === Subject: Re: help! solving matrix equations ... >realistic problems, I forgot to limit that y cannot be 0. and in fact, I >want ||w||=1... the norm of w = 1. >Sorry I forgot this practicality constraint... >Under this new added condition, what can be a good solution? Ah, that changes things considerably. So now you want to minimize || P a - C w || subject to ||w|| = 1. Using a Lagrange multiplier, we can write the Lagrangian as F(a,w,lambda) = (P a - C w)^T (P a - C w) - lambda (w^T w - 1) We look for a stationary point, where the gradients with respect to a and w are both 0. I get C^T P a = (C^T C - lambda) w P^T P a = - P^T C w Assuming P^T P is invertible, you want a = - (P^T P)^(-1) P^T C w, and then C^T (I + P (P^T P)^(-1) P^T) C w = lambda w i.e. w should be an eigenvector of C^T (I + P (P^T P)^(-1) P^T) C for eigenvalue lambda. Moreover, I then get ||P a - C w||^2 = lambda ||w||^2 so you want to take a normalized eigenvector for the lowest eigenvalue. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: help! solving matrix equations ... >realistic problems, I forgot to limit that y cannot be 0. and in fact, I >want ||w||=1... the norm of w = 1. >Sorry I forgot this practicality constraint... >Under this new added condition, what can be a good solution? > Ah, that changes things considerably. So now you want to > minimize || P a - C w || subject to ||w|| = 1. Using a Lagrange > multiplier, we can write the Lagrangian as > F(a,w,lambda) = (P a - C w)^T (P a - C w) - lambda (w^T w - 1) > We look for a stationary point, where the gradients with respect to > a and w are both 0. I get > C^T P a = (C^T C - lambda) w > P^T P a = - P^T C w > Assuming P^T P is invertible, you want a = - (P^T P)^(-1) P^T C w, and > then > C^T (I + P (P^T P)^(-1) P^T) C w = lambda w > i.e. w should be an eigenvector of C^T (I + P (P^T P)^(-1) P^T) C > for eigenvalue lambda. Moreover, I then get > ||P a - C w||^2 = lambda ||w||^2 > so you want to take a normalized eigenvector for the lowest > eigenvalue. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada Just in case your silence is a result of you being completely overwhelmed, I'd be happy to contribute an informa discussion of the absolute basics, starting with Robert Israel's original suggestion to solve the equation [P|C]*x=nullvector Let me know. Reinhard === Subject: Re: Why Do Americans Call It Math? > I used to say Maths, as a Brit, I still do sometimes in the UK, but I gave > putting the s on the end makes it awkward to lisp out. Math avoids the > horrible triple consonant 'ths' > That's a double consonant, and occurs in lots of plurals. It's easy enough. Indeed; one can only assume that he doesn't take baths. And I bet he doesn't have any diphthongs in his catchphrase! Phil -- ... one Marine noticed one of the prisoners was still breathing. A Marine can be heard saying on the pool footage provided to Reuters Television: He's ing faking he's dead. He faking he's ing dead. The Marine then raises his rifle and fires into the man's head. The pictures are too graphic for us to broadcast, Sites said. === Subject: Re: Why Do Americans Call It Math? > > I used to say Maths, as a Brit, I still do sometimes in the UK, but I gave > putting the s on the end makes it awkward to lisp out. Math avoids the > horrible triple consonant 'ths' > > That's a double consonant, and occurs in lots of plurals. It's easy enough. > Indeed; one can only assume that he doesn't take baths. > And I bet he doesn't have any diphthongs in his catchphrase! But he does play to his strengths. Paul -- Hanging on in quiet desperation is the English way. The time is gone, the song is over. Thought I'd something more to say. === Subject: Re: Why Do Americans Call It Math? > > > I used to say Maths, as a Brit, I still do sometimes in the UK, but I gave > putting the s on the end makes it awkward to lisp out. Math avoids the > horrible triple consonant 'ths' > > That's a double consonant, and occurs in lots of plurals. It's easy enough. When I was in fourth grade, our teacher was saying that is was uncommon to have a word with three consecutive consonants. I instantly raised my hand to say I knew a word with five consecutive consonants: thousandths. She nicely informed that was only four, because th was a single sound. David Ames > > Indeed; one can only assume that he doesn't take baths. > > And I bet he doesn't have any diphthongs in his catchphrase! > But he does play to his strengths. > Paul === Subject: Re: Why Do Americans Call It Math? > horrible triple consonant 'ths' > > That's a double consonant, and occurs in lots of plurals. It's easy enough. >> Indeed; one can only assume that he doesn't take baths. >> And I bet he doesn't have any diphthongs in his catchphrase! >But he does play to his strengths. I guess someone will have to add a postscript about his hardscrabble offspring's jockstrap heading downstream to his birthplace, first lengthwise then in a corkscrew manner. I saw it in a truly earthshaking filmstrip. === Subject: Re: Why Do Americans Call It Math? > horrible triple consonant 'ths' > > That's a double consonant, and occurs in lots of plurals. It's easy enough. >> >> Indeed; one can only assume that he doesn't take baths. >> >> And I bet he doesn't have any diphthongs in his catchphrase! >But he does play to his strengths. > I guess someone will have to add a postscript about his hardscrabble > offspring's jockstrap heading downstream to his birthplace, first > lengthwise then in a corkscrew manner. I saw it in a truly > earthshaking filmstrip. How abstract. I assume it's set in his birthplace Christchurch, and starring Frenchmen in their nightclothes, I assume? Such heartthrobs tug at my heartstrings. === Subject: Re: Why Do Americans Call It Math? >> And I bet he doesn't have any diphthongs in his catchphrase! >But he does play to his strengths. > I guess someone will have to add a postscript about his hardscrabble > offspring's jockstrap heading downstream to his birthplace, first > lengthwise then in a corkscrew manner. I saw it in a truly > earthshaking filmstrip. > How abstract. I assume it's set in his birthplace Christchurch, and > starring Frenchmen in their nightclothes, I assume? Such heartthrobs > tug at my heartstrings. I wish this thread was in danish. I would have loved to use the word angstskrig. Asger. === Subject: Re: Why Do Americans Call It Math? >> And I bet he doesn't have any diphthongs in his catchphrase! >But he does play to his strengths. > I guess someone will have to add a postscript about his hardscrabble > offspring's jockstrap heading downstream to his birthplace, first > lengthwise then in a corkscrew manner. I saw it in a truly > earthshaking filmstrip. > How abstract. I assume it's set in his birthplace Christchurch, and > starring Frenchmen in their nightclothes, I assume? Such heartthrobs > tug at my heartstrings. > I wish this thread was in danish. I would have loved to use > the word angstskrig. Use it often enough in English contexts and there's a chance it may be accepted in the language. English is notorious for taking words from other languages and, usually, mispronouncing them. Recent examples loaded with consonant clusters include ersatz, perestroika and smorgasbord. Note that we don't hold with those funny mutilations to the vowels which foreigners seem to use. 8-) Paul -- Hanging on in quiet desperation is the English way. The time is gone, the song is over. Thought I'd something more to say. === Subject: Re: Why Do Americans Call It Math? spose I better answer really... some respondent was right, I find 'baths' hard to say. At least Thistle has an 'i' seprating the h and the s. Since this is a Math newsgroup, I'll end it now. Have a 1004 digit prime twin for bedtime, 1^1004+189770491,93. Systematic search of first 1600 due any time real soon now (as Bill G. used to say) gould luck and gould day what's a dipthong...? don't bother Richard Miller >> And I bet he doesn't have any diphthongs in his catchphrase! >But he does play to his strengths. > I guess someone will have to add a postscript about his hardscrabble > offspring's jockstrap heading downstream to his birthplace, first > lengthwise then in a corkscrew manner. I saw it in a truly > earthshaking filmstrip. > How abstract. I assume it's set in his birthplace Christchurch, and > starring Frenchmen in their nightclothes, I assume? Such heartthrobs > tug at my heartstrings. > I wish this thread was in danish. I would have loved to use > the word angstskrig. > Asger. === Subject: Re: Principle = Principal Curvatures of a Hypersurface In R3 the prototypical situation is that of the surface z = Axx/2 + Bxy + Cyy/2 at O = (x, y) = (0, 0). According to whether BB - AC > 0, =0 or < 0 it is a hyperbolic paraboloid, a parabolic cylinder or an elliptic paraboloid. The principal curvatures at O are the eigenvalues of the symmetric 2x2 matrix belonging in the standard way to this quadratic form. A transformation to principal axes yields z = L.x'x'/2 + M.y'y'/2. The coefficients L and M are the pricipal curvatures. In Rn one can play exactly the same game: Establish a Cartesian coordinate system P-X1-X2-...-Xn at point P of the hypersurface S with axis Xn along the normal to S and axes X1, X2, ..., X(n-1) in the tangent hyperplane to S at P. In specific cases this change of coordinate system may be easy or difficult. No general theory exists to tell this in advance. If S is indeed C2-smooth at P then S is representable in these local coordinates by a function xn = f(x1, x2,...,x(n-1)). Its Taylor series expansion starts with a quadratic form. The coefficients are the pure and mixed 2nd-order partial derivatives of f. The eigenvalues of the symmetric matrix belonging to this form are the principal curvatures. A transformation to principal axes yields xn = L1.x1'.x1' / 2 + ... + Ln-1.x(n-1)'.x(n-1)' / 2 + o(x^2). The answers to your questions: (1) no. (2) no in general; yes only after transformation to principal axes. (3) For n > 4: no, even if the tangent hyperplane is not parallel to the original Xn axis; one has to solve an n-th-degree polynomial equation to find the eigenvalues. Johan E. Mebius >Here's one that has me stumped ... >I'm trying to find the n-1 principle curvatures of a hypersurface >f(x_1,x_2...x_n) in an n-dimensional Hilbert space. >How do I go about doing this? All the texts I have referred to only >consider surfaces in R3 with simple (not to mention obvious) >parametrizations. >Note that in my case, I dont have a handy parametrization of the form >x_j=x_j(p_1,p_2...p_n-1) , j=1,n (1) >where p_1,p_2 ... p_n-1 are the n-1 parameters characterizing the >n-dimensional hypersurface. >My questions boil down to >(1) Is there a general method of determining a parametrization of the >form (1) for any given hypersurface? >(2) Are the principle curvatures then merely the vector norms of the >diagonal elements of the rank n-1 tensor of 2nd derivatives of the >position vectors defined by (1)? >(3) Can the principle curvatures be determined without recourse to >such a parametrization? >Lets assume that the hyperfurface f has continuous 1st and 2nd partial >derivatives w.r.t each of x_1,x_2 ...x_n. >-Sharat === Subject: Re: Principle = Principal Curvatures of a Hypersurface >(1) Is there a general method of determining a parametrization of the >form (1) for any given hypersurface? I have no proper answers, trying to extrapolate from 3D, I stand to correction for any inaccuracy or error. No, even in 3D, parameterization exploits symmetry for simplification in particular cases of surfaces of revolution, helical surfaces, torses etc. Geodesics in hyperspace upto an arbitrary parameter of obliqueness using geodesic polar coordinates could be considered... a lot of this is from GR. >(2) Are the principle curvatures then merely the vector norms of the >diagonal elements of the rank n-1 tensor of 2nd derivatives of the >position vectors defined by (1)? >(3) Can the principle curvatures be determined without recourse to >such a parametrization? In 3D, the product of principal curavatures comes out purely from the metric related to Tensor product R1212/g. Gauss Egregium theorem states that it is an isometric mapping invariant, a pure class product of first fundamental form metric coefficients, second fundamental form coefficients L,M, and N get eliminated in final result only of highest order _products_of curvatures, but not individual curvatures. This is also true in n-dimensional Riemannian geometry. Flatlanders need not know a priori how their land is bent or twisted in the embedded surrounding hyperspace, knowing the corresponding multiple curvature product hyper-invariant. === Subject: shapes and circle help Hello Are there any good methods to solve the 2 excercises below relating to shapes and circles The wheel of a wheelbarrow rotates 60 times when it is pushed a distance of 50 m calculate the radius of the wheel and the small circle has an areaof 16piecm2 the larger circle has a circumference of 18 pie cm calculate the SHADED area give your answer in terms of pie for this you have to imagine a small circle in a big circle. === Subject: Re: shapes and circle help > Are there any good methods to solve the 2 excercises below relating > to shapes and circles? Yes. Read a few more examples. After learning the formulas for perimeter and area of circles, watch how they are being usefully applied in those examples. === Subject: Root finder XI Root Finder xi. by Jon Giffen. It is found that the roots to the polynomial, a[0]+a[1]t+a[2]t^2+...+a[n]t^n where T=(t,t^2,t^3,..,t^n) and N=(a[1],a[2],a[3],...,a[n]) are given by, (D*N)|S|^2 D + (S*N)|D|^2 S T = (-a[0]){---------------------------} (D*N)^2|S|^2 + (S*N)^2|D|^2 where D=(1,2,3,...,n) S=(a[1],2a[2],3a[3],...,na[n]) Solve the 6th degree polynomial, t^6 + t - 10=0 a[0] = -10 N=(1,0,0,0,0,1) D=(1,2,3,4,5,6) S=(1,0,0,0,0,6) D*N=7 S*N=7 |D|^2 = 91 |S|^2 = 37 37(1,2,3,4,5,6)+91(1,0,0,0,0,6) T = (10)------------------------------- 7(37+91) (1280,740,1110,1480,1850,7680) = ------------------------------ = (t,t^2,t^3,t^4,t^5,t^6) 896 since t and t^6 need only be considered, t = 1280/896 = 10/7 t^6 = 7680/896 = 60/7 60/7 + (60/7)^(1/6) - 10 = 0.002 which is almost zero. (10/7)^6 + 10/7 - 10 = -0.07 which is also close Solve the 6th degree polynomial, t^6 - t - 10=0 t = -(60/7)^(1/6) from the prior example f(t) = t^7 + t^3 + t - 20 = 0 a[0] = -20 N=(1,0,1,0,0,0,1) D=(1,2,3,4,5,6,7) S=(1,0,3,0,0,0,7) D*N = 11 S*N = 11 |S|^2 = 59 |D|^2 = 138 59(1,2,3,4,5,6,7)+138(1,0,3,0,0,0,7) T = (20)------------------------------------ 11(59+138) t^7 = 27580/2167 t=1.4382 f(1.4382) = -2.8597 applying this to Newton's Method, 1.4382 - (-2.857)/[7(1.4382^6)+3(1.4382^2) + 1] = 1.4795 f(1.4795)=0.2361 1.4795 - (0.2361)/[7(1.4795^6)+3(1.4795^2) + 1] = 1.4766 f(1.4766)=0.000092 ~ 0 f(t) = t^7 + t^3 + t + 20 = 0 t = -1.4766 from last example f(t) = 2t^4 + 3t^3 + 2t^2 + t - 13 = 0 T*(1,2,3,2)-13 = 0 a[0]= -13 N=(1,2,3,2) D=(1,2,3,4) S=(1,4,9,8) |D|^2=30 |S|^2 = 162 D*N=22 S*N=52 13[22(162)(1,2,3,4)+52(30)(1,4,9,8)] T = -------------------------------------- 162(22^2) + 30(52^2) t^4 = 347568/159520 = 2.1787 t = 1.2149 f(1.2149) = 0.9038 applying Newton's Method, 1.2149 - 0.9038/[8(1.2149^3)+9(1.2149^2)+4(1.2149)+1] = 1.1879 f(1.1879) = 0.0213 ~ 0 f(t)=t^6 - t^5 + 4t^4 - 5t^3 + t^2 - t - 100 = 0 (-1,1,-5,4,-1,1)*T - 100 = 0 dividing the negatives from the positives, (0,1,0,4,0,1)*T - (1,0,5,0,1,0)*T = 100 (0,1,0,4,0,1)*T - 100p = (1,0,5,0,1,0)*T + 100(1-p) = 0 p is some ratio Applying the formula to each partition of N, (D*N)|S|^2 D + (S*N)|D|^2 S T = (-a[0]){---------------------------} (D*N)^2|S|^2 + (S*N)^2|D|^2 where D=(1,2,3,...,n) S=(a[1],2a[2],3a[3],...,na[n]) for (0,1,0,4,0,1)*T - 100p , a[0]=-100p D*N=2+16+6=24 |S|^2=4+16^2+36=296 S*N=2+16+6=24 |D|^2=1+4+9+16+25+36=91 for (1,0,5,0,1,0)*T + 100(1-p) , a[0]=100(1-p) D*N=1+15+5=21 |S|^2=1+15^2+25=251 S*N=1+15+5=21 |D|^2=91 296(1,2,3,4,5,6)+91(0,2,0,16,0,6) (100p)---------------------------------- 24(296+91) 251(1,2,3,4,5,6)+91(1,0,15,0,5,0) =100(p-1)--------------------------------- 21(251+91) Taking the magnitude of both sides |296(1,2,3,4,5,6)+91(0,2,0,16,0,6)| = |(296,774,888,2640,1480,2322)|=4003.36 |251(1,2,3,4,5,6)+91(1,0,15,0,5,0)| = |(342,502,2118,1004,1710,1506|=3324.91 4003.36 3324.91 p--------- = (p-1)------- 9288 7182 (0.9310)p = p-1 1 = (1-0.9310)p p = 14.5 then f(t)=t^6 - t^5 + 4t^4 - 5t^3 + t^2 - t - 100 = 0 t^6 = 1450/4 =362.5 t=2.67 f(2.67 )=239.38 t^5 = 858400/9288= 92.420 t=2.1263 f(2.1263)=-15.2200 Selecting -15.2200 for Newton's Method, f(t)=t^6 - t^5 + 4t^4 - 5t^3 + t^2 - t - 100 = 0 2.1263+15.22/ [6(2.1263^5)-5(2.1263^4)+16(2.1263^3)-15(2.1263^2)+2(2.1263)-1] t=2.1877 f(2.1877)= 1.3927 The development of this is given at, http://mypeoplepc.com/members/jon8338/polynomial/id7.html Jon Giffen === Subject: Root Finder 12 Root Finder 12 by Jon Giffen. This solution was so simple that I couldn't believe it. But I tried it, and it works. It is found that the roots to the polynomial, a[0]+a[1]t+a[2]t^2+...+a[n]t^n=0 where T=(t,t^2,t^3,..,t^n) and N=(a[1],a[2],a[3],...,a[n]) are given by, -a[0] T= -----D D*N D=(1,2,3,4,...,n) Solve the 6th degree polynomial, f(t)=t^6 + t - 10=0 a[0] = -10 N=(1,0,0,0,0,1) D=(1,2,3,4,5,6) a[0]=-10 D*N=7 -a[0] 10 T = -----D = ---(1,2,3,4,5,6) D*N 7 t =10/7 t=1.42835 f(1.42835)=-0.071 t^6=60/7 t=1.43056 f(1.43056)= 0.002 Solve the 49th degree polynomial, f(t)=t^49 + t^16 + 40t^5 - 6000 = 0 a[0]=-6000 N=(1,0,0,0,40,0,..,1,0,...,1) D=(1,2,3,..,49) D*N=1+200+16+49=266 -a[0] 6000 T = -----D = ------(1,2,3,..,49) D*N 266 t^49=294000/266=1105.26 t=1.15375 f(1.15375)=-3575.99 Applying Newton's Method, t=1.15375-(-3575)/[49(1.15375^48)+16(1.15375^15)+200(1.15375)^4] =1.15375 again, so the answer must depend on distant decimal places. mixed signs, no pattern f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t - 100 = 0 a[0]=-100 N=(-1,1,5,4,-1,1) D=(1,2,3,4,5,6) D*N=-1+2+15+16-5+6=33 -a[0] 100 T = -----D = ---(1,2,3,4,5,6) D*N 33 t^6=600/33=18.1818 t=1.62158 f(1.62158)=-43.0453 too low t^5=500/33=15.1515 t=1.72223 f(1.72223)=-27.0815 too low f^4=400/33=12.1212 t=1.86589 f(1.86589)= 2.16509 pretty close Applying Newton's Method, 1.86589- 2.16509 ------------------------------------------------------------------ 6(1.86589^5)-5(1.86589^4)+16(1.86589^3)+15(1.86589^2)+2(1.86589)-1 t=1.856637 f(1.856637)=0.0192 t^2 + 2t - 3 = 0 a[0]=-3 N=(2,1) D=(1,2) D*N=4 -a[0] 3 T = -----D =---(1,2) t^2=3/4 t=0.866 ~ 1 D*N 4 The lengthy development of this is given at, http://mypeoplepc.com/members/jon8338/polynomial/id7.html Jon Giffen === Subject: Root Finder 13 Root Finder 13 Jon Giffen Another approach is considered, along with a possibility for finding the root to an Infinite Series. It is discovered that the property of the nth degree polynomial, a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 where N=(a[1],a[2],a[3],...,a[n]) T=(t.t^2.t^3,t^4,..., t^n ) Is, |C|^2 |T|^2=(-a[0])^2{-------------------} |N|^2|C|^2-(N*C)^2 where C=(a[1],2a[2],3a[3],4a[4],...,na[n]) f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t - 100 = 0 a[0]=-100 N=(-1,1,5,4,-1,1) C=(-1,2,15,16,-5,6) |C|^2=547 |N|^2=48 N*C=153 |N|^2 |C|^2 - (N*C)^2 = 48(547)-153^2=2847 |C|^2 D T =(-a[0]){----------------------}^(1/2) --- |N|^2 |C|^2 - (N*C)^2 |D| where D=(1,2,3,4,5,6) and 100 547 T = -------- {-----}^(1/2) (1,2,3,4,5,6) = 4.594933(1,2,3,4,5,6) 91^(1/2) 2847 t^6=27.5696 t=1.738097 t^5=22.9746 t=1.871758 ---------- t=1.856637 is the correct root Notice that the root to a polynomial that is so long, that it is virtually an infinite Power Series; is found by using the solution to the Geometric Series, |C|^2 |T|^2=t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{------------------} |N|^2|C|^2-(N*C)^2 adding 1 to both sides, |C|^2 1+t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{------------------}+1 |N|^2|C|^2-(N*C)^2 then the sum S=1/(1-t^2) but |C|^2 S=(-a[0])^2{------------------}+1 (1-t^2)=1/S t^2=1-1/S and |N|^2|C|^2-(N*C)^2 t ={1 - 1/S}^(1/2) where N=(a[1],a[2],a[3],...,a[n]) T=(t,t^2,t^3,t^4,..., t^n ) C=(a[1],2a[2],3a[3],4a[4],...,na[n]) to the nth degree power series, a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 Development Suppose the polynomial, a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 Is expressed as, a[0]+a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n=0 Where c is almost 1 then dividing the two, a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n= -a[0] --------------------------------------------------- a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n = -a[0] Suppose K=(a[1]c,a[2]c^2,a[3]c^3,...a[n]c^n) then simply, T*(K-N)=0 so (K-N) is orthogonal to T. Consequently, N*(K-N) T is parallel to N - ------- K and |K-N|^2 N*(K-N) [m(N - -------- K) - Q]*N=0 solve this for m. Then |K-N|^2 N*(K-N) T= m(N - -------- K) |K-N|^2 Substitute m in the above and take the square of the magnitude of both sides. Then (K-N)*(K-N) 0 |T|^2=Lim (-a[0])^2 ---------------------------- = --- c->1 |N|^2(K-N)*(K-N)-[N*(K-N)]^2 0 applying L'Hopital two times with respct to c, |C|^2 |T|^2=(-a[0])^2{-------------------} |N|^2 |C|^2-(N*C)^2 where d C = Lim ---- K = =(a[1],2a[2],3a[3],4a[4],...,na[n]) c->1 dc E.O.P. Jon Giffen http://mypeoplepc.com/members/jon8338/polynomial/id7.html === Subject: Re: Root Finder 12 > Root Finder 12 > by Jon Giffen. > This solution was so simple that I couldn't believe it. > But I tried it, and it works. > It is found that the roots to the polynomial, > a[0]+a[1]t+a[2]t^2+...+a[n]t^n=0 where > T=(t,t^2,t^3,..,t^n) and > N=(a[1],a[2],a[3],...,a[n]) are given by, > -a[0] > T= -----D > D*N > D=(1,2,3,4,...,n) Ok then, attempt to construct the quadratic formula for your method: at^2 + bt + c = 0 T = (t, t^2) N = (b, a) D = (1, 2) T = -c/(b+2a) * (1,2) = (t, t^2) t = -c/(b+2a) t^2 = -2c/(b+2a) Clearly you are dead wrong, since the correct t's are t = (-b + sqrt[b^2-4ac])/(2a) and t = (-b - sqrt[b^2-4ac])/(2a) If you can not reconstruct the quadratic formula, then you are wrong. Any polynomials with a root approximated by your method are just coincidences. === Subject: Re: Root Finder 12 ex. t^2+2t-3=0 N=(2,1) |N|^2=5 Q=(3/5)(2,1) D=(1,2) |D|^2=5 Q*D=12/5 (mD-Q)*D=0 m|D|^2=Q*D m=(Q*D)/|D|^2 = 12/25 T=mD = (12/25)(1,2) t^2=24/25 ~ 1 >>Root Finder 12 >>by Jon Giffen. >>This solution was so simple that I couldn't believe it. >>But I tried it, and it works. >>It is found that the roots to the polynomial, >>a[0]+a[1]t+a[2]t^2+...+a[n]t^n=0 where >>T=(t,t^2,t^3,..,t^n) and >>N=(a[1],a[2],a[3],...,a[n]) are given by, >> -a[0] >>T= -----D >> D*N >>D=(1,2,3,4,...,n) > Ok then, attempt to construct the quadratic formula for your method: > at^2 + bt + c = 0 > T = (t, t^2) > N = (b, a) > D = (1, 2) > T = -c/(b+2a) * (1,2) = (t, t^2) > t = -c/(b+2a) > t^2 = -2c/(b+2a) > Clearly you are dead wrong, since the correct t's are > t = (-b + sqrt[b^2-4ac])/(2a) > and > t = (-b - sqrt[b^2-4ac])/(2a) > If you can not reconstruct the quadratic formula, then you are wrong. > Any polynomials with a root approximated by your method are just > coincidences. === Subject: Re: Root Finder 12 > Root Finder 12 > by Jon Giffen. > This solution was so simple that I couldn't believe it. > But I tried it, and it works. > [...] > Solve the 6th degree polynomial, > f(t)=t^6 + t - 10=0 > a[0] = -10 N=(1,0,0,0,0,1) D=(1,2,3,4,5,6) > a[0]=-10 D*N=7 > -a[0] 10 > T = -----D = ---(1,2,3,4,5,6) > D*N 7 > t =10/7 t=1.42835 f(1.42835)=-0.071 10/7 is not a root of t^6 + t - 10 = 0. The Rational Root Test says that any rational solutions to this polynomial are +/-1, +/-2, +/5, or +/10. It can only be an approximation. Strike Twelve. (The Rational Root Theorem -- a.k.a. the Rational Zero Theorem -- can be found at http://mathworld.wolfram.com/RationalZeroTheorem.html . > Solve the 49th degree polynomial, > f(t)=t^49 + t^16 + 40t^5 - 6000 = 0 > a[0]=-6000 N=(1,0,0,0,40,0,..,1,0,...,1) D=(1,2,3,..,49) > D*N=1+200+16+49=266 > -a[0] 6000 > T = -----D = ------(1,2,3,..,49) > D*N 266 > t^49=294000/266=1105.26 t=1.15375 f(1.15375)=-3575.99 Once again, the Rational Root Theorem says that the only possible rational roots are integers. > Applying Newton's Method, Ah, so. You aren't finding roots after all, only approximations to them. You've been told repeatedly that this isn't the same as finding the roots. > t=1.15375-(-3575)/[49(1.15375^48)+16(1.15375^15)+200(1.15375)^4] > =1.15375 again, so the answer must depend on distant decimal > places. The problem here is you don't have enough precision to make Newton's Method work. > [...] > t^2 + 2t - 3 = 0 > a[0]=-3 N=(2,1) D=(1,2) D*N=4 > -a[0] 3 > T = -----D =---(1,2) t^2=3/4 t=0.866 ~ 1 > D*N 4 What? Your method can't even solve a quadratic equation? That's when you know it's really bad. -- Christopher Heckman === Subject: Re: Root Finder 12 ex. t^2+2t-3=0 N=(2,1) |N|^2=5 Q=(3/5)(2,1) D=(1,2) |D|^2=5 Q*D=12/5 (mD-Q)*D=0 m|D|^2=Q*D m=(Q*D)/|D|^2 = 12/25 T=mD = (12/25)(1,2) t^2=24/25 ~ 1 ex. at^2+bt+c=0 N=(b,a) |N|^2=b^2+a^2 Q=(-c/[b^2+a^2])(b,a) D=(1,2) |D|^2=5 Q*D=(-c/[b^2+a^2])(b+2a) (mD-Q)*D=0 m=(Q*D)/|D|^2 = (1/5)(-c/[b^2+a^2])(b+2a) T=mD=(1/5)(-c/[b^2+a^2])(b+2a)(1,2) t^2 =(2/5)(-c/[b^2+a^2])(b+2a) b+/-{b^2-4ac}^(1/2) t ={(2/5)(-c/[b^2+a^2])(b+2a)}^(1/2)=-------------------- 2a solve and find the required correction >>Root Finder 12 >>by Jon Giffen. >>This solution was so simple that I couldn't believe it. >>But I tried it, and it works. >>[...] >>Solve the 6th degree polynomial, >>f(t)=t^6 + t - 10=0 >>a[0] = -10 N=(1,0,0,0,0,1) D=(1,2,3,4,5,6) >>a[0]=-10 D*N=7 >> -a[0] 10 >>T = -----D = ---(1,2,3,4,5,6) >> D*N 7 >>t =10/7 t=1.42835 f(1.42835)=-0.071 > 10/7 is not a root of t^6 + t - 10 = 0. The Rational Root Test says > that any rational solutions to this polynomial are +/-1, +/-2, +/5, > or +/10. It can only be an approximation. > Strike Twelve. > (The Rational Root Theorem -- a.k.a. the Rational Zero Theorem -- > can be found at http://mathworld.wolfram.com/RationalZeroTheorem.html . >>Solve the 49th degree polynomial, >>f(t)=t^49 + t^16 + 40t^5 - 6000 = 0 >>a[0]=-6000 N=(1,0,0,0,40,0,..,1,0,...,1) D=(1,2,3,..,49) >>D*N=1+200+16+49=266 >> -a[0] 6000 >>T = -----D = ------(1,2,3,..,49) >> D*N 266 >>t^49=294000/266=1105.26 t=1.15375 f(1.15375)=-3575.99 > Once again, the Rational Root Theorem says that the only possible rational > roots are integers. >>Applying Newton's Method, > Ah, so. You aren't finding roots after all, only approximations to them. > You've been told repeatedly that this isn't the same as finding the roots. >>t=1.15375-(-3575)/[49(1.15375^48)+16(1.15375^15)+200(1.15375)^4] >> =1.15375 again, so the answer must depend on distant decimal >>places. > The problem here is you don't have enough precision to make Newton's > Method work. >>[...] >>t^2 + 2t - 3 = 0 >>a[0]=-3 N=(2,1) D=(1,2) D*N=4 >> -a[0] 3 >>T = -----D =---(1,2) t^2=3/4 t=0.866 ~ 1 >> D*N 4 > What? Your method can't even solve a quadratic equation? That's when > you know it's really bad. > -- Christopher Heckman === Subject: Re: Root Finder 12 I thought you already posted the last word on this subject. Are you suffering from some kind of attention deficit disorder, or are you being deliberately misleading? If there's a third possibility, I'd welcome your explanation. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Root Finder 12 You people are savage, and there's no call for it. I suppose you are one of the anal perfectionist obsessed with keeping on course down to the Angstrom to offset the Gudermanian. Newton's Method is an invention of genius. Why not use it? Newton just didn't come up with the approximations to plug into it... or did he? > I thought you already posted the last word on this subject. Are you > suffering from some kind of attention deficit disorder, or are you > being deliberately misleading? If there's a third possibility, I'd > welcome your explanation. > -- > There are two things you must never attempt to prove: the unprovable > -- and the obvious. > -- > Democracy: The triumph of popularity over principle. > -- > http://www.crbond.com === Subject: Re: Root Finder 12 There's always room for improvement > I thought you already posted the last word on this subject. Are you > suffering from some kind of attention deficit disorder, or are you > being deliberately misleading? If there's a third possibility, I'd > welcome your explanation. > -- > There are two things you must never attempt to prove: the unprovable > -- and the obvious. > -- > Democracy: The triumph of popularity over principle. > -- > http://www.crbond.com === Subject: Re: Root finder XI > Root Finder xi. > by Jon Giffen. > It is found that the roots to the polynomial, > a[0]+a[1]t+a[2]t^2+...+a[n]t^n where > [...] > f(t) = t^7 + t^3 + t - 20 = 0 > [...] > applying this to Newton's Method, You've been told about this before. There are already good places to start off Newton's Method, so you haven't provided anything new. Strike Eleven. -- Christopher Heckman === Subject: Re: Root finder XI > Root Finder xi. You told us that ix was the final one. === Subject: Re: Root finder XI > > Root Finder xi. > > You told us that ix was the final one. Based on his posts, did you really think you could believe him? -- Christopher Heckman === Subject: 20th Century obsession with self defeating statements Why are mathematicians focussed on the impossible? What impossible entities have they proven? A program that examines itself, if it halts then continues? A number that is different to all other listed numbers, only by an explicit infinite clause that concatinates the opposing digit of all listed infinite numbers. A statement that asserts you can't prove me? A *finite* sized algorithm that gives the maximum output length of *any* sized algorithm? A set that doesn't contain itself, yet contains all sets that don't contain themselves? Mathematical history is full of tragedy, read up on any great mathematician and you will find his life ended either early or in sorrow. Be objective with the proofs handed to you, most of the above are just a narrowly defined domain and a self referential negative clause defined on naive interpretations of that domain, specifically the defining clauses of the domain are just twisted. Cardinal representation is a shorthand for numbers, a digit by some power of 10, the implied connotation is that digit is selected from a total set of digits, Cantors specifically defined opposing digit is obscure. Herc === Subject: Re: Simple Group Theory Question One more comment makes things pretty easy. If you know about commutators, use a^2 = 1 for all a to calculate the commutators, which gives the thm Arturo Magidin speaks of. Van === Subject: Daryl answer the question please....... How many digits d is D matched to on any new infinite random list? I'm not constructing a list given your number, you can put your number in a black box and hold the key. THEN I generate a countable infinite random list. THEN you hand over the key. THEN we check how many digits of your number I matched on my INDPENDANT list. How many digits d is D matched to on any *new* infinite *random* list? Herc >How many digits d is D matched to on any *new* infinite *random* list? I don't know. -- Josh Purinton Does anyone know? Is it AA/ 0 A/ 1 B/ >1 C/ <10 D/ >10 E/ >1,000,000 F/ unlimited G/ infinite H/ all of them digits a countable random list will match any given number? Do any people who follow Cantors proof know how many digits of your NEW sequence will be matched? === Subject: Re: Daryl answer the question please....... HERC777 says... >How many digits d is D matched to on any new infinite random list? >I'm not constructing a list given your number, you can put your number >in a black box and hold the key. >THEN I generate a countable infinite random list. >THEN you hand over the key. >THEN we check how many digits of your number I matched on my INDPENDANT >list. >How many digits d is D matched to on any *new* infinite *random* >list? Okay, assume that we're producing our digits by some truly random physical process (such as radioactive decay) so that each digit is equally likely and is statistically independent of all previously generated digits. Let R(i) be random real number i, where R(0) is my real. Then what's likely to be the case is the following: 1. For every n > 0, there exists a number k(n) such that R(0) agrees with R(k(n)) in the first n decimal places. 2. For every k > 0, there exists a number n(k) such that R(0) disagrees with R(k) at position n(k). So there would be no bound on the number of decimal places that match, but none of your reals would be exactly equal to my real. -- Daryl McCullough Ithaca, NY === Subject: Re: Daryl answer the question please....... 1 -> for every digit place, there is a person from the infinite list that matches up to that digit 2 -> for every person, there exists a digit on his sequence that is different why does 2 hold? are you saying 2 random processes never have the same output? are you working backwards from what's likely, assuming the infinite set is incomplete? Herc === Subject: Re: Daryl answer the question please....... HERC777 says... >1 -> for every digit place, there is a person from the infinite list >that matches up to that digit >2 -> for every person, there exists a digit on his sequence that is >different >why does 2 hold? are you saying 2 random processes never have the same >output? Right. With probability 1, no two random processes produce the same (infinite) output. The probability that two randomly generated sequences of digits will agree on the first n digits is 10^{-n}. So the probability that they will agree *everywhere* is the limit as n --> infinity of 10^{-n}, which is 0. >are you working backwards from what's likely, assuming the >infinite set is incomplete? Well, I'm assuming that if you have countably many events, each of which is probability 0, then the probability of all of them together is still probability 0. -- Daryl McCullough Ithaca, NY === Subject: Re: Daryl answer the question please....... Why would you use a statistical method for finite sets here? With infinite outcomes, each individual oucome is P=0. P=0 means possible in this context. If HHHHHHHHH.. is in the range of outputs, and infinite trials are performed, there is no limit to the number of heads. Logical conclusion, all heads. Herc === Subject: Re: Daryl answer the question please....... says... >How many digits d is D matched to on any new infinite random list? >I'm not constructing a list given your number, you can put your number >in a black box and hold the key. >THEN I generate a countable infinite random list. >THEN you hand over the key. >THEN we check how many digits of your number I matched on my INDPENDANT >list. >How many digits d is D matched to on any *new* infinite *random* >list? >Herc >>How many digits d is D matched to on any *new* infinite *random* list? >I don't know. >Josh Purinton >Does anyone know? >Is it >AA/ 0 >A/ 1 >B/ >1 >C/ <10 >D/ >10 >E/ >1,000,000 >F/ unlimited >G/ infinite >H/ all of them >digits a countable random list will match any given number? >Do any people who follow Cantors proof know how many digits of your >NEW sequence will be matched? === Subject: David Ullrich answer the questions please........... An infinite number of people each toss a coin infinite times. Can you guarantee a new sequence of heads and tails? A ____ You hand the sequence to me in a locked box and keep the key. Then I generate a second infinite random list of H&T sequences. Then you hand me the key. How many digits of your sequence did I match the second time? (the list is independant of the sequence) A_____ Herc === Subject: Re: David Ullrich answer the questions please........... >An infinite number of people each toss a coin infinite times. Can you >guarantee a new sequence of heads and tails? Well, since you said please: I don't have any idea what you're asking. What do you mean, a new sequence? (My best guess is you mean a sequence that's never been seen before. But that makes no sense, because _no_ infinite sequence has ever been seen.) >A ____ >You hand the sequence to me in a locked box and keep the key. Ah, would that we could. >Then I generate a second infinite random list of H&T sequences. >Then you hand me the key. >How many digits of your sequence did I match the second time? >(the list is independant of the sequence) If both sequences are generated randomly and the two sequences are independent then of course it's impossible to say with certainty how many matches there are - random is like that. But with probability one there will be infinitely many matches, in fact the set of places where the two sequences match will have asymptotic density 1/2. >A_____ >Herc ************************ David C. Ullrich === Subject: Re: David Ullrich answer the questions please........... <4cs8q0tn63pf5p4p5g33e75ieqruat41s8@4ax.com> 9 results for new number ullrich cantor [Herc] An infinite number of people each toss a coin infinite times [DU] because _no_ infinite sequence has ever been seen. 20 people understood the question so far, do you want a clause on the second sentence saying within the domain of the 1st sentence? Daryl has this problem too, P=0 so an infinite sequence is impossible to make... unless you use the diagonal. I'll rephrase in case you can answer. When many many people all toss coins many many times, can you toss your coin in such a way that it makes a new sequence of heads and tails that no other person has made with their coin? If there are n people, by the binomial distribution and with 90% confidence interval, you will have to toss your coin approx. log(n) times. What happens as n->oo? I'm amazed such a large group of people here in sci.math don't recoil from the obvious error in uncountable theory and see the logically obvious, that with infinite people, the sequences all covered are infinite in length. Herc you don't think that Klingon trick will really work do you? === Subject: Re: David Ullrich answer the questions please........... >9 results for new number ullrich cantor >[Herc] >An infinite number of people each toss a coin infinite times >[DU] >because _no_ infinite sequence has ever been seen. >20 people understood the question so far, do you want a clause on the >second sentence saying within the domain of the 1st sentence? Daryl >has this problem too, P=0 so an infinite sequence is impossible to >make... unless you use the diagonal. >I'll rephrase in case you can answer. >When many many people all toss coins many many times, can you toss your >coin in such a way that it makes a new sequence of heads and tails that >no other person has made with their coin? That's a repharasing of An infinite number of people each toss a coin infinite times. Can you guarantee a new sequence of heads and tails? ? many many is a rephrasing of infinite? Also in the first question you asked whether I could guarantee a sequence with a certain property, now you ask whether I _can_ toss it in a certain way... The answer to the second version of the question depends on (i) how many people toss a coin, and how many times (ii) whether you're asking (a) can I be certain that my sequence will be different or (b) is it possible that my sequence is different, if I'm allowed to control my sequence. >If there are n people, by the binomial distribution and with 90% >confidence interval, you will have to toss your coin approx. log(n) >times. >What happens as n->oo? >I'm amazed such a large group of people here in sci.math don't recoil >from the obvious error in uncountable theory and see the logically >obvious, that with infinite people, the sequences all covered are >infinite in length. Yeah, that _is_ hard to understand... >Herc >you don't think that Klingon trick will really work do you? ************************ David C. Ullrich === Subject: Re: David Ullrich answer the questions please........... <4cs8q0tn63pf5p4p5g33e75ieqruat41s8@4ax.com> <3d1aq0hcipbkk03i9scg3u6vfoprm94co8@4ax.com> OK, I'm going to see if you can answer this new question, because I fully know you are mistaken that you can find a new (different) sequence when that sequence is fully within_the_range of an effectively identical person in a set of infinite people all trying to copy you. Try to get the gist of it and work out my intended meaning if there is ambiguity. Given 1/ an infinite sequence D 2/ a set S of s infinite lists of infinite sequences Given s is sufficiently large, what portion of S contain a finite maximum to the initial length of D that is matched? Say D = Say S1 = { ... } Assume S1 due to rare random fluctuations does not contain infinite H.. Then it has some finite limit, it could be 3 by the example above. Basically, what is the confidence interval that an infinite list does not contain some given sequence? Will it tend to always match it to infinite digits? Will it tend to contain some finite limit? 99% of the time, any given sequence [WILL] / [WILL NOT] be matched to infinite precision on a random infinite list. Herc === Subject: Re: David Ullrich answer the questions please........... >OK, I'm going to see if you can answer this new question, because I >fully know you are mistaken that you can find a new (different) >sequence when that sequence is fully within_the_range of an effectively >identical person in a set of infinite people all trying to copy you. Beezarre. You keep changing the question - now the others are trying to copy me? _Previously_ they went first. You really need to make up your mind what the question is... >Try to get the gist of it and work out my intended meaning if there is >ambiguity. >Given >1/ an infinite sequence D >2/ a set S of s infinite lists of infinite sequences Is S countable? >Given s is sufficiently large, what portion of S contain a finite >maximum to the initial length of D that is matched? This question makes no sense unless I assume that s was a typo for S. Assuming that, the question makes no sense. What initial length are you talking about? What the heck is a finite maximum to an initial length? Oh. Maybe you mean to ask what portion of S contains an initial segment of maximal length matching D. It's obviously impossible to answer this question without be that none of the sequences in S match D at _all_. Or it could be that all the sequences in S are exactly the same as D. >Say D = Say S1 = { >... >Assume S1 due to rare random fluctuations does not contain infinite H.. >Then it has some finite limit, Huh? I have no idea what it means to say S1 has some finite limit. (Hint: this is because it makes no sense to say that.) >it could be 3 by the example above. >Basically, what is the confidence interval that an infinite list does >not contain some given sequence? >Will it tend to always match it to infinite digits? >Will it tend to contain some finite limit? >99% of the time, any given sequence [WILL] / [WILL NOT] be matched to >infinite precision on a random infinite list. >Herc ************************ David C. Ullrich === Subject: Re: David Ullrich answer the questions please........... <4cs8q0tn63pf5p4p5g33e75ieqruat41s8@4ax.com> <3d1aq0hcipbkk03i9scg3u6vfoprm94co8@4ax.com> >OK, I'm going to see if you can answer this new question, because I >fully know you are mistaken that you can find a new (different) >sequence when that sequence is fully within_the_range of an effectively >identical person in a set of infinite people all trying to copy you. Beezarre. You keep changing the question - now the others are trying to copy me? _Previously_ they went first. You really need to make up your mind what the question is... [Herc] [read the OP, the 1st question is you try to be different to the infinite set, the 2nd question is another infinite set tries to copy you] >Try to get the gist of it and work out my intended meaning if there is >ambiguity. >Given >1/ an infinite sequence D >2/ a set S of s infinite lists of infinite sequences Is S countable? [Herc] [Yes, S is a finite set. s e N. S = SET s = size] >Given s is sufficiently large, what portion of S contain a finite >maximum to the initial length of D that is matched? This question makes no sense unless I assume that s was a typo for S. [Herc] [Given the size (s) of the set (S) is large enough to get an expected random distrution] Assuming that, the question makes no sense. What initial length are you talking about? What the heck is a finite maximum to an initial length? Oh. Maybe you mean to ask what portion of S contains an initial segment of maximal length matching D. [Herc - yes] It's obviously impossible to answer this question without be that none of the sequences in S match D at _all_. Or it could be that all the sequences in S are exactly the same as D. [Herc] [That's why S is actually a set of S1, S2, S3... Ss Assuming s is large enough, what is the _typical_ behaviour for any Sx?] >Say D = Say S1 = { >... >Assume S1 due to rare random fluctuations does not contain infinite H.. >Then it has some finite limit, Huh? I have no idea what it means to say S1 has some finite limit. (Hint: this is because it makes no sense to say that.) [Herc] [S1 contains an initial segment of maximum length matching D. I meant bound not limit. Note that S1 may contain an initial segment of maximum length matching D, and S2 may contain D] >it could be 3 by the example above. >Basically, what is the confidence interval that an infinite list does >not contain some given sequence? >Will it tend to always match it to infinite digits? >Will it tend to contain some finite limit? >99% of the time, any given sequence [WILL] / [WILL NOT] be matched to >infinite precision on a random infinite list. >Herc ************************ David C. Ullrich [Herc] === Subject: Re: please help me >>A baseball is popped straight up with an initial velocity of 32 >>feet per second, >Giving KE = 1/2 m v^2 = 1/2 m * (32fps)^2 >>What is the maximum height >>reached by this baseball? >It is the height at which PE=KE, and PE+KE=k throughout the trajectory. No, it is the height at which velocity is 0, thereby forcing KE to be 0 also. >PE = mgh = m*32fpsps*h >1/2 m * (32 ft / sec)^2 = m * (32 ft/sec^2) * h >m factors out >1/2 * 1024 ft^2/sec^2 = 32 ft/sec^2 * h(ft) >512 ft^2/sec^2 / 32 ft/sec^2 = h >16 ft = h. >It's an energy balance. >Now apply KE=PE throughout the trajectory to determine >> its height above the ground is a function of >>time >I tolerance everything and tolerate everyone. >I love: Dona, Jeff, Kim, Kimmie, Mom, Neelix, Tasha, and Teri, alphabetically. >I drive: A double-step Thunderbolt with 657% range. >I fight terrorism by: Using less gasoline. <> === Subject: ? solving linear eqn Given A*x = b, here x and b are vectors and A is a matrix, square or not. I saw the following: 1) given A, b and then solve for b but how about the following: 2) given x and b, solve for A? by Cheng Cosine Nov/23/2k4 Ut === Subject: Re: ? solving linear eqn > Given A*x = b, here x and b are vectors and A is a matrix, square or not. > I saw the following: > 1) given A, b and then solve for b > but how about the following: > 2) given x and b, solve for A? Someone suggested to extend (2) into more general form that all A, x, and b are matrix. Then I can easily analysis A*x = b as what textbook usually taught to do for A*x = b as usual. But this approach just remind me another linear equation below. A*x+x*B = C Here all are matrices. A is m-by-m and B is n-by-n, while both x and C are m-by-n. I read in some linear system books called this as Lyapunov equation, and some interesting theorems are given. However, I don't see how to analysize linear problems of this kind. To be more specific, in analyzing A*x = b, I read ppl use the eigenvector or spectral expansion or more generally the SVD. Then one can see when the solution exist and unique. But I don't see any way to do analysis for A*x+x*B = C. Any suggestions? by Cheng Cosine Nov/24/2k4 UT === Subject: Re: ? solving linear eqn > Given A*x = b, here x and b are vectors and A is a matrix, square or not. > I saw the following: > 1) given A, b and then solve for b > but how about the following: > 2) given x and b, solve for A? > Someone suggested to extend (2) into more general form that > all A, x, and b are matrix. Then I can easily analysis A*x = b > as what textbook usually taught to do for A*x = b as usual. > But this approach just remind me another linear equation below. > A*x+x*B = C > Here all are matrices. A is m-by-m and B is n-by-n, while both > x and C are m-by-n. > I read in some linear system books called this as Lyapunov equation, > and some interesting theorems are given. However, I don't see how > to analysize linear problems of this kind. To be more specific, in analyzing > A*x = b, I read ppl use the eigenvector or spectral expansion or more > generally the SVD. Then one can see when the solution exist and unique. > But I don't see any way to do analysis for A*x+x*B = C. > Any suggestions? > by Cheng Cosine > Nov/24/2k4 UT Cheng, that someone is quite right. Let us first consider how to solve an equation in square matrices (1) A * X = B, With A and B given, provided A is of maximum rank, in other words det(A) is unequal 0 (but for Christ's sake never bother to compute a determinant, use the methods suggested below instead), the solution X will then be found by finding the inverse of A, call it inv(A), multiply both sides with inv(A) on the left and we get, since inv(A) * A = E (the unity matrix with the diagonal elements =1, all other elements =0), E*X=inv(A)*B, and since of course the unity matrix multiplied with any matrix leaves that matrix unchanged, we find X=inv(A)*B. What your original question started from, albeit with a matrix A but a vector b and an unknown vector x, was to turn things aroung and ask for A, given x and b. Guess what, in terms of matrices the new viewpoint is not so drastically different from (1), we merely now talk about a matrix equation (2) X * A = B, and again, provided A [ has maximum rank/has a non-zero determinant/is invertable] - you guessed it, the three conditions stated in the square brackets are equivalent, provided the condition is met, the solution is obtained by forming inv(A), multiplying both sides (2) with inv(A) but on the right this time, and get X=B*inv(A). The main computational task here obviously is to find the inverse of A. Using the time-honoured methods first developed by C.F.Gauss you adjoin the unity matrix E to the right of A, call that structure (A|E) and through the repeated application of elementary row operations bring the A part of that adjoined structure to diagonal form. Call that transformed diagonal matrix t(A), and the simultaneously transformed unity matrix t(E). If t(A) is not maximum rank, ie. it has 0 elements in the diagonal, i.e. its determinant is 0, rejoice since (1), or (2), have no solution and your work is done. If however t(A) has maximum rank you now have a structure (t(A)|t(E)) where t(A) is a diagonal matrix and t(E) certainly no longer looks like a unity matrix. By taking the column vectors of t(E) one by one, call them c and solving for vectors x in (3) t(A) * x = c you have obtained inv(A) as the successive x vectors are none other than the columns of inv(A). For 3x3 or 4x4 matrices the entire 's computation can be done on a notepad and need not take longer than 60 seconds. With a bit of practice your lecturer will accuse you cheating! At long last to your particular problem. Given vectors a and b solve for a matrix X so that (4) X * a = b Simple. Transform a to a matrix A by interpreting a as the first column vector of A, keep adding columns until you have a square matrix, A. A will need to have an inverse, so you have to keep checking. For 2- or 3- vectors this is trivial, but for larger structures I am sure one could find an algorithm of the successive application of elementary row operations a la C.F.Gauss so you don't just randomly form A from a to then find that the result does not have maximum rank. Let us now assume you have adjoined a to a suitable A. You may now adjoining any columns at all to b to transform this vector into a square matrix B and BINGO, the solution will be: (5) X = B * inv(A) as per the exercise we did for (2). So, ZVK has given you the professional mathematician's answer to your problem, I have provided the market gardener's version, what more could you ask for? Actually, a few very important things require an answer in connection with (4), and I leave these for you as an exercise: (a) are there vectors a and b for which an X cannot be found? SIMPLE, just look at (5) and ask yourself what a would make it impossible for A to have an inverse; (b) obviously if there is one solution X there is an infinity of them, and it would certainly be desirable to express this infinite solution space in terms of a space, ie. show its dimension and perhaps write down elements that allow the general solution to be described as a linear combination of these elements. Good luck. Reinhard PS. === Subject: Re: ? solving linear eqn > Given A*x = b, here x and b are vectors and A is a matrix, square or not. > I saw the following: > 1) given A, b and then solve for b > but how about the following: > 2) given x and b, solve for A? > by Cheng Cosine > Nov/23/2k4 Ut In 1), you probably meant ... then solve for x. Question 2), provided x is not the zero vector, may have infinitely many solutions. Just play around with A being 1-by-2 unknown [u, v], x a 2-by-1 column vector with 1's as entries, b a 1-by-1 matrix [1].) The question was somewhat playful anyway, but without any effort to see what's to be expected. Want to see more? (If not, skip the rest. You will need something like normed linear algebra to understand it.) Serious treatment comes under the name backward error analysis: Suppose one has obtained an approximate solution x of the equation C * x = b with given C and b. A question can be asked is x an exact solution of a system A*x=b where A is close to C? Write A=C+H, then we are looking for a small H so that (C+H)*x=b. That means H*x=b-C*x (b-C*x is called the residual). If the norms of matrices and vectors are compatible (submultiplicative), then norm(H) * norm(x) >= norm(b-C*x), so that norm(H) >= norm(b-C*x) / norm(x). Can the lower bound be achieved? Actually yes, and I will give the construction for Euclidean norm: set H = (b-C*x)*x^transposed / norm(x)^2. Your question, to summarize, has an answer A = C + H. For other norms, one would have to play with the dual norm, and a finite-dimensional version of Hahn-Banach Theorem. Sorry you asked? :-) === Subject: Re: The simplest possible TOE > We all exist as artifacts in the mind of God or Mother Nature I believe the single simplest TOE would be as follows: Because. A slightly more useful version might be: what is, is; what exists, exists. A. === Subject: Re: The simplest possible TOE you are on the right track. You and your teachers must be the the world's smartest philosopher. > Lord of Chaos(Suresh Devanathan) >> We all exist as artifacts in the mind of God or Mother Nature > I believe the single simplest TOE would be as follows: > Because. > A slightly more useful version might be: what is, is; what exists, exists. > A. === Subject: Re: The simplest possible TOE > We all exist as artifacts in the mind of God or Mother Nature 1) How can you have an or in The simplest possible TOE, idiot? 2) What predictions can you make with your infantile TOE, idiot? 3) What technologies spring forth from your infantile TOE, idiot? India: 1.1 billion assholes and 1.1 million flush toilets. The miracle is that there are no lines. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: The simplest possible TOE idiot, just because u fail to realize that there are infinite number of axiomatic systems that are simulateneously made true, is not my fault. Idiot, you ramble, without paying deep attention to what is being said. >> We all exist as artifacts in the mind of God or Mother Nature > 1) How can you have an or in The simplest possible TOE, idiot? > 2) What predictions can you make with your infantile TOE, idiot? > 3) What technologies spring forth from your infantile TOE, idiot? > India: 1.1 billion assholes and 1.1 million flush toilets. The > miracle is that there are no lines. > -- > Uncle Al > http://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals) > http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: The simplest possible TOE >>We all exist as artifacts in the mind of God or Mother Nature > 1) How can you have an or in The simplest possible TOE, idiot? > 2) What predictions can you make with your infantile TOE, idiot? > 3) What technologies spring forth from your infantile TOE, idiot? > India: 1.1 billion assholes and 1.1 million flush toilets. The > miracle is that there are no lines. Who needs a flush toilet when there is the Ganges. Bob Kolker === Subject: Re: The simplest possible TOE hello whitey >We all exist as artifacts in the mind of God or Mother Nature >> 1) How can you have an or in The simplest possible TOE, idiot? >> 2) What predictions can you make with your infantile TOE, idiot? >> 3) What technologies spring forth from your infantile TOE, idiot? >> India: 1.1 billion assholes and 1.1 million flush toilets. The >> miracle is that there are no lines. > Who needs a flush toilet when there is the Ganges. > Bob Kolker === Subject: Re: The simplest possible TOE > hello whitey I beg your pardon? Bob Kolker === Subject: Re: The simplest possible TOE i m done with u. >> hello whitey > I beg your pardon? > Bob Kolker === Subject: Re: The simplest possible TOE You and Uncle Al must be the most mannered person in the world. Lord of Chaos(Suresh Devanathan) >i m done with u. > hello whitey >> I beg your pardon? >> Bob Kolker === Subject: Re: The simplest possible TOE The technology that springs forth, is the universe itself. >> We all exist as artifacts in the mind of God or Mother Nature > 1) How can you have an or in The simplest possible TOE, idiot? > 2) What predictions can you make with your infantile TOE, idiot? > 3) What technologies spring forth from your infantile TOE, idiot? > India: 1.1 billion assholes and 1.1 million flush toilets. The > miracle is that there are no lines. > -- > Uncle Al > http://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals) > http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: The simplest possible TOE Prediction: God does whatever he wants to do. >> We all exist as artifacts in the mind of God or Mother Nature > 1) How can you have an or in The simplest possible TOE, idiot? > 2) What predictions can you make with your infantile TOE, idiot? > 3) What technologies spring forth from your infantile TOE, idiot? > India: 1.1 billion assholes and 1.1 million flush toilets. The > miracle is that there are no lines. > -- > Uncle Al > http://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals) > http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: Automorphisms of abelian p-groups: Aut(Z_p+Z_{p^2}) >> What is the automorphism group of Z_p+Z_{p^2} for p prime? > This is what I have so far: Each automorphism must fixes the class of >> elements of order p. So, it must fixes the subgroup Z_p+Z_p. But I'm >> not sure how to continue. Can someone please lead me a little? >G = C_p x C_{p^2}, with presentation , clearly >has p^2*(p-1) elements of order p^2 -- each cyclic subgroup of >order p^2 has p(p-1) such elements, and there are p such cyclic >subgroups: , , ..., -- and p(p-1) elements of >order p that aren't powers of elements of order p^2 -- the >non-identity elements in , , ..., -- so >|Aut(G)| = p^3*(p-1)^2. >automorphisms for Aut(G) = , with A^{p(p-1)} = 1, >B^p = 1, C^{p-1} = 1, D^p = 1: >A(x) = x^n, A(y) = y, where n generates (Z_{p^2},*) =~ C_{p(p-1)} >B(x) = xy, B(y) = y >C(x) = x, C(y) = y^m, where m generates (Z_p,*) =~ C_{p-1} >D(x) = x, D(y) = x^p y >Filling in the details of the presentation of Aut(G), i.e., >finding sufficient relations among the generators, is left as an >exercise for the reader. > You might have added that Aut(G) has order p^3 (p-1)^2, See the last clause of my first, admittedly not very clearly written, paragraph. > with a normal > extraspecial group of order p^3 (generated by A^(p-1), B, D, with A^(p-1) > central), and the quotient group is isomorphic to Z_{p-1} x Z_{p-1}. above, rather than discovering it later while playing around with relations among the generators of Aut(G). (Why do you think I left that as an exercise for the reader? :-) -- Jim Heckman === Subject: Re: Determination of sine frequency >does anybody know a convenient method for the determination of the frequency >of a sine wave when exactly four aquidistant samples are known (A0, A1, A2, >A3). The tricky point is that these four samples do not cover one complete >period (this would be easy determined by the discrete fourier transform) but >only one part of one period. If f(x) = k * sin(x*2*pi*f), f''(x) = -k * (2*pi*f)^2 * sin(x*2*pi*f) You can estimate f''(A1) = (f(A2)-f(A0))/t^2 Where t is the interval between samples. This assumes that the data is normalized to the range [-1..1] and probably some other things I didn't consider. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Quantum Gravity Hypothesis The physicist Lee Smolin explains that space is relational, very analogous to being inside the words of a sentence: The geometry of a universe is very like the grammatical structure of a sentence. Just as a sentence has no structure and no existence apart from the relationships between the words, space has no existence apart from the relationships that hold between the things in the universe. Coordinates are a convienient means of mathematical modeling, ...but reality actually defines itself with events. Events don't happen in a fixed absolute background of space, or time; events characterize the evolution OF space-time. The metric of space-time becomes defined by events, such, that there is no space-time if there are no events. A metric field can be defined by the primary substratum of events. Thus the intrinsic geometrical structure of spacetime is predicated on the pseudo-Riemannian spaces via the affine relationships Ö all physical events are fully reducible to manifestations of the substratum i. e. the event density generating a metric field. The overlap of events as ripples- being the wave functions - circular conic 2D cross sections, generates new smaller ripples that are contained in the outer past ripples-cross sections, thus the locality principle is not violated and describes non-paradoxically, what appears as non-local transfer of information. [event_2^0_[[[_[_[event 2^n_]___]_]]]]] Intersections[the overlapping] of event boundaries also provides a better definition for bits of physical information, where the information density of the universe is continually increasing. Time is defined as an iterative sequence of outer[past] events including all inner[future] events. http://www.iomas.com/gina/ultrahiq/Mega-Society/NoesisMay/SupernovaCL.asp http://www.martinelli.org/rexpansion/ === Subject: Re: Quantum Gravity Hypothesis Russell E. Rierson schrieb > The physicist Lee Smolin explains that space is relational, very > analogous to being inside the words of a sentence: The geometry of > a universe is very like the grammatical structure of a sentence. Just > as a sentence has no structure and no existence apart from the > relationships between the words, space has no existence apart from the > relationships that hold between the things in the universe. Nice example. The point is that all real sentences we hear or read have some other structure. For example, this sentence has also the structure of a stream of bits. Acoustic sentences have the structure of sound waves. Ilja === Subject: Re: Quantum Gravity Hypothesis > The physicist Lee Smolin explains that space is relational, very > analogous to being inside the words of a sentence: The geometry of > a universe is very like the grammatical structure of a sentence. Just > as a sentence has no structure and no existence apart from the > relationships between the words, space has no existence apart from the > relationships that hold between the things in the universe. http://www.edge.org/3rd_culture/smolin/smolin_p3.html We see that, at least naively time has completely disappeared from the formalism. This has led to what is called the problem of time in quantum cosmology, which is how to [] provide an interpretation according to which time is not part of a fundamental description of the world, but only reappears in an appropriate classical limit. If he progresses to substituting expression for description he will have turned an important corner, not just regarding the concept time but also observer. Of course, this too is the essential problem with space-time. It connotes a clumsy (in a complex world), primal concept of observer - which is consistent with the general primal understanding of the nature of Nature as empirical. http://www.effectuationism.com/forum/messages/27/27.html?1071620499 With further development: Indefinite and dynamic Man/Person- -Ground in tension with moving animal/object - Time- -Being. Peter Kinane http://www.effectuationism.com/ === Subject: Re: Quantum Gravity Hypothesis > The physicist Lee Smolin explains that space is relational, very > analogous to being inside the words of a sentence: The geometry of > a universe is very like the grammatical structure of a sentence. Just > as a sentence has no structure and no existence apart from the > relationships between the words, space has no existence apart from the > relationships that hold between the things in the universe. ...[trim]... Also take the time to check Cahill's work in this field: www.mountainman.com.au/process_physics The geometry of the universe might be generated out of randomness. === Subject: Re: Quantum Gravity Hypothesis > The physicist Lee Smolin explains that space is relational, very > analogous to being inside the words of a sentence: The geometry of > a universe is very like the grammatical structure of a sentence. Just > as a sentence has no structure and no existence apart from the > relationships between the words, space has no existence apart from the > relationships that hold between the things in the universe. That is just another way of stating that we use mathematical MODELS of the world. Inherently such models are not real. > Coordinates are a convienient means of mathematical modeling, Hmmm. Say manifolds rather than coordinates, because that's what modern physical theories actually use for this modeling. > ...but > reality actually defines itself with events. Events don't happen in > a fixed absolute background of space, or time; events characterize the > evolution OF space-time. The metric of space-time becomes defined > by events, Sure. See above. > such, that there is no space-time if there are no events. But in any interesting physical situation there are always events. Here's an example list: Object A exists at proper time t0 Object A exists at proper time t1 Object A exists at proper time t2 ... Object B ... Clearly for a classical theory in which proper time is continuous the events are not countable as this particular enumeration suggests. > [... excursion into never-never land] Tom Roberts tjroberts@lucent.com === Subject: Re: Quantum Gravity Hypothesis > The physicist Lee Smolin explains that space is relational, very > analogous to being inside the words of a sentence: The geometry of > a universe is very like the grammatical structure of a sentence. Just > as a sentence has no structure and no existence apart from the > relationships between the words, space has no existence apart from the > relationships that hold between the things in the universe. > Coordinates are a convienient means of mathematical modeling, ...but > reality actually defines itself with events. Events don't happen in > a fixed absolute background of space, or time; events characterize the > evolution OF space-time. The metric of space-time becomes defined > by events, such, that there is no space-time if there are no events. > A metric field can be defined by the primary substratum of events. > Thus the intrinsic geometrical structure of spacetime is predicated on > the pseudo-Riemannian spaces via the affine relationships ? all > physical events are fully reducible to manifestations of the > substratum i. e. the event density generating a metric field. I think the wording is sloppy. An event is a point in 4 coordinates x,y,z,t. The term event density *might* refer to spacetime field density, but do I have to guess? > The overlap of events as ripples- being the wave functions - circular > conic 2D cross sections, generates new smaller ripples that are > contained in the outer past ripples-cross sections, These so-called ripples imply an occurance. An occurance is much more complex than an event. An occurance needs dx,dy,dz,dt,de (e=energy) at some fuzzy location x,y,z,t. The de, usually a photon, proves the existance of matter, which in turn, from the PoV of GR tells spacetime where it is. Hence we survey spacetime using photons, as a result of the occurances. Bilge and I had a discussion about this before, and Bilge pointed out that we should not screw with the definition of what an event is, leave it as x,y,z,t and I agree. >thus the > locality principle is not violated and describes non-paradoxically, > what appears as non-local transfer of information. > [event_2^0_[[[_[_[event 2^n_]___]_]]]]] > Intersections[the overlapping] of event boundaries You see, the term event boundaries is poor vocabulary, (hey I used it and was rightly corrected), and should be termed occurance boundaries, because dx,dy,dz,dt,de is fuzzy. Occurances continually over-lap. >also provides a > better definition for bits of physical information, where the > information density of the universe is continually increasing. A non-zero divergence of information. No wonder the universe will always be smarter than us! > Time is defined as an iterative sequence of outer[past] events > including all inner[future] events. I think the ISU has defined physical time very well, if one intends to introduce new definitions of time, they must be able to transform that definition to the ISU standard or justify a change in that standard. Did Lee Smolin really say this in all seriousness? Ken S. Tucker === Subject: Re: Quantum Gravity Hypothesis > The physicist Lee Smolin explains that space is relational, very > analogous to being inside the words of a sentence: The geometry of > a universe is very like the grammatical structure of a sentence. Just > as a sentence has no structure and no existence apart from the > relationships between the words, space has no existence apart from the > relationships that hold between the things in the universe. > > Coordinates are a convienient means of mathematical modeling, ...but > reality actually defines itself with events. Events don't happen in > a fixed absolute background of space, or time; events characterize the > evolution OF space-time. The metric of space-time becomes defined > by events, such, that there is no space-time if there are no events. > > A metric field can be defined by the primary substratum of events. > Thus the intrinsic geometrical structure of spacetime is predicated on > the pseudo-Riemannian spaces via the affine relationships ? all > physical events are fully reducible to manifestations of the > substratum i. e. the event density generating a metric field. > I think the wording is sloppy. An event is a point > in 4 coordinates x,y,z,t. The term event density > *might* refer to spacetime field density, but do I > have to guess? The book Gravitation - by Misner, Thorne, and Wheeler[1st chapter], says that points in spacetime are characterized by what happens there. Give a point in spacetime the name event. World lines of Worldlines fill up spacetime. Events relate to other events. > The overlap of events as ripples- being the wave functions - circular > conic 2D cross sections, generates new smaller ripples that are > contained in the outer past ripples-cross sections, > These so-called ripples imply an occurance. > An occurance is much more complex than an event. > An occurance needs dx,dy,dz,dt,de (e=energy) at > some fuzzy location x,y,z,t. > The de, usually a photon, proves the existance > of matter, which in turn, from the PoV of GR tells > spacetime where it is. Hence we survey spacetime > using photons, as a result of the occurances. Yes, an event is a happening or an occurance. http://dictionary.reference.com/search?q=event&r=67 QUOTE: event Something that takes place; an occurrence. A significant occurrence or happening. See Synonyms at occurrence. A social gathering or activity. The final result; the outcome. Sports. A contest or an item in a sports program. Physics. A phenomenon or occurrence located at a single point in space-time, regarded as the fundamental observational entity in relativity theory. END OF QUOTE > Bilge and I had a discussion about this before, > and Bilge pointed out that we should not screw > with the definition of what an event is, leave > it as x,y,z,t and I agree. Tensors are coordinate independent. In the real world, events[occurances] relate to other events[occurances]. The real world doesn't require a superimposing of Cartesian, or other coordinates, on it. >thus the > locality principle is not violated and describes non-paradoxically, > what appears as non-local transfer of information. > > [event_2^0_[[[_[_[event 2^n_]___]_]]]]] > > Intersections[the overlapping] of event boundaries > You see, the term event boundaries is poor > vocabulary, (hey I used it and was rightly corrected), > and should be termed occurance boundaries, > because dx,dy,dz,dt,de is fuzzy. The terms Event and occurance are interchangable for this purpose. > Occurances continually over-lap. >also provides a > better definition for bits of physical information, where the > information density of the universe is continually increasing. > A non-zero divergence of information. No wonder > the universe will always be smarter than us! > Time is defined as an iterative sequence of outer[past] events > including all inner[future] events. > I think the ISU has defined physical time very > well, if one intends to introduce new definitions > of time, they must be able to transform that definition > to the ISU standard or justify a change in that standard. The above definition doesn't contradict the ISU standards. > Did Lee Smolin really say this in all seriousness? > Ken S. Tucker Yes. http://books.guardian.co.uk/reviews/scienceandnature/0%2C6121%2C438833%2C00. html QUOTE But in Einstein's theory, space is something else altogether. It arises out of the relationships between objects - planets, galaxies and so on. Think of a sentence, says Smolin; it is not simply a container into which one puts words. Without any words, there would be no sentence; similarly, in Einstein's theory, space has no existence apart from the objects that move within it. END OF QUOTE === Subject: Re: Quantum Gravity Hypothesis ... Russell, I think you did good post. > A metric field can be defined by the primary substratum of events. > Thus the intrinsic geometrical structure of spacetime is predicated on > the pseudo-Riemannian spaces via the affine relationships ? all > physical events are fully reducible to manifestations of the > substratum i. e. the event density generating a metric field. > > I think the wording is sloppy. An event is a point > in 4 coordinates x,y,z,t. The term event density > *might* refer to spacetime field density, but do I > have to guess? > The book Gravitation - by Misner, Thorne, and Wheeler[1st chapter], > says that points in spacetime are characterized by what happens > there. Give a point in spacetime the name event. World lines of > Worldlines fill up spacetime. Events relate to other events. Yes, the universe has many photons following worldlines for reference, but when you actually measure one, by aborption or emission, Heisenburg makes us limit our accuracy to dx,dy,dz,dt,de, which I term an occurance. In basic relativity we *simplify* the occurance to be an event at x,y,z,t, but then we leave the real world of measurement and go into an imaginary world of continuous manifolds independent of energy and matter, that is inconsistent with GR and QT. > The overlap of events as ripples- being the wave functions - circular > conic 2D cross sections, generates new smaller ripples that are > contained in the outer past ripples-cross sections, > > These so-called ripples imply an occurance. > An occurance is much more complex than an event. > An occurance needs dx,dy,dz,dt,de (e=energy) at > some fuzzy location x,y,z,t. > The de, usually a photon, proves the existance > of matter, which in turn, from the PoV of GR tells > spacetime where it is. Hence we survey spacetime > using photons, as a result of the occurances. > Yes, an event is a happening or an occurance. > http://dictionary.reference.com/search?q=event&r=67 > QUOTE: > event > > Something that takes place; an occurrence. > A significant occurrence or happening. See Synonyms at occurrence. > A social gathering or activity. > The final result; the outcome. > Sports. A contest or an item in a sports program. > Physics. A phenomenon or occurrence located at a single point in > space-time, regarded as the fundamental observational entity in > relativity theory. Close but if you want to invite Webster to lecture on physics, well you know what I mean, > END OF QUOTE > Bilge and I had a discussion about this before, > and Bilge pointed out that we should not screw > with the definition of what an event is, leave > it as x,y,z,t and I agree. > Tensors are coordinate independent. In the real world, > events[occurances] relate to other events[occurances]. The real world > doesn't require a superimposing of Cartesian, or other coordinates, on > it. Basically that's true. But a tensor must be able to be specialized as a set of measurements wrt some arbituary CS. So somewhere in the process one will need to be *able* to define an event at x,y,z,t, by an occurance, dx...de, at the point of measurement. >thus the > locality principle is not violated and describes non-paradoxically, > what appears as non-local transfer of information. > > [event_2^0_[[[_[_[event 2^n_]___]_]]]]] > > Intersections[the overlapping] of event boundaries > > You see, the term event boundaries is poor > vocabulary, (hey I used it and was rightly corrected), > and should be termed occurance boundaries, > because dx,dy,dz,dt,de is fuzzy. > The terms Event and occurance are interchangable for this purpose. I'm not sure. It's my impression Smolin is presenting an analogy for a description of quantizing GR, not an easy thing to do, understandably. But I object to *bastardizing* (no offense intended), the definition of event. His ripples in spacetime are better descibed as the result of occurances because they involve energy. > Occurances continually over-lap. > >also provides a > better definition for bits of physical information, where the > information density of the universe is continually increasing. > > A non-zero divergence of information. No wonder > the universe will always be smarter than us! > > Time is defined as an iterative sequence of outer[past] events > including all inner[future] events. > > I think the ISU has defined physical time very > well, if one intends to introduce new definitions > of time, they must be able to transform that definition > to the ISU standard or justify a change in that standard. > The above definition doesn't contradict the ISU standards. OK, but it seems like a complicated way to describe a Cesium clock, unless he has something different in mind. > Did Lee Smolin really say this in all seriousness? > Ken S. Tucker > Yes. > http://books.guardian.co.uk/reviews/scienceandnature/0%2C6121%2C438833%2C00. h tml > QUOTE > But in Einstein's theory, space is something else altogether. It > arises out of the relationships between objects - planets, galaxies > and so on. Think of a sentence, says Smolin; it is not simply a > container into which one puts words. Without any words, there would be > no sentence; similarly, in Einstein's theory, space has no existence > apart from the objects that move within it. > END OF QUOTE Ken S. Tucker === Subject: Re: Quantum Gravity Hypothesis > The physicist Lee Smolin explains that space is relational, very > analogous to being inside the words of a sentence: The geometry of > a universe is very like the grammatical structure of a sentence. Just > as a sentence has no structure and no existence apart from the > relationships between the words, space has no existence apart from the > relationships that hold between the things in the universe. > Coordinates are a convienient means of mathematical modeling, ...but > reality actually defines itself with events. Events don't happen in > a fixed absolute background of space, or time; events characterize the > evolution OF space-time. The metric of space-time becomes defined > by events, such, that there is no space-time if there are no events. Snip to make my point. But we know there is empty space-time because we can move into it Russell. We also know that empty space-time curves because of gravity. The metric is independent of any events there. The patterns of curved space-time around mass are absolute. Albert Einstein Einstein asked not about what atoms are; he said; but instead he wanted to know what was inbetween them. You may question what empty space-time means but not its existence. Mitch Raemsch -- Light Falls -- === Subject: Re: Quantum Gravity Hypothesis > The physicist Lee Smolin explains that space is relational, very > analogous to being inside the words of a sentence: The geometry of > a universe is very like the grammatical structure of a sentence. Just > as a sentence has no structure and no existence apart from the > relationships between the words, space has no existence apart from the > relationships that hold between the things in the universe. > Coordinates are a convienient means of mathematical modeling, ...but > reality actually defines itself with events. Events don't happen in > a fixed absolute background of space, or time; events characterize the > evolution OF space-time. The metric of space-time becomes defined > by events, such, that there is no space-time if there are no events. > A metric field can be defined by the primary substratum of events. > Thus the intrinsic geometrical structure of spacetime is predicated on > the pseudo-Riemannian spaces via the affine relationships - all > physical events are fully reducible to manifestations of the > substratum i. e. the event density generating a metric field. > The overlap of events as ripples- being the wave functions - circular > conic 2D cross sections, generates new smaller ripples that are > contained in the outer past ripples-cross sections, thus the > locality principle is not violated and describes non-paradoxically, > what appears as non-local transfer of information. > [event_2^0_[[[_[_[event 2^n_]___]_]]]]] > Intersections[the overlapping] of event boundaries also provides a > better definition for bits of physical information, where the > information density of the universe is continually increasing. > Time is defined as an iterative sequence of outer[past] events > including all inner[future] events. Reading the above I am reminded of the cartoon where you see a physicist, systems analyst, mathematician , engineer or whatever with all these arcane symbols leading to a box with a sign that says 'and then a miracle occurs'. Someone comes along and looks at it and says - good work but you might want to tighten your reasoning up here - pointing to the bit that says 'and then a miracle occurs'. For some reason when I read the above I am reminded of the cartoon and feel an irresistible urge to say - you might like to tighten you reasoning up here. The only difference is at least you understand what - 'and then a miracle occurs' - means. AFAICS the above is basically unintelligible gibberish. But then again what the heck would I know - the physics books I read usually express their ideas in mathematical language not words whose meaning is as obscure as the book of revelations. Bill > http://www.iomas.com/gina/ultrahiq/Mega-Society/NoesisMay/SupernovaCL.asp > http://www.martinelli.org/rexpansion/ === Subject: Re: Deep Thoughts # 17: Liar Paradox is a Formal Metamathematical Theorem >>You _really_ need to work on the irony detector. Truth >>and provability are _different_ things - when you define >>one to be the other you exhibit amazing ignorance of the >>most basic issues. >************************ >David C. Ullrich >>We call a formula F(v1,. . .,vn) correct if for every n-tuple (a1,. . >>.,an), the sentence F(a1,. . .,an) is true. We let N be the >>first-order system whose axioms are all the correct formulas (this >>includes all logically valid formulas.) Thus, the provable formulas >>of N are nothing more than the axioms of N. >Recursion Theory for Metamathematics - Raymond M. Smullyan, 1993 > he's not defining provability to be the > same as truth. (He's setting up a _specific_ formal system in > which the two coincide. There's a big difference.) That doesn't make any sense. How can you define provability (or anything else) without being in a system? Smullyan is setting up a specific formal system and you're not? So what are you talking about if it's not a formal system? How does that work? > ************************ > David C. Ullrich === Subject: Re: Deep Thoughts # 17: Liar Paradox is a Formal Metamathematical Theorem >> >You _really_ need to work on the irony detector. Truth >and provability are _different_ things - when you define >one to be the other you exhibit amazing ignorance of the >most basic issues. >************************ >David C. Ullrich >We call a formula F(v1,. . .,vn) correct if for every n-tuple (a1,. . >.,an), the sentence F(a1,. . .,an) is true. We let N be the >first-order system whose axioms are all the correct formulas (this >includes all logically valid formulas.) Thus, the provable formulas >of N are nothing more than the axioms of N. >Recursion Theory for Metamathematics - Raymond M. Smullyan, 1993 >> he's not defining provability to be the same as truth. (He's setting >> up a _specific_ formal system in which the two coincide. There's a >> big difference.) > That doesn't make any sense. It does. Really. Given a language: * Truth (defined in the model theory) is a semantic relation that holds between a sentence of the language and an interpretation of the language. * Provability (defined in the proof theory) is a syntactic relation that holds between a set of sentences in the language and a given sentence of the language. It *follows* trivially from the definitions of N, truth, and provability that N |- A iff A is true in the standard model of the language of arithmetic iff A is an axiom of N. That's a simple theorem. It isn't a definition of truth or provability. Makes good sense. HTH. === Subject: get me a printout on that Data! [Spock] Captain the Klingons are attempting to hijack our computer. [Kirk] Scotty! Full power dammit! [Spock] Captain, they've locked their quantum drive computer into our infinite encryptable login mainframe. [Kirk] Those pesky Klingons, if their quantum computer tried EVERY COMBINATION in parallel we'll be goners! [Spock] Don't worry captain, I've identified their quantum device, its an Enter-prizonator 3000, I'm diagonalising now... [Kirk] Great Spock! You should be a lecturer! [Spock] I've reprogrammed our computer security Captain, the infinite login sequence is now set to the inverted diagonal of the Enterprizonator 3000, its spockandthepointyearedchick... [Kirk] Don't tell me, its a secret remember, good work crew, warp factor 9! Herc === Subject: Re: get me a printout on that Data! > [Spock] > Captain the Klingons are attempting to hijack our computer. > [Kirk] > Scotty! Full power dammit! > [Spock] > Captain, they've locked their quantum drive computer into our infinite > encryptable login mainframe. > [Kirk] > Those pesky Klingons, if their quantum computer tried EVERY COMBINATION > in parallel we'll be goners! > [Spock] > Don't worry captain, I've identified their quantum device, its an > Enter-prizonator 3000, I'm diagonalising now... > [Kirk] > Great Spock! You should be a lecturer! > [Spock] > I've reprogrammed our computer security Captain, the infinite login > sequence is now set to the inverted diagonal of the Enterprizonator > 3000, its spockandthepointyearedchick... > [Kirk] > Don't tell me, its a secret remember, good work crew, warp factor 9! > Herc Herc your stuff is getting better! === Subject: Re: get me a printout on that Data! Herc === Subject: US usage of the terms college and university Can some US reader enlighten me as to the usage of the terms college and university in your country? Does, for example, a college math course have the same conotation as a university math course? They are quite different here in Canada. Dan Toronto, Canada === Subject: Re: US usage of the terms college and university >Can some US reader enlighten me as to the usage of the terms college >and university in your country? Does, for example, a college math >course have the same conotation as a university math course? As others have noted, there is no standard distinction. (As they say, The nice thing about standards is that there are so many to choose from!) To add another usage: the departments of my university are organized into seven colleges: the College of Liberal Arts and Sciences, the College of Education, the College of Law, etc. You might find this useful: http://www.math.niu.edu/~rusin/teaching-math/usa for non-USAns of the US school system. I have heard surprised foreign visitors compare our College Algebra course to College Alphabet. Sad to say, at least a quarter of US post-secondary students need to take this course. dave === Subject: Re: US usage of the terms college and university >>Can some US reader enlighten me as to the usage of the terms college >>and university in your country? Does, for example, a college math >>course have the same conotation as a university math course? > As others have noted, there is no standard distinction. (As they say, > The nice thing about standards is that there are so many to choose from!) > To add another usage: the departments of my university are organized into > seven colleges: the College of Liberal Arts and Sciences, the > College of Education, the College of Law, etc. > You might find this useful: > http://www.math.niu.edu/~rusin/teaching-math/usa > for non-USAns of the US school system. > I have heard surprised foreign visitors compare our College Algebra > course to College Alphabet. Sad to say, at least a quarter of US > post-secondary students need to take this course. > dave It's sad. Heck, it was a shock to me. I spent my undergraduate career at a school where the lowest level math class available was a calculus class intended for people who never took HS calc. So when I went to grad school and saw many sections of a class called Algebra and Trigonometry I was rather taken aback. -Ron === Subject: Re: US usage of the terms college and university 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: > Can some US reader enlighten me as to the usage of the terms college > and university in your country? Does, for example, a college math > course have the same conotation as a university math course? They > are quite different here in Canada. My understanding is that in most cases the difference is that a college has only undergraduate programs, a university has both undergraduate and graduate programs. But it is also not uncommon for a university to have subdivisions called colleges, representing either academic units (UCI College of Medicine) or housing units (Kresge College at UCSC). Anyway, I wouldn't expect much difference between the two math course phrases you describe. -- David Eppstein Computer Science Dept., Univ. of California, Irvine http://www.ics.uci.edu/~eppstein/ === Subject: Re: US usage of the terms college and university >> Can some US reader enlighten me as to the usage of the terms college >> and university in your country? Does, for example, a college math >> course have the same conotation as a university math course? They >> are quite different here in Canada. >My understanding is that in most cases the difference is that a college >has only undergraduate programs, a university has both undergraduate and >graduate programs. But it is also not uncommon for a university to have >subdivisions called colleges, representing either academic units (UCI >College of Medicine) or housing units (Kresge College at UCSC). >Anyway, I wouldn't expect much difference between the two math course >phrases you describe. What's the difference between Harvard College and Harvard University? Harvard College is the undergraduate program at Harvard. It is part of the Faculty of Arts and Sciences and offers programs in the liberal arts. Harvard University refers to the entire educational institution, including the undergraduate college, the graduate and professional schools, research centers, administration, and affiliates. === Subject: Re: US usage of the terms college and university > Can some US reader enlighten me as to the usage of the terms college > and university in your country? There is no standardized distinction between college and university in the United States. Many Americans think that there is, but there is not. > Does, for example, a college math > course have the same conotation as a university math course? The connotation of college math course is math course encountered in higher education (that is, tertiary education) rather than what an American would call a high school math course, that is math course encountered in secondary education. Of course, some people in the United States repeat in higher education courses that they ought to have learned in secondary education. Moreover, much of the precalculus mathematics that some Americans don't take until college/university is actually a junior-high subject in many other parts of the world, e.g., Taiwan. > They are quite different here in Canada. It's a puzzlement to people from many countries that Americans don't draw a sharp terminological distinction between colleges and universities, but it is, alas, a fact. Hope this helps! When I write to international audiences, such as this newsgroup, I usually try to remember to say university when referring to my alma mater (which was indeed denominated a university) just so that I don't confuse readers from other countries. But when speaking to other Americans, I might simply say, e.g., When I went to college . . . to refer to that same institution of higher education. -- Karl M. Bunday P.O. Box 1456, Minnetonka MN 55345 Learn in Freedom (TM) http://learninfreedom.org/ remove .de to email === Subject: Re: US usage of the terms college and university A college can be a junior college, also called a community college; or it may be a university. In The USA, a university is a college but a college is not always a university. A university is a college. A university permits the study to earn an bachelor of arts degree, complete some certificate program, or earn master of arts or master of science degree. Some universities have programs for earning PhD. A university is a 4 year institution, the frequent amount of time expected for earning bachelor of arts or bachelor of science degrees. A university is a college, and offers a higher range of study than community colleges (which are also called junior colleges). A community college or junior college permits study to earn a lower degree, called associate in arts degree; also for studying in some certificate programs. A community college is a 2 year institution. The A.A. degree can be earned in about two years. Be aware, sometimes a student will use more than 2 years of study at the community college. Students change their major field of study, sometimes enroll in fewer courses during a semester; maybe take time off to do other things. The same variations happen by students at universities. Some students take more than 4 years at a university to earn a Bachelor degree. In summary, distinction between college an university is more adequately characterised as so: Community College or Junior College: 2 year college; AA degree, some certificate programs, courses for personal interest, vocational training programs. University( also a college): 4 year college; BA & BS Degrees, Masters degree, at some, PhDs. Some certificate programs G C === Subject: Re: US usage of the terms college and university > Can some US reader enlighten me as to the usage of the terms college > and university in your country? Does, for example, a college math > course have the same conotation as a university math course? They > are quite different here in Canada. Both refer to tertiary, not secondary, institutions. A college has only undergraduate, a university both undergraduate and graduate, programs. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: US usage of the terms college and university >> Can some US reader enlighten me as to the usage of the terms college >> and university in your country? Does, for example, a college math >> course have the same conotation as a university math course? They >> are quite different here in Canada. >Both refer to tertiary, not secondary, institutions. >A college has only undergraduate, a university both >undergraduate and graduate, programs. This may or may not be the case. When Michigan State College was upgraded to Michigan State University in title, this was the only change. It was already a university in all but name. And there are officially named universities which do not give graduate courses, or give only a few professional courses, which cannot be used for graduate degrees. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: US usage of the terms college and university >Both refer to tertiary, not secondary, institutions. >A college has only undergraduate, a university both >undergraduate and graduate, programs. > This may or may not be the case. When Michigan State > College was upgraded to Michigan State University in > title, this was the only change. It was already a > university in all but name. The sequence was Michigan State College of Agriculture and Applied Science (1925) -> Michigan State University of Agriculture and Applied Science (1955) -> Michigan State University (1964). === Subject: Re: US usage of the terms college and university There was some Japanese movie a few years ago. The subtitles said a certain person would be starting college ... but it was really more like what we call middle school in the US. I don't know if there is a Japanese term college, or if the subtitle writer was from some third country where college has this meaning. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: US usage of the terms college and university > Can some US reader enlighten me as to the usage of the terms college > and university in your country? Does, for example, a college math > course have the same conotation as a university math course? They > are quite different here in Canada. > Both refer to tertiary, not secondary, institutions. > A college has only undergraduate, a university both > undergraduate and graduate, programs. But Tertiary's what I did at 15-17 before I went to Uni. Phil -- ... one Marine noticed one of the prisoners was still breathing. A Marine can be heard saying on the pool footage provided to Reuters Television: He's ing faking he's dead. He faking he's ing dead. The Marine then raises his rifle and fires into the man's head. The pictures are too graphic for us to broadcast, Sites said. === Subject: Re: US usage of the terms college and university > Can some US reader enlighten me as to the usage of the terms college > and university in your country? Does, for example, a college math > course have the same conotation as a university math course? They > are quite different here in Canada. > Both refer to tertiary, not secondary, institutions. > A college has only undergraduate, a university both > undergraduate and graduate, programs. An interesting opinion... I have seen good colleges (complete with graduate departments) become universities simply by ordering new stationery. === Subject: Re: US usage of the terms college and university >> Can some US reader enlighten me as to the usage of the terms college >> and university in your country? Does, for example, a college math >> course have the same conotation as a university math course? They >> are quite different here in Canada. >> Both refer to tertiary, not secondary, institutions. >> A college has only undergraduate, a university both >> undergraduate and graduate, programs. >An interesting opinion... >I have seen good colleges (complete with graduate departments) become >universities simply by ordering new stationery. And I have seen (many) bad-to-middling colleges become universities simply by creating new (bad-to-bad) graduate programs (for instance, in hospitality). And, of course, ordering new stationery and signage. Lee Rudolph === Subject: Witzzle Pro Mail In Math Contest Kaidy has a new mail-in contest for you and your children. It is a unique individual or class contest. The Fall Witzzle Pro Mail-In Math and/or teachers will be notified in January, 2005. Witzzle Pro Games have been the basis for almost 10 years of math contests. Students up to grade 8 may enter. Teachers may enter whole classes in one step. Each winner and teacher, if school based entry with school email, gets a prize. It is a great game that lets students play to learn while they learn to play! Visit http://mathfun.com/WitzzleProMailInContest.html for the contest information! Kaidy and MathFun.com are dedicated to promoting math awareness, learning and success for all children! Check back in February 2005 for our Spring contest. === Subject: Re: solid angle >Now that I come to think of it over my evening meal, there was indeed >a way of seeing this without doing the integrals. The area of the unit >sphere outside the cone is equal to that of a cylinder of identical >height and radius (=1). Now the height simply is equal to the 2 * >sinus(arctan(A/D). >Frank That's true of course, but to show an area on a sphere = the area of its projection on the cylinder is itself a calculus (integration) problem, no? --Lynn === Subject: Re: solid angle > That's true of course, but to show an area on a sphere = the area of > its projection on the cylinder is itself a calculus (integration) > problem, no? > --Lynn Yes, I believe that is true, although I have a notion that Archimedes had figured this out already - but I do not know whether that is correct. Anyway, I only wanted to indicate that one could understand the equation using this knowledge. But I remain very happy indeed that you have explained the analytic underpinnings of these connections. Frank === Subject: Re: solid angle >> That's true of course, but to show an area on a sphere = the area of >> its projection on the cylinder is itself a calculus (integration) >> problem, no? >> --Lynn >Yes, I believe that is true, although I have a notion that Archimedes >had figured this out already - but I do not know whether that is >correct. Archimedes did figure it out, and by using the ideas of integration, which have nothing to do with differential calculus. Archimedes and Euclid and the Greeks educated in Euclidean geometry understood limits and what we call the Riemann integral, although they could not often calculate it. Consider the surface area of the frustrum of a cone which comes close to matching a small slice of the sphere by planes perpendicular to the cylinder tangent to the sphere. Make sure the slope of the cone is close to that of the that of the sphere at points of the sector. Then the ratio of the area of the frustrum to that cut off on the cylinder is close to one. The rest follows using limits. One might say that this proof is not correct by modern standards, and this can be argued, but not very well. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: solid angle >>Now that I come to think of it over my evening meal, there was indeed >>a way of seeing this without doing the integrals. The area of the unit >>sphere outside the cone is equal to that of a cylinder of identical >>height and radius (=1). Now the height simply is equal to the 2 * >>sinus(arctan(A/D). >>Frank >That's true of course, but to show an area on a sphere = the area of >its projection on the cylinder is itself a calculus (integration) >problem, no? >--Lynn No, Now that it's clear you are looking for the solid angle of a sphere with its polar caps lopped off, you can use Pappus' theorem which says to get the area, multiply the arc length by 2piR, the equatorial path, so you get a = atan(A/D) arc = 2aR Area = 2piR*2aR (and with R = 1), = 4 pi atan(A/D) steradians If you include sinus you get the cylindrical area equivalent, not reducible to steradians. If you have something to say, write an equation. If you have nothing to say, write an essay === Subject: Re: solid angle >>That's true of course, but to show an area on a sphere = the area of >>its projection on the cylinder is itself a calculus (integration) >>problem, no? >>--Lynn >No, Now that it's clear you are looking for the solid angle of a >sphere with its polar caps lopped off, you can use Pappus' theorem >which says to get the area, multiply the arc length by 2piR, the >equatorial path, so you get > a = atan(A/D) > arc = 2aR > Area = 2piR*2aR (and with R = 1), > = 4 pi atan(A/D) steradians > >If you include sinus you get the cylindrical area equivalent, not >reducible to steradians. >If you have something to say, write an equation. >If you have nothing to say, write an essay I'm not sure whether your response is directed at Frank or myself, but I will stick by my statement that to show an area on a sphere = the area of its projection on the cylinder is itself a calculus (integration) problem. Even though the case of this particular area can be calculated via Pappus' theorem, I would guess, not having looked at the proof recently, that you are just substituting one calculus result with another by using Pappus' theorem. And who would know whether the original poster knew Pappus' theorem? --Lynn === Subject: Re: solid angle >That's true of course, but to show an area on a sphere = the area of >its projection on the cylinder is itself a calculus (integration) >problem, no? >--Lynn >>No, Now that it's clear you are looking for the solid angle of a >>sphere with its polar caps lopped off, you can use Pappus' theorem >>which says to get the area, multiply the arc length by 2piR, the >>equatorial path, so you get >> a = atan(A/D) >> arc = 2aR >> Area = 2piR*2aR (and with R = 1), >> = 4 pi atan(A/D) steradians >> >>If you include sinus you get the cylindrical area equivalent, not >>reducible to steradians. >>If you have something to say, write an equation. >>If you have nothing to say, write an essay >I'm not sure whether your response is directed at Frank or myself, but >I will stick by my statement that to show an area on a sphere = the >area of its projection on the cylinder is itself a calculus >(integration) problem. Even though the case of this particular area >can be calculated via Pappus' theorem, I would guess, not having >looked at the proof recently, that you are just substituting one >calculus result with another by using Pappus' theorem. And who would >know whether the original poster knew Pappus' theorem? >--Lynn I goofed. I need to retract the Pappus' schema; it needs the centroid of the arc to traverse the path. So I (cough, cough) integrated Int(2piR^2cos a da) = sin atan(A/D)*2pi *2 (R = 1) But, ding, ding, sin(atan) can be simplified so we get Angle = 4pi/sqrt(1 + D^2/A^2) This seems more appealing. John Polasek If you have something to say, write an equation. If you have nothing to say, write an essay === Subject: quantifier conundrum i am reading d. velleman's 'how to prove it' and have come to an an exercise from the book below will illustrate my point. where && is conjunction, || is disjunction, ! is not, E is the existential quantifier and A is the universal quantifier. the book gives: let T(x,y) mean x teaches y ExEy [ T(x,y) && !EuEv( T(u,v) && ( u != x || v != y ) ) ] ExEy [ T(x,y) && AuAv!( T(u,v) && ( u != x || v != y ) ) ] ExEy [ T(x,y) && AuAv( !T(u,v) || !( u != x || v != y ) ) ] ExEy [ T(x,y) && AuAv( !T(u,v) || ( u = x && v = y ) ) ] and several transformations later which i hope are correct, we have: ExEy [ T(x,y) && AuAv( T(u,v) -> ( u = x && v = y ) ) ] which i'd read as: there exist one or more x's and one or more y's such that x teaches y and for all people u and all people v, if u teaches v then u is x and v is y. is that the right way to read this statement? anyway the main problem i'm having is that i'm still quite murky on exactly what x and y represent in this statement, because the existential quantifier guarantees that there be one or more things in existence, so where it says u = x && v = y, which x and y is this referring to? in other words there may exist several x's and y's where x teaches y, so if for all u and v if u teaches v does that mean that u = the several x, and v the several y? === Subject: Re: quantifier conundrum |ExEy [ T(x,y) && AuAv( T(u,v) -> ( u = x && v = y ) ) ] [...] |anyway the main problem i'm having is that i'm still quite murky on |exactly what x and y represent in this statement, because the |existential quantifier guarantees that there be one or more things in |existence, so where it says u = x && v = y, which x and y is this |referring to? Read the formula from the inside out. The subformula T(x,y) && AuAv( T(u,v) -> ( u = x && v = y ) ) has two free variables in it, x and y, so it describes a property of two given individuals x and y. Namely, it says that x teaches y, and that x is the only teacher and y is the only student (to paraphrase a little). In relation to this subformula, x and y both refer simply to some given individuals. To get a definite statement one would have to supply the two individuals. Now apply the existential quantifier Ey. You get a formula with just one free variable, x, which says something about x. Namely, it says that x has (one or more) student(s) who have the relationship described in the previous paragraph to x. But of course there can be only one such student y. So it still means that x is the only teacher, and x has only one student. Now apply the existential quantifier Ex. The resulting sentence is closed, i.e. no free variables, and says that one or more people have the property described in the previous paragraph. But of course there can be only one such teacher x. So it says there's a unique teacher, and that the one teacher has only one student. Keith Ramsay === Subject: Re: quantifier conundrum > where && is conjunction, || is disjunction, ! is not, E is the > existential quantifier and A is the universal quantifier. > let T(x,y) mean x teaches y > ExEy [ T(x,y) && !EuEv( T(u,v) && ( u != x || v != y ) ) ] > ExEy [ T(x,y) && AuAv!( T(u,v) && ( u != x || v != y ) ) ] > ExEy [ T(x,y) && AuAv( !T(u,v) || !( u != x || v != y ) ) ] > ExEy [ T(x,y) && AuAv( !T(u,v) || ( u = x && v = y ) ) ] > and several transformations later which i hope are correct, we have: > ExEy [ T(x,y) && AuAv( T(u,v) -> ( u = x && v = y ) ) ] Nicely done. > which i'd read as: There's a x and a y such that T(x,y) and for all u,v if T(u,v) then u = x, v = y There's a x and a y such that T(x,y) and for all u,v for which T(u,v), u = x, v = y. There's a x and a y such that T(x,y) and any other u,v with T(u,v) are x and y. There's a x and a y such that T(x,y) and x and y are the only such pair. There is a unique x and y such that T(x,y) > there exist one or more x's and one or more y's such that x teaches y > and for all people u and all people v, if u teaches v then u is x and > v is y. > is that the right way to read this statement? Very clumsy. The expression there's one or more isn't preferred except perhaps for lax minds. Better and clearer is simply there is a. In contrast, there's exactly one. There is one however may have some ambiguity about it. > anyway the main problem i'm having is that i'm still quite murky on > exactly what x and y represent in this statement, because the > existential quantifier guarantees that there be one or more things in > existence, so where it says u = x && v = y, which x and y is this > referring to? in other words there may exist several x's and y's where > x teaches y, so if for all u and v if u teaches v does that mean that u > = the several x, and v the several y? x and y are elements in the range of the quantifiers. The statement says there is a x and a y such that T(x,y) and ... The and ... part says that if there's a u and a v with T(u,v), then they're x and y. That one or more stuff I consider over complex, distracting from, rather than adding to clarity. I suggest you think (Ex) there is a x and (E!x) there is exactly one x or there is a unique x and avoid thinking (Ex) there is one x as mildly ambiguous and (Ex) there is or are, one or more than one, x or x's to give it's full and rigorous grammatical due, as abstrusely absurd. ;-) Do you have a dollar? Yes I have a dollar. Do you have five? No, only two. There you see, a already means one or more. === Subject: help! vector differentiation! Hi all, Can anybody help me on how to differentiate function of vectors with respect to a vector? For example, I know d(w'*A*w)/dw=2w'*A where w is a column vector, A is symmetrical matrix, w' denotes the transpose. w'*A*w is a scalar function of vector w. The derivative of w'*A*w w.r.t. w = 2w' * A The second derivative of w'*A*w w.r.t. w = 2 A ----------------------------------- But what should be the first and second derivative of (w'*A*w)*w'*B? The reason I ask this because I want to find second derivative for (w'*B*w) / (w'*A*w) === Subject: Re: help! vector differentiation! > Hi all, > Can anybody help me on how to differentiate function of vectors with respect > to a vector? > For example, > I know d(w'*A*w)/dw=2w'*A > where w is a column vector, A is symmetrical matrix, w' denotes the > transpose. > w'*A*w is a scalar function of vector w. > The derivative of w'*A*w w.r.t. w = 2w' * A > The second derivative of w'*A*w w.r.t. w = 2 A > ----------------------------------- > But what should be the first and second derivative of (w'*A*w)*w'*B? > The reason I ask this because I want to find second derivative for > (w'*B*w) / (w'*A*w) This result and many others are easily derived by using Ricci notation, instead of the cumbersome matrix notation. This particular result is in a short note on my website: http://www.numerical-algorithms.com/ in the PDF dpocument: http://www.numerical-algorithms.com/notes/ortho.pdf equation (58). In the usual matrix notation and a fixed font this works out to: xT.A.x a F = ------ = - xT.B.x b Fkl = a [ 2 2 2 2 T T 2 2 T ] - [ - A - - B - - - [(Ax)(Bx)+(Bx)(Ax)] + 2 - - (Bx)(Bx) ] b [ a b a b b b ] T a = x.A.x T b = x.B.x This by the way is a case of a matrix (A-lambda*B) perturbed by an outer product of vectors whose inverse can be expressed in terms of the inverse of (A-B). Jentje Goslinga === Subject: Re: help! vector differentiation! >> Hi all, >> Can anybody help me on how to differentiate function of vectors with >> respect to a vector? >> For example, >> I know d(w'*A*w)/dw=2w'*A >> where w is a column vector, A is symmetrical matrix, w' denotes the >> transpose. >> w'*A*w is a scalar function of vector w. >> The derivative of w'*A*w w.r.t. w = 2w' * A >> The second derivative of w'*A*w w.r.t. w = 2 A >> ----------------------------------- >> But what should be the first and second derivative of (w'*A*w)*w'*B? >> The reason I ask this because I want to find second derivative for >> (w'*B*w) / (w'*A*w) > This result and many others are easily derived by using Ricci > notation, instead of the cumbersome matrix notation. > This particular result is in a short note on my website: > http://www.numerical-algorithms.com/ > in the PDF dpocument: > http://www.numerical-algorithms.com/notes/ortho.pdf > equation (58). > In the usual matrix notation and a fixed font this works out to: > xT.A.x a > F = ------ = - > xT.B.x b > Fkl = > a [ 2 2 2 2 T T 2 2 T ] > - [ - A - - B - - - [(Ax)(Bx)+(Bx)(Ax)] + 2 - - (Bx)(Bx) ] > b [ a b a b b b ] > T > a = x.A.x > T > b = x.B.x > This by the way is a case of a matrix (A-lambda*B) perturbed by an > outer product of vectors whose inverse can be expressed in terms > of the inverse of (A-B). > Jentje Goslinga Hi Jentje, That's interesting... I'd like to learn Ricci notation if it outweighs the cubersom matrix-vector notation... But the paper on your website is not openable by my Acroreader 6.0, what is the problem? === Subject: Re: help! vector differentiation! > Hi Jentje, > That's interesting... I'd like to learn Ricci notation if it outweighs > the cubersom matrix-vector notation... Well, there isn't much to learn about it, try calculating the simple result you wanted yourself starting with: Q = (akl xk xl)/(bmn xm xn) = a/b (no free indices) Qi = d/dxi Q Qi = [ail xl + aki xk]/b - (a/b^2) [bmi xm + bin xn] = [ail xl + aik xk]/b - (a/b^2) [bim xm + bin xn] = 2 (a/b) [(aik xk)/a - (bik xk)/b] (because of the symmetry of A and B) Qij = d/dxj Qi Qij = 2 (a/b) [aij/a - bij/b - (aik xk)/a^2 (2 ajk xk)... ] + 2 [(aik xk)/a - (bik xk)/b] d/dxj (a/b) Qij = 2 (a/b) [aij/a - bij/b - (aik xk)/a^2 (2 ajk xk)... ] + 2 [(aik xk)/a - (bik xk)/b] * 2(a/b)[(ajk xk)/a - (bjk xk)/b] (subst from previous) and so on. You can use the Indicial Tensor package in Maxima or Macsyma for matrices without too many problems. Some of the results in my little note on orthogonal matrices I managed to derive by machine in this fashion. Sorry, the result I gave you is for symmetric matrices which I think is more or less implied by your question, at least the denominator matrix (usually called B) in the generalized eigenvalue problem is normally assumed to be symmetric and positive definite, some kind of a norm. For non-symmetric matrices you just retain aik xk and aki xk separately throughout the calculation and translate them back as Ax and ATx. > But the paper on your website is not openable by my Acroreader 6.0, what is > the problem? No idea; I tried it, it works fine for me with Acro 4.05. No problem, glad to have been of help, Jentje Goslinga === Subject: Re: Jac(O), O = integers in a number field : Assume K/Q is a number field, and let O denote the integers in this : extension. : Of course, O is a Noetherian, integrally closed, domain of (Krull) : dimension 1, aka Dedekind domain. In particular, every nonzero prime : ideal is maximal and so the only nonmaximal prime ideal in O is the : zero ideal. : My question: what is the Jacobson radical, Jac(O)? Use the fact that the intersection of all prime ideals of a commutative ring is equal to the ideal of nilpotent elements in the ring. Ted Hwa === Subject: Re: Jac(O), O = integers in a number field > : Assume K/Q is a number field, and let O denote the integers in this > : extension. > : Of course, O is a Noetherian, integrally closed, domain of (Krull) > : dimension 1, aka Dedekind domain. In particular, every nonzero prime > : ideal is maximal and so the only nonmaximal prime ideal in O is the > : zero ideal. > : My question: what is the Jacobson radical, Jac(O)? > Use the fact that the intersection of all prime ideals of a commutative > ring is equal to the ideal of nilpotent elements in the ring. > Ted Hwa I do not see how this helps--although what you say is certainly correct, it applies to all commutative rings, even those where the Nilradical is *strictly* contained in the Jacobson radical. Here, however, I wish to prove that the Nilradical is equal to the Jacobson radical. Can you give some more details please? Jenny === Subject: Re: Jac(O), O = integers in a number field >> >> : Assume K/Q is a number field, and let O denote the integers in this >> : extension. >> >> : Of course, O is a Noetherian, integrally closed, domain of (Krull) >> : dimension 1, aka Dedekind domain. In particular, every nonzero prime >> : ideal is maximal and so the only nonmaximal prime ideal in O is the >> : zero ideal. >> >> : My question: what is the Jacobson radical, Jac(O)? >> Use the fact that the intersection of all prime ideals of a commutative >> ring is equal to the ideal of nilpotent elements in the ring. >> Ted Hwa >I do not see how this helps--although what you say is certainly >correct, it applies to all commutative rings, even those where the >Nilradical is *strictly* contained in the Jacobson radical. Here, >however, I wish to prove that the Nilradical is equal to the Jacobson >radical. Can you give some more details please? I think you can solve your original problem as follows. Let a be a nonzero element of O. Choose a prime number p (in Z) not dividing N(a), and let P be a prime ideal of O containing

. Then the norm of P (i.e. the index of P in O) is a power of p, whereas the index of in O is coprime to p, so cannot be contained in P, and hence a is not in P. Derek Holt. === Subject: Re: Ring of continuous real functions > The expression 'algebra' is non-descriptive, even misleading, > while a proper description 'vector ring' is clear and helpful. Not true. The expression 'algebra' is overloaded, I give you that:) However, it is hardly ever misleading in a given context. OTOH, the term 'vector ring' would lead to confusion: what do you mean? an algebra? Mathematics is full of overloaded terminology. While this is not an ideal state of affairs, the alternative would be to use words bearing no relation whatsoever to the vocabulary of the natural languages. Count the various (mathematical) meanings of the words 'degree' and 'normal' that you are familiar with. Any confusions? Hope not! Enjoy your algebra, and learn to live with overloaded terminology! Jyrki Lahtonen, Turku, Finland === Subject: Re: Ring of continuous real functions Count the various (mathematical) meanings of the words > 'degree' and 'normal' that you are familiar with. Any confusions? > Hope not! Oops! May be 'degree' isn't a particularly useful example. In my native (mathematical) Finnish we use the same word for both 'degree' and 'rank' (in pretty much all the meanings of mathematical English). Now there's some serious overloading:) Jyrki === Subject: Re: Ring of continuous real functions Oops! May be 'degree' isn't a particularly useful example. In my > native (mathematical) Finnish we use the same word for both > 'degree' and 'rank' (in pretty much all the meanings of mathematical > English). Now there's some serious overloading:) What's the correct spelling for this sentence. In English the word tu has three different spellings. ;-) A train arrived at two minuts to two and departed two minute after two. Another train also arrived at two minuts to two and departed two minutes after two. The first train was at the station from two to two to two two. The second train was at the station from two to two to two two too. Were those two trains too overloaded? ;-) === Subject: Re: Ring of continuous real functions [snipped joke] :) Subtle (and not so subtle) puns are mostly wasted on me (especially when I'm down with a flu) so I only figured out your point late last night. Ok, help me out, please. What's the correct word? overladen, overlaid, something else? Jyrki === Subject: Re: Ring of continuous real functions Ok, help me out, please. What's the correct word? > overladen, overlaid, something else? overburdened, overlaid. === Subject: Re: Ring of continuous real functions > overburdened, overlaid. Jyrki === Subject: Re: Ring of continuous real functions > overburdened, overlaid. I'm not quite sure what point William is trying to make here. The verb to overload is indeed the usual English term for what you were describing, Jyrki, especially in a technical context. -- Jim Heckman === Subject: Re: Ring of continuous real functions > I'm not quite sure what point William is trying to make here. > The verb to overload is indeed the usual English term for what > you were describing, Jyrki, especially in a technical context. surprisingly (?) didn't have the verb to overload, so I got rather worried. My copy of Webster's Thesaurus does have it, and it describes the words meaning the same way as you do. My other worry was due to the fact that I had misremembered a (related) term overlay used in a computer programming context. It's been 14 years since I lived in an English speaking country, but I still feel like I sort of know English. Therefore an occasional reminder that I'm not at the level of an educated native speaker is needed and, indeed, good for me:) Jyrki === Subject: Re: Ring of continuous real functions === > Subject: Re: Ring of continuous real functions [...] >>Remember? Oh yea, linear algebra gives 'algebra' a parochial >> meaning. Properly then isn't an 'algebra' just an inner product >> space? > ??? No. Completely different things. Most inner product spaces > don't have amultiplication. The 'inner product' of two vectors > is a scalar for an inner product space. For algebras, the product > of two vectors is another vector. > An algebra A is a vector space A with a vector product? Not just any vector product, but a *bilinear* vector product. I.e., x(ay + bz) = a(xy) + b(xz) and (ax + by)z = a(xz) + b(yz) for all x,y,z in A and a,b in F. I already went over this with you in alt.math.undergrad some time ago. > R^3 with the cross product is an algebra? > An algebra with a non-associative, anti-commutative product? Of course. R^3 with the cross product is just the Lie algebra so(3). A Lie algebra is one satisfying x^2 = 0 and the Jacobi identity x(yz) + y(zx) + z(xy) = 0. > Or is an algebra A is a vector space A with a vector product that is > both sides distributive over addition? In other words a vector ring. Ring usually implies associativity, which is not part of most people's default definition of an algebra, although when dealing with associative algebras many authors will say something like associative algebra. > Four dimensional vectors with matrix multiplication is an example > of a non-commutative vector ring. > For vectors x,y and scalar a, a needed axiom is > a(xy) = (ax)y = x(ay) ? Just a special case of bilinearity. (See definition above.) > Is a (vector) multiplicative identity required? No. But again many authors, when dealing with such algebras, multiplicative identity [or 1]. > The expression 'algebra' is non-descriptive, even misleading, Not if it's well defined, which it is. > while a proper description 'vector ring' is clear and helpful. See above. -- Jim Heckman === Subject: Re: Ring of continuous real functions === Subject: Re: Ring of continuous real functions William Elliot An algebra A is a vector space A with a vector product? > Not just any vector product, but a *bilinear* vector product. > I.e., x(ay + bz) = a(xy) + b(xz) and (ax + by)z = a(xz) + b(yz) > for all x,y,z in A and a,b in F. I already went over this with > you in alt.math.undergrad some time ago. You did? No did find. > R^3 with the cross product is an algebra? > Of course. R^3 with the cross product is just the Lie algebra > so(3). A Lie algebra is one satisfying x^2 = 0 and the Jacobi > identity x(yz) + y(zx) + z(xy) = 0. A noncommutative, nonassociative algebra. > Or is an algebra A is a vector space A with a vector product that > is both sides distributive over addition? In other words a vector > ring. > Ring usually implies associativity, which is not part of most > people's default definition of an algebra, although when dealing > with associative algebras many authors will say something like > associative algebra. So distributivity is the only assured ring property. > For vectors x,y and scalar a, a needed axiom is > a(xy) = (ax)y = x(ay) ? > Just a special case of bilinearity. (See definition above.) Ok, how's this? Have I got it straight? An algebra is a vector space with a bilinear vector product. That is, for all vectors x,y,z, scalars a, a(xy) = (ax)y = x(ay) x(y + z) = xy + xz, (x + y)z = xz + yz Matrices for example are an associative, noncommutative algebra with identity. If F is a field, then F, F[x], F[[x]], F(x) are F-algebras with scalars F. They are also algebras with scalars the prime subfield or any subfield of F. ---- === Subject: Re: Ring of continuous real functions [...] > Ok, how's this? Have I got it straight? > An algebra is a vector space with a bilinear vector product. > That is, for all vectors x,y,z, scalars a, > a(xy) = (ax)y = x(ay) > x(y + z) = xy + xz, (x + y)z = xz + yz Yes. > Matrices for example are an > associative, noncommutative algebra with identity. *Square* matrices over a *field*, yes. The algebra of nxn matrices over the field F, often denoted by M_n(F) or M(n,F), is extremely important in the theory of associative algebras, since it can shown that every n-dimensional associative algebra A over F is isomorphic to a subalgebra of M(n,F) if A has an identity, or of M(n+1,F) if not. > If F is a field, then F, F[x], F[[x]], F(x) are F-algebras with scalars F. > They are also algebras with scalars the prime subfield or any subfield of > F. Yes. -- Jim Heckman === Subject: Ring of continuous real functions > An algebra is a vector space with a bilinear vector product. > That is, for all vectors x,y,z, scalars a, > a(xy) = (ax)y = x(ay) > x(y + z) = xy + xz, (x + y)z = xz + yz > Matrices for example are an > associative, noncommutative algebra with identity. > *Square* matrices over a *field*, yes. The algebra of nxn > matrices over the field F, often denoted by M_n(F) or M(n,F), is > extremely important in the theory of associative algebras, since > it can shown that every n-dimensional associative algebra A over > F is isomorphic to a subalgebra of M(n,F) if A has an identity, > or of M(n+1,F) if not. So either way it's isomorphic to a subalgebra of M(n+1,F) A subalgebra is a + and * closed subset of the algebra with the same scalars and of course closed by scalar multiplication. Examples of infinite dimensional associative algebras are C(R) and { f:R -> R } with pointwise *. === Subject: Re: This Week's Finds in Mathematical Physics (Week 209) > This Week's Finds in Mathematical Physics - Week 209 > John Baez > Time flies! This June, Peter May and I organized a workshop on > n-categories at the Institute for Mathematics and its Applications: Do you consider category theory to be a branch of physics??? Or where is the physics in This Week's Finds? Arnold Neumaier === Subject: Re: This Week's Finds in Mathematical Physics (Week 209) >> This Week's Finds in Mathematical Physics - Week 209 >> John Baez >> Time flies! This June, Peter May and I organized a workshop on >> n-categories at the Institute for Mathematics and its Applications: > Do you consider category theory to be a branch of physics??? > Or where is the physics in This Week's Finds? Who cares? Category Theory is good mathematics, so please keep it coming on sci.math. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Parametrizations of U(2) >U(2) (the group of unitary 2x2 matrices) is a smooth manifold of >dimension 4 (over the real numbers), ie. there exist an atlas of (local) >diffeomorphisms mapping some subset of R^4 into U(2), such that U(2) is >completely covered. >For example the subgroup SU(2) can be parametrized with >{{z_1, z_2},{-(z_2)*,z_1*}} s.t. |z_1|^2 + |z_2|^2 = 1, >where z_1 and z_2 are complex numbers. I'm looking for parametrizations >of the remaining unitary matrices in U(2)SU(2) with four real parameters. > What's wrong with the definition? A matrix M = {{z_1, z_2},{z_3,z_4}} > lies in U(2) iff M* M = I, i.e. > (z_1)* (z_1) + (z_3)* (z_3) = 1 > (z_1)* (z_2) + (z_3)* (z_4) = 0 > (z_2)* (z_1) + (z_4)* (z_3) = 0 > (z_2)* (z_2) + (z_4)* (z_4) = 1 > The middle two equations say the same thing: roughly, that the ratios > (z_4)/(z_1)* and -(z_2)/(z_3)* are equal, to r , say. That is, > z_4 = r (z_1)*, z_2 = -r (z_3)* . The first and last equations together > then say that |r| = 1. So now you have a parameterization > {{z_1, -r (z_3)*},{z_3, r(z_1)*}} s.t. |z_1|^2 + |z_2|^2 = 1, |r| = 1. > Your parameterization of SU(2) is contained in this; SU(2) is the > set of matrices with r = 1. Dave, Andor === Subject: Re: the modesty of Mr Wolfram. On the Foundation of mathematics and mathematica > [ ... ] I mean, _I_ could easily have > written Mathematica if I'd felt like it [ ... ] No, you couldn't. Han de Bruijn === Subject: A restatement of the anti-Wolfram argument > [ ... ] I mean, _I_ could easily have > written Mathematica if I'd felt like it [ ... ] > No, you couldn't. Han has a good point. If you tried to write something remotely similar to MATLAB or Mathematica from scratch you would wind up spending long years implementing various numerical and symbolic algorithms, designing a language, UI, etc. It would take years, really, even for the best programmer on earth. That is a counter-argument to Wolfram's work. Chaitin praises Mathematica, he says it's a substantial AI for mathematics. I agree with Chaitin in that Mathematica knows a good deal about mathematics, but it probably lacks anything that could be called mathematical *reasoning*. The problem is Wolfram's own argument that reality is in fact based on very simple algorithms. Why is Mathematica so complex then? Maybe, it's because mathematics is no simple thing? And if so, since mathematics is part of human reality, then it turns out that Wolfram was wrong in his argument for universal simplicity, that complexity is an illusion. Complexity is a real thing that our minds are made to deal with. Our brains aren't there for nothing. And this is easy to see: the shortest program that can implement something like Mathematica on a barebones universal computer is still quite long! -- Eray Ozkural === Subject: Re: A restatement of the anti-Wolfram argument , ...... then it turns out that Wolfram was wrong in his argument for universal simplicity, that complexity is an illusion. In no part of A New Kind of Science, Wolfram suggests that complexity is an illussion. His thesis is that complexity arises from the interaction of simple laws. In the same sense that the mechanics of the Solar System is governed by the 3 laws of Galileo and the 'Newton's law of Gravitation. His book is a disprover of the false Chaitin's mesure of complexity: The complexity of a sequence of numbres is mesured by the length of the minimum program that can reporduce the sequence Here is a program that contradicts that thesis: T = -3.14 : D = .01 : Y = 0 DO WHILE T < 3.14 X = T + SIN(5*Y) Y = T - COS(2*X) DRAW(X,Y) T = T + D LOOP This is a curve perfectly simetric and periodic, that is of low complexity. But if you change the Y for X in the third line and de X foy in the fourth, it results a chaotic curve without periodicity. The length of the two programs are the same but not its complexity. Refer. G. CHAITIN Randomnes ans Mathematical Proof Scientific American 1975 Exploring Randomness Springer 2001 I. STEWART Does God Play Dice? Blackwell 1989 === Subject: Re: A restatement of the anti-Wolfram argument > The problem is Wolfram's own argument that reality is in fact based on > very simple algorithms. Why is Mathematica so complex then? Maybe, > it's because mathematics is no simple thing? Wolfram's point seems to be that all that complex mathematics is redundant: in order to effectively understand and manipulate the real world, much simpler ideas might suffice, or even be more useful. (again: W's point, not mine). And if so, since > mathematics is part of human reality, Perhaps -human- reality is not what W. is talking about. then it turns out that Wolfram > was wrong in his argument for universal simplicity, that complexity is > an illusion. Complexity is a real thing that our minds are made to > deal with. Our brains aren't there for nothing. And this is easy to > see: the shortest program that can implement something like > Mathematica on a barebones universal computer is still quite long! Perhaps Mathematica is complex only because -humans- have made their own lives and thoughts too complex (in other words, because mathematics is unnecessarily complex). (Again: attempts to present W's pov, not my pov). -- Herman Jurjus === Subject: Re: A restatement of the anti-Wolfram argument > The problem is Wolfram's own argument that reality is in fact based on > very simple algorithms. Why is Mathematica so complex then? Maybe, > it's because mathematics is no simple thing? > Wolfram's point seems to be that all that complex mathematics is > redundant: in order to effectively understand and manipulate the real > world, much simpler ideas might suffice, or even be more useful. (again: > W's point, not mine). > And if so, since > mathematics is part of human reality, > Perhaps -human- reality is not what W. is talking about. > then it turns out that Wolfram > was wrong in his argument for universal simplicity, that complexity is > an illusion. Complexity is a real thing that our minds are made to > deal with. Our brains aren't there for nothing. And this is easy to > see: the shortest program that can implement something like > Mathematica on a barebones universal computer is still quite long! > Perhaps Mathematica is complex only because -humans- have made their own > lives and thoughts too complex (in other words, because mathematics is > unnecessarily complex). (Again: attempts to present W's pov, not my pov). Yes, I understand W.s point, but I also think he's overlooking the enormous amount of complexity created by billions of years of evolution. I don't think Wolfram understands that the life processes are very detailed and complex processed that cannot really be dumbed down to 1D CAs. Every few years, we find new stuff in neuroscience and molecular biology that suggests that we should multiply the complexity of some things we thought were simple by ten. Regardless of experimental findings, if W. were right, then it should have been the case that Mathematica is really only seemingly complex, and its real complexity, program-size complexity should have been extremely small. But by any stretch of imagination, you can't compress, with current human knowledge, all of Mathematica to less than a hundred kilobytes, and that is a remarkable complexity. (Maybe the absolute compressibility is much larger, probably so, but it's not a Rule 110 or anything like that) Also for the brain, W. seems to suggest that the brain is the result of a very very tiny program, which I find myself in sheer disagreement with. I am certain that it is optimized very well, but it's not 10 bits. And one other point, surely a {0,1}* generator can create all the complexity in the world. But that takes time and space! An inordinate amount of time and space! It is COMPLETELY IRRELEVANT to claim that since there is such a generator, COMPLEXITY is an ILLUSION. I would not write a volume to explain the previous two sentences! What a waste of resources! -- Eray === Subject: Re: A restatement of the anti-Wolfram argument Discussion, linux) >> [ ... ] I mean, _I_ could easily have >> written Mathematica if I'd felt like it [ ... ] >> No, you couldn't. > Han has a good point. He might have a good point, if anyone with two brain cells to rub together thought David was serious. -- Jesse F. Hughes [Iota]'s the smallest infinitesimal, Russell, there are smaller infinitesimals. -- Ross Finlayson === Subject: Re: A restatement of the anti-Wolfram argument >> >> [ ... ] I mean, _I_ could easily have >> written Mathematica if I'd felt like it [ ... ] >> >> No, you couldn't. > Han has a good point. > He might have a good point, if anyone with two brain cells to rub > together thought David was serious. Well, then let's see what David says. That statement was taken out of quote, he was really poking fun about some other issue, but there is no reason why he should not be half-serious. -- Eray Ozkural === Subject: Re: A restatement of the anti-Wolfram argument <878y8pvhx2.fsf@phiwumbda.org> Discussion, linux) > > [ ... ] I mean, _I_ could easily have > written Mathematica if I'd felt like it [ ... ] > > No, you couldn't. > Han has a good point. >> He might have a good point, if anyone with two brain cells to rub >> together thought David was serious. > Well, then let's see what David says. That statement was taken out of > quote, he was really poking fun about some other issue, but there is > no reason why he should not be half-serious. Maybe it's simply a bit of cultural knowledge, but I believe any construction of the form I could easily have do [some monumental task] if I'd felt like it is a joke. Unless James Harris says it. Of course, we all have our suspicions about James S Harris == David C. Ullrich, so he might surprise me here. -- I arrest anybody I think needs arresting, Mr. Carter, and I'm not in the habit of explaining why. There's a law about that --- You're in Dodge, Mr. Carter. -- Gunsmoke radio show (A Bush favorite) === Subject: Re: the modesty of Mr Wolfram. On the Foundation of mathematics and mathematica ~^>Pn0&%&Ux8>1=w8P?^q%:g?%]2+oVLC;x!s,~MYjl!j>x`k>b9B5_NaM'4_X:z Zw76-- > I asked here once, does Wolfram have any important theorem whatsoever > about cellular automata, or all he has is a handful of CA runs? The > important theorems I remember were found by some other people, maybe > it's my memory that does not serve me right. > I'm asking this, because I want to decide if I should buy the book, > and that is the litmus test for me. If he has any significant > theorems, then the book would be worth reading. Otherwise, it's good > for knocking out your roommate. I saw it. It's thick, and it has > pretty pictures, but that's not sufficient for me. If your criteria for buying NKS is it needs to have some new important theorem in it, then you shouldn't buy it. But this criteria would eliminate a fairly sizable portion of quite useful books. I can think of a couple of good reasons to buy the book. First, if you have an interest in cellular automata, NKS covers a wide variety of cellular automata and shows examples of how they can be related to a fairly wide variety of problems. So, NKS is a quite useful reference for anyone interested in cellular automata. Second, whether you accept the concepts presented in NKS as a new kind of science or not they clearly represent a different way of thinking about a variety. And it is quite useful to having a fresh viewpoint whether you agree with the viewpoint or not. Simply thinking about why an argument is or is not valid is quite useful. But if you can't get past Wolfram's style in this book, can't stand Mathematica or insist on rigorous proof of all concepts presented then you will definitely not be happy with this book. -- To reply via email subtract one hundred nine === Subject: Re: the modesty of Mr Wolfram. On the Foundation of mathematics and mathematica > I'm asking this, because I want to decide if I should buy the book, > and that is the litmus test for me. If he has any significant > theorems, then the book would be worth reading. Otherwise, it's good > for knocking out your roommate. I saw it. It's thick, and it has > pretty pictures, but that's not sufficient for me. > If your criteria for buying NKS is it needs to have some new important > theorem in it, then you shouldn't buy it. But this criteria would > eliminate a fairly sizable portion of quite useful books. > I can think of a couple of good reasons to buy the book. First, if you > have an interest in cellular automata, NKS covers a wide variety of > cellular automata and shows examples of how they can be related to a > fairly wide variety of problems. So, NKS is a quite useful reference for > anyone interested in cellular automata. Yes, I'm interested in CAs. > Second, whether you accept the concepts presented in NKS as a new kind > of science or not they clearly represent a different way of thinking > about a variety. And it is quite useful to having a fresh viewpoint > whether you agree with the viewpoint or not. Simply thinking about why > an argument is or is not valid is quite useful. In fact, I agree that a computational POV is useful. > But if you can't get past Wolfram's style in this book, can't stand > Mathematica or insist on rigorous proof of all concepts presented then > you will definitely not be happy with this book. I started reading the online version but quickly grew suspicious of the content, and I couldn't continue reading because the online version hurt my eyes. I think, from such a title, I would be expecting a really major result or demonstration of some computational phenomena that would make me rethink everything. I might even agree that a purely experimental approach can be useful. Maybe he should just compress the relevant ideas minus the ranting to a 40-page paper.Compression is a sign of intelligence. -- Eray Ozkural === Subject: Re: the modesty of Mr Wolfram. On the Foundation of mathematics and mathematica >Say, now that people are talking about it, has anyone ever seen Mr.W >and JSH in the same room at the same time? Hm... > Hmm, indeed. Some might suggest that JSH's bitterness about the fact > that mathworld refuses to mention his work shows that your unstated > conjecture is full of beans. But it's clear to me that that's all > just a smokescreen. > Enhances one's respect for Wolfram. I mean, _I_ could easily have > written Mathematica if I'd felt like it, but I can't imitate JSH's > writing style - I've tried many times. Yeah right. You're not fooling anyone David. I mean is anyone stupid enough to believe that JSH is for real? Now we need someone with the mathematical and technical knowledge to become JSH. Also this person has to have been hanging around net^H^H^Hsci.math for long enough. Now consider: David Ullrich Started posting Feb 1995 James Harris Started posting Feb 1996 Both are still around. What more evidence do we need? -William Hughes === Subject: Re: Is absolute integral really the neccessary and sufficient condition for a system to be stable? >> For Bounded-In-Bounded-Output system, >> Is the absolute integrability the neccessary and sufficient condition for >> a >> system to be stable? >> That's to say: Integrate(abs(h(t)), t from -inf to inf) < +inf >> Any proof? >> I also want to know if this is the condition for LTI system only, or it >> is >> the condition for Linear systems or Time Invariant systems. >> Absolute integral is difficult to check, does it have any alternative >> equivalent condtions? > The very word impulse response, h(t) exists only for LTI systems. so > your formula for stability applies to LTI. The equivalent is :finding > zero input response of system i.e characteristic roots and modes. I > think thats universal i.e for stability of any system, charateristic > mode should --->0 when t--->inf. > A. Kumar Hi Kumar, modes for a system? and for (non)linear system? === Subject: question about stablility of systems Hi all, About the criteria of systems stablity: 1) h(t) absolutely integrable, for LTI systems. 2) the eigenvalues having negative real part(continous system). (for LTI systems only???) 3) the eigenvalues having magnitude less than 1(discrete system). (for LTI systems only???) what else criteria do we have? I am also wondering which of these critiria are applicable to LTI system only, which are applicable to Linear system, and which are applicable to non-linear systems... === Subject: Re: question about stablility of systems I did not mean that Lyapunov is not used for Linear Systems. I mean to say is, there is no need to take Lyapunov Function for Small Linear Systems. I appreciate you for suggestion Lyapunov Criteria. > Hi all, > About the criteria of systems stablity: > 1) h(t) absolutely integrable, for LTI systems. > 2) the eigenvalues having negative real part(continous system). (for LTI > systems only???) > 3) the eigenvalues having magnitude less than 1(discrete system). (for LTI > systems only???) > what else criteria do we have? > I am also wondering which of these critiria are applicable to LTI system > only, which are applicable to Linear system, and which are applicable to > non-linear systems... === Subject: Re: question about stablility of systems 4) It should be BIBO stable( Bounded Input and Bounded Output) says, for finte input the systme output should be finite. Athreya > Hi all, > About the criteria of systems stablity: > 1) h(t) absolutely integrable, for LTI systems. > 2) the eigenvalues having negative real part(continous system). (for LTI > systems only???) > 3) the eigenvalues having magnitude less than 1(discrete system). (for LTI > systems only???) > what else criteria do we have? > I am also wondering which of these critiria are applicable to LTI system > only, which are applicable to Linear system, and which are applicable to > non-linear systems... === Subject: Re: question about stablility of systems Negative real parts correspond to exponential decay, e^(-at). Any function with such a characteristic will result in transient changes dying away. Positive real parts correspond to exponential growth, e^(+at). Any function with such a characteristic will result in transient changes building up. This is a characteristic of all systems. However, if a non-linear system has bounds (BIBO) then any instability resulting from positive real will be limited also. > 2) the eigenvalues having negative real part(continous system). (for LTI > systems only???) === Subject: Re: matrix diagolization > In the diagonalization operation of a matrix A. > [Q Gamma] = eig(A) > Q= > 1.0000 0 -1.0000 0 > 0 1.0000 0 -1.0000 > 0 0 0.0000 0 > 0 0 0 0.0000 > Gamma= > -0.0113 0 0 0 > 0 -0.0253 0 0 > 0 0 -0.0113 0 > 0 0 0 -0.0253 > It seems that the eigenvalues are not different and the diagolization is > actually failed. > However, I was expecting the matix diagonalizable. I am wondering whether > there is any techniques dealing with the > matrix that can NOT be diagonalizable to simplify the math reasoning? > -- > -- > Yan ZHANG > http://www.geocities.com/iam_yanzhang/ Please show the matrix A, so that there is something to analyze. Assuming that the software was MATLAB: the singular or near-singular Q suggests repeated or near-repeated eigenvalues of A. But I repeat, what is A? === Subject: Re: matrix diagonalization > When a matrix A is diagonalizable, we usually write it as the following form > for the sake of mathematical reasoning. > A = P^(-1) Q P > I am hoping to know, when a SQUARE matrix B is NOT diagonalizable, what > technique should we usually employ in mathematical simplification? Single > value decomposation (SVD) or other techniques? > -- > Yan ZHANG > http://www.geocities.com/iam_yanzhang/ Schur decomposition. === Subject: approximate entropy I would appreciate any answers to me directly (george@netvision.net.il), george I am a math and science writer for the Swiss newspaper Neue Entropy measure. (Pincus, PNAS Vol. 88. 1991. See for example Ivar Peterson's (http://www.maa.org/mathland/mathtrek_10_11_04.html ) In the course of my research I have been hearing some criticism about ApEn. I would be very happy to receive anybody's comments. 1) ApEn seems like a reasonable measure of randomness or complexity. But one of the first tests of its applicability turned out to be counter-intuitive: the ApEn's of transcendental and algebraic numbers are quite intermingled. (Pincus and Kalmann, PNAS Vol. 94., 1997) At that point one option would have been to say that ApEn produces unreasonable results and should be thrown out. Rather, an artificially created puzzle remained in the literature. 2) Following that, many authors published papers which compute ApEn of various physiological variables (heart rhythm, EEG, hormone levels). The papers seem to be purely descriptive and non-refutable. The authors write up the results and leave it to future research to figure out why ApEn is high in certain instances and low in others. When computing a measure - any measure - one will always get some numerical results. 3) Recently, ApEn was applied to financial time-series (Pincus and Kalmann, be totally anecdotal: (A) two one-week periods of the Dow Jones index are picked out and their ApEn's are declared significantly different (without any indication of how significant). A more comprehensive comparison of all low ApEn periods versus high ApEn periods would have maybe given an indication of what is going on, but that was not done. (B) The evidence on the Hang-Seng index describes one episode other periods with high ApEn can be gleaned with no crash following, and vice versa. So? Furthermore, the evidence ends just after the crash in (C) The section on the Black and Scholes model seems to be in error since it claims that the B-S model implies Brownian motion of the underlying stock prices. But Brownian motion is an assumption of the B-S model, not an implication. -- -------- George Szpiro Neue Z.9frcher Zeitung (Switzerland) POB 6278 Jerusalem 91060 Israel === Subject: Re: approximate entropy Your post interested me, so I read up on a program in Mathematica to do the calculations.( there is an Matlab version available in a book by Edward Beltrami). I applied it to Hofstader's sequence, the last digits of the primes and Pi's digits and the Approximate entropy came out larger in that order. My program is really slow, but it does seem to give the ApEn function as defined in the paper. It is a lot like a Lyapunov Largest exponent in the way I've calculated it, but it more a probability measure on the variables than a direct result of the variables. It is also much harder and takes longer than a Lyapunov since it has two distinct sums in it. It is more closely related to correlation dimension than Kaplan-York dimension in it's method of calculation. My conclusion is that it is a good tool for understanding time series/ sequences and how random/ rough they are. It generally agrees with other methods in it's results. Clear [f,n,d,c,Phi,ApEn,a,i,j,k,r,m,g,digits] (*Steven M. Pincus,Approximate entropy as a measure of system complexity, PNAS,vol 88,pp2297-2301,March 1991,Mathematics*) digits=100 $MaxExtraPrecision =digits f[n_]:=Floor[Mod[10^n*Pi,10]] (* approximate Entropy for Pi digits sequence*) d[i_,j_,m_,n_]:=Max[Table[Abs[f[i+k-1]-f[j+k-1]],{k,1,m-1}]] c[i_,r_,m_,n_]:=N[Sum[If[d[i,j,m,n]True] y=Fit[a,{1,x},x] gb=Plot[y,{x,1,digits}] Show[{ga,gb}] >I would appreciate any answers to me directly (george@netvision.net.il), >george > I am a math and science writer for the Swiss newspaper Neue >Entropy measure. (Pincus, PNAS Vol. 88. 1991. See for example Ivar >Peterson's >(http://www.maa.org/mathland/mathtrek_10_11_04.html ) In the course of my >research I have been hearing some criticism about ApEn. I would be very >happy to receive anybody's comments. >1) ApEn seems like a reasonable measure of randomness or complexity. But >one of the first tests of its applicability turned out to be >counter-intuitive: the ApEn's of transcendental and algebraic numbers are >quite intermingled. (Pincus and Kalmann, PNAS Vol. 94., 1997) At that point >one option would have been to say that ApEn produces unreasonable results >and should be thrown out. Rather, an artificially created puzzle remained in >the literature. >2) Following that, many authors published papers which compute ApEn of >various physiological variables (heart rhythm, EEG, hormone levels). The >papers seem to be purely descriptive and non-refutable. The authors write up >the results and leave it to future research to figure out why ApEn is high >in certain instances and low in others. When computing a measure - any >measure - one will always get some numerical results. >3) Recently, ApEn was applied to financial time-series (Pincus and Kalmann, >be totally anecdotal: > (A) two one-week periods of the Dow Jones index are picked out >and their ApEn's are declared significantly different (without any >indication of how significant). A more comprehensive comparison of all low >ApEn periods versus high ApEn periods would have maybe given an indication >of what is going on, but that was not done. > (B) The evidence on the Hang-Seng index describes one episode >other periods with high ApEn can be gleaned with no crash following, and >vice versa. So? Furthermore, the evidence ends just after the crash in > (C) The section on the Black and Scholes model seems to be in >error since it claims that the B-S model implies Brownian motion of the >underlying stock prices. But Brownian motion is an assumption of the B-S >model, not an implication. -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : alternative email: rlbtftn@netscape.net URL : http://home.earthlink.net/~tftn === Subject: Re: (probably stupid) question on Fermat decidability > In sci.math, Peter Webb > . So if it's undecidable then it must be true, which > contradicts it being undecidable. So you get a contradiction, > therefore it cannot be undecidable. > Would anyone care to point out where I'm going wrong here? > The word undecidable is a translation of unentscheidbare, thus the >> undecidability must be Goedel's undecidability that is, formal >> undemonstrability >> within a system of axioms. A counterexample is not a formal >> demosntration. > Huh? > A counter example is most definitely a formal demonstration that something > is false. > If somebody shows me that > 15^235 + 126^235 = 127^235 > That establishes that FLT is untrue. > Sorry, can't oblige. > 15^235 + 126^235 - 127^235 = -2090269352984120073921722601405239849756 > 9870115681397780853008649734764326885583515335140989537921571270545556 > 9604711364290726616189649060384898822338675269590920961620286373126245 > 7900319179143756931647521906559845674970685046408230891391743862001266 > 1458414228168068171186039437154027547644676324833775228907190508149387 > 6230595833929891310280939770288390843783350735273423166543492008228388 > 7629825926635922080687553561414046616794750931246133230719835685424633 > 75358215282629964552579522143894992 > :-) > [rest snipped] Undecidability is its own conclusion. Cam === Subject: Re: Josh Purinton ANSWER THE QUESTION! === > Subject: Re: Cantor's Theory: Mathematical creationism > say we modify Tail to TRUE and Head to FALSE. > Then we can simplify the diag proof further. > Let G be a sequence of {TRUE, FALSE} such that: > G(i) = ! P_i(i) > Since G(i) = ! P_i(i), the sequence G differs from each sequence P_i > Is that also the major component of a proof that you can > find a new sequence of heads and tails if infinite people > were tossing coins infinite times each? If the coin tosses are numbered i=1,2,... and the people are numbered 1,2,... then this is a proof that there exists a sequence that is different from all P_i. - Randy === Subject: Maximal number of distinct factors Given a number n, is there a tight upper bound on the number of its distinct factors (need NOT be prime), d(n)? Specifically, is d(n) = O(logn)? abey Notes: 1. The average number of distinct prime factors, omega(n), is known to be ~loglogn [see http://mathworld.wolfram.com/DistinctPrimeFactors.html ]. So average d(n) is probably theta(logn). 2. When n is power of a prime number, then d(n) is trivally theta(logn). 3. When n is square-free i.e. it is a product of distinct primes, average omega(n) is known to be ~ logn/(loglogn) [see url above]. If this is also an upper bound, then maximal d(n) is ~ n^(1/loglogn). 4. I expect that maximal d(n) would occur for a distribution of prime factors of n that is a combination of the cases 2 and 3 above. ps: I offer my apologies if this happens to be a repost. === Subject: Re: Maximal number of distinct factors > Given a number n, is there a tight upper bound on the number of its > distinct factors (need NOT be prime), d(n)? > Specifically, is d(n) = O(logn)? No > ps: I offer my apologies if this happens to be a repost. It is, and so I post here my recent s.m.r. reply (hasn't yet turned up there ) See Hardy & Wright. In the 5-th edition: Theorem 314 states that d(n) is not O(log(n)^k) for any k. Theorem 315 states that d(n) is O(n^k) for any k > 0. Theorem 317 states that limsup (log d(n) log log n/log n) = log 2, and so for k > 0 d(n) < 2^{(1+k)log n/log log n} for all large n, and d(n) > 2^{(1-k)log n/log log n} for infinitely many n. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Maximal number of distinct factors thank you, Robin abey === Subject: Follow-up: Is this one a t-distribution? Hi all, this is a follow-up to a recent post of mine. I was asking wether f(x,y) = 1/pi * prod_(m=1)^L ( 1 / (1 + r_m*x^2 + r_m*y^2) ) * sum_(n=1)^L r_n / (1 + r_n*x^2 + r_n*y^2) is the density of a kind of bivariate Student-t distribution. To make things a bit simpler, it might also help to know if f(x,y) = C(a,b) * (1 + a*(x^2+y^2))^(-1) * (1 + b*(x^2+y^2))^(-1) represents a special form of a bivariate Student-t distribution? Above, a,b>0 and C(a,b) is just the normalization constant. For the special case a=b, it it easily seen that f(x,y) is bivariate t (with 2 degrees of freedom); ...but what if a is different from b, is it still some sort of t-distribution ?? Any help is welcome. Christian. === Subject: Answer the Logarithm problem I am posting here the website i have developed in orderso get something new in field of mathematics http://ece.nsit.ac.in/website/vipin/Ideas.htm === Subject: A question in matrix analysis Given n N-dimensional linearly independent vectors (set V1) (where N>n) I choose additional (N-n) vectors (set V2) and build the matrix M (of size NxN) such that: Its first n column vectors are the set V1. Its next n+1:N column vectors are the set V2. I require that the union set of V1 and V2 is linearly independent - thus the matrix M is invertible. Given the set V1 what choices of V2 are allowed such that the inverse of M consists from column vectors that non of them have zeros in their first n components? (some restrictions may also be given on V1 - I'm not sure - perhaps they can't be orthogonal - although they are linearly independent). === Subject: Re: question for math teachers >>rphenry@home.com comments on the sequence of jr.hi and hischool math courses: >The way geometry is taught, you need to know some algebra. > This is a bad idea. Euclid's students knew no algebra; it > had not yet been invented. I have to disagree- I worked with my daughter (and several of her friends) through last year's geometry class. Other than a few solve for x problems involving the Pythagorean Theorem (light pole this tall, shadow this long, how tall is the man casting the shadow sort of stuff) there was no algebra. There were, however, a lot of theorems, but really understanding *why* was not (to my mind) adequately taught. That's what I spent a lot of time with these girls on. As a result, I believe, they all scored directly at the top of the class. (as a result of course of their understanding, not just my tutoring and help). >The way algebra 2 is taught, you need to know some geometry. >Etc. while currently this is true, there is no reason I can think of that it needs to be. As for the forgetting of things over the intervening year- I disagree that it doesn't mean that they never understood it in the first place. I am talking about kids who, unlike me and several other nerds of then-high-school-age, do not spend their summers and/or free time on science and math; they spend it on basketball and socializing and camp. The fact that my daughter forgot how to exactly use exponent rules, how to simplify alebraic fractions, etc- as soon as she did a few examples, she recalled the way things work, yes. But I can bet that next year, when she starts trig, she'll have forgotten the sine-cosine relations, the rules for angles, etc- because she's not using them this year at all. For example- I do a lot of math in my school and work life. However, yesterday I had to derive an expression for a general form of deformation of columns and beams, and in integrating ended up with a ODE , second order, nonhomogeneous. I had to think for *quite a while* to recall the part about setting my particular equation equal to Ax + B, before solving it became simple again. Not because I didn't understand Diff Eq's the years ago when I took that class-but because I hadn't *used* that skill in a while. I think that happens to most people. -k wallace > This is an indication that those involved in designing the > elementary and high school programs have no understanding > of mathematics. >>Actually, Geometry in high school requires at least introductory algebra >>knowledge. > It should not. > Also, the proof-based Euclidean Geometry in high school using has >>returned, and very strongly. > Not really. Students mainly memorize theorems and proofs, > instead of getting an understanding of what is involved. > Of course, any decent mathematical program would have > taught the most important part of algebra in first grade, > or at the latest second, and proofs in general in > elementary school. But we cannot teach it to most of > the teachers, and it is quite possible that the students, > taught to memorize definitions and facts, and taught how > to solve specific types of problems, can no longer learn > it unless they have somehow or other kept the mental > abilities alive after the educationists have done their > damage. >>The usual sequence of course, sometimes switching the order of >>Algebra-Intermediate and Geometry, is necessary because the subject matter >>build as one studies each successive course. Strong algebra skill is used in >>Trigonometry; > Algebra SKILL, not any understanding of the fundamentals > of algebra. With any understanding, the knowledge of > what methods are needed is there, and the rest is just > practice. > strong algebra skill and some details of Trigonometry are used in >>Calculus. > Not if someone is going to understand calculus. > The basic concept of limit should be introduced no > later than infinite decimals. >>The Geometry is a foundation course that may help in understanding of >>trigonometry, > You mean five or six facts? > but most students will forget most of so many different theorems, >>and such forgetting seems not to be a serious problem in learning trigonometry, >>since the numeric sense of algebra secures the students progress. > Numeric sense? Numeric understanding can be important, > particularly induction and the structure of integers, > but nothing computational is more than a minor asset. > Most >>students will find trigonometry to be easier and more fun to study than >>Euclidean geometry. > Because it is not needed. But unless one is going to > spend a lot of time on computational tricks, there is > little in the course. > We are never going to progress as long as those who think > that one can get understanding from computation and > examples, or even that those help in understanding, are > teaching anything of mathematics. The educationists have > no understanding of concepts or how they can be learned. === Subject: Re: question for math teachers >rphenry@home.com comments on the sequence of jr.hi and hischool math courses: >>The way geometry is taught, you need to know some algebra. >> This is a bad idea. Euclid's students knew no algebra; it >> had not yet been invented. >I have to disagree- I worked with my daughter (and several of her >friends) through last year's geometry class. Other than a few solve for >x problems involving the Pythagorean Theorem (light pole this tall, >shadow this long, how tall is the man casting the shadow sort of stuff) >there was no algebra. There were, however, a lot of theorems, but >really understanding *why* was not (to my mind) adequately taught. >That's what I spent a lot of time with these girls on. As a result, I >believe, they all scored directly at the top of the class. (as a result >of course of their understanding, not just my tutoring and help). I see no disagreement that algebra is needed to understand geometry. Euclid's students would have been able to do those problems. The key part of algebra is the use of variables; this belongs early, as mathematical language, not as a mechanical means to solving problems. >>The way algebra 2 is taught, you need to know some geometry. >>Etc. >while currently this is true, there is no reason I can think of that it >needs to be. >As for the forgetting of things over the intervening year- I disagree >that it doesn't mean that they never understood it in the first place. Details may be forgotten, but concepts not. However, the current emphasis on objective testing makes it difficult, and in some cases impossible, to evaluate the learning of concepts. However, details can be looked up, but not concepts. I >am talking about kids who, unlike me and several other nerds of >then-high-school-age, do not spend their summers and/or free time on >science and math; they spend it on basketball and socializing and camp. So what? >The fact that my daughter forgot how to exactly use exponent rules, how >to simplify alebraic fractions, etc- as soon as she did a few examples, >she recalled the way things work, yes. But I can bet that next year, >when she starts trig, she'll have forgotten the sine-cosine relations, >the rules for angles, etc- because she's not using them this year at all. So what? The details are easy to relearn, if the concepts are there. You are making my point for me. >For example- I do a lot of math in my school and work life. However, >yesterday I had to derive an expression for a general form of >deformation of columns and beams, and in integrating ended up with a ODE > , second order, nonhomogeneous. I had to think for *quite a while* to >recall the part about setting my particular equation equal to Ax + B, >before solving it became simple again. Not because I didn't understand >Diff Eq's the years ago when I took that class-but because I hadn't >*used* that skill in a while. I think that happens to most people. Again, so what? You could also have gone to Mathematica or Maple or Matlab or Maxima, and not knowing the trick to solving that particular equation is not of great importance. Knowing how to use the concepts to derive the equations is important, but knowing the tricks of solving is not, even for most mathematicians. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: question for math teachers <-9Cdneped6eP8j7cRVn-oQ@comcast.com> at 12:28 PM, hrubin@odds.stat.purdue.edu (Herman Rubin) said: >Details may be forgotten, but concepts not. I find that the details quickly come back if I remember the concepts. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: question for math teachers >>rphenry@home.com comments on the sequence of jr.hi and hischool math courses: >The way geometry is taught, you need to know some algebra. >This is a bad idea. Euclid's students knew no algebra; it >had not yet been invented. >>I have to disagree- I worked with my daughter (and several of her >>friends) through last year's geometry class. Other than a few solve for >>x problems involving the Pythagorean Theorem (light pole this tall, >>shadow this long, how tall is the man casting the shadow sort of stuff) >>there was no algebra. There were, however, a lot of theorems, but >>really understanding *why* was not (to my mind) adequately taught. >>That's what I spent a lot of time with these girls on. As a result, I >>believe, they all scored directly at the top of the class. (as a result >>of course of their understanding, not just my tutoring and help). > I see no disagreement that algebra is needed to understand > geometry. Euclid's students would have been able to do > those problems. > The key part of algebra is the use of variables; this belongs > early, as mathematical language, not as a mechanical means > to solving problems. you are right, I posted in the wrong spot. I was disagreeing with the post portion right above yours, that states the way geometry is taught, you need to know some algebra. >The way algebra 2 is taught, you need to know some geometry. >Etc. >>while currently this is true, there is no reason I can think of that it >>needs to be. >>As for the forgetting of things over the intervening year- I disagree >>that it doesn't mean that they never understood it in the first place. > Details may be forgotten, but concepts not. However, the > current emphasis on objective testing makes it difficult, > and in some cases impossible, to evaluate the learning of > concepts. However, details can be looked up, but not > concepts. yes, you are right. I >>am talking about kids who, unlike me and several other nerds of >>then-high-school-age, do not spend their summers and/or free time on >>science and math; they spend it on basketball and socializing and camp. > So what? The point is, I guess, that the lack of intense interest doesnt' keep the details fresh. YOu are correct, though, in your assesment of the fact that details are not concepts...unfortunately, details are often taught to the exclusion, it seems, of a good basis in concepts. I think that is what most of my complaint is regarding math curriculum; however, I still think it makes more sense to teach alg.1 then alg. 2, then geometry and then trig. >>The fact that my daughter forgot how to exactly use exponent rules, how >>to simplify alebraic fractions, etc- as soon as she did a few examples, >>she recalled the way things work, yes. But I can bet that next year, >>when she starts trig, she'll have forgotten the sine-cosine relations, >>the rules for angles, etc- because she's not using them this year at all. > So what? The details are easy to relearn, if the concepts > are there. You are making my point for me. yes, ok. Perhaps I have fallen victim in some way to the misapprehension that knowing lots of details means understanding concepts. I'll have to examine that one. Odd, I do enough tutoring of 1-200 level math (at our local CC, my other job ) that I should keep that nugget of knowledge much more firmly in the forefront. I run into both types quite often- the ones who have memorized details, but can't see the forest for the trees, and the ones who say I can *SEE* exactly what needs to happen,I just don't know what rules to use to GET it there. -kwallace >>For example- I do a lot of math in my school and work life. However, >>yesterday I had to derive an expression for a general form of >>deformation of columns and beams, and in integrating ended up with a ODE >> , second order, nonhomogeneous. I had to think for *quite a while* to >>recall the part about setting my particular equation equal to Ax + B, >>before solving it became simple again. Not because I didn't understand >>Diff Eq's the years ago when I took that class-but because I hadn't >>*used* that skill in a while. I think that happens to most people. > Again, so what? You could also have gone to Mathematica or > Maple or Matlab or Maxima, and not knowing the trick to solving > that particular equation is not of great importance. Knowing > how to use the concepts to derive the equations is important, > but knowing the tricks of solving is not, even for most mathematicians. === Subject: Re: Parametrizations of U(2) >U(2) (the group of unitary 2x2 matrices) is a smooth manifold of >dimension 4 (over the real numbers), ie. there exist an atlas of (local) >diffeomorphisms mapping some subset of R^4 into U(2), such that U(2) is >completely covered. >For example the subgroup SU(2) can be parametrized with >{{z_1, z_2},{-(z_2)*,z_1*}} s.t. |z_1|^2 + |z_2|^2 = 1, >where z_1 and z_2 are complex numbers. I'm looking for parametrizations >of the remaining unitary matrices in U(2)SU(2) with four real parameters. What's wrong with the definition? A matrix M = {{z_1, z_2},{z_3,z_4}} lies in U(2) iff M* M = I, i.e. (z_1)* (z_1) + (z_3)* (z_3) = 1 (z_1)* (z_2) + (z_3)* (z_4) = 0 (z_2)* (z_1) + (z_4)* (z_3) = 0 (z_2)* (z_2) + (z_4)* (z_4) = 1 The middle two equations say the same thing: roughly, that the ratios (z_4)/(z_1)* and -(z_2)/(z_3)* are equal, to r , say. That is, z_4 = r (z_1)*, z_2 = -r (z_3)* . The first and last equations together then say that |r| = 1. So now you have a parameterization {{z_1, -r (z_3)*},{z_3, r(z_1)*}} s.t. |z_1|^2 + |z_2|^2 = 1, |r| = 1. Your parameterization of SU(2) is contained in this; SU(2) is the set of matrices with r = 1. You can interpret this construction as a demonstration of the fibration U(n-1) --> U(n) -> S^(2n-1) when n=2. dave === Subject: Re: axioms, sry by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iANHFw922175; Gitte, please STOP using the domain HULU.DK, you do not own the rights for this mail, i do.. Sorry to be writing here, but had to get a hold of you somewhere.. /Lars U === Subject: Re: Modern censorship > finding a polynomial-time prime finding algorithm. Polynomial in what variable, the value of the prime itself, or the number of digits in the prime (i.e. logarithm of the value of the prime)? By the way, in the months immediately after Ron Rivest got his public-key cryptosystem published in Scientific American, I developed an algorithm for directly generating prime numbers in any desired range (except if the range is too narrow) and automatically providing a mathematical proof they are really prime. For example, if you want to generate a product of two similar-size primes whose leading decimal digits are your telephone number, for example 18005551212xxxxxxxxxxxxx, first you use my algorithm to generate a prime with appx. half the digits of that desired product, i.e. any prime in the range [10**11, 10**12] in that case, then divide the desired range of the product [180055512120000000000000, 180055512129999999999999] by that first prime to get the desired range for the second prime and apply my algorithm again to get that second prime. The product is published, and easily identifiable as belonging to you, while the primes can't be guessed easily. (Of course you'd use more than 24 digits. I used only 24 here only to illustrate the idea.) === Subject: Re: quaternions in part: >the only thing of interest about >them is their relationships among each other... completely arbitrary >to me. Ah, but those relationships a) yield a division algebra, and b) correspond to vector algebra in three dimensions, so they're not _completely_ arbitrary. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: Good intro to Lie Groups & Algebras > I am looking for a good book and/or online literatures that provides a > good introduction to Lie groups and algebras for someone who has not > gone through a course in differential geometry. > I have a background in advanced calculus, point set topology, real and > functional analysis, and basic complex analysis. Unfortunately, I have > not taken a course in diff. geom. nor understand the concept of manifolds. > If someone knows of a book that is a good intro to diff. geom. that > covers Lie groups/algebras and manifolds, that would be great. > --john > PS: Please post responses to newsgroup. Do not email to me directly. I concur with a previous suggestion that you read Warner's book, but I really think that you should also familiarize yourself with just the rudiments of differential geometry. You'll have to do it even to know what a Lie group is, and at least in my own experience it is pretty futile to say I am going to learn X and expect to pick up the necessary tools as I go along. The tools are heavy and better handled when studied independently. -- Ryan Reich ryanr@uchicago.edu === Subject: notion wanted Hi NG, I am looking for a notion I can't remember at the moment. How do you call a topological space which can be represented by (at most) countably many compact subsets? [Exhaustion method] J. Please reply to my email-address, as well. Thx. === Subject: Re: Problem from Atiyah-McDonald > I have a problem with Exercise 11 from Chapter 2. It states a question: > (Let A be a commutative ring with 1, f - A-module homomorphism (what > is important - not necessarily a ring homomorphism)) > If f: A^m -> A^n is injective is it always the case that m <= n ? > Yes. Suppose that m > n. We can assume that m = n + 1. > We have a map f: A^m -> A^n. We have to show it has a non-trivial kernel. > It is represented by a matrix of size n+1 by n. Call this matrix M. > Then we can write down a kernel element. Let M_j be M with row j deleted. > Then (det M_1, -det M_2, det M_3, ....) is in the kernel. > Of course this might be zero. But then all the M_j have zero determinant. > So we reduce the problem to showing that a map g: A^n -> A^n > represented by a zero-determinant matrix N has a non-trivial kernel. > Unless N is zero (nothing to prove then), then N has a largest > submatrix with nonzero determinant. We can suppose that it is the > top left k by k matrix. Let P be the top left k+1 by k+1 submatrix of N. > Consider the first row of the adjugate of P. If we add n - k - 1 zeros > to the end of this we get a nonzero element in the kernel of g. > I'm sure all this can be more succintly expressed using exterior algebra. > -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 > Francis Wheen, _How Mumbo-Jumbo Conquered the World_ Once i learned it like this: Let be Z the subring given by 'the integers in A'. Then R:=Z[a_ij] is noetherian and f|R: R^m -> R^n is injective given by the matrix (a_ij). So one can assume A is noetherian. Now take a minimal prime ideal in the noetherian R and localize - which preserves monomorphisms, so one can assume A is artinian. Then use the 'length' (A & McD, Ch 6) which is 'additive' to get m = n + length(cokernel) <= n. -- use mail for mail not no nonail === Subject: Re: Distance between two points in a cube OK so lets start with the 2-dimensional case to review: the distance between 2 points, (x1,y1) and (x2,y2), is like the length of the hypotenuse of a right triangle. We find the distance traveled in the x direction, square it, and add it to the distance in the y direction squared. The square-root of this sum is the total distance. Ex) (from your question) (290,135) and (25,149). distance= sqrt [ (25-290)^2 + (149-135)^2] d= sqrt[265^2 + 14^2] d= sqrt[70225 + 196] d= sqrt[70421] = 265.37.. This is the total distance in 2 dimensions. In the three-dimensional case, the only difference is the change in the z direction (you cant use the same variable for more than one dimension!), so you use the pythagorean theory again, by simply adding the squared change in z before taking the square root. So in your problem, > x : y : x > 290 : 135 : 8 (point 1) > 25 : 149 : 123 (point 2) we have to find the difference travelled in each direction, square them, add them together and finally take the square root. d= sqrt[(25-290)^2 + (149-135)^2 + (123-8)^2] d= sqrt[70225 + 196 + 13225] d= 289.216 Hope this helps. I've never posted something like this before so i don't know the proper way to show notation, but i guess you wont mind. === Subject: Re: Distance between two points in a cube > calculate the difference between two? set points in(on?) a cube > x : y : (z) ? ; 290 : 135 : 8 (point 1) ; 25 : 149 : 123 (point 2) If the points are in a cube or sphere it does n't matter to find direct distance. But if they are On a cube,then.. Is the side of cube (a) known? First ensure that the points in fact stay on a cube. To stay on an enclosing cube [corner points are (0,0,0),(a,a,a)] << x y z (a-x)(a-y)(a-z)=0 >> holds. After this, d^2 = (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2. === Subject: Re: any smart way of proving |sinc(x)| integrates to infinity? >Integrate(|sinc(x)|, from -inf to +inf) = infinity. >> Sure. It's a periodic function, so the integral is equal to the >> integral of one period times the number of periods. >Except it's not periodic. sinc(x) = sin(x) / x. It goes up and down but the >amplitude also tapers off. You're absolutely correct. In mitigation, I plead that I'm so used to missing and misplaced parentheses that I assumed the poster actually meant sin(cx). But I definitely should have said so. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if you're afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: any smart way of proving |sinc(x)| integrates to infinity? >I want to know if there is any smart way of proving >Integrate(|sinc(x)|, from -inf to +inf) = infinity. > Since it's symetric about the y-axis, you can reduce the problem to 0 to > +inf if you wish. The zero-lobe is finite, so you can safely ignore it. > One tactic you could use is to prove that the area of each lobe > (x=(2*n-1)*pi/2 to (2*n+1)*pi/2)) is greater than the nth term in > some series known to be infinite, say (a_constant)/n. >does knowing that >Integrate(sinc(x), from -inf to +inf) = 1 and >Integrate(sinc^2(x), from -inf to +inf) = 1 >help me solve the problem? > I don't think so. And I think you mean sinc(pi*x), at least in the first > case. Concerning your last comment: The OP probably meant sinc(x), but there are two different standard definitions for the sine cardinal function. According to one of those definitions, sinc(x) = ( 1 if x = 0, ( sin(pi*x)/(pi*x) otherwise. David === Subject: Re: PDE help >I'm trying to solve this 2nd order linear PDE. Actually, even if I >could find out whether or not a solution exists may be good enough. I >couldn't convince Maple/Mathematica to solve this for me. >(1-x^2)*(1-y^2)*diff(H(x,y,z),x,y)+(1-x^2)*(1-z^2)*diff(H(x,y,z),x,z)+(1-z^ 2)*(1-y^2)*diff(H(x,y,z),z,y)=(1-x^2)*(1-y^2)*(1-z^2)*H(x,y,z) >And I want the solution to be such that >H(-1,y,z)=H(x,-1,z)=H(x,y,-1)=0. For convenience, start with a change of variables H(x,y,z) = h(x+1,y+1,z+1) so the boundary conditions on h are at x=0, y=0 and z=0. The pde becomes x*(-2+x)*y*(-2+y)*diff(h(x,y,z),x,y) + x*(-2+x)*z*(-2+z)*diff(h(x,y,z),x,z) +z*(-2+z)*y*(-2+y)*diff(h(x,y,z),y,z) = -x*(-2+x)*y*(-2+y)*z*(-2+z)*h(x,y,z) Suppose you had a solution given by a power series in x,y,z. The boundary conditions say that all nonzero terms are of degree at least 1 in each variable. If the terms of lowest total degree have total degree d, say sum_{i+j+k=d} c_{i,j,k} x^i y^j z^k, then taking the terms of lowest degree from the pde we find that sum_{i+j+k=d} c_{i,j,k} x^i y^j z^k (ij+ik+jk) = 0 But since ij+ik+jk can't be 0, that would imply c_{i,j,k} = 0. So there are no nontrivial solutions that are analytic in a neighbourhood of (0,0,0) (for the transformed equation), or (-1,-1,-1) for the original. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: November 25 is Infinite Clause day!! -Not to mention HERC777's understanding. The question is not whether -the diagonal number's digits are within the list (most likely, -they are this is your 1st error, you overlook the fact all digit sequences to infinite length are computable the conclusion of Cantor's proof relies on the existence of a new sequence of digits which is clearly non existent. -We also need to construct Diag. This is fairly simple, -and one can have many variants, so we need just pick one: - -if List(N,N) = 4 then Diag(N) = 5 else Diag(N) = 4 this is your second error, your construction can never be realised. its like saying construct the set of all sets that don't contain themselves, if its a member of itself its not a member, if its not a member of itself its a member. you can theorise the construction but you cannot refute your basic premises of set theory because of one theorem, regardless if a contradiction is formed. this is all you are doing. let R = 1, 2, 3, 4... let p <> R1, p <> R2, p <> R3... therefore R is incomplete. your claim that p is a valid construction but its not rigorous. Its like your Godel proof, you are trying to prove ! (proof(X) <-> X) proof(X) <=> Exists a proof of X. so you allow this theorem, G <-> ! proof(G) but you don't allow this meta-consistency theorem X <-> proof(X). Either of these theorems will work but they are mutually exclusive, no one here has shown a PROOF that ! (proof(X) <-> X). In AI the meta-constistency theorem is called Truth Maintenance, only proven facts are allowed and never need to be revoked. The alternate system is called belief revision. Godel's proof only works in one automated theorem system. Same with the set of reals, assume all possible combinations of digits are present in a countable list, with UTM(n, 0) they actually are. Now examine the construction of diag, it is a flawed construction! QED. -Does an N exist such that for every positive M in J, - Diag(M) = List(N,M)? - -The answer clearly is no (if M=N Diag(N) != List(N,N) by construction). This is your 3rd error. Why is it when I write 'obviously' it gets picked up yet everyone uses 'clearly' around here. What you all actually mean by 'clearly' is clearly it's in the text book! Your construction breaks the premise of a complete list, it doesn't contradict it. An infinite number of people toss a coin RANDOMLY an infinite number of times. After 10 people have tossed a coin 5 times each, the 1st 3 places in the sequence are (tend to be) all covered. hhh.. hht.. hth.. htt.. thh.. tht.. tth.. ttt.. there are 2 duplicates due to random spread. After several million people have tossed a coin 30 times each, by the binomial distribution, all combinations have been covered up to 15 tosses. 000000000000001..010101010 000000000000010..010101010 000000000000011..010101010 000000000000100..010101010 000000000000101..010101010 000000000000110..010101010 000000000000111..010101010 000000000001000..010101010 ... 111111111111110..010101010 111111111111111..010101010 <- - - - - - - ->|<- - - - -> all combinations still random covered The number of coins that get covered increases without bound, logarithmically. The length of 'covered combinations' is approx. log(people doing coin tosses). As the countable list approaches infinite length.. As #people -> oo, number of digits covered -> log(oo). Therefore an infinite list contains all sequences of {H, T} of infinite length. 1 HTHTHTHTHT 2 HHHHHHHHH 3 TTTTTTTTTTT 4 THTHTHTHTHT .. Take the diagonal! HHTH... Invert it! TTHT.. ALL SEQUENCES ARE PRESENT, TTHT.. is not a new sequence. INFINTIE people toss coins infinite times each, can you with 100% certainty form a NEW {H, T} sequence? CLUE : NO!! Inverting the diagonal is not a new sequnce of tosses, and modifying the real diagonal does not a new real make. Herc === Subject: Re: November 25 is Infinite Clause day!! > Barb Knox, George Green, Jesse James, John Savard, Dave B, you, > Ullrich, Will and dozens of others have all been vocally opposed to me > in their defence of Cantors diag proof. Now they are all silent? is > it apathy? No, it's knowing an idiot when we see one. > This is the bait : > An infinite number of people toss a coin infinite times each. This is already ambiguous. There is more than one kind of infinity. One question: is there a LAST one of these infinite number of people? If so, is there a LAST toss in any (or each) of their individual strings of infinite tosses? > Can you guarantee a new sequence of Heads and Tails? This is simply meaningless. There are AN INFINITE NUMBER OF sequences of heads and tails being generated here (one for each tosser). They could all be the same. They could all be different. The fact that the sample space is as big as it is means (among other things) that usual notions of probability are IRRELEVANT to can. If I ask you to pick a natural number at random, the probability that you will pick any PARTICULAR rational number is 0. This question is simply STUPID. > If the answer is NO, which I think most if not many people will agree > it is, They don't know . And YOU don't know what YOU mean by can you guarantee. In the first place, there are not now nor will there ever be an infinite number of people in the world. In the second place, even if there WERE an infinite number of people in the world, the number of sequences-of-flips is a much BIGGER infinity than that, so the probability of any particular sequence occurring AT ALL (as 1 flipper's) IS 0. Guarantee a new sequence?? EVERY INDIVIDUAL flipper's sequence is new! It will CERTAINLY, EVENTUALLY, be a sequence that NOBODY HAS EVER EVEN THOUGHT ABOUT before! === Subject: Re: November 25 is Infinite Clause day!! <9kq772-t9i.ln1@sirius.athghost7038suus.net> <96c872-j7j.ln1@sirius.athghost7038suus.net> > An infinite number of people toss a coin infinite times each. > Can you guarantee a new sequence of Heads and Tails? the smarter people of the group can't see this is a well formed question but 20 others have answered it. > Can you guarantee a new sequence of Heads and Tails? > This is simply meaningless. There are AN INFINITE NUMBER OF sequences > of heads and tails being generated here (one for each tosser). > They could all be the same. They could all be different. The then assume the worst case, because you HAVE to make a unique sequence, that is, you cannot rely on some pattern forming in the set that allows you to find a unique sequence. You have a problem that the output is unkown, and you have a problem understanding why you need a guarantee. Guarantee means for ANY possible set. >In the first place, there are not now >nor will there ever be an infinite number of people in the world. >In the second place, even if there WERE an infinite number of people >in the world, the number of sequences-of-flips is a much BIGGER >infinity than that, so the probability of any particular sequence Just read it as, assume an infinite list of binary number expansions from 0.000... to 0.11111.. >occurring AT ALL (as 1 flipper's) IS 0. Guarantee a new sequence?? >EVERY INDIVIDUAL flipper's sequence is new! It will CERTAINLY, >EVENTUALLY, be a sequence that NOBODY HAS EVER EVEN THOUGHT ABOUT before! That is the fake infinity that Cantors rubbish forces you to believe. There is no logical reason why 2 different people can't flip the same side of a coin again and again. You say any sequence is P=0 then in the next breath say they certainly exist. When 100 people have tossed coins for a while, to get a complete new sequence it must be longer than log(100) tosses, because HHHHHH someone got it already HHHHHT someone got it already HHHHTH someone got it already HHHHTT someone got it already HHHTHH someone got it already HHHTHT someone got it already HHHTTH someone got it already HHHTTT someone got it already HHTHHH someone got it already HHTHHT someone got it already HHTHTH someone got it already HHTHTT someone got it already HHTTHH someone got it already HHTTHT someone got it already HHTTTH someone got it already HHTTTT someone got it already HTHHHH someone got it already HTHHHT someone got it already HTHHTH someone got it already HTHHTT someone got it already HTHTHH someone got it already HTHTHT someone got it already HTHTTH someone got it already HTHTTT someone got it already HTTHHH someone got it already HTTHHT someone got it already HTTHTH someone got it already HTTHTT someone got it already HTTTHH someone got it already HTTTHT someone got it already HTTTTH someone got it already HTTTTT someone got it already THHHHH someone got it already THHHHT someone got it already THHHTH someone got it already THHHTT someone got it already THHTHH someone got it already THHTHT someone got it already THHTTH someone got it already THHTTT someone got it already THTHHH someone got it already THTHHT someone got it already THTHTH someone got it already THTHTT someone got it already THTTHH someone got it already THTTHT someone got it already THTTTH someone got it already THTTTT someone got it already TTHHHH someone got it already TTHHHT someone got it already TTHHTH someone got it already TTHHTT someone got it already TTHTHH someone got it already TTHTHT someone got it already TTHTTH someone got it already TTHTTT someone got it already TTTHHH someone got it already TTTHHT someone got it already TTTHTH someone got it already TTTHTT someone got it already TTTTHH someone got it already TTTTHT someone got it already TTTTTH someone got it already TTTTTT someone got it already When 1,000,000 people have all tossed coins for a while, your UNIQUE SEQUENCE must be even longer, around 15 digits long for a high confidence of it being unique. As the number of people -> oo, to toss a unique sequence it must be longer_than_log(S) as S->oo. Sure you're up to it? Herc === Subject: Re: help! solving matrix equations ... >Suppose I have two matrices, P is 15x3, an C is 15 x 10, >how can I find two vectorx, a: 3 x 1 and w: 10 x 1, >such that P*a=C*w, >or P*a and C*w as near each other as possible in Mean square error >sense(min. mean square error)... Ummm, there's a _very_ simple solution. >Is there any systematic way of doing this? For a less trivial problem, you might want P a - C w = b for some given vector b. Consider the block matrix M = [P | C] (i.e. the matrix whose first 3 columns are P and whose last 10 are C), and use the standard least squares approach to solving M x = b. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: help! solving matrix equations ... >>Suppose I have two matrices, P is 15x3, an C is 15 x 10, >>how can I find two vectorx, a: 3 x 1 and w: 10 x 1, >>such that P*a=C*w, >>or P*a and C*w as near each other as possible in Mean square error >>sense(min. mean square error)... > Ummm, there's a _very_ simple solution. >>Is there any systematic way of doing this? > For a less trivial problem, you might want P a - C w = b for some > given vector b. > Consider the block matrix M = [P | C] (i.e. the matrix whose first 3 > columns are P and whose last 10 are C), and use the standard least > squares approach to solving M x = b. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada Hi Robert, Following your advice, I put M=[P | C], which is 15x13, and y=[a', -w']' which is an expanded column vector of size 13x1, I want to find M*y=0, but M is 15x13 and it has no null space. So I have to find the best appoximate solution to M*y=0, however, the LS solution gives 0, y=pinv(M)*0, however, I really want the best such non-zero y, am I right? === Subject: Re: help! solving matrix equations ... >Suppose I have two matrices, P is 15x3, an C is 15 x 10, >how can I find two vectorx, a: 3 x 1 and w: 10 x 1, >such that P*a=C*w, >or P*a and C*w as near each other as possible in Mean square error >sense(min. mean square error)... >> Ummm, there's a _very_ simple solution. >Following your advice, I put M=[P | C], which is 15x13, and y=[a', -w']' >which is an expanded column vector of size 13x1, You didn't follow my first advice... >I want to find M*y=0, >but M is 15x13 and it has no null space. Every matrix has a null space. >So I have to find the best >appoximate solution to M*y=0, however, the LS solution gives 0, y=pinv(M)*0, >however, I really want the best such non-zero y, am I right? Who said anything about y being nonzero? What would be your criterion for a best nonzero y? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: help! solving matrix equations ... >>Suppose I have two matrices, P is 15x3, an C is 15 x 10, >>how can I find two vectorx, a: 3 x 1 and w: 10 x 1, >>such that P*a=C*w, >>or P*a and C*w as near each other as possible in Mean square error >>sense(min. mean square error)... > Ummm, there's a _very_ simple solution. >>Following your advice, I put M=[P | C], which is 15x13, and y=[a', -w']' >>which is an expanded column vector of size 13x1, > You didn't follow my first advice... >>I want to find M*y=0, >>but M is 15x13 and it has no null space. > Every matrix has a null space. >>So I have to find the best >>appoximate solution to M*y=0, however, the LS solution gives 0, >>y=pinv(M)*0, >>however, I really want the best such non-zero y, am I right? > Who said anything about y being nonzero? > What would be your criterion for a best nonzero y? > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada realistic problems, I forgot to limit that y cannot be 0. and in fact, I want ||w||=1... the norm of w = 1. Sorry I forgot this practicality constraint... Under this new added condition, what can be a good solution? === Subject: Re: logic of the Cantorian followers mind > Both Cantor's theory and Creationism theory are founded on the > proposition that we must acknowledge the existence of some abstract > infinite entity lying beyond what we can observe in order to understand > the reality that we do observe. Mathematics is a study of abstract entities, both finite and infinite. These necessarily lie beyond what we can observe. Have you ever observed a real number, or even an integer? Mathematics isn't about the reality that we do observe, although the study of these abstract entities can certainly help us to understand that reality. I understand that you're bothered by non-computable real numbers. That's okay. I suppose different people have different tolerances for abstraction. Just don't try to impose your view on mathematics as a whole. The mainstream accepts the existence of objects that can't be finitely described. This mainstream mathematics produces many beautiful and useful results, and is just as internally consistent as the more restrictive system you seem to prefer. You can rail against it for disagreeing with your intuitions of what exists, but that's not an objectively valid reason for rejecting it. It works. You can like it or dislike it, but you would do well at least to understand it. === Subject: Re: logic of the Cantorian followers mind The > mainstream accepts the existence of objects that can't be finitely > described. Exactly what objects are those? === Subject: Re: logic of the Cantorian followers mind : :> The :> mainstream accepts the existence of objects that can't be finitely :> described. : Exactly what objects are those? Most of the real numbers. Most of the sets of integers. Most of the languages over any alphabet. Stephen === Subject: Re: logic of the Cantorian followers mind >> Both Cantor's theory and Creationism theory are founded on the >> proposition that we must acknowledge the existence of some abstract >> infinite entity lying beyond what we can observe in order to understand >> the reality that we do observe. >Mathematics is a study of abstract entities, both finite and infinite. >These necessarily lie beyond what we can observe. Have you ever >observed a real number, or even an integer? Mathematics isn't about the >reality that we do observe, although the study of these abstract >entities can certainly help us to understand that reality. >I understand that you're bothered by non-computable real numbers. It's even worse than that. Herc even refuses to accept that the full set of COMPUTABLE real numbers is not itself computable (i.e. recursively enumerable). He objects to the use of a diagonal argument there too. >That's >okay. I suppose different people have different tolerances for abstraction. >Just don't try to impose your view on mathematics as a whole. The >mainstream accepts the existence of objects that can't be finitely >described. This mainstream mathematics produces many beautiful and useful >results, and is just as internally consistent as the more restrictive >system you seem to prefer. You can rail against it for disagreeing with >your intuitions of what exists, but that's not an objectively valid >reason for rejecting it. It works. You can like it or dislike it, but you >would do well at least to understand it. -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: logic of the Cantorian followers mind Hi Barb, so you too think that you can come up with a new sequence of heads and tails after countable infinite many people toss coins infinite times each? Herc === Subject: Re: logic of the Cantorian followers mind > If you studied logic you'd have learnt what propositions are, > statements that are either true or false. > Cantors definition of the diag number is typically: > D(i) = L(i,i) mod 9 + 1 ? I guess you meant mod 10, does this matter ? So the set up is Start proposition M for any function L from N^2 to {0,1,..,9}. For each i, (L(i,j) : j in N) represents a real R(i) via the following convergent sum: sum_j L(i,j)/(10^j) Then the claim is that D := sum_k D(i)/(10^j) is not in {R(i) : i in N} > We can represent this as an equivalent infinite clause. > D1 = L(1,1) mod 9 + 1 AND > D2 = L(2,2) mod 9 + 1 AND > D3 = L(3,3) mod 9 + 1 AND ... > this results in the proposition > D1 <> L(1,1) AND D2 <> L(2,2) AND ... > I defined 4 propositions. > C (Cantor) = there are uncountable numbers Erm, perhaps there are uncountably many real numbers > P (Property of Diag) = Diag <> Real1 and Diag <> Real2 and Diag <> Real3 ... SEE ABOVE Ok, I used R(i) instead of Reali > A (Argument) = All sequences of digits are computable, and already > present on the list of computable reals I see, you are calling this proposition argument, rather than claiming it is an argument. > R (Rebuttal) = the set 0.1, 0.2, 0.3...0.9, 0.01, 0.02, 0.03... also > contains every sequence of digits, but using strictness analysis it > does not contain any irrationals as all numbers have finite length. Well, I still don't know what strictness analysis is, but R is clearly false if I am supposed to read it that the 10n+m th element in the sequence written is m/(10^{n+1}) > Your argument is at the beginner Cantorian level NOT(A) and is not > covered in this topic. > Herc So you are assuming A, is this right? I agree that, given A, C is false. You haven't given me any reason to beleive A though. As C follows from ZF, if C is false, then ZF is inconsistent. === Subject: Re: logic of the Cantorian followers mind the usual construction is like this : D(i) = 5 iff L(i.i) =/=4 = 4 otherwise The reason you don't use L(i,i) mod 10 + 1 is because it doesn't result in a digit. You can't use (L(i,i) + 1) mod 10 because you may convert the diagonal 0.099999.. to 0.100000 which is the same number and nullifies the argument. The topic is the psychology of Cantorian followers, not Yet Another Formal Demonstration of the idiot proof. Strictness analysis is what Cantor follower fall back on when it is demonstrated that all permutations are computable. It's type checking in compilers. ODD + ODD = EVEN EVEN + EVEN = EVEN EVEN + ODD = ODD Using these formula, can you determine the correctness of this sum? 438948748774843897 + 49874387487438743 = 9384848748747841 Though in Cantors proof, it's an attack on incompleteness of all finite permutations because even though they go to infinity each element is finite. I'm not sure why you can't parse this pseudo sentence. > What is obviously wrong is that A = All sequences of digits are > computable, and already present on the list of computable reals. Brief induction will prove it, any length sequence is computable. R is true, ask Ghost, ZF doesn't prove C you don't know how to use it. Herc === Subject: Re: Why Do Americans Call It Math? > I used to say Maths, as a Brit, I still do sometimes in the UK, but I gave > putting the s on the end makes it awkward to lisp out. Math avoids the > horrible triple consonant 'ths' That's a double consonant, and occurs in lots of plurals. It's easy enough. > Math or Mathematics is easier to say... one has to bow to simplicity A fine sentiment for a mathematician, I don't think! -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: how to maximize f(x) over g(x), where f(x) grows following the same trend as g(x)? >I am wondering what kind general techniques can be applied to >maximize (f(x)/g(x)) >where f(x) and g(x) grows following the similar trend, ie. both grow large >and large , etc... >so when f(x) increase, g(x) increases too... >How to find such maximization? Any thoughts? >Esp. when x is a vector? How to do maximization? The same techniques as in any other maximization problem over an unbounded domain. You can write just about any positive objective h(x) in this way: just take f(x) = g(x) h(x) where g(x) grows rapidly enough that f(x) also grows. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: solid angle >Here you can find a schematic drawing of a PET camera with a patients head inside. >http://www.fmrib.ox.ac.uk/~stuart/thesis/chapter_3/image3_2.gif >Here is a real picture of a camera >http://pet.radiology.uiowa.edu/webpage/Overview/scanner.jpg >And this is a nice 3D view of the detector ring > >http://www.cineactive.com/portfolio/nobel/images/hi_res/pet_2_Thumbz_hi_res . jpg >Yes, I am sure that D was the diameter, not the radius. >Frank > http://math.asu.edu/~kurtz/images/PET.jpg > which shows a cross section of the detector. The part of the sphere > that the detector would see is the part outside of the cone with > central angle phi_0, which I have indicated with a thicker line on the > sphere cross section. The definition of a solid angle is the surface > area of a unit sphere that the angle subtends. So if you imagine this > figure as being rotated, we just need to compute the volume of the > unit sphere outside the cone. Since the area of a unit sphere is 4Pi, > we can subtract the sphere area inside the cone. Using spherical > coordinates, remembering the spherical dS is rho^2 sin(phi) dphi > dtheta with rho = 1, the area is: > Area = 4 Pi - 2*Int[0..2Pi] Int[0..Phi_0] sin(phi) dphi dtheta > = 4 Pi + 4 Pi *(cos(phi)[phi = 0..Phi_0] > = 4 Pi + 4 Pi ( cos(Phi_0) - 1) > = 4 Pi cos (Phi_0) = 4 Pi sin (alpha) > = 4 Pi sin(arctan(A/D) > --Lynn Now that I come to think of it over my evening meal, there was indeed a way of seeing this without doing the integrals. The area of the unit sphere outside the cone is equal to that of a cylinder of identical height and radius (=1). Now the height simply is equal to the 2 * sinus(arctan(A/D). Frank === Subject: Re: solid angle >Hello. >I read somewhere the following equation for the solid angle covered by >a PET-camera: >solid angle = 4 * pi * sin(arctg(A/D)) >A PET camera is a device used to detect and image positron emitters. >Nowadays, it consists of a ring detector, into which the patient is >brought. >A in the equation above refers to the depth of the ring detector, ie. >the dimension parallel to the ring axis. >D stands for the diameter of the ring. >I cannot see how to get to the equation. Using a plane through the >axis, I would believe that in such a plane the angle covered by the >detectors would equal 4 * arctg(A/D)). But where does the sinus come >from? >So I looked up the definition of solid angle in Mathworld and tried to >make a calculation. >I used as the origin the point on the ring axis halfway between its >borders. The x-axis was placed on the ring axis. >Then, using the definition of solid angle and the fact that for all >points on the detector y^2 + z^2 = R^2 because of the ring geometry (R >being the radius of the ring detector), I arrived at the following >integral: >Int (-R -- +R) Int (-A/2 -- +A/2) (Sqrt(R^2-y^2) / (x^2 + R^2)^3/2) >dx dy >I was unable to compute this integral (even with the help of >Mathematica), and cannot see how it could lead to an equation as the >one cited above. So I probably made several mistakes. I would >appreciate any help to sort out this mess. >Frank > Frank, this may clear things up. > Clearly A/2 should be the radius of the ring and D the depth, but not > clear whether D is to the plane of the ring or to the full depth. > Assume to the plane of the ring. > > Then you have a half angle > th = atan(A/2D) > Using an unknown spherical radius R, the area of the spherical segment > is > A = 2 pi R h = 2 pi R^2(1 - cos th) > For the solid angle, divide R^2 out: > solid A = 2 pi (1 - cos th) steradians > So for half-angle th = 90 deg (half sphere), A = 2pi and for th = 180 > (full sphere) A = 4pi. > John Polasek > If you have something to say, write an equation. > If you have nothing to say, write an essay === Subject: Re: solid angle > Hello. > > I read somewhere the following equation for the solid angle covered by > a PET-camera: > solid angle = 4 * pi * sin(arctg(A/D)) > A PET camera is a device used to detect and image positron emitters. > Nowadays, it consists of a ring detector, into which the patient is > brought. > A in the equation above refers to the depth of the ring detector, ie. > the dimension parallel to the ring axis. > D stands for the diameter of the ring. > I could n't figure out all, but this helps you in parts: A sphere of > diameter sqrt(D^2+A^2)=2 R is cut centrally by a cylinder/ring > diameter D. The length of intersected PET patient's tube is A. Then > surface area is simply = 2 pi R A and solid angle =2 pi R A /R^2 = 2 > pi A/R = 4 pi A/D = 4 pi sin(alpha) , where alpha is angle between > planes of the ring at opening and sphere. Formula 4 pi > sin(arctan(A/D) is confusing, 4 pi A/D was quite OK. > There is one interesting result that I can't help giving, even if not > related to your question : Remaining volume is simply pi*A^3/6 . Frank === Subject: Re: solid angle >Here you can find a schematic drawing of a PET camera with a patients head inside. >http://www.fmrib.ox.ac.uk/~stuart/thesis/chapter_3/image3_2.gif >Here is a real picture of a camera >http://pet.radiology.uiowa.edu/webpage/Overview/scanner.jpg >And this is a nice 3D view of the detector ring > >http://www.cineactive.com/portfolio/nobel/images/hi_res/pet_2_Thumbz_hi_res . jpg >Yes, I am sure that D was the diameter, not the radius. >Frank > http://math.asu.edu/~kurtz/images/PET.jpg > which shows a cross section of the detector. The part of the sphere > that the detector would see is the part outside of the cone with > central angle phi_0, which I have indicated with a thicker line on the > sphere cross section. The definition of a solid angle is the surface > area of a unit sphere that the angle subtends. So if you imagine this > figure as being rotated, we just need to compute the volume of the > unit sphere outside the cone. Since the area of a unit sphere is 4Pi, > we can subtract the sphere area inside the cone. Using spherical > coordinates, remembering the spherical dS is rho^2 sin(phi) dphi > dtheta with rho = 1, the area is: > Area = 4 Pi - 2*Int[0..2Pi] Int[0..Phi_0] sin(phi) dphi dtheta > = 4 Pi + 4 Pi *(cos(phi)[phi = 0..Phi_0] > = 4 Pi + 4 Pi ( cos(Phi_0) - 1) > = 4 Pi cos (Phi_0) = 4 Pi sin (alpha) > = 4 Pi sin(arctan(A/D) > --Lynn Especially, it became clear where that sinus comes from. Being a (nuclear) physician (using PET cameras amongst other things), I could not have figured this out on my own. I am always amazed at the quality of the answers one gets on the Usenet and at the kind of people who are answering. I find it incredible kind of people of your status to be prepared to get your hands dirty on problems that are probably way Frank === Subject: Simple Group Theory Question I can't find the correct way to proving the following: Let G be a non-abelian group of order 8, prove that G has a member of order 4. I know that G can only be either an isomorphism of D_4 or the Quaternion, but I need to prove this directly without relying on that. === Subject: Re: Simple Group Theory Question days. My association with the Department is that of an alumnus. >I can't find the correct way to proving the following: >Let G be a non-abelian group of order 8, prove that G has a member of >order 4. >I know that G can only be either an isomorphism of D_4 or the >Quaternion, but I need to prove this directly without relying on that. Every element in a group of order 8 must be of order 1, 2, 4, or 8 (as a consequence of Lagrange's Theorem). The only element of order 1 is the identity. If there were an element of order 8, say x, then would have 8 elements, hence =G; but if G is cyclic, then G is abelian. So G does not have any elements of order 8. If all elements of G were of order 1 or 2, you'd be sunk. So you want to show that not all elements are of order 1 or 2. That is, that not all elements satisfy x^2 = 1. If you can do that, then there must be some element not of order 2; it can't be of order 8, so it must be of order 4. You can prove something stronger: THEOREM. Let H be a group where x^2 = 1 for all x in H. Then H is abelian. Hint for the proof: Use the fact that (xy)^2 = 1 = 1*1 = x^2y^2 to show that xy = yx for all x. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Simple Group Theory Question > I can't find the correct way to proving the following: > Let G be a non-abelian group of order 8, prove that G has a member of > order 4. The orders of elements of G can only be 1, 2, 4 or 8. If you have order 8 then you must also have order 4. The only other possibility is all have order 1 and 2. In that case a^2 = 1 for all a in G. What do you know about groups with this property? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Simple Group Theory Question I can't find the correct way to proving the following: Let G be a non-abelian group of order 8, prove that G has a member of order 4. I know that G can only be either an isomorphism of D_4 or the Quaternion, but I need to prove this directly without relying on that. === Subject: Re: max-min problem of a set of points in space? >You are saying that for M=13 in 3 dimensional space, you end up puting one >at the center? Yes. If you arrange them in layers of 3,7,3 in a tetrahedral packing, the minimum distance between any 2 points will be the radius of the large sphere. >I was thinking of putting all 13 on the unit sphere equally spaced with the >constraint that their grativity center must be at 0. I think you will get a minimum distance less than the radius if you put 13 points on the surface of a 3-D sphere. >For M=3, you agree that putting all 3 on the unit sphere equally spaced with >the constraint that their grativity center must be at 0 is the best, right? Yes, an equilateral triangle formation. In fact, I think you can generalize that for any M <= N+1 the best packing is the (M-1)-dimensional object with all M points an equal distance from each other, and all on the surface of the hypersphere. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: topological lemma Suppose that we are given a directed system of compact Hausdorff spaces indexed by N_0 X_0 --> X_1 --> X_2 --> ... so that the image of a space is closed in the following space and that the direct limit X is itself compact. Then what additional condition is needed to ensure that there is an N such that X=X_N? The general case is false by taking X_k = {0} cup [1/k,1]. TIA, Tobias -- everyone who casts a shadow seems to stand in the sun reverse my forename for mail! - saibot === Subject: Re: topological lemma > Suppose that we are given a directed system of compact Hausdorff spaces > indexed by N_0 > X_0 --> X_1 --> X_2 --> ... > so that the image of a space is closed in the following space and that the > direct limit X is itself compact. > Then what additional condition is needed to ensure that there is an N such > that X=X_N? The general case is false by taking X_k = {0} cup [1/k,1]. Is the direct limit compact? (I suppose one may think it's [0,1] but in the dircet limit surely {0} is an open set.) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: topological lemma >> Suppose that we are given a directed system of compact Hausdorff spaces >> indexed by N_0 >> X_0 --> X_1 --> X_2 --> ... It may be assumed additionally that the maps are homeomorphisms onto their (closed) image. >> so that the image of a space is closed in the following space and that >> the direct limit X is itself compact. >> Then what additional condition is needed to ensure that there is an N >> such that X=X_N? The general case is false by taking X_k = {0} cup >> [1/k,1]. > Is the direct limit compact? > (I suppose one may think it's [0,1] but in the dircet limit surely > {0} is an open set.) direct limit should be the disjoint union of (0,1] with a point which is not compact... So the general case could still be correct, but how to prove it? For my needs, it would suffice to prove some weaker case because the X_i really are manifolds with strictly increasing dimension. -- everyone who casts a shadow seems to stand in the sun reverse my forename for mail! - saibot === Subject: Re: topological lemma > Suppose that we are given a directed system of compact Hausdorff spaces > indexed by N_0 > > X_0 --> X_1 --> X_2 --> ... > > It may be assumed additionally that the maps are homeomorphisms onto their > (closed) image. > so that the image of a space is closed in the following space and that > the direct limit X is itself compact. > > Then what additional condition is needed to ensure that there is an N > such that X=X_N? The general case is false by taking X_k = {0} cup > [1/k,1]. >> Is the direct limit compact? >> (I suppose one may think it's [0,1] but in the dircet limit surely >> {0} is an open set.) > direct limit should be the disjoint union of (0,1] with a point which is > not compact... > So the general case could still be correct, but how to prove it? For my > needs, it would suffice to prove some weaker case because the X_i really > are manifolds with strictly increasing dimension. I should add that in my case X is not compact, but that is essentially what I want to show with this argument. OK, to be more concrete: I have a sequence of compact Hausdorff spaces (manifolds of strictly increasing dim) X_0 --> X_1 --> X_2 --> ... so that every space is homeomorphic to its image which is closed in the next space; the direct limit X need not be compact. Now I have a map f : K --> X from some compact Hausdorff X. The problem is to show that f(K) is already contained in some X_N. By taking everywhere the intersection with f(K), the direct limit is preserved: take Y_i = f(K) / X_i and Y = f(K). So Y is the direct limit of the Y_i and is compact, which gives the above situation. -- everyone who casts a shadow seems to stand in the sun reverse my forename for mail! - saibot === Subject: Re: topological lemma > >> Suppose that we are given a directed system of compact Hausdorff spaces >> indexed by N_0 >> >> X_0 --> X_1 --> X_2 --> ... >> >> It may be assumed additionally that the maps are homeomorphisms onto >> their (closed) image. >> so that the image of a space is closed in the following space and that >> the direct limit X is itself compact. >> >> Then what additional condition is needed to ensure that there is an N >> such that X=X_N? The general case is false by taking X_k = {0} cup >> [1/k,1]. > > Is the direct limit compact? > (I suppose one may think it's [0,1] but in the dircet limit surely > {0} is an open set.) > >> direct limit should be the disjoint union of (0,1] with a point which is >> not compact... >> So the general case could still be correct, but how to prove it? For my >> needs, it would suffice to prove some weaker case because the X_i really >> are manifolds with strictly increasing dimension. > I should add that in my case X is not compact, but that is essentially > what I want to show with this argument. > OK, to be more concrete: > I have a sequence of compact Hausdorff spaces (manifolds of strictly > increasing dim) > X_0 --> X_1 --> X_2 --> ... > so that every space is homeomorphic to its image which is closed in the > next space; the direct limit X need not be compact. > Now I have a map f : K --> X from some compact Hausdorff X. The problem is > to show that f(K) is already contained in some X_N. > By taking everywhere the intersection with f(K), the direct limit is > preserved: take Y_i = f(K) / X_i and Y = f(K). So Y is the direct limit > of the Y_i and is compact, which gives the above situation. How about this. if f(K) is compact and not in some X_n there are infinitely many n with f(K) intersect X_{n+1} not contained in X_n. Pick a point in each of these. The set of these points is infinite, closed (interscetion with each X_j is finite so closed) and contained in a compact so compact. But it also has the discrete topology: impossible! -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: topological lemma > How about this. if f(K) is compact and not in some X_n there > are infinitely many n with f(K) intersect X_{n+1} not contained > in X_n. Pick a point in each of these. The set of these points is > infinite, closed (interscetion with each X_j is finite so closed) and > contained in a compact so compact. But it also has the discrete topology: > impossible! Hey, great! Damn, been thinking about this problem for nearly uncountably many hours now, and then it's so easy... -- everyone who casts a shadow seems to stand in the sun reverse my forename for mail! - saibot === Subject: Re: Integers Into Grid Of Coprimality: Game > Here is a simple grid game based on coprimality. > Any number of players. > Start with an n-by-n grid drawn on paper. > (I suggest an n of at least 5 or 6.) > Players take turns writing integers into any of the grid's empty > squares. > Each integer must have not been used before in the game, > each integer must be >= 1 and <= 2*n^2, > and each integer must be coprime to every other integer in its row > and > column. > The last player to be able to move is the winner. > What would be a good strategy for this game? > Simply always write a prime number? > A 10x10 grid only requires 100 primes and there are 2262 available > <=20000, so the game always ends in a draw. > Also, I suggest, especially for large n, that a list of the integers > 1 > through 2*n^2 be kept, and each integer crossed off as it is used. > > > > Leroy Quet You must have misread my game's rules. :/ You want the integers to not be bigger than 2*n^2. When n=10, the integers must each be <=200, not 20000. So there are far fewer primes to work with. Leroy Quet === Subject: Re: Integers Into Grid Of Coprimality: Game > Here is a simple grid game based on coprimality. > Any number of players. > Start with an n-by-n grid drawn on paper. > (I suggest an n of at least 5 or 6.) > Players take turns writing integers into any of the grid's empty > squares. > Each integer must have not been used before in the game, > each integer must be >= 1 and <= 2*n^2, > and each integer must be coprime to every other integer in its row > and > column. > The last player to be able to move is the winner. What would be a good strategy for this game? > Simply always write a prime number? > A 10x10 grid only requires 100 primes and there are 2262 available > <=20000, so the game always ends in a draw. > Also, I suggest, especially for large n, that a list of the integers > 1 > through 2*n^2 be kept, and each integer crossed off as it is used. Leroy Quet > You must have misread my game's rules. > :/ > You want the integers to not be bigger than 2*n^2. > When n=10, the integers must each be <=200, not 20000. > So there are far fewer primes to work with. Duh! No wonder it seemed too easy. Ok, only 47 out of 100 squares (46 primes + 1) are no-brainers. That leaves 53 composites. Maybe I'll have another shot at it. > Leroy Quet === Subject: Abstract Algebra Back again with another abstract algebra proglem. Hope I can get the english terms somewhat correct. I didnt know how to do the text an symbols here in a newsgroup so I put it on a website. I did the symbols in MS Word so I just cross my fingers that people can see it. Url to page: http://w1.351.telia.com/~u35111589/abs.htm -- Stefan === Subject: Re: Abstract Algebra days. My association with the Department is that of an alumnus. >Back again with another abstract algebra proglem. Hope I can get the >english terms somewhat correct. >I didnt know how to do the text an symbols here in a newsgroup so I put >it on a website. I did the symbols in MS Word so I just cross my fingers >that people can see it. >Url to page: >http://w1.351.telia.com/~u35111589/abs.htm Suggestion for typing it in text: Let S_n be the permutation group on the set X={1,2,...,n}, and let s (or sigma) be an element of S_n. Define a relation ~ on X by x ~ y if and only if (sigma)^m(x) = y for some integer m. Show that ~ is an equivalence relation on X. there, because that would be interpreted as for every integer m, (sigma)^m(x)=y). You want to show that ~ is an equivalence relation. That means showing three things: (i) That ~ is reflexive; that is, for all x in X, x ~ x. (ii) That ~ is symmetric: that is, for all x and y in X, if x~y, then y~x. (iii) That ~ is transitive: that is, for all x, y, and z in X, if x~y and y~z, then x~z. So, let's take it one at a time. (i) Let x in X. We want to show that there is some integer m for which sigma^m(x)=x. Well... do you know whether there is some integer for which sigma^m is the identity? How about using that one? Or, what is sigma^0? (ii) Let x and y in X, and assume that x~y. That means that there exists m in Z such that (sigma)^m(x) = y. We want to show that there exists an n in Z such that (sigma)^n (y) = x. How about sigma^{-m}? (iii) We are assuming that there are integers n and m such that sigma^n(x)=y and sigma^m(y)=z. You want to show that there is an integer k such that sigma^k(x)=z. What is sigma^m(sigma^n(x))? Can you use that? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Abstract Algebra [SNIP] > Suggestion for typing it in text: > Let S_n be the permutation group on the set X={1,2,...,n}, and let s > (or sigma) be an element of S_n. > Define a relation ~ on X by > x ~ y if and only if (sigma)^m(x) = y for some integer m. > Show that ~ is an equivalence relation on X. > You want to show that ~ is an equivalence relation. That means showing > three things: > (i) That ~ is reflexive; that is, for all x in X, x ~ x. > (ii) That ~ is symmetric: that is, for all x and y in X, if x~y, then > y~x. > (iii) That ~ is transitive: that is, for all x, y, and z in X, if x~y > and y~z, then x~z. > So, let's take it one at a time. > (i) Let x in X. We want to show that there is some integer m for which > sigma^m(x)=x. Well... do you know whether there is some integer > for which sigma^m is the identity? How about using that one? Or, > what is sigma^0? > (ii) Let x and y in X, and assume that x~y. That means that there > exists m in Z such that (sigma)^m(x) = y. We want to show that > there exists an n in Z such that (sigma)^n (y) = x. How about > sigma^{-m}? > (iii) We are assuming that there are integers n and m such that > sigma^n(x)=y and sigma^m(y)=z. You want to show that there is an > integer k such that sigma^k(x)=z. What is sigma^m(sigma^n(x))? > Can you use that? I will look at that and hopefully I will get it. -- Stefan === Subject: Re: Abstract Algebra > [SNIP] > Suggestion for typing it in text: > > Let S_n be the permutation group on the set X={1,2,...,n}, and let s > (or sigma) be an element of S_n. > > Define a relation ~ on X by > > x ~ y if and only if (sigma)^m(x) = y for some integer m. > > Show that ~ is an equivalence relation on X. > You want to show that ~ is an equivalence relation. That means showing > three things: > > (i) That ~ is reflexive; that is, for all x in X, x ~ x. > (ii) That ~ is symmetric: that is, for all x and y in X, if x~y, then > y~x. > > (iii) That ~ is transitive: that is, for all x, y, and z in X, if x~y > and y~z, then x~z. > > So, let's take it one at a time. > > (i) Let x in X. We want to show that there is some integer m for which > sigma^m(x)=x. Well... do you know whether there is some integer > for which sigma^m is the identity? How about using that one? Or, > what is sigma^0? > > (ii) Let x and y in X, and assume that x~y. That means that there > exists m in Z such that (sigma)^m(x) = y. We want to show that > there exists an n in Z such that (sigma)^n (y) = x. How about > sigma^{-m}? > > (iii) We are assuming that there are integers n and m such that > sigma^n(x)=y and sigma^m(y)=z. You want to show that there is an > integer k such that sigma^k(x)=z. What is sigma^m(sigma^n(x))? > Can you use that? > I will look at that and hopefully I will get it. There is another way to look at this problem.If this is from a course you are studying I would be interested in what level and where it is given.Perhaps abstract algebra in a university or a bright highschool student studying on his own.You may e-mail a reply to this retired mathematician. for your question ,dynamical systems people look at it as follows.let S be any set (eg S={1.2.3....n) ) and let f :S->S be any one-one (fx=fy -> x=y) and onto S (every y in S can be written y=fx for some x in S) f^n(x) is the result of applying f n-times starting from x :1_x =f(x) 2_x =f(1_x) .............n_x =f((n-1)_x),for n a posiive integer ,f^-1(x) =y means f(y) =x and -1_x=f^-1(x) , -2_x=f^-1(-1_x) ...-n_x=f^-1((-n+1)_x) .0_x =f^0(x) =x but you know all this. For any x in S ,the set {n_x: over all integers n=0,1,2,3,...,-1,-2,-3....) is called the orbit determined by x,and is written orb(x).It can be checked by you that(1) if y is in the orbit of x then orb(y) is the same set as orb(x) .It follows that (2)if y is not in orb(x) then orb(x) and orb(y) have no elements in common .Also note that x=0_x is in orb(x) .This means the set S is a union of disjoint orbits,namely the the subsets E of S that can be written E=orb(x) for some xin S (notuniquely of course-any other member of orb(x) gives the same E .x~y means n_x=y for some integer n(your definition ) which means exactly that x and y belong to the same orbit . The properties defining an equivalence relation are now trivial viz- x is in the same orbit as x ,if x is in the same orbit as y then y is in the same orbit as x .if x is in the same orbit as y ,say E and y and and z belong to the same orbit ,say F then E=F since both orbits contain y (and different orbits have no elements in common.).So x and z belong to the same orbit E (or F) .So you must carefully show (1) and (2) above . The pair (S,f)is called a discrete dynamical system in the catagory of sets.This is one of the major interpretation s of equivalence relation.SMN === Subject: Re: ALL the groups of order 24 > D_12 = C_12 x| C_2 = (C_4 x C_3) x| C_2 > I think this is a little clearer than D_12 = C_4 x| D_3 above, > But C_12 x| C_2 doesn't even specify a group. There are three > non-trivial homomorphisms C_2 -> Aut(C_12), giving three > non-isomorphic C_12 x| C_2 groups. > By constrast, C_4 x| D_3 is unambiguous, given that we wouldn't > write x| if we meant x. A good point. C_12 x| C_2 can have G = ; x^12 = 1 = y^2 yxy = x^r ; with r = =/- 1, +/- 5 ; so if x| includes x, this could be 4 cases, 1 of which is Abelian. I guess its best to give the presentation. (I was using the action by C_2 = to be yxy = x^(-1), but that isn't what the notation means.) Van === Subject: Re: ALL the groups of order 24 A few more notes on the groups of order 24. Its still useful to write D_n = C_n x| C_2 = x^n = 1 = y^2, yxy^(-1) = x^(-1) Look at the center Z and G/Z, and perhaps one of the normal subgroups H and G/H, which tell of something of the subgroup structure of the groups, assuming we know about the structure of the groups of order <=15. C_2 x A_4 is self evident assuming we know the structure of A_4, which I like to think of as A_4/N = C_3. The center Z = C_2, and N =~ C_2 x C_2 is the normal subgroup of A_4, so that H = C_2 x N = C_2 x C_2 x C_2 is a normal subgroup of G, consisting of the 7 elements or order 2, and G/Z = A_4, G/H = A_4/N = C_3, so all the element of this group are either or order 2 or 3. Consider the case; |x| = 12, |y| = 2, yxy = x^r ; then r is in Z*12 = (-1, 5, 7). Clearly r = - 1 ==> D_12 = C_12 x| C_2 with Z = = C_2, Normal subgroup H = C_12, so G/Z = D_6 and G/H = C_2. Consider yxy = x^5 and yxy = x^7 (r = 5,7). We can see what is going on here by looking at 5Z mod 12 and 7Z mod 12; r = 5 has Z = = C_4 ; r = 7 has Z = = C_6. Also we see that the powers of x are permuted by r = 5; (1 5) (2 10) (4 8) (7 11) r = 7; (1 7) (3 9) (5 11) r = 7 corresponds to C_3 x D_4 x^{12}, y^2, yxyx^5 in the above post, r = 5 corresponds to C_4 x D_3 x^{12}, y^2, yxyx^{-5}, in the above. r = 5 G/Z = D_3 ; r = 7 has G/Z = C_2 x C_2 . This shows that all 3 groups r = -1, 5, 7 are non isomorphic. Van Van === Subject: Re: ALL the groups of order 24 Another group is G = C_2 x D_6 = Dih(C_2 x C_6) (from above--I don't know what Dih(A x B) means. But it has center Z = C_2 x C_2, and since for n = 2m with m odd, D_2m = C_2 x D_m, then G/Z = D_3. The above presentation is strange, and I haven't checked it. I prefer x^6 = 1 = y^2 = z^2, with D_6 = and z commuting with x and y. Next, C_3 x Q with Z = C_6, and G/Z = C_2 x C_2, and Q x| C_3 with Q = , x_3 = x_1*x_2, |x_i| = 4, and the (outer) auto z(x_i) =x_(i+1), cyclic permutation of the x_i in Q with order 3. This is SL(2,3), discussed above. Also I just noticed A_4 x C_2 = (C_2 x C_2 x C_2) x| C_3 from above, which I prefer. S_4 = PGL_2(3) = Hol(C_2 x C_2). I don't know what Hol(C_2 x C_2) is, but we know that PGL_2(3) must be S_4 as PSL(2,3) = A_4 is easy by looking at a few elements, and Z*_3 can be thought of as +/-1, and if det = -1, the elements are odd, while +1 are the even elements, so det A = sgn(A) if A is in S_4 or PGL_2(3), and sgn(A) is the parity of A as a permutation. I don't know what case Dic(C_12) corresponds to, but I believe it is what I call r = 5 or 7 and x^12 = 1 = y^2 in the above post. I am not familiar with any of these either. Q_6 x^{-2}y^6, x^{-2}(xy)^2 tilde{A_4} x^{-3}y^3, x^{-2}yxy noname1 x^4, y^6, (xy)^2, (x^{-1}y)^2 noname2 x^{-2}y^2, x^{-2}(xy)^3 C_2 x Q_3 x^6, y^4, y^{-1}xyx ; What do Q_3 and Q_6 refer to? -------- C_3 x| C_8 with x^3 = 1 = y^8 and yxy^(-1) = x^2 has Z = = Z_4, and G/Z = D_3. C_3 x| D_4 (where the kernel of the action is non-cyclic) has D_4 = ; x^4 = 1 = y^2 = z^3 and xzx^(-1) = z^2, with the center = kernel of the action = C_2 x C_2 is not cyclic. If y acted on both x and z we would get D_12, and Z = C_2, which was considered above. Van === Subject: Re: ALL the groups of order 24 > Another group is G = C_2 x D_6 = Dih(C_2 x C_6) (from above--I don't > know what Dih(A x B) means. Dih(G), where G is any abelian group, is the generalized dihedral group G x| C_2, where the non-trivial element of C_2 maps elements to their inverses. > S_4 = PGL_2(3) = Hol(C_2 x C_2). I don't know what Hol(C_2 x C_2) is, Hol(G), where G is any group, is the holomorph of G, that is, G x| Aut(G), where Aut(G) has its natural action. > I don't know what case Dic(C_12) corresponds to, but I believe it > is what I call r = 5 or 7 and x^12 = 1 = y^2 in the above post. No, those are D_3 x C_4 and D_4 x C_3. Dic(C_12) is the dicyclic group of order 24. See http://en.wikipedia.org/wiki/Dicyclic_group > I am not familiar with any of these either. > tilde{A_4} x^{-3}y^3, x^{-2}yxy This is the double cover of A_4. It's the same as SL_2(3). > noname1 x^4, y^6, (xy)^2, (x^{-1}y)^2 This was number 8 in my list. > noname2 x^{-2}y^2, x^{-2}(xy)^3 C_3 x| C_8. > C_2 x Q_3 x^6, y^4, y^{-1}xyx ; > What do Q_3 and Q_6 refer to? Q_n is Dic(C_2n). This conflicts with the usual meaning of Q_8. === Subject: Re: ALL the groups of order 24 > Another group is G = C_2 x D_6 = Dih(C_2 x C_6) (from above--I don't > know what Dih(A x B) means. > Dih(G), where G is any abelian group, is the generalized dihedral > group G x| C_2, where the non-trivial element of C_2 maps elements > to their inverses. I see. Makes sense. > S_4 = PGL_2(3) = Hol(C_2 x C_2). I don't know what Hol(C_2 x C_2) is, > Hol(G), where G is any group, is the holomorph of G, that is, > G x| Aut(G), where Aut(G) has its natural action. I saw this in some notes somewhere a while ago, and was planning to archives built into a larger reference, which is one reason I add all these posts. Another reason is to get feedback like this. > I don't know what case Dic(C_12) corresponds to, but I believe it > is what I call r = 5 or 7 and x^12 = 1 = y^2 in the above post. > No, those are D_3 x C_4 and D_4 x C_3. Dic(C_12) is the dicyclic group > of order 24. See http://en.wikipedia.org/wiki/Dicyclic_group > I am not familiar with any of these either. > tilde{A_4} x^{-3}y^3, x^{-2}yxy > This is the double cover of A_4. It's the same as SL_2(3). > noname1 x^4, y^6, (xy)^2, (x^{-1}y)^2 > This was number 8 in my list. > noname2 x^{-2}y^2, x^{-2}(xy)^3 > C_3 x| C_8. > C_2 x Q_3 x^6, y^4, y^{-1}xyx ; >What do Q_3 and Q_6 refer to? > Q_n is Dic(C_2n). This conflicts with the usual meaning of Q_8. I saw this somewhere also, which I hope to find. I thought Q_n had |x| = 2^n and y^2 = x^2(n-1) ; something like this. I will have to look it up. Van === Subject: Re: ALL the groups of order 24 Dic(C_6) x C_2 = C_6 x| C_4 ; I don't know which of Pawel Gladki's groups this corresponds to, but x^6 = 1 = y^4, with yxy^(-1) = x^(-1) is the only possible non-Abelian group. Conjugation by y fixes x^3, and y^2 x = x y^2, so Z = x = C_2 x C_2, and G/Z = D_3. The last thing is to verify that D_4 x C_3 = C_3 x (C_4 x| C_2) = C_4 x| C_6 = (C_2 x C_2) x| C_6 which I don't quite understand, since D_4 = C_4 x| C_2, is clear, acting by conugation, with yxy^(-1) = x^(-1), so the action squared is the identity. But A(C_2 x C_2) = D_3 has actions of order both 2 and 3. I suppose we could choose the action of order 2, interchanging the generators of C_2 x C_2, and ignore the action of order 3, which is the cyclic permutation of the 3 non-identity elements of C_2 x C_2, but it does not seem like this would give a group isomorphic to the first, though I haven't checked it. I guess these groups must be isomorphic, as this is the last group. If anyone has comments on this last group I would welcome them. Van === Subject: Re: ALL the groups of order 24 > Dic(C_6) x C_2 = C_6 x| C_4 ; I don't know which of Pawel Gladki's > groups this corresponds to, C_2 x Q_3. > D_4 x C_3 = C_3 x (C_4 x| C_2) = C_4 x| C_6 = (C_2 x C_2) x| C_6 > But A(C_2 x C_2) = D_3 has actions of order both 2 and 3. Yes, it's ambiguous, so I shouldn't have included it. The action of order 2 (kernel of order 3) was the one intended. The other action gives A_4 x C_2. === Subject: Re: The Cairo tile [was: Euclid's construction of the dodecahedron] [reply address is baloglouAToswego.edu] **************************************************************************** ** > *I spoke too soon: further web searches (on Cairo tiling rather than > Cairo tile) led to http://www.geocities.com/williamwchow/java/j8.htm, > where the very name Cairo tiling is attributed to F.S. Hill Jr. and a > reference is made to Doris Schattschneider's M. C. Escher: Visions of > Symmetry, where the tiling appears -- misclassified as a pgg rather > than p4g** -- in pages 226-7 and 318 (with a photo of a 'Cairo-tiled' > column in Baarn's New Lyceum) ... and it is also mentioned that the > tiling is also found in Japan (weakening my Arab theory somewhat***). There is no misclassification, Schattschneider is 100% right and I was ... 10% wrong: the reflections and 90-degree rotation are destroyed due to 'interlacing' (as she *does* point out, but I read that note a bit too carelessly and the figure in *that* page is a bit too small)! **************************************************************************** * Another way to destroy everything but the diagonal (pgg) glide reflections is coloring that renders the other isometries inconsistent with color: see http://www.oswego.edu/~baloglou/103/cairo-3-pgg.html (using 3 colors) and http://www.oswego.edu/~baloglou/103/cairo-5-pgg.html (using 5 colors); see also http://www.oswego.edu/~baloglou/103/cairo-5-p4.html for an example where everything but the p4 90-degree rotation fails to work (consistently with color) and http://www.oswego.edu/~baloglou/103/cairo-4a-p4g.html , http://www.oswego.edu/~baloglou/103/cairo-4b-p4g.html for two colorings (in 4 colors) that preserve the p4g structure in its entirety. baloglouAToswego.edu === Subject: Re: the modesty of Mr Wolfram. On the Foundation of mathematics and mathematica > I'm asking this, because I want to decide if I should buy the book, You can read it free online. http://www.wolframscience.com/nksonline/toc.html You can read many reviews online as well. e.g. http://www.maa.org/reviews/wolfram.html === Subject: Re: the modesty of Mr Wolfram. On the Foundation of mathematics and mathematica says... >> It's eerily reminiscent of JSH. He sounds like JSH with brains. > For all us outsiders, who IS JSH anyway? -- Phil Viton Ohio State University === Subject: Re: the modesty of Mr Wolfram. On the Foundation of mathematics and mathematica > For all us outsiders, who IS JSH anyway? A mathematically incompetent nuisance who periodically claims to have an elementary proof of FLT. Everytime he is shown to be in error he comes back and tries again, just as badly as prior. Bob Kolker === Subject: Re: the modesty of Mr Wolfram. On the Foundation of mathematics and mathematica > For all us outsiders, who IS JSH anyway? > A mathematically incompetent nuisance who periodically claims to have an > elementary proof of FLT. He hasn't been talking about an elementary proof of FLT lately, has he? Anyway, a more adequate description of JSH is that he is a genius, although not at mathematics, but at making contributors to sci.math jump through hoops. === Subject: Re: logic of the Cantorian followers mind === > David Petry Subject: Cantor's Theory: Mathematical creationism > Cantor's theory (classical set theory) has the same relationship to > the mathematical sciences as Creationism theory has to the physical > sciences. They are similar in content and similar in origin. Cantor's > theory is essentially a creation myth. Not so. > Both Cantor's theory and Creationism theory are founded on the > proposition that we must acknowledge the existence of some abstract > infinite entity lying beyond what we can observe in order to understand > the reality that we do observe. Not so. _Assume_ that ZF is consistent, then Cantors theorem follows. This is perfectly normal in Mathematics. If you don't want to work in ZF, then don't. If ZF is inconsistent, then this inconsistency is provable. Go and prove it. > - - - - - - - - - - - - - - - - - - - - - - - - - - > I cofounded this theory just a couple of days ago. The reason I am > having a problem debunking Cantor is because it is faith based. They > invent assume > an INVISIBLE, imaginary entitiy which is impossible to disprove > no matter what is your substance. I repeat: if it is inconsistent, then it is possible to prove that. > This is the reasoning behind Cantor supporters. > 4894389.. > 4389439.. > 4378498.. > 4894894.. > .. > Diag0 is 4374.. > Diag1 is 5485.. This is not reasoning. This is just some symbols with no explaination as to how they should be interpreted. > Intuitively this what? > is a new sequence, but that is not a *property* of the > number, I thought you said it was a sequence, now it's a number. There is a difference. > its merely DEFINED AS DIFFERENT. You haven't defined anything. > P = Diag1 <> Real1 and Diag1 <> Real2 and Diag1 <> Real3 ... > P is not a result of the proof, its the *definition* of Diag1 just in > propositional form, ?? > but its a poor definition. Correct! I'd go further: what you have written makes no sence. > What is obviously wrong is that A = All sequences of digits are > computable, and already present on the list of computable reals. Is this a definition? > Although A intuitively disproves P, Cantor followers use R, rebutal to > the argument A A is an argument then? Really? > R = the set 0.1, 0.2, 0.3...0.9, 0.01, 0.02, 0.03... also contains > every sequence of digits, but using strictness analysis it does not > contain any irrationals as all numbers have finite length. Well, this seems to be a statement. Ok, it would be if you made it clear what you meant by sequence of digits, and strictness analysis. > P -> C (Cantors proof, there are uncountable numbers) P is meaningless, so can not imply anything. > A -> !P (All sequences are computable, so Cantors premise of a new > sequence is flawed) > R -> !(A -> !C) (Listing all sequences of digits does not imply you > contain all reals) > They skip over the fact P is disproven, A seems to be debunked just > because sets of rationals that contain all sequences are incomplete. > This is why A still holds: > ------------------------------------------------------------ > An infinite number of people each toss a coin infinite times, can you > guarantee a new sequence of heads and tails? > ------------------------------------------------------------ If countably many people, each toss a coin countably many times, then there is a sequence of coin tosses that is realised by their tossing. This is clear by Cantors diagonal argument. > The answer is clearly no, you cannot come up with something new that > infinite people have done before. You seem to be the one believing things withouot proof. > Cantor supporters should try to > answer this question. > Herc I refer you to: http://www.math.ucla.edu/~asl/bsl/0401/0401-001.ps It's a very well written paper that trys to understand the misunderstandings that people have with Cantors diagonal argument. It contains a simple exposition of that argument too. === Subject: Re: logic of the Cantorian followers mind If you studied logic you'd have learnt what propositions are, statements that are either true or false. Cantors definition of the diag number is typically: D(i) = L(i,i) mod 9 + 1 We can represent this as an equivalent infinite clause. D1 = L(1,1) mod 9 + 1 AND D2 = L(2,2) mod 9 + 1 AND D3 = L(3,3) mod 9 + 1 AND ... this results in the proposition D1 <> L(1,1) AND D2 <> L(2,2) AND ... I defined 4 propositions. C (Cantor) = there are uncountable numbers P (Property of Diag) = Diag <> Real1 and Diag <> Real2 and Diag <> Real3 ... SEE ABOVE A (Argument) = All sequences of digits are computable, and already present on the list of computable reals R (Rebuttal) = the set 0.1, 0.2, 0.3...0.9, 0.01, 0.02, 0.03... also contains every sequence of digits, but using strictness analysis it does not contain any irrationals as all numbers have finite length. Your argument is at the beginner Cantorian level NOT(A) and is not covered in this topic. Herc === Subject: Re: Periods of exponential generators >Given only a^n + b^n = c^n, 0(a^n + b^n) == 0 mod c, >(c^n - a^n) == 0 mod b, and >(c^n - b^n) == 0 mod a. We could argue a+b mod c != 2, because a^0 + b^0 mod c = 2. The degenerate case. In which case a^n+b^n==2 mod c for all n, and so a^n+b^n!==0 for any n. However, 3+4 mod 5 = 2, and 3^2+4^2 mod 5 = 0, so that isn't right. Repeated values.... Hm. 2,2,0... Perhaps an analogy to the linear recurrence generator is valid. For 1st order recurrence generators such as linear congruential, multiplicative, and exponential generators, if a value repeats, it is terminal. It's an absorbing state, but this isn't a Markov chain. For 2nd order generators, there is a memory of the previous *two* values, and sequences like 2,2,0 then become possible. It seems it may be the case that for each dual generator associated with FLT there is a *well studied* 2nd order recurrence generator with associated matrix. As Galois field theory is applicable only for prime bases, and we have only about it in her thesis, which I have had bound but can barely read due to nystagmus from sleeping pills this summer. x.2 = a.0*x.0 + a.1*x.1 (mod c) initial conditions: x.0 = 2; x.1 = 2; c = 5. desired result: x.2 = 0 associated generator: 3^n + 4^n mod 5 a.0 = ? a.1 = ? I got more than one solution when I tried this last but that was about two months ago. However, at this point I can work with properties instead of values, to some limited extent. I tolerance everything and tolerate everyone. I love: Dona, Jeff, Kim, Kimmie, Mom, Neelix, Tasha, and Teri, alphabetically. I drive: A double-step Thunderbolt with 657% range. I fight terrorism by: Using less gasoline. === Subject: zeta(2) I read somewhere that the results pi^2/6 = Sum(1/n^2, n = 1 .. Infinity) and pi^2/12 = - Sum((-1)^n/(n^2), n = 1 .. Infinity) are equivalent. I don't see it, though. How do you go from one of them to the other? Am I missing something obvious? -- mpr === Subject: Re: zeta(2) > I read somewhere that the results > pi^2/6 = Sum(1/n^2, n = 1 .. Infinity) > and > pi^2/12 = - Sum((-1)^n/(n^2), n = 1 .. Infinity) This is 1 - 1/4 + 1/9 - 1/16 + 1/25 - ... = 1 + 1/4 + 1/9 + 1/16 + 1/25 + ... - 2(1/4 + 1/16 + 1/36+ ...) = 1 + 1/4 + 1/9 + 1/16 + 1/25 + ... - (1/2)(1 + 1/4 + 1/9 + ...) =( 1/2)(1 + 1/4 + 1/9 + 1/16 + 1/25 + ...) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: ALL the groups of order 24 One can easily see this since the centers are C_4 and C_2, repectively. Also, one can write for any n, D_n = C_n x| C_2, so that D_12 = C_12 x| C_2 = (C_4 x C_3) x| C_2 I think this is a little clearer than D_12 = C_4 x| D_3 above, though of course D_3 = C_3 x| C_2 ==> D_12 = C_4 x| (C_3 x| C_2) = (C_4 x C_3) x| C_2, since when (n,m) = 1, the only possibility is that the semi-direct product = direct product, so C_4 x| C_3 = C_4 x C_3. === Subject: Re: ALL the groups of order 24 > D_12 = C_12 x| C_2 = (C_4 x C_3) x| C_2 > I think this is a little clearer than D_12 = C_4 x| D_3 above, But C_12 x| C_2 doesn't even specify a group. There are three non-trivial homomorphisms C_2 -> Aut(C_12), giving three non-isomorphic C_12 x| C_2 groups. By constrast, C_4 x| D_3 is unambiguous, given that we wouldn't write x| if we meant x. === Subject: Re: Invariant Galilean Transformations (FAQ) On All Laws >: >>message >: >> Since Eleaticus seems to have problems accepting t' = t as part >of the >: >> Galilean Transformation, and since Eleaticus does not have the >: >> mathematical sophistication to comprehend the validity of that >: >> particular >: >> transformation of t within the Galilean Transformation, this is >: >> attempt >: >> to explain what is happening without introducing t' in the >: >> Transformation. >: >>: >> Eleaticus has explicitly demonstrated what many would have >suspected for >: >> a >: >> long time: that he is completely ignorant of multivariable >calculus. He >: >> has no familiarity with, or concept of, the Chain Rule in >multivariable >: >> calculus. Take, for example, his much beloved Galilean >Transformation: >: >>: >> x' = x - vt, >: >Two rulers, side my side. >: >>: >0'-----1'-----2'-------x' t'= 0 >: >0------1-----2--------x t = 0 >: >>: >One moves to the right a distance vt. >: >-----------------------0'-----1'-----2'-------x' >: >0------1-----2--------x<----------vt------->: >x' = x+ vt. >: >>: >I prefer going the other way, you see. >: >>: >> Assuming that v is positive, the correct equation is x' = x-vt. >: >Is it? Oh.... yes, I see what you mean. >: Observing the diagram again, I notice that the way that you have >drawn it >: does not indicate either that x' = x+vt or that x' = x-vt, so we >were >: both mistaken. Also, it indicates that you diagram was badly drawn. >: Your diagram SHOULD have looked like >: <-----------vt-------->0------1------2--------x' >: 0------1-----2------------------x >: then x' = x-vt would be the appropriate equation. >Ok, if you insist on show the displacement between the respective >origins >o and o' I'm happy with that. It looks like it should be the same >displacement >between x and x', but no problem. > <-----------vt-------->0------1------2--------x' >0------1-----2------------------x<-----vt----->: >> You have >: >> the same situation for which the same equation is appropriate, >but you >: >> have associated the wrong equation with the situation which you >have >: >> described. Here is a hint. In your second picture, the values >of x' are >: >> less than the corresponding values of x, >: >Well, yes, they are, I see that now. >: >I should have drawn >: >-----------------------0-----1-----2-------x >: >0'-----1'-----2'------x'<-------+vt------->: >x' = x+ vt. >: Again, your diagram is badly drawn. It should have looked like >: <------vt-------->0-----1-----2-------x >: 0------1------2-------x' >: THAT is the appropriate diagram for x' = x + vt. >Ah, I see. Well, the zero end of the ruler is more important anyway. >: >Or maybe >: >0'-----1'-----2'-----3'----4' >: >---------------|<--+vt--->: >---------------0-----1-----2 >: >4' = 2+vt. So v = 1, t = 2. Better? >: No. Because again, your diagram is wrong. Also, for some reason >which I >: cannot even begin to fathom, you are denoting points on the x' axis >with >: 0', 1', 2', 3', 4'. Why on earth do you want to put dashes on the >numbers >: 0, 1, 2, 3, 4, when dashes don't belong there???? >They are called primes, not dashes, as I'm sure even an >incompetent >relativist will tell you. >The reason for putting them there is for clarity. Garbage. The 0', 1', 2', 3', 4', are already on an axis marked x'. The marking of x' is clear enough without you burdening the 0, 1, 2, 3, 4, with extraneous and unnecessary primes. It also does not justify your ridiculous usage of 4' in your formula 4' = 2+vt. Your usage of 4' in that formula suggests that you really think that the prime on the 4 has some significance, which it does not. >When the two rulers >are side by side at t=0, x' = x-v0 = x. Writing x = x-vt doesn't make >much sense, does it? Of course it doesn't, but what does that have to do with the fact that the primes do not belong on 0, 1, 2, 3, 4, on the x' axis? I make a statement that 0, 1, 2, 3, 4, on the x' axis should not have primes on them, and you defend doing so by bringing the completely irrelevant fact that one must distinguish between x and x'. What does it say about your way of doing things if you defend doing something silly by raising a completely irrelevant point? >You do realize vt is a distance, I trust? Yes, of course I do. Stop being so sarcastic and condescending. You do not have the educational background on this subject to allow you any right to be sarcastic or condescending. >Since >you do not like my diagram, preferring yours, I'll do it this way. ><------------vt----------->0'---------x' >0--------x >Now it should be easy for you to see 0' = 0-vt It is not true that 0 = 0-vt unless vt = 0. >AND x' = x-vt. >Right? Except for the fact that you are using 0' on the x' axis where you should be using 0. x' is a variable in its own right, and its values are real, e.g., 0, 1, 2, etc. x' is not a function of x whose value at 0 is denoted by 0', at 1 is denoted by 1', etc, but a variable. >: >> whereas, if your equation had >: >> been right (with v and t both positive), the values of x' would >be GREATER >: >> than the corresponding values of x. >: >>: >> Substituting -v for v, and interchanging primed and unprimed >coordinates >: >> in x' = x - vt, we get x = x' + vt, which is equivalent to x' = >x - vt. >: >Well done. Now, are you saying v cannot be negative without >interchanging >: >the frames? >: No, I didn't. I don't know whether your misinterpretation of what I >I don't know that about you, either, but I'm fair and try to give you >the >benefit of the doubt. No you are not. You are trying to put words into my mouth, and pretend that I said them. But why suggest it in the first place? >Since you didn't say that, we'll make v >negative, ok? >x'-------------0'<--(-vt)--->---------------x--------------0 >Now we have x' = x+vt, 0' = 0+vt, right? Wrong. x' is not a function of x whose value at 0 is denoted by 0', at 1 is denoted by 1', at 2 is denoted by 2', etc. x' is a variable in its own right, and its values are real, so that it can take values 0, 1, 2, 3, etc. The formula x' = 0 is meaningful. The expression x' = 0' is not. The answer to your question depends on the sense in which you measure x and x'. Is the positive direction to the left or to the right? Is v positive or negative? Is t positive or negative? You have left these questions unanswered. The convention must be established before we can relate the variables x, x', v, t. >It should be, it's your >dagram >mirrored. >: It looked like you wanted vt >: positive, so that required v positive and t positive or v negative >and t >: negative. >No no, t is not ever negative. Wrong. t must be measured from a specific datum. Any time before that datum has a negative value for t. t is a coordinate in spacetime which can take negative values as well as positive values. >v can be, though. Cars do usually have >a reverse gear as well as forward gears. That is irrelevant. In linear motion along the x axis, v is positive if x increases as t increases, and v is negative if x decreases as t increases. If we have a road in the north-south direction, and x increases as you move north, then all cars on the northbound side (and facing north) moving forwards have positive v. All cars on the northbound side (and facing north) moving in reverse have negative v. All cars on the southbound side (and facing south) moving forwards have negative v. All cars on the southbound side (and facing south) moving in reverse have positive v. It is not simply the case that v is positive when a car moves forward and v is negative when a car moves in reverse. >: I chose the former (t positive) as the more likely. Of >: course the equations work just as well with v negative as with v >positive, >: and I resent you suggesting that I intended otherwise. >Resentful because I asked you a question? No, I am not resentful because of that. I am resentful because of the hidden implication regarding my intentions and assertions, and also because of the attempt to attribute a statement to me that I never made. >A. Excuse me, sir, but could you direct me to city hall? >D. I resent you suggesting I do not know where city hall is. >Not a very bright reply, is it? And only an idiot would consider it to be an analogous situation. You questioned my intentions and assertions. You attempted to attribute a claim to me that I never made. None of this is true in the specious example that you made in an attempt to belittle my genuine concerns that a statement would be attributed to me that I never made. The two cases have nothing in common with each other. >: >For the Andersen Transforms, I get >: >xi = (x')/sqrt(1/v^2/c^2), >: >tau = (t + vx/c^2)/sqrt(1/v^2/c^2), >: These transformations require that xi = 0 iff x' = 0. >Well yes, of course. What is so terrible about that? I did not say that that was terrible. I just made an observation. What is terrible is that you have no consistency regarding your independent variables. You used x' in one of your transformation equations, and you used x and t in the other one. >It is also true of the Lorentz equations. Have you forgotten them? No. The Lorentz transformation law for xi is xi = (x-vt)/sqrt(1-v^2/c^2). It is not xi = x'/sqrt(1-v^2/c^2). >xi = (x-vt)/sqrt(1/v^2/c^2), >tau = (t - vx/c^2)/sqrt(1/v^2/c^2), >and since x' - x-vt, Where is it implicit that x' = x-vt. When Einstein used x' = x-vt, he was entitled to do so since he explicitly defined x' in this way. You have not so explicitly defined x'. >xi = (x')/sqrt(1/v^2/c^2), so that xi = 0 iff x' = x-vt = 0. Only x' = x-vt. You have not made such a definition. Einstein explicitly defined x' = x-vt when he first used the variable x'. >: Are you sure that >: that was what you meant? >Yes, I am. Why would you question it? I looks fine to me. Because you had not taken the trouble to define x'. >: I ask this particularly in view of the fact you >: have provided the expression for tau in terms of x and t, and so you >: should have provided the expression for xi in terms of x and t as >well, >: otherwise your equations are useless. >Me? No, I'm just copying from Einstein's text. But I'll agree the >equations are useless. Garbage. The difference from Einstein's usage of the equations in that form is that HE HAD ALREADY EXPLICITLY DEFINED x' IN TERMS OF x AND t. You had not. Because x' was ALREADY meaningful in Einstein's paper, then his equations make sense, and are useful. You have not given any such definition of x', and THAT is the reason why YOUR equations were useless. In other words, the difference between Einstein's usage of the equations and yours is that Einstein took care to establish the correct context for the equations, which was something that you could not be bothered to do so. >: >and the faster you go, the later you arrive. >: >I think t' = t in ANY transform. >: Nonsense. t' = t in the Galilean transformation, where time has >been >: assumed absolute. For the Lorentz Transformation, >: x' = (x - vt)/sqrt(1 - v^2/c^2), >Oh... I see. >x' = (x-vt) >AND >x' = (x-vt) / sqrt(1-v^2/c^2) ? No. >Perhaps you should actually read what Einstein said. Here it is: > http://www.fourmilab.ch/etexts/einstein/specrel/www/ >x' = x-vt >xi = (x')/sqrt(1-v^2/c^2) >See? Stop being so bloody condescending. It is particularly galling when you are simultaneously doing such a good job of flaunting your own ignorance. The notation has changed in the intervening years. Einstein used (x,t) as the coordinates for one frame of reference, and (xi,tau) as the coordinates in the second frame of reference. He also defined the quantity x' as x' = x-vt. In present day notation, (x',t') are the coordinates in the second frame of reference. In other words, x' now means what Einstein denoted by xi, and t' now means what Einstein denoted by tau. A person has to be particularly perverse in order to take x' as it is used at present (i.e. as I used it), and expect it to mean exactly the same thing that Einstein did in his paper of 1905. This is particularly so since I never even mentioned his paper of 1905. What I find particularly interesting is your assertion that I don't know and I also find it interesting that for some reason, you expect the world at large to use the same notation that Einstein did in 1905, even if the 1905 paper has nothing to do with the discussion. Your deliberate misinterpretation of my equation x' = (x-vt)/sqrt(1-v^2/c^2) is not appreciated. If you have an objection of my using that equation for the Lorentz Transformation Law, I would expect you to come up with something more substantial than just pathetically pointing out that Einstein used x' to mean something completely different in his paper of 1905. >: t' = (t - vx/c^2)/sqrt(1 - v^2/c^2). >: >> y' = y, >: >>: >> This becomes y = y', when substituting -v for v, and >interchanging primed >: >> and unprimed coordinates. y = y' is equivalent to y' = y. >: >>: >> z' = z. >: >>: >> This becomes z = z', which is equivalent. >: >Yeah... >: >and x' = x when v = 0, which it does in Einstein's equation, >: Which is what you would expect. So what. >So I've made my point, that's what. But your point was completely irrelevant to the conversation at that point. The So what was because what you had written was completely irrelevant to what was being discussed. It still is. >The equations are ing useless, as you now agree. No, I didn't. You are twisting my words to mean what you want them to mean. If you are going to enter this discussion, then the very least that I am entitled to expect from you is honesty and integrity. If you can't be bothered being honest, then there is no point to the discussion. >: I ask this particularly in view of the fact (Einstein has] >: provided the expression for tau in terms of x and t, and so >[Einstein] >: should have provided the expression for xi in terms of x and t as >well, >: otherwise [Einstein's] equations are useless. >Is there anything else I can help you with? More condescension on your part, which you have no entitlement to give. If you quote with the intention of deliberately deceiving, then you demonstrate that people can safely ignore your dishonesty with impunity, as your dishonesty makes your contributions worthless. >[Final bit snipped due to irrelevance and impertinence of McAnally's >idiocy.] Talk about the pot calling the kettle black. If there's anybody here with a lack of intellect, it is you. After all, you write meaningless drivel. I note that you have cut the other parts where I quite rightly pointed out that you kept answering my points with complete and utter irrelevancies. That seems to be one of your techniques. If you can't respond directly, then you will make a completely irrelevant comment, and then pretend that it is a constructive contribution to the discussion. Perhaps it would help if you actually made the attempt to understand what was being written, and what people were actually saying, and then respond to what they are actually saying, and not just responding to what you want to *pretend* that they are saying, which is what you are doing now. But then, I suppose that is too much like hard work for somebody like you, and you would rather just fly off the handle spewing insults, and making observations without caring one jot about whether what you are writing is cogent to the situation at hand. ----- === Subject: Re: Invariant Galilean Transformations (FAQ) On All Laws To Androcles McAnally, shadows with the pseudonym Speicher, enjoy your tour into the psychotic world of ultra kook dom, the're trolls, by any other name. Ken [........] === Subject: Re: Google SCHOLAR! > Is it just me or does it seem to not have any ACS publications in the > search results? > May be just you. Although it doesn't actually sem to have the pubs.acs.org > pages - maybe they're hidden by the ACS. See eg > 22> (or ), yet this doesn't have the copy of the > first paper that's on the nottingham.ac.uk webserver > Stephan > -- Looks like an interesting paper - shame it was published in a closed access journal. Darren. === Subject: Re: This Week's Finds in Mathematical Physics (Week 209) > 24) John Baez, Space and state, spacetime and process. > Notes available at http://www.ima.umn.edu/categories/#wed2 I looked in 24 for buzzwords and found symmetries of spacetime IR^4, symmetric monoidal function, symmetry group, Feynman diagram, and finally positively oriented point. came across with exactly the same ideas dating back at least to 2001. Just the figures are more instructive now. They reminded me of plumbers' work. Who can tell me why John did not yet understand what I am suggesting. Inspired by Pauli, I would like to say that elapsed time is not just not symmetrical but also not even asymmetrical. Eckard === Subject: Re: This Week's Finds in Mathematical Physics (Week 209) > Inspired by Pauli, I would like to say that elapsed time is > not just not symmetrical but also not even asymmetrical. More of your customary amathematical gibbering :-( -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: (probably stupid) question on Fermat decidability >> The same conclusion (undecidable in the system -> true in the >> standard model) holds for statements of the specific kind like FLT or >> Goldbach, where if they were false, then there would be a verifiable >> counterexample. >No!! I am confused! Everything you write below seems correct, but I do not see Derek Holt. >Either one of A or ~A, where A is a formally undecidable statement >within a model, can be added as an axiom to that model to form a new >and larger model which remains consistent. If, for example, A is the >Goldbach conjecture, then adding ~A to the model for arithmetic as it >is generally understood does not lead to a contradiction. All it says >is that a counterexample can not be found by any finite number of >steps within the model which omits the axiom. >Let's see if a simpler example would help. >Within the natural numbers, consider the statement there is no natural >number x such that x*x=2. Let's pretend for the sake of illustration >that it is formally undecidable and then add it as an axiom that there >exists an x such that x*x=2. A completely consistent theory is the >result. The new theory includes the theory of natural numbers as a >sub-theory (and all true derivable theorems within that model would >remain derivable in the larger model), but there is still no *natural* >number x such that x*x=2 which can be found by any finite number of >deductions from the Peano axioms. >It's a subtle point, perhaps, but a formal system and an >interpretation of a formal system are very different things. In the >example given, one can no longer interpret a variable x as a natural >number but is forced to interpret it as an irrational number, which >contains the interpretation as a natural number as merely a special >case. >Also as the example shows, adding underivable statements as axioms can >lead to interesting mathematics. >Paul >-- >Hanging on in quiet desperation is the English way. >The time is gone, the song is over. >Thought I'd something more to say. === Subject: JSH: Better now I don't know if you noticed but I had a tremendous drop in confidence concomittant with a dramatic grip of existential crisis. But I feel much better now and my confidence is restored, and I've learned to handle better the implications of my own research. I guess there's no way any of you would understand what it's like to end up with some research result far bigger than you ever wanted, finding something you didn't want to find, where that makes everything so much harder. It would have been easier for me in many ways for there not to be this problem with algebraic integers which is so HUGE where it impacts so many areas in mathematics and extends out into the physics world through group theory--though not so much in a bad way! It's like the screw-up in the math world might have helped physics as group theory got developed, so what if the underlying mathematical ideas were slightly off? Hey, in physics, if it works, it works. In any event, for me, there was that lagging doubt that maybe somewhere along the way someone would find something just WRONG with what I had, and why would that be a surprise or not a good thing? But a lot of people having looked over my work, with a lot of criticisms considered lead me to finally just have to accept that it is correct, despite my severe misgivings. You know that technique of non-polynomial factorization may lead to new physics theories that can probe *into* a finer mathematical ever even dreamed might exist? Sigh. Of course you don't. That's ok though as it'd shock the hell out of me if any of you had even a minor grasp of how big all of this is, as that would mean you are, um, different in important ways than what I've gathered over years of making these posts and watching the effects. Now the wait is on a math journal. I'm still not sufficiently surrogate factoring got a nice paper, but not for us, please send to another journal from a journal in the top 5, while Combinatorica passed on my paper deriving the prime counting function, which is kind of ok, as it had a minor, sort of error, which they didn't notice. I kind of wonder about this peer review thing. Hey, NONE of them have given me copies of comments from reviewers. When do you get those? Southwest Journal of Pure and Applied Mathematics didn't either. Is there some kind of weird rule going on where I don't get comments from reviewers? Maybe from fear I'll just post them? James Harris === Subject: Re: JSH: Better now !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > I don't know if you noticed but I had a tremendous drop in confidence > concomittant with a dramatic grip of existential crisis. > But I feel much better now and my confidence is restored, and I've > learned to handle better the implications of my own research. > I guess there's no way any of you would understand what it's like to > end up with some research result far bigger than you ever wanted, > finding something you didn't want to find, where that makes everything > so much harder. Well, don't despair. Even some far bigger heap of elephant poop than you ever wanted can be swept away eventually. Anyway, you are drunk again, right? Or hast thou eaten on the insane root that takes the reason prisoner? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: Better now > I don't know if you noticed but I had a tremendous drop in confidence > concomittant with a dramatic grip of existential crisis. > But I feel much better now and my confidence is restored, and I've > learned to handle better the implications of my own research. Unfortunately, James, what you think of as better appears to the world at large as worse. That existential crisis is reality seeping in. === Subject: Re: JSH: Better now > Now the wait is on a math journal. I'm still not sufficiently > surrogate factoring got a nice paper, but not for us, please send to > another journal from a journal in the top 5, while Combinatorica > passed on my paper deriving the prime counting function, which is kind > of ok, as it had a minor, sort of error, which they didn't notice. I > kind of wonder about this peer review thing. I wonder if JSH is on to something ... not about math, but about journals. === Subject: Re: JSH: Better now [snippage] > I don't know if you noticed but I had a tremendous drop in confidence > concomittant with a dramatic grip of existential crisis. I do hope you're managing to avoid the logical fallacy known as cum hoc, ergo propter hoc there. > It would have been easier for me in many ways for there not to be this > problem with algebraic integers What problem? > which is so HUGE where it impacts so > many areas in mathematics Such as? > and extends out into the physics world > through group theory--though not so much in a bad way! Examples? > It's like the screw-up in the math world might have helped physics as > group theory got developed, so what if the underlying mathematical > ideas were slightly off? Is that supposed to be English? > Hey, in physics, if it works, it works. > In any event, for me, there was that lagging doubt that maybe > somewhere along the way someone would find something just WRONG with > what I had, and why would that be a surprise or not a good thing? Well, it wouldn't be a surprise (groogle for oops author:jstevh@msn.com for evidence). > But a lot of people having looked over my work, with a lot of > criticisms considered lead me to finally just have to accept that it > is correct, despite my severe misgivings. What are your misgivings? > You know that technique of non-polynomial factorization may lead to > new physics theories that can probe *into* a finer mathematical > ever even dreamed might exist? 'May' in the sense that any given event 'may' lead to any other given event, or in some stronger sense which you will now demonstrate? btw, my physics isn't what it once was, what's the mathematical structure of an electron? > Now the wait is on a math journal. > I'm still not sufficiently > surrogate factoring got a nice paper, but not for us, please send to > another journal from a journal in the top 5, They were politely telling you to sod off, you do realise that James? > while Combinatorica > passed on my paper deriving the prime counting function, which is kind > of ok, as it had a minor, sort of error, which they didn't notice. Submitting papers containing errors? How out of character. > I > kind of wonder about this peer review thing. > Hey, NONE of them have given me copies of comments from reviewers. > When do you get those? When anyone takes your work remotely sseriously. > Southwest Journal of Pure and Applied > Mathematics didn't either. > Is there some kind of weird rule going on where I don't get comments > from reviewers? Maybe from fear I'll just post them? Established maths journals have been dealing with cranks for a lot longer than you've been around, James. They have a special file for papers such as yours. > James Harris PS any chance you're gonna re-examine your FLT work, in the light of your new polynomial factorization techniques? That was some of your best material, I think. -- Larry Lard Replies to group please === Subject: Re: JSH: Better now > Is there some kind of weird rule going on where I don't get comments > from reviewers? Maybe from fear I'll just post them? > James Harris I pointed out to you a few weeks ago how comments on your paper from a peer reviewer were likely to start. Do you really think a busy Prof. somewhere is going to wade through your paper making pointing all of these things out, and trying to guess what you mean? If he did, would you dismiss these comments as a style critique The point is, if you don't define things, and don't write things clearly, people can not judge whether the claim is correct or not. P.S. don't take these things personally, the last peer review I got started by noting: p1 l 15: word field has been omitted after residue and there were various such corrections. I was very grateful for the help. === Subject: Re: JSH: Better now > I don't know if you noticed but I had a tremendous drop in confidence > concomittant with a dramatic grip of existential crisis. ??? > But I feel much better now and my confidence is restored, and I've > learned to handle better the implications of my own research. Read: 'arrogance' for 'confidence'. > In any event, for me, there was that lagging doubt that maybe > somewhere along the way someone would find something just WRONG with > what I had, and why would that be a surprise or not a good thing? > But a lot of people having looked over my work, with a lot of > criticisms considered lead me to finally just have to accept that it > is correct, despite my severe misgivings. Apparently you do NOT read or understand your critics when they present iron-clad refutations of your claims or counter-examples to your results. > You know that technique of non-polynomial factorization may lead to > new physics theories that can probe *into* a finer mathematical > ever even dreamed might exist? Pigs may have wings, too. (More likely, in fact!) -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Better now >I don't know if you noticed but I had a tremendous drop in confidence >concomittant with a dramatic grip of existential crisis. >But I feel much better now and my confidence is restored, and I've >learned to handle better the implications of my own research. >[...] >In any event, for me, there was that lagging doubt that maybe >somewhere along the way someone would find something just WRONG with >what I had, and why would that be a surprise or not a good thing? Uh, two things to say about that: (i) LOL. Plenty of people _have_ found things that are simply WRONG. (ii) You know, this is one of many reasons why nobody ever believes anything you say. You never _express_ any doubts about the validity of your work - it's clear, simple, anyone who denies it is lying or doesn't understand simple algebra, etc etc etc. But every once in a while you admit that you _previously_ _had_ some doubts. It happens periodically that you tell us you had doubts at time t, when at time t you _said_ there was no doubt at all! See, this means every once in a while you _admit_ that you were _lying_ previously... >[...] >Now the wait is on a math journal. I'm still not sufficiently >surrogate factoring got a nice paper, but not for us, please send to >another journal from a journal in the top 5, while Combinatorica >passed on my paper deriving the prime counting function, which is kind >of ok, as it had a minor, sort of error, which they didn't notice. I >kind of wonder about this peer review thing. >Hey, NONE of them have given me copies of comments from reviewers. >When do you get those? This is very curious. I always get a copy of the referee's comments when I submit a paper - that happens when the paper is accepted, accepted subject to suggested revisions, or (gasp) rejected. >Southwest Journal of Pure and Applied >Mathematics didn't either. >Is there some kind of weird rule going on where I don't get comments >from reviewers? Maybe from fear I'll just post them? People have explained this. Why do you keep asking the same question if you're just not going to believe the answer? It seems quite likely that most editors are not sending your papers out for review, because it's obvious at a glance that it's just trash, and the editor wouldn't want to waste a reviewer's time. If an editor does send it to a reviewer, the reviewer will explain that it's just nonsense. So why do the editors not tell you that it's all nonsense, instead replying politely that maybe you should consider a different journal? Possibly because they like to be polite. Possibly because they realize that if they explain what they actually think about the paper that will lead to a pointless but heated exchange. (Answer the following for yourself: in cases where they _have_ explained more fully, like the comments you posted here about how the Prime Counting Function had no theoretical content, how did you react? You sent them nasty letters in reply, didn't you? Yes you did, admit it. Why should an editor willingly get into that sort of thing?) >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Better now Discussion, linux) > Hey, NONE of them have given me copies of comments from reviewers. > When do you get those? Southwest Journal of Pure and Applied > Mathematics didn't either. > Is there some kind of weird rule going on where I don't get comments > from reviewers? Maybe from fear I'll just post them? Maybe the decisions to reject your submissions have come from the editors. He is under no obligation to send it out to reviewers, after all. If the editor tells you that the paper isn't right for his journal, he likely didn't solicit reviewers. I thought the fact you never saw the original reviews for the Southwest Journal was odd. But when I commented on this in <87hdt7thl7.fsf@phiwumbda.org>, both David Ullrich and Arturo Magidin said that they've had papers published without seeing referee reports. -- If .999... = 1 then (.999...)/1 should equal 1 let's see (.999...)/1 = .999... [Therefore] .999... still=/= 1 -- An astonishing proof by S. Enterprize === Subject: Re: JSH: Better now days. My association with the Department is that of an alumnus. >I thought the fact you never saw the original reviews for the >Southwest Journal was odd. But when I commented on this in ><87hdt7thl7.fsf@phiwumbda.org>, both David Ullrich and Arturo Magidin >said that they've had papers published without seeing referee reports. I have also received rejections without an explicit referee report. In one case I was simply told that the referee thought the paper was not of sufficiently wide interest to warrant publication in our journal; in another I was again just told of a suggestion by the referee, rather than receive a full referee report. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: Better now > In any event, for me, there was that lagging doubt that maybe > somewhere along the way someone would find something just WRONG with > what I had, and why would that be a surprise or not a good thing? Perhaps you've missed all the times people have pointed out the numerous things that are just WRONG with what you had. > Hey, NONE of them have given me copies of comments from reviewers. > When do you get those? Southwest Journal of Pure and Applied > Mathematics didn't either. > Is there some kind of weird rule going on where I don't get comments > from reviewers? Maybe from fear I'll just post them? My guess is most of the reviewers felt it was unreadable. Others probably didn't see the connection to Galois Theory mentioned in the introduction. If your paper is returned with comments of those type, where there is no perceived hope for it, they probably don't bother. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Automorphisms of abelian p-groups: Aut(Z_p+Z_{p^2}) >> What is the automorphism group of Z_p+Z_{p^2} for p prime? >> This is what I have so far: Each automorphism must fixes the class of >> elements of order p. So, it must fixes the subgroup Z_p+Z_p. But I'm >G = C_p x C_{p^2}, with presentation , clearly >has p^2*(p-1) elements of order p^2 -- each cyclic subgroup of >order p^2 has p(p-1) such elements, and there are p such cyclic >subgroups: , , ..., -- and p(p-1) elements of >order p that aren't powers of elements of order p^2 -- the >non-identity elements in , , ..., -- so >|Aut(G)| = p^3*(p-1)^2. >automorphisms for Aut(G) = , with A^{p(p-1)} = 1, >B^p = 1, C^{p-1} = 1, D^p = 1: >A(x) = x^n, A(y) = y, where n generates (Z_{p^2},*) =~ C_{p(p-1)} >B(x) = xy, B(y) = y >C(x) = x, C(y) = y^m, where m generates (Z_p,*) =~ C_{p-1} >D(x) = x, D(y) = x^p y >Filling in the details of the presentation of Aut(G), i.e., >finding sufficient relations among the generators, is left as an >exercise for the reader. You might have added that Aut(G) has order p^3 (p-1)^2, with a normal extraspecial group of order p^3 (generated by A^(p-1), B, D, with A^(p-1) central), and the quotient group is isomorphic to Z_{p-1} x Z_{p-1}. Derek Holt. === Subject: Re: EZ Root > This shows the steps on how to find roots to the nth degree > polynomial. An example of a 6th degree polynomial is provided, > and two of its solutions are calculated. OK, I have a little test for this. > by Jon Giffen > The polynomial > a[0]+a[1]t+a[2]t^2+a[3]t^3+ ... + a[n] = 0 > Has the roots, > |Q| > (t,t^2,t^3,..t^n)= T = --- C + Q > |C| > Where > C=(D*N)S+(S*N)D and > N=(a[1], a[2], a[3],....,a[n]) > D=( 1 , 2 , 3 ,...,n ) > S=(a[1],2a[2],3a[3],...,na[n]) > Q=(-a[0]/|N|^2)N > |N|^2=a[1]^2+a[2]^2+a[3]^2+..+a[n]^2 > |Q| > T[0] = Q + --- C > |C| > L[1]*C=0 > L[1]*Q=0 Solve for L[1] > |T[0]| > T[1] = T[0] + -------- L[1] > |L[1]| > L[2]*L[1]=0 > L[2]*T[0]=0 Solve for L[2] > |T[1]| > T[2] = T[1] + ------ L[2] > |L[2]| > L[3]*L[1]=0 > L[3]*L[2]=0 > L[3]*T[1]=0 Solve for L[3] > . |T[i-1]| > T[i] = T[i-1] + -------- L[i] > |L[i]| > t^6+t-10=0 Try using the polynomial created when translating the given polynomial 1 unit to the right: (t-1)^6+(t-1)-10 = t^6-6t^5+15t^4-20t^3+15t^2-5t-10 = 0. > N=(1,0,0,0,0,1) > |N|^2=2 > Q=(10/2)(1,0,0,0,0,1)=5(1,0,0,0,0,1) > |Q|=(5)2^(1/2) > D=( 1 , 2 , 3 ,...,n )=(1,2,3,4,5,6) > S=(a[1],2a[2],3a[3],...,na[n])=(1,0,0,0,0,6) > S*N=(1,0,0,0,0,6)*(1,0,0,0,0,1)=1+6=7 > D*N=(1,2,3,4,5,6)*(1,0,0,0,0,1)=1+6=7 > C=(S*N)D+(D*N)S > =7[(1,2,3,4,5,6)+(1,0,0,0,0,6)] > =7(2,2,3,4,5,12) > |C|=(7)202^(1/2) > T=Q+C|Q|/|C| > =5(1,0,0,0,0,1)+7(2,2,3,4,5,12)[(5)2^(1/2)]/[(7)202^(1/2)] > =5(1,0,0,0,0,1)+5(2,2,3,4,5,12)/101^(1/2) > t^6 = 5 + 5(12)/101^(1/2) = 10.9702 t=1.4906 > 10.9702+1.49-10=2.46 > t=1.4906-2.46/(6(1.4906^5)+1)=1.4361 N.M. > t^6=8.7721 t=1.491 > t^5=6.1083 t=1.436 > t^4=4.2534 t=1.436 > t^3=2.9618 t=1.436 > t^2=2.0624 t=1.436 > T[0]=(1.4361,2.0624,2.9618,4.2534,6.1083,8.7721) > |T[0]|=12.1425 > L*Q=(a,b,0,0,0,c)*(1,0,0,0,0,1)=0 a=-c > L*T[0]=(-c,b,0,0,0,c)*T[0] > = -1.43617c + 8.7721c + 2.0624b=0 c=1 b=-3.5570 > L=(-1,-3.5570,0,0,0,1) |L|=3.8278 > T[1]=T[0]+L|T[0]|/|L| > = (1.4361, 2.0624,2.9618,4.2534,6.1083,8.7721) > +( -1,-3.5570, 0, 0, 0, 1)(12.1425/3.8278) > ----------------------------------------------------------------- > = (1.4361, 2.0624,2.9618,4.2534,6.1083,8.7721) > +( -1,-3.5570, 0, 0, 0, 1)(3.1728) > ----------------------------------------------------------- > = ( 1.4361, 2.0624,2.9618,4.2534,6.1083,8.7721) > +(-3.1728,-11.2858, 0, 0, 0,3.1728) > ----------------------------------------------------------- > (-1.7367, -9.2234,2.9618,4.4534,6.1083,11.9449) > t= -1.7367, 3.304i ,1.436 ,1.452 ,1.436 ,1.5119 > t^6=8.7721+3.1728=11.9449 t=-1.5119 > 11.9449-1.5119-10=0.433 > t=-1.5119-0.433/(6(-1.5119^5)+1)=-1.5025 N.M. > (-1.5025)^6-1.5025-10=0.0056~0 === Subject: Re: Why Do Americans Call It Math? > I used to say Maths, as a Brit, I still do sometimes in the UK, but I gave > putting the s on the end makes it awkward to lisp out. Math avoids the > horrible triple consonant 'ths' > Math or Mathematics is easier to say... one has to bow to simplicity > Richard Miller >

How do you pronounce schedule?
>If an American pronouce it like British people,
>do you think he's snobbish?
If an American pronounced schedule like the BritĒs he would be > speaking correct English! If a Brit pronounced it like an American he would > be considered uneducated. But able to beat the Germans ... === Subject: Re: General solutions to (dr/dt)^2=k >> I _gave_ you a general mathematical description of these curves! >> Any differentiable curve at all, parametrized by arc length. >> ************************ >> David C. Ullrich >Is this an unsolved problem? Huh? >I think someone have proved that arc >length parametrization of anything except a straight line is >impossible. Huh??? Find a book on calculus of several variables. Very informally, given any curve you can simply _define_ a parameter by arclength, and that gives you your parametrization by arclength. Officially: Say c(t) is a differentiable curve - inserting all the fine print, suppose that c'(t) is continuous and never vanishes. Let L(t) = int_0^t |c'(s)| ds, so L(x) is the length of the curve from t=0 to t=x. Now L is a continuous strictly increasing function, so it has a continuous inverse G(t). Note that L(G(t)) = t plus the chain rule show that L'(G(t)) G'(t) = 1, so G'(t) = 1/L'(G(t)). Now define C(t) = c(G(t)). Then C is just a reparamatrization of the original curve c. And C is parametrized by arclength: The chain rule shows that C'(t) = c'(G(t)) G'(t) = c'(G(t))/L'(G(t)), and the fundamental theorem of calulus shows that L'(t) = |c'(t)|; hence L'(G(t)) = |c'(G(t)|, and putting that in above gives |C'(t)| = |c'(G(t))/|c'(G(t))|| = 1. (And hence in your original somwwhat informal notation, (dC/dt)^2 = 1. So: that condition is satisfied by a suitable reparametrization of more or less any curve.) >Now, I know there are several algorithms for approximation >but I'm interested in the theoretical aspect, i.e. existence and >uniqueness of solutions. >Mike ************************ David C. Ullrich === Subject: Re: General solutions to (dr/dt)^2=k >> >> I _gave_ you a general mathematical description of these curves! >> Any differentiable curve at all, parametrized by arc length. >> >> ************************ >> >> David C. Ullrich >Is this an unsolved problem? > Huh? >I think someone have proved that arc >length parametrization of anything except a straight line is >impossible. > Huh??? Find a book on calculus of several variables. > Very informally, given any curve you can simply > _define_ a parameter by arclength, and that gives > you your parametrization by arclength. Officially: > Say c(t) is a differentiable curve - inserting all the > fine print, suppose that c'(t) is continuous and never > vanishes. Let > L(t) = int_0^t |c'(s)| ds, > so L(x) is the length of the curve from t=0 to t=x. > Now L is a continuous strictly increasing function, > so it has a continuous inverse G(t). Note that > L(G(t)) = t plus the chain rule show that > L'(G(t)) G'(t) = 1, > so G'(t) = 1/L'(G(t)). > Now define > C(t) = c(G(t)). > Then C is just a reparamatrization of the original > curve c. And C is parametrized by arclength: The > chain rule shows that > C'(t) = c'(G(t)) G'(t) = c'(G(t))/L'(G(t)), > and the fundamental theorem of calulus shows that > L'(t) = |c'(t)|; hence L'(G(t)) = |c'(G(t)|, and putting > that in above gives > |C'(t)| = |c'(G(t))/|c'(G(t))|| = 1. > (And hence in your original somwwhat informal notation, > (dC/dt)^2 = 1. So: that condition is satisfied by a suitable > reparametrization of more or less any curve.) >Now, I know there are several algorithms for approximation >but I'm interested in the theoretical aspect, i.e. existence and >uniqueness of solutions. >Mike > ************************ > David C. Ullrich those elliptical integrals but I guess you answered the question on the existence of solutions to the problem. Mike === Subject: Re: General solutions to (dr/dt)^2=k > > I _gave_ you a general mathematical description of these curves! > Any differentiable curve at all, parametrized by arc length. > > ************************ > > David C. Ullrich >Is this an unsolved problem? >> Huh? >>I think someone have proved that arc >>length parametrization of anything except a straight line is >>impossible. >> Huh??? Find a book on calculus of several variables. >> Very informally, given any curve you can simply >> _define_ a parameter by arclength, and that gives >> you your parametrization by arclength. Officially: >> Say c(t) is a differentiable curve - inserting all the >> fine print, suppose that c'(t) is continuous and never >> vanishes. Let >> L(t) = int_0^t |c'(s)| ds, >> so L(x) is the length of the curve from t=0 to t=x. >> Now L is a continuous strictly increasing function, >> so it has a continuous inverse G(t). Note that >> L(G(t)) = t plus the chain rule show that >> L'(G(t)) G'(t) = 1, >> so G'(t) = 1/L'(G(t)). >> Now define >> C(t) = c(G(t)). >> Then C is just a reparamatrization of the original >> curve c. And C is parametrized by arclength: The >> chain rule shows that >> C'(t) = c'(G(t)) G'(t) = c'(G(t))/L'(G(t)), >> and the fundamental theorem of calulus shows that >> L'(t) = |c'(t)|; hence L'(G(t)) = |c'(G(t)|, and putting >> that in above gives >> |C'(t)| = |c'(G(t))/|c'(G(t))|| = 1. >> (And hence in your original somwwhat informal notation, >> (dC/dt)^2 = 1. So: that condition is satisfied by a suitable >> reparametrization of more or less any curve.) >>Now, I know there are several algorithms for approximation >>but I'm interested in the theoretical aspect, i.e. existence and >>uniqueness of solutions. >Mike >> ************************ >> David C. Ullrich >those elliptical integrals but I guess you answered the question on >the existence of solutions to the problem. ??? I have no idea what problem you're referring to, nor what solutions to (dr/dt)^2 = k have to do with elliptic integrals. Don't know much about elliptic integrals either, except that at least typically there _is_ no closed-form expression for them... >Mike ************************ David C. Ullrich === Subject: Ring of continuous real functions === Subject: Re: Ring of continuous real functions Should I also send you email response? >> 2) If p:C(X) -> R is a nonzero algebra homomorphism then there's >> an unique a in X such that p(f) = f(a) for each f in C(X). > Remember that homomorphisms are assumed to be linear, so a > non-zero one will be surjective. >>Remember? Oh yea, linear algebra gives 'algebra' a parochial >> meaning. Properly then isn't an 'algebra' just an inner product >> space? > ??? No. Completely different things. Most inner product spaces > don't have amultiplication. The 'inner product' of two vectors > is a scalar for an inner product space. For algebras, the product > of two vectors is another vector. An algebra A is a vector space A with a vector product? R^3 with the cross product is an algebra? An algebra with a non-associative, anti-commutative product? Or is an algebra A is a vector space A with a vector product that is both sides distributive over addition? In other words a vector ring. Four dimensional vectors with matrix multiplication is an example of a non-commutative vector ring. For vectors x,y and scalar a, a needed axiom is a(xy) = (ax)y = x(ay) ? Is a (vector) multiplicative identity required? The expression 'algebra' is non-descriptive, even misleading, while a proper description 'vector ring' is clear and helpful. ---- === Subject: Re: Ring of continuous real functions === > Subject: Re: Ring of continuous real functions > Should I also send you email response? > 2) If p:C(X) -> R is a nonzero algebra homomorphism then there's > an unique a in X such that p(f) = f(a) for each f in C(X). >> Remember that homomorphisms are assumed to be linear, so a >> non-zero one will be surjective. >Remember? Oh yea, linear algebra gives 'algebra' a parochial > meaning. Properly then isn't an 'algebra' just an inner product > space? >> ??? No. Completely different things. Most inner product spaces >> don't have amultiplication. The 'inner product' of two vectors >> is a scalar for an inner product space. For algebras, the product >> of two vectors is another vector. > An algebra A is a vector space A with a vector product? > R^3 with the cross product is an algebra? > An algebra with a non-associative, anti-commutative product? a noncommutative nonassociative algebra :-) Even better, R^3 with the cross product is a Lie algebra ! -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Partitioned Partitions Assume you have to partition the letters AAAAABBC (=5,2,1) to three boxes holding 4,2 and 2 letters. Is there some general formula how many possibilities this gives? -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn === Subject: Re: Partitioned Partitions > Assume you have to partition the letters AAAAABBC (=5,2,1) > to three boxes holding 4,2 and 2 letters. Is there > some general formula how many possibilities this gives? ^^^^^^^ Hi Hauke! It's the coefficient of a^5*b^2*c^1 * x^4*y^2*z^2 in the taylor expansion of 1/[ (1-a*x)*(1-a*y)*(1-a*z)*(1-b*x)*(1-b*y)*(1-b*z)*...*(1-c*z) ]. In Maple code: 1/mul(mul(1-i*j,i=[a,b,c]),j=[x,y,z]); subs({seq}(i=0,i=[a,b,c,x,y,z]),diff(%,[a$5,b$2,c$1,x$4,y$2,z$2])); Result: 368640 For practical computations, it might be faster to derive recursion formulas. === Subject: Re: Schwartz'n'eleaticus Paradox > SR is consistent? > Yes, it is consistent. > Your example assumes that simultaneity is absolute; which is > inconsistent with SR. You have assumed your conclusion, which is > falacious. > Why go so far out of your way to prove your corrupt cretinism? > Non-absolute simultaneity has no utility when discussing human sized objects > in impact. If you think it does, then make it ant sized objects. If you could do the math, or if you even had an inkling about SR, you would realize that your statement above is idiotic. I award you five wilsons for it. Paul Cardinale === Subject: Re: the modesty of Mr Wolfram. On the Foundation of mathematics and mathematica >Mathematica is a tool. Can you give me a citation for the original paper on >the hammer? > That's Hammer, and I believe the original paper is here: > Say, now that people are talking about it, has anyone ever seen Mr.W > and JSH in the same room at the same time? Hm... I have met Mr. W, and seen JSH's picture on his web site. You can tell the difference. Oh, I guess die-hards will say, Are you sure that picture really was JSH? I think Mr. W has a big ego. He also has a big accomplishment to his credit: founding and running Wolfram Research, Inc. He does not display any of the bitterness that is so prominent in the postings of JSH. -- Chris Henrich God just doesn't fit inside a single religion. === Subject: Re: the modesty of Mr Wolfram. On the Foundation of mathematics and mathematica >>Mathematica is a tool. Can you give me a citation for the original paper on >>the hammer? >That's Hammer, and I believe the original paper is here: >Say, now that people are talking about it, has anyone ever seen Mr.W >and JSH in the same room at the same time? Hm... Hmm, indeed. Some might suggest that JSH's bitterness about the fact that mathworld refuses to mention his work shows that your unstated conjecture is full of beans. But it's clear to me that that's all just a smokescreen. Enhances one's respect for Wolfram. I mean, _I_ could easily have written Mathematica if I'd felt like it, but I can't imitate JSH's writing style - I've tried many times. >dave ************************ David C. Ullrich === Subject: Re: Daryl McCullough ANSWER THE QUESTION === >Subject: Re: Cantor's Theory: Mathematical creationism >------------------------------------------------------------ >An infinite number of people each toss a coin infinite times, can you >guarantee a new sequence of heads and tails? >------------------------------------------------------------ >The answer is clearly no, you cannot come up with something new that >infinite people have done before. Cantor supporters should try to >answer this question. You cannot guarantee a new sequence, but with probability one, all of them will be different, as will the sequence tossed by another person. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Daryl McCullough ANSWER THE QUESTION HERC777 says... >------------------------------------------------------------ >An infinite number of people each toss a coin infinite times, can you >guarantee a new sequence of heads and tails? >------------------------------------------------------------ Yes. The sequence C(i) defined by if person number i tosses a heads on toss number i, then C(i) = tails, otherwise C(i) = heads. -- Daryl McCullough Ithaca, NY === Subject: Re: Daryl McCullough ANSWER THE QUESTION new sequence of H&T missing from infinite random attempts. Josh finally bit his tongue, can you answer this question: >How many digits d is D matched to on any new infinite random list? It's easy to construct a list L such that, for all d, the first d digits of L's inverted diagonal are the first d digits of some element of L (that is, for all d, there exists an n such that for all k < d, L_n(k) = !L_k(k)). But even such a list will not necessarily have its own inverted diagonal as an element. Take the list such that the nth entry consists of n 1s followed by an unending sequence of 0s: 0 0 0 . . . 1 0 0 0 . . . 1 1 0 0 0 . . . 1 1 1 0 0 0 . . . Here, every finite prefix of L's inverted diagonal 111... is a finite prefix of some element of L (in fact, for all d, for all k < d, L_d(k) = !L_k(k)). But, since each element of L has only a finite number of 1s followed by an infinite number of 0s, L's inverted diagonal is not actually an element of L. -- Josh Purinton You completely ignored the question and branched off into the topic logic of the Cantorian followers mind. I'm not constructing a list given your number, you can put your number in a black box and hold the key. THEN I generate a countable infinite random list. THEN you hand over the key. THEN we check how many digits of your number I matched on my INDPENDANT list. How many digits d is D matched to on any *new* infinite *random* list ? Herc >How many digits d is D matched to on any *new* infinite *random* list? I don't know. -- Josh Purinton Does anyone know? Is it AA/ 0 A/ 1 B/ >1 C/ <10 D/ >10 E/ >1,000,000 F/ unlimited G/ infinite H/ all of them digits a countable random list will match any given number? Do any people who follow Cantors proof know how many digits of your NEW sequence will be matched? === Subject: Re: Daryl McCullough ANSWER THE QUESTION Discussion, linux) > HERC777 says... >>------------------------------------------------------------ >>An infinite number of people each toss a coin infinite times, can you >>guarantee a new sequence of heads and tails? >>------------------------------------------------------------ > Yes. The sequence C(i) defined by > if person number i tosses a heads on toss number i, > then C(i) = tails, otherwise C(i) = heads. But this is a new question, isn't it Herc? Before, you asked whether it was *possible* to flip a sequence not on the list (i.e., that it is possible the list is incomplete). Now you ask whether one can *guarantee* there's a sequence not on the list (i.e., that the list is necessarily incomplete). As it turns out, both answers are yes. But the questions are still different. -- Just because you're ... in a Ph.d program it does not mean that you're up to the challenge of being a real mathematician. Only those who have a purity of mind and dedication to the truth as the highest ideal have a chance. --James Harris, as Sir Galahad the Pure. === Subject: Re: Daryl McCullough ANSWER THE QUESTION <87is7wh1qy.fsf@phiwumbda.org> I'm surprised your feeble mental powers detected a change. Yes, to stop ignoramuses interpreting an extreme case from is it possible FOR YOU to form a new sequence I changed it to guarantee a new sequence. you don't even have to toss a coin to get it, you can run your finger down the diagonal and proclaim your infinite newness. Herc >>------------------------------------------------------------ >>An infinite number of people each toss a coin infinite times, can you >>guarantee a new sequence of heads and tails? >>------------------------------------------------------------ I bet school kids get it right. I bet 3rd year stats / probability students get it right. === Subject: Re: Daryl McCullough KILLFILE THE TROLL stuff that's wasted on trolls. Phil === Subject: Re: solid angle by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iANFdPL11151; >> Hello. >> I read somewhere the following equation for the solid angle covered by >> a PET-camera: >> solid angle = 4 * pi * sin(arctg(A/D)) >> A PET camera is a device used to detect and image positron emitters. >> Nowadays, it consists of a ring detector, into which the patient is >> brought. >> A in the equation above refers to the depth of the ring detector, ie. >> the dimension parallel to the ring axis. >> D stands for the diameter of the ring. Sorry, typo(duplicate label) is corrected. >A sphere(imagined for calculation facility,Dsph corner to corner) of >diameter sqrt(D^2+A^2)=2 R = Dsph is cut centrally by a cylinder/ring >diameter D. The length of intersected PET patient's tube is A. Then >surface area is simply = 2 pi R A and solid angle =2 pi R A /R^2 = 2 >pi A/R = 4 pi A/Dsph = 4 pi sin(alpha), where alpha is angle between >planes of the ring at opening and sphere. Formula 4 pi >sin(arctan(A/D) is confusing, 4 pi A/Dsph is quite OK. === Subject: one linear algebra problem Hi all, I am unable to solve the following problem. Any help is welcome. PROBLEM: Let V be an n dimensional vector space over R. Does there exists an (n-1)-dimensional subspace W of E^2(V) Such that dimension{w in W| w#b = 0}=1 for all b in V{0} ?? NOTATIONS: Notation: We work in the exterior powers of an n-dimensional vector space V over R. E^k(V) will be the k-th exterior power, and # will denote exterior product. --------------------------- for n=3, i know there does not exist any. for arbitrary n i dont know how to proceed. akash === Subject: Principle Curvatures of a Hypersurface Here's one that has me stumped ... I'm trying to find the n-1 principle curvatures of a hypersurface f(x_1,x_2...x_n) in an n-dimensional Hilbert space. How do I go about doing this? All the texts I have referred to only consider surfaces in R3 with simple (not to mention obvious) parametrizations. Note that in my case, I dont have a handy parametrization of the form x_j=x_j(p_1,p_2...p_n-1) , j=1,n (1) where p_1,p_2 ... p_n-1 are the n-1 parameters characterizing the n-dimensional hypersurface. My questions boil down to (1) Is there a general method of determining a parametrization of the form (1) for any given hypersurface? (2) Are the principle curvatures then merely the vector norms of the diagonal elements of the rank n-1 tensor of 2nd derivatives of the position vectors defined by (1)? (3) Can the principle curvatures be determined without recourse to such a parametrization? Lets assume that the hyperfurface f has continuous 1st and 2nd partial derivatives w.r.t each of x_1,x_2 ...x_n. -Sharat === Subject: Re: Principle Curvatures of a Hypersurface have a look at the book, Elementary Topics in Differential Geometry by J.A. Thorpe, Springer, 1979. It develops differential geometry in the n-dimensional space. Chap. 21 covers curvatures of surfaces. Ciao Karl > I'm trying to find the n-1 principle curvatures of a hypersurface > f(x_1,x_2...x_n) in an n-dimensional Hilbert space. > How do I go about doing this? All the texts I have referred to only > consider surfaces in R3 with simple (not to mention obvious) > parametrizations. === Subject: Re: Principle Curvatures of a Hypersurface Hi Sharat, > Here's one that has me stumped ... > I'm trying to find the n-1 principle curvatures of a hypersurface > f(x_1,x_2...x_n) in an n-dimensional Hilbert space. > How do I go about doing this? All the texts I have referred to only > consider surfaces in R3 with simple (not to mention obvious) > parametrizations. > Note that in my case, I dont have a handy parametrization of the form > x_j=x_j(p_1,p_2...p_n-1) , j=1,n (1) > where p_1,p_2 ... p_n-1 are the n-1 parameters characterizing the > n-dimensional hypersurface. > My questions boil down to > (1) Is there a general method of determining a parametrization of the > form (1) for any given hypersurface? Unlikely, you can only approximate locally using geodesic coordinates, a nonlinear (quadratic I believe) transformation of space. You can try to integrate the constrained geodesics emanating from a point on a geodesic circle but the result will not be exact. > (2) Are the principle curvatures then merely the vector norms of the > diagonal elements of the rank n-1 tensor of 2nd derivatives of the > position vectors defined by (1)? No, see below. > (3) Can the principle curvatures be determined without recourse to > such a parametrization? Yes, because the second derivative information of the surface may be found locally by using a linear transformation of space. Suppose we have found a point on the hypersurface (a manifold of order n-1 embedded in an n-dimensional space) and computed the unit normal vector to the surface at this point. We can now perform a linear transformation of space by using the Householder reflection or some other orthogonal rotation matrix to rotate the surface normal into the vector En = [0,0,0,..,0,1]. This rotation can then be applied to the second derivative matrix of the surface itself, essentially transforming it to a coordinate system whose first n-1 coordinates are within the surface. It is then an easy matter to project onto the surface by taking the principal minor of order n-1. The eigenvalues and eigenvectors of the reduced Hessian are the curvatures and principal directions respectively. Finally, we pad the eigenvectors with zeroes and transform them back. This procedure can be applied for illustration to a surface in 3D given as z=F(x,y) or even G(x,y,z)=0 showing that the principal directions and principal curvatures may be computed without having a parametric form of the surface and wholly from the embedding space without recourse to the differential geometry of a surface. In the general case of a manifold (m > 1), the block reflector as described by Schreiber and Parlett [1] may be used to construct an orthogonal transformation which maps the (generally not mutually orthogonal) normals to the constraint surfaces into orthogonal unit vectors Ek (k=n-m...n). This is described in the paper under The standard task. An orthogonal basis within the manifold is easily constructed the most obvious one being trivial Ek k=1,n-m vectors padded with m zeroes: [1,0,0...][0,0,0] [0,1,0...][0,0,0] This orthogonal basis of vectors is then mapped back to the full n-dimensional space using the inverse of the mapping. In the case of a reflector it is symmetric so the inverse is equal to itself. In summary, given a set of m independent vectors the block reflector may be used to construct a complete orthogonal decomposition of space into m mutually orthogonal coordinates and n-m mutually orthogonal coordinates in its orthogonal complement. The block reflector based on a set of vectors yk (k=1..m) reverses vectors in the range of yk and leaves vectors in its orthogonal complement unchanged. The problem is that while its construction from a given set of vectors is trivial, its construction based on the required mapping of a set of vectors into another set [subject to isometry constraints as with an ordinary Householder transformation] is much harder. To find the principal directions and curvatures of a hypersurface we may use the Householder reflector. For a manifold (m > 1), we may use the block reflector to effect a transformation of space to the manifold but the principal directions and curvatures are governed by composite quadratic forms involving all constraint surfaces. To optimize some objective function subject to a set of m constraints [a manifold] we compute its Hessian, then compute the block reflector for the normals to the constraint surfaces and transform the Hessian using the block reflector. Then we project by taking the principal minor of order n-m and compute its eigensystem. Finally, the eigenvectors are padded with m zeroes and transformed back using the block reflector. The n-m back transformed eigenvectors are the principal directions of the objective function within the manifold but expressed in coordinates of the embedding space and they are orthogonal to the normals to the constraint surfaces. In my paper on the constrained geodesic I treat the general case of an n-dimensional space with m constraint surfaces resulting in an n-m dimensional manifold. I have tried to explain this in my manuscript on the constrained geodesic: http://www.numerical-algorithms.com/ I have had few reactions on this paper so maybe it is not so clear... I have actually used this approach in a numerical setting in the notorious Sphere Problem where objective functions of configurations of points on the surface of a sphere are optimized. In the case of p points on a sphere, the Riemannian space has 3p dimensions and there are p unit norm constraints and 3 rotation constraints so the manifold has dimension n=2p-3. The key paper (difficult) on the block reflector is: Block Reflectors: Theory and Computation Robert Schreiber and Beresford Parlett Siam Journal of Numerical Analysis Vol 25 No.1, February 1988 Unfortunately, the block reflector may have been presented or perceived as a tool for wholesale matrix block row annihilation but it has much wider application as a tool for subspace manipulation. [The block reflector may be too expensive for ordinary applications involving block row operations because its computation is expensive, requiring an SVD of the order of the block. This type of problem goes back to Kronecker who studied quadratic forms with linear constraints and tried to say something about the eigenvalues of the constrained quadratic form. In the simplest case think of an ellipsoid centered at the origin in 3D and aligned with the axes which is cut with a plane passing through the origin and think of what happens to the main axes. Good luck, Jentje Goslinga > Lets assume that the hyperfurface f has continuous 1st and 2nd partial > derivatives w.r.t each of x_1,x_2 ...x_n. > -Sharat === Subject: Re: .999... stillstill=/=1 >... = oo >.999... = .999 + oo = indeterminate => oo = indeterminate - 0.999 = indeterminate (1) >Lim >10^n ( .999...) =/= 1000... >n-->oo 1000... = 1000 + oo = 1000 + indeterminate = indeterminate lim {n --> oo} (10^n ( .999...)) = = lim {n --> indeterminate} ((10^n) * indeterminate) = = lim {n --> indeterminate} indeterminate = something != indeterminate Is this correct? If so, since something is not indeterminate is it determinate? Also ...... = ? (maybe oooo?) Remark: in the context of floating-point programming arithmetic with indeterminate value (NaN) is not meaningless for some applications. === Subject: Simple question: Taylor expansion of eigenvalues of a large matrix Hello! I want to make a Taylor series expansion around a parameter that appears in the eigenvalues from a 12x12 matrix. Even though half of the matrix entries are zero, the eigenvalues are still the solutions of a large polynomial. Is there anyway to get the Taylor series approximation with respect to my parameter without knowing the analytical form of the full eigenvalues? Also, I would prefer not to have to make any assumptions about the magnitude of the eigenvalue itself, since its magnitude will vary strongly with a second parameter (not the parameter which I want to make the series expansion around). I put the matrix in Mathematica and it doesn't seem to be able to handle it... === Subject: Re: Simple question: Taylor expansion of eigenvalues of a large matrix > I want to make a Taylor series expansion around a parameter >that appears in the eigenvalues from a 12x12 matrix. Even though half >of the matrix entries are zero, the eigenvalues are still the >solutions of a large polynomial. Is there anyway to get the Taylor >series approximation with respect to my parameter without knowing the >analytical form of the full eigenvalues? Also, I would prefer not to You certainly shouldn't need an explicit expression for the eigenvalues; everything can be done implicitly. >have to make any assumptions about the magnitude of the eigenvalue >itself, since its magnitude will vary strongly with a second parameter >(not the parameter which I want to make the series expansion around). >I put the matrix in Mathematica and it doesn't seem to be able to >handle it... Perhaps you mean something like this? This is Maple, but I assume Mathematica has something similar. # Make a random sparse matrix with entries containing parameter p > with(LinearAlgebra): M:= RandomMatrix(12,12, density=1/3) + p*RandomMatrix(12,12, density=1/3); [76 + 32 p , 0 , 93 , -53 , -83 p , 0 , 0 , 72 + 43 p , 0 , 0 , 0 , 0] [-62 , 28 , 0 , 0 , -84 , 0 , 0 , -59 , 40 + 21 p , 0 , -81 p , 49 - 24 p] [0 , 0 , 0 , 0 , -82 , -62 , 0 , -68 p , 0 , 0 , 9 - 93 p , -28 + 75 p] [-35 , -86 + 85 p , 50 , -23 , 0 , 54 , 0 , 0 , -51 + 8 p , 94 - 62 p , 0 , 4 + 65 p] [52 p , 0 , 36 p , -67 p , -91 - 17 p , 93 p , -36 p , -79 p , 0 , 88 + 8 p , 38 - 90 p , 0] [-76 + 60 p , -4 p , 0 , 52 p , 0 , -53 p , -90 p , 0 , 49 , 0 , 8 p , -34 + 73 p] [-15 p , 0 , 46 p , 0 , 9 - 67 p , 0 , -61 , 0 , -18 p , 0 , 0 , 0] [0 , -68 , -24 p , 68 , 0 , 0 , 0 , 0 , 49 p , 0 , 0 , 0] [53 , 10 p , 67 p , -94 - 82 p , 0 , 0 , 0 , -58 p , 46 p , 0 , 60 p , 27 p] [0 , -18 , -37 , -93 , 79 , -67 , 0 , -95 p , -93 , 0 , 0 , 0 ] [0 , 0 , 0 , 40 , -97 p , 14 p , 1 , 47 p , 0 , 8 , 0 , -32] [0 , 0 , 0 , 83 , 50 + 81 p , -5 p , 41 , 0 , 0 , -91 , 0 , 0 ] > P:= CharacteristicPolynomial(M,t); > series(RootOf(P,t), p=0, 3); Output is a rather complicated series involving a root of the polynomial obtained from P by substituting p=0. I wouldn't want to expand this to much higher order, as things will rapidly get very complicated, I think. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: summation of combinations upto logn What is the value/closest approximation of summation of n choose i for i varying from 1 to logn ? Edmond === Subject: Re: summation of combinations upto logn >What is the value/closest approximation of summation of n choose i for i >varying from 1 to logn ? A first-order approximation for the sum for i <= j, j small compared to sqrt(n), is (n choose j)/(1 - j/n) -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: summation of combinations upto logn >What is the value/closest approximation of summation of n choose i for i >varying from 1 to logn ? I assume by logn you mean floor(ln(n)). The sum from 1 to m is, in Maple's terminology, F(n,m) = 2^n - 1 - binomial(n,m+1) hypergeom([1,m+1-n],[m+2],-1) but that's probably not much help to you. If floor(ln(n)) = m, we have as n -> infinity m = ln(n) + O(1), F(n,m) = (n choose m) (1 + m/(n-m+1) + m(m-1)/((n-m+1)(n-m+2)) + ...) = (n choose m) (1 + O(m/n)) = exp((ln(n))^2 - ln(n) ln(ln(n)) + O(ln(n)) = n^(ln(n) - ln(ln(n)) + O(1)) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: summation of combinations upto logn > What is the value/closest approximation of summation of n choose i for i > varying from 1 to logn ? Just some hints: first, Stirling's formula will give you a good approximation of n choose k where k is the largest integer less than log n. Also, n choose k-1 is appoximately (log n)/n times (n choose k) n choose k-2 is approximately (log n)^2/n^2 times (n choose k) etc., so the sum is fairly well approximated by the sum of the last few terms. === Subject: Re: Joel shifting ignored [SNIP] > Ken S. Tucker > > Let us not forget the Tucker Automobile, > that occupies Spaceport Three. > Ooont Billy Joel Das singin Uptown girl dah dah dah diddy doo daaaah Must have left the radio on in the Tucker, someone check Spaceport Three, before the Goon Squad haps!! === Subject: Jac(O), O = integers in a number field Assume K/Q is a number field, and let O denote the integers in this extension. Of course, O is a Noetherian, integrally closed, domain of (Krull) dimension 1, aka Dedekind domain. In particular, every nonzero prime ideal is maximal and so the only nonmaximal prime ideal in O is the zero ideal. My question: what is the Jacobson radical, Jac(O)? Recall Jac(O) is defined to be the intersection of all the maximal ideals in O. Equivalently, Jac(O) is the set of all x in O such that 1 - xy is a unit in O, for all y in O. Almost certainly, Jac(O) is the zero ideal. But I do not know how to prove this. Some things that came to my mind: - Krull's intersection theorem - bigcap_{n=1}^{infty} Jac(O)^n = zero ideal - Nakayama's Lemma - IM=M implies M = 0, where M is a finitely generated O-module and I is an ideal of O contained in Jac(O) - any ideal in O is generated by (at most) two elements - if I and J are coprime ideals of O, then IJ = I intersect J Try as I might, I was not able to make any of this work to my advantage. Any hints/comments will be much appreciated. Jennifer N.B. I am aware that there is a powerful theory to Dedekind domains, in particular, all nonzero ideals in a Dedekind domain have an essentially unique factorization into prime ideals. This may or may not help. Either way, I think it should be possible to work out what Jac(O) is w/o using any of this, i.e. in more elementary terms. === Subject: The simplest possible TOE We all exist as artifacts in the mind of God or Mother Nature === Subject: Re: The simplest possible TOE > We all exist as artifacts in the mind of God or Mother Nature I believe the single simplest TOE would be as follows: Because. A slightly more useful version might be: what is, is; what exists, exists. A. === Subject: Re: The simplest possible TOE you are on the right track. You and your teachers must be the the world's smartest philosopher. > Lord of Chaos(Suresh Devanathan) >> We all exist as artifacts in the mind of God or Mother Nature > I believe the single simplest TOE would be as follows: > Because. > A slightly more useful version might be: what is, is; what exists, exists. > A. === Subject: Re: The simplest possible TOE > We all exist as artifacts in the mind of God or Mother Nature 1) How can you have an or in The simplest possible TOE, idiot? 2) What predictions can you make with your infantile TOE, idiot? 3) What technologies spring forth from your infantile TOE, idiot? India: 1.1 billion assholes and 1.1 million flush toilets. The miracle is that there are no lines. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: The simplest possible TOE idiot, just because u fail to realize that there are infinite number of axiomatic systems that are simulateneously made true, is not my fault. Idiot, you ramble, without paying deep attention to what is being said. >> We all exist as artifacts in the mind of God or Mother Nature > 1) How can you have an or in The simplest possible TOE, idiot? > 2) What predictions can you make with your infantile TOE, idiot? > 3) What technologies spring forth from your infantile TOE, idiot? > India: 1.1 billion assholes and 1.1 million flush toilets. The > miracle is that there are no lines. > -- > Uncle Al > http://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals) > http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: The simplest possible TOE >>We all exist as artifacts in the mind of God or Mother Nature > 1) How can you have an or in The simplest possible TOE, idiot? > 2) What predictions can you make with your infantile TOE, idiot? > 3) What technologies spring forth from your infantile TOE, idiot? > India: 1.1 billion assholes and 1.1 million flush toilets. The > miracle is that there are no lines. Who needs a flush toilet when there is the Ganges. Bob Kolker === Subject: Re: The simplest possible TOE hello whitey >We all exist as artifacts in the mind of God or Mother Nature >> 1) How can you have an or in The simplest possible TOE, idiot? >> 2) What predictions can you make with your infantile TOE, idiot? >> 3) What technologies spring forth from your infantile TOE, idiot? >> India: 1.1 billion assholes and 1.1 million flush toilets. The >> miracle is that there are no lines. > Who needs a flush toilet when there is the Ganges. > Bob Kolker === Subject: Re: The simplest possible TOE > hello whitey I beg your pardon? Bob Kolker === Subject: Re: The simplest possible TOE i m done with u. >> hello whitey > I beg your pardon? > Bob Kolker === Subject: Re: The simplest possible TOE You and Uncle Al must be the most mannered person in the world. Lord of Chaos(Suresh Devanathan) >i m done with u. > hello whitey >> I beg your pardon? >> Bob Kolker === Subject: Re: The simplest possible TOE The technology that springs forth, is the universe itself. >> We all exist as artifacts in the mind of God or Mother Nature > 1) How can you have an or in The simplest possible TOE, idiot? > 2) What predictions can you make with your infantile TOE, idiot? > 3) What technologies spring forth from your infantile TOE, idiot? > India: 1.1 billion assholes and 1.1 million flush toilets. The > miracle is that there are no lines. > -- > Uncle Al > http://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals) > http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: The simplest possible TOE Prediction: God does whatever he wants to do. >> We all exist as artifacts in the mind of God or Mother Nature > 1) How can you have an or in The simplest possible TOE, idiot? > 2) What predictions can you make with your infantile TOE, idiot? > 3) What technologies spring forth from your infantile TOE, idiot? > India: 1.1 billion assholes and 1.1 million flush toilets. The > miracle is that there are no lines. > -- > Uncle Al > http://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals) > http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: The simplest possible TOE > We all exist as artifacts in the mind of God or Mother Nature Mother Nature is retarded. === Subject: I need some help proving these simple equations I have to prove these equations, but i cant see how i shall start. 1 - | x/y | = | x | / | y | , x = {R} and y {R} / {0} 2 - x (1 - x) <= 1/4 , x = {R} Does anyone have a idea ? /Jens Larsen === Subject: Re: I need some help proving these simple equations >2 - x (1 - x) <= 1/4 , x = {R} It's a polynomial, so it is continuous and differentiable everywhere. Take the derivative to find the local maxima and minima of the function. Hopefully you will find a local maximum. Then check to see if it is an absolute maximum by comparing the function's value there to values for other x. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: I need some help proving these simple equations >2 - x (1 - x) <= 1/4 , x = {R} > It's a polynomial, so it is continuous and differentiable everywhere. > Take the derivative to find the local maxima and minima of the function. > Hopefully you will find a local maximum. Then check to see if it is an > absolute maximum by comparing the function's value there to values for other > x. > --Keith Lewis klewis {at} mitre.org > The above may not (yet) represent the opinions of my employer. The task is not to discover the inequality but to prove an already given inequality. So, the use of Calculus is like using a sledgehammer where a hammer would do. Down to earth: What happens if you subtract x*(1-x) from both sides and recognize a square on the right-hand side? === Subject: Re: I need some help proving these simple equations days. My association with the Department is that of an alumnus. >I have to prove these equations, but i cant see how i shall start. >1 - | x/y | = | x | / | y | , x = {R} and y {R} / {0} There are 5 cases: x<0 and y<0; x<0 and y>0; x=0 and y anything; x>0 and y<0; x>0 and y>0. You need to deal with these cases because the definition of absolute value depends on whether the number is positive or negative. Check that in each case you have equality, and you are done. For example: if x<0 and y<0, then x/y is positive. So |x/y| = x/y. On the other hand, since x<0, |x|=-x; and since y<0, |y|=-y. So |x|/|y| = (-x)/(-y) = x/y. Thus, |x/y| = x/y = (-x)/(-y) = |x|/|y|, proving that the equality holds when both x and y are negative. Now check the other cases. >2 - x (1 - x) <= 1/4 , x = {R} >Does anyone have a idea ? x(1-x) <= 1/4 if and only if x(1-x) - (1/4) <= 0 x(1-x) - (1/4) = x - x^2 - (1/4), and Now notice that x^2 - x + (1/4) = (x-(1/2))^2. So x(1-x) <= 1/4 if and only if (x-(1/2))^2 >=0. Now what? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Finding unique sums.