mm-36 === Subject: 1. What is considered to be the toughest field of mathematics at thegraduate level?2. Which schools are considered to be the best schools for graduatelevel mathematics? === Subject: I guess that is going to be how one defines toughest.As for top schools, it will probably come as no surprise:These schools are always in the top 20:Harvard, Princeton, M.I.T., Duke, Cal @ Berkeley, U of Michigan, U of Wisc@Madison, U of Minnesota@ Twin Cities, etc...> 1. What is considered to be the toughest field of mathematics at the> graduate level?> 2. Which schools are considered to be the best schools for graduate> level mathematics? === Subject: Don't know about toughest. If you ask which one is hottest, it will have to be algebraic geometry, which is continuously expanding and absorbing other fields of mathematics.> 1. What is considered to be the toughest field of mathematics at the> graduate level?> 2. Which schools are considered to be the best schools for graduate> level mathematics? === Subject: > 1. What is considered to be the toughest field of mathematics at the> graduate level?assumption of easiest to toughestall others >> stats >> abstractHerc === Subject: okay, how about we define toughest as in the toughest to get an A in.I mean a significant number of people will make A's in calculus anddifferential eqns etc. but are there certain courses out there thatare extremely difficult to make A's in? === Subject: > 1. What is considered to be the toughest field of mathematics at the> graduate level?There is no such thing. It's a personal thing.> 2. Which schools are considered to be the best schools for graduate> level mathematics?Again, for graduate level there is no such thing. What counts is theadvisor and the critical mass (the number) of the faculty working inthe field you are interested in. Jan Bielawski === Subject: > Don't know about toughest. If you ask which one is hottest, it will have > to be algebraic geometry, which is continuously expanding and absorbing > other fields of mathematics.Oh, come on. Jan Bielawski === Subject: I think getting an A would primarily depend on your professor/instructor.Acheiving an A in a maths course doesn't always mean you have mastered thesubject, nor does getting a B or C mean you haven't mastered the subject.Like it or not, no matter what course you take at university, subjectivityreigns supreme in the grading arena. Albeit, some subjects are worse thanothers. I think the term toughest may prove difficult to define in thiscontext. Any subject within the realm of Advanced maths is difficult andrequires a lot of work to understand it thoroughly.I think toughest may turn out to be a matter of personal opinion.Everyone thinks and learns differently; so, some people may catch on to onesubject faster than another subject.LurchLurch> 1. What is considered to be the toughest field of mathematics at the> graduate level?> 2. Which schools are considered to be the best schools for graduate> level mathematics? === Subject: >okay, how about we define toughest as in the toughest to get an A in.>I mean a significant number of people will make A's in calculus and>differential eqns etc. but are there certain courses out there that>are extremely difficult to make A's in?That will certainly differ enormously from school to school, andfrom instructor to instructor within a school. What a particularperson finds tough depends mainly on that person's level of preparationand abilities.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Start with a closed polygonal path in R^n, constructed by choosing the generate another closed path by drawing segments between the midpoints of adjacent segments in the first path. Repeat this transformation for a while.What can you say about the shape of the path after a large number of iterations?-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: > Start with a closed polygonal path in R^n, constructed by choosing the > generate another closed path by drawing segments between the midpoints > of adjacent segments in the first path. Repeat this transformation for > a while.> What can you say about the shape of the path after a large number of > iterations?I'm not sure if this is a question or a problem, but anyway,here's a quick sketch of what I think the solution is - just in case,I've inserted a spoiler warning...S P O I L E R||V||V||V||V||V||VWe are iterating the linear transformationT(v1,v2,...,vn) = ((v1+v2)/2, (v2+v3)/2, ..., (vn+v1)/2)where the vectors vi indicate the vertices of the polygonal path. Sothe behaviour after a large number of iterations can easily bedetermined by considering the eigenvalues of the circulant matrix1/2 1/2 0 0 ... 0 0 0 1/2 1/2 0 ... 0 0 . . . 0 0 0 0 ... 1/2 1/21/2 0 0 0 ... 0 1/2which are 0, 1, and others which are all of absolute value < 1. So thepowers of this matrix quickly converge to (a conjugate of) an all-zeromatrix with a single 1 on the main diagonal - and the vectors willconverge to a single value which is their arithmetic mean(corresponding to the all-ones eigenvector).I did this very quickly - hope I got it right...- EM === Subject: > Start with a closed polygonal path in R^n, constructed by choosing the > generate another closed path by drawing segments between the midpoints > of adjacent segments in the first path. Repeat this transformation for > a while.> What can you say about the shape of the path after a large number of > iterations?I may be wrong, but if what I figure is correct, then my reply mightbe a start to what you want.I think we can consider each dimension separately, reducing theproblem to simply repeatedly averaging adjacent (modulo something)terms of a (1-dimensional) sequence.If the original terms of the sequence are {a(k)}, indexed from 1 to M(M = number of vertices), then leta(0,k) = a(k),and a(n,k) = (a(n-1,{k+1})+a(n-1,k))/2,for n = a positive integer = the number of iterations upon theoriginal sequence. And {k+1} is k(mod M) +1.I get, ignoring the mod part,a(n,m) = (1/2^n) sum{k=0 to n} binomial(n,k) * a(m-k),for M >= m > n.I guess that if we simply define a(-m), where -m is negative, asa(-m(mod(M)), where 1 <= -m(mod M) <= M, then I think thisclosed form in terms of the original sequence works.(Add a multiple of M to the -m of the index until you get an integerfrom 1 to M.)So, m may be <= n in the sum.I do not know how this will answer the questions you MIGHT still have,though, or if I have made a mistake.Leroy Quet === Subject: >> Start with a closed polygonal path in R^n, constructed by choosing the >> generate another closed path by drawing segments between the midpoints >> of adjacent segments in the first path. Repeat this transformation for >> a while. >> What can you say about the shape of the path after a large number of >> iterations?>We are iterating the linear transformation>>T(v1,v2,...,vn) = ((v1+v2)/2, (v2+v3)/2, ..., (vn+v1)/2)>>where the vectors vi indicate the vertices of the polygonal path. So>the behaviour after a large number of iterations can easily be>determined by considering the eigenvalues of the circulant matrix>>1/2 1/2 0 0 ... 0 0> 0 1/2 1/2 0 ... 0 0> .> .> .> 0 0 0 0 ... 1/2 1/2>1/2 0 0 0 ... 0 1/2>>which are 0, 1, and others which are all of absolute value < 1. So the>powers of this matrix quickly converge to (a conjugate of) an all-zero>matrix with a single 1 on the main diagonal - and the vectors will>converge to a single value which is their arithmetic mean>(corresponding to the all-ones eigenvector). Well, yes, that says the path shrinks down to a point, but what aboutthe _shape_ of the path as it shrinks? That will depend on theeigenvalues that are next in absolute value and their eigenvectors(as well as the original v_i).Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: > Start with a closed polygonal path in R^n, constructed by choosing the > generate another closed path by drawing segments between the midpoints > of adjacent segments in the first path. Repeat this transformation for > a while.> What can you say about the shape of the path after a large number of > iterations?The polygon will become planar. One can notice this by observing thatthe sum of the angles approaches that of a planar polygon. === Subject: Got this puzzle the other day, but can't figure it out. Impossible?First, let a friend constructs a right-angled triangle with *integer*sides a and b (the shorter sides, that is)Then he throws away the integers, but hands you the area, A.Question: Can you find two integers, a and b, to recreate aright-angled triangle with area A ??? === Subject: > Got this puzzle the other day, but can't figure it out. Impossible?> First, let a friend constructs a right-angled triangle with *integer*> sides a and b (the shorter sides, that is)> Then he throws away the integers, but hands you the area, A.> Question: Can you find two integers, a and b, to recreate a> right-angled triangle with area A ???Sure. If all else fails, an approach of trial and error will work.You know that A = 1/2 ab, so ab=2A. Start counting up until you find integers a and b that have product 2A. You may not find what your friend started with, but you will find something.For Example: Given area=6if a=1, b=12if a=2, b=6if a=3, b=4This gives you three possible choices for a and b.Your solution won't be unique unless A=1/2 or A=p/2 where p is prime.-- Will Twentyman === Subject: > Got this puzzle the other day, but can't figure it out. Impossible?> First, let a friend constructs a right-angled triangle with *integer*> sides a and b (the shorter sides, that is)> Then he throws away the integers, but hands you the area, A.> Question: Can you find two integers, a and b, to recreate a> right-angled triangle with area A ???How long are you willing to wait?-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: >> Got this puzzle the other day, but can't figure it out. Impossible?>> First, let a friend constructs a right-angled triangle with *integer*> sides a and b (the shorter sides, that is)>> Then he throws away the integers, but hands you the area, A.>> Question: Can you find two integers, a and b, to recreate a> right-angled triangle with area A ???Are you sure the problem was stated this way?You can always take a=1 and b=2AArea = a*b/2 = 1*2A/2 = ADirk Vdm === Subject: Pythagorean Triangles (right triangles whose sides are all integers) haveinteger area.We can sometimes patch two PTs together, either additively or subtractively,and get a non-right triangle whose sides and area are all integers. Forexample, the PT (5,12,13) and the PT (9,12,15) can be aligned along their 12side to make either the triangle (13,14,15) or the triangle (4,13,15).Prove that, given a triangle D whose sides and area are integers, either Dis a PT or there is some positive integer multiple of D which can be patchedfrom two PTs, as described above. === Subject: Will Self escribi.97 en elmensaje|nvhb9teb9tblk53@corp.supernews.com:> Pythagorean Triangles (right triangles whose sides are all integers)> have integer area.>> We can sometimes patch two PTs together, either additively or> subtractively, and get a non-right triangle whose sides and area are> all integers. For example, the PT (5,12,13) and the PT (9,12,15) can> be aligned along their 12 side to make either the triangle (13,14,15)> or the triangle (4,13,15).>> Prove that, given a triangle D whose sides and area are integers,> either D is a PT or there is some positive integer multiple of D> which can be patched from two PTs, as described above.If the triangle isn't rectangle and its sides are integers, by cosine lawthe three cosines are rationals. If also the area is integer, the threeheights are rationals and then also the three sines. Therefore, the tangentsof the three angles are rationals.Then any height, interior to triangle, divides it in two right triangles,whose acute angles have rational tangents, being its hypotenuses integers.Multiplying the triangle sides by the gcd od the denominators of thetangents of its two sub-triangle acute angles, we get two PythagoreanTriangles. If the height is exterior, the triangle id the difference of tworight triangles with rational sides and, multiplying by a suitable factor asbefore, integers.-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: > Will Self escribi.97 en el> mensaje|nvhb9teb9tblk53@corp.supernews.com:> Pythagorean Triangles (right triangles whose sides are all integers)> have integer area.>> We can sometimes patch two PTs together, either additively or> subtractively, and get a non-right triangle whose sides and area are> all integers. For example, the PT (5,12,13) and the PT (9,12,15) can> be aligned along their 12 side to make either the triangle (13,14,15)> or the triangle (4,13,15).>> Prove that, given a triangle D whose sides and area are integers,> either D is a PT or there is some positive integer multiple of D> which can be patched from two PTs, as described above.>> If the triangle isn't rectangle and its sides are integers, by cosine law> the three cosines are rationals. If also the area is integer, the three> heights are rationals and then also the three sines. Therefore, thetangents> of the three angles are rationals.>> Then any height, interior to triangle, divides it in two right triangles,> whose acute angles have rational tangents, being its hypotenuses integers.> Multiplying the triangle sides by the gcd od the denominators of the> tangents of its two sub-triangle acute angles, we get two Pythagorean> Triangles. If the height is exterior, the triangle id the difference oftwo> right triangles with rational sides and, multiplying by a suitable factoras> before, integers.> -- >> Ignacio Larrosa Ca.96estro> A Coru.96a (Espa.96a)> ilarrosaQUITARMAYUSCULAS@mundo-r.com>Golly, that was fast. Well done, Ignacio.BTW for readers, when Ignacio says If the triangle isn't rectanglehe means If the triangle isn't a right triangle. Just to dispel anyconfusion. (Spanish would be a much better International Language :-) === Subject: both of which are integers. Of course given time one can find the twointegers that A came from initially--something the original writer did notrequest.>> Got this puzzle the other day, but can't figure it out. Impossible?>> First, let a friend constructs a right-angled triangle with *integer*> sides a and b (the shorter sides, that is)>> Then he throws away the integers, but hands you the area, A.>> Question: Can you find two integers, a and b, to recreate a> right-angled triangle with area A ???>> How long are you willing to wait?>> --> Dave Seaman> Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.> === Subject: A hemispherical bubble is placed on a spherical bubble of radius 1. Asmaller hemispherical bubble is then placed on the first one. This processis continued until n chambers, including the sphere, are formed. Usemathematical induction to prove that the maximum height of any bubble towerwith n chambers is 1 + sqrt(n).The solution of this problem requires a little mathematical discovery. === Subject: > A hemispherical bubble is placed on a spherical bubble of radius 1. A> smaller hemispherical bubble is then placed on the first one. This process> is continued until n chambers, including the sphere, are formed. Use> mathematical induction to prove that the maximum height of any bubbletower> with n chambers is 1 + sqrt(n).>> The solution of this problem requires a little mathematical discovery.This is a good problem, and the formulation in terms of bubbles isnicely phrased, an elegant touch. Also, the little mathematical discoverymay come as a pleasant surprise. === Subject: > A hemispherical bubble is placed on a spherical bubble of radius 1. A> smaller hemispherical bubble is then placed on the first one. This process> is continued until n chambers, including the sphere, are formed. Use> mathematical induction to prove that the maximum height of any bubble tower> with n chambers is 1 + sqrt(n).>> The solution of this problem requires a little mathematical discovery.>Discovery: Reduce the problem to a unit circle with ever smallersemi-circles being place upon the next larger. === Subject: I was listening to the radio a while ago and they were talking about a> set theory. The theory basically said that everything belongs in a> set, because if an element does not belong in a set, it belongs in the> set of elements that do not belong in a set (does that make sense?).Yes, the inference if not-p then q . . . . 1 if q then p . . . . . . 2 therefore p . . . . . . . . . . . 3is valid. So the _form_ of the argument makes sense. But whethereverything belongs to a set depends on the set theory, some set theorieshave a universal set (and its true of those set theories) but some settheories do not have a universal set (and it isn't true of those settheories).Above, given any fixed x, not-p is not (x belongs to some set A) andq is x belongs to some set B. The truth of premiss 2 is seen bytaking A to be B. Unfortunately premiss 1 is not true of all settheories.GC> What is this theorem called? Does anyone have any resources on the> web for it?> Taras> PS: I hope what I have written above is correct/makes sense :)-- === Subject: > No, I don't agree that lim h-->0 ( |h|/h) = 1. This limit dne!>> How to compute this limit:> lim |z + h|/h> h -> 0>> where h is a real number and z is a complex number>> Let z = x + iy> |z + h|/h = ((x+h)^2 + y^2)/h> = (x^2 + y^2)/h + 2x + h>> lim = oo if z /= 0> = 1 if z = 0 Top posting is very confusing.> Let's consider lim h->0 (|h|/h).> Suppose h is really really small, like 1/zillion. Then |h| is 1/zillion > and h is 1/zillion and |h|/h is 1. This works out the same no matter > how small h is. Yes, and when h = -1/zillion, |h|/h = -1.> When evaluating this limit, it makes no difference whether |h|/h is > defined for h=0. All that matter is what happens when h is CLOSE TO 0. > And for any value of h close to 0, |h|/h is 1. [...]See above.-- Paul SperryColumbia, SC (USA) === Subject: > No, I don't agree that lim h-->0 ( |h|/h) = 1. This limit dne!>> How to compute this limit:>> lim |z + h|/h> h -> 0>> where h is a real number and z is a complex number>> Let z = x + iy> |z + h|/h = ((x+h)^2 + y^2)/h> = (x^2 + y^2)/h + 2x + h> lim = oo if z /= 0> = 1 if z = 0> Top posting is very confusing.> Let's consider lim h->0 (|h|/h).> Suppose h is really really small, like 1/zillion. Then |h| is 1/zillion > and h is 1/zillion and |h|/h is 1. This works out the same no matter > how small h is. > Yes, and when h = -1/zillion, |h|/h = -1.> When evaluating this limit, it makes no difference whether |h|/h is > defined for h=0. All that matter is what happens when h is CLOSE TO 0. > And for any value of h close to 0, |h|/h is 1. [...]> See above.oops. === Subject: > How to compute this limit:> lim |z + h|/h> h -> 0>> where h is a real number and z is a complex number>> Let z = x + iy> |z + h|/h = ((x+h)^2 + y^2)/hOk, I agree, I made mistake.let h = u + iv = e^r e^it|z + h|/h = sqr((x+u)^2 + (y+v)^2) / (u + iv) = sqr((x+u)^2 + (y+v)^2) (u - iv) / (u^2 + v^2)when z = 0|z + h|/h = (u - iv) / sqr(u^2 + v^2) = e^r e^-it / e^r = e^-itlim(h->0) |h|/h = e^-it = any value on the complex unit circleby just keeping t constant while r -> -oo hence, limit doesn't existwhen z /= 0lim(h->0) |z+h|/h = |z| lim(h->0) (u - iv) / (u^2 + v^2) = |z| lim(h->0) e^r e^-it / e^2r = |z| lim(h->0) e^-r e^-it = the infinity of the complex planeno matter what t does while r -> -oo === Subject: An unending argument over math depends on hidden issues,misunderstandings, and logical errors. What I'm doing in this post isgiving an exposition which should remove all of the above, as I'llpoint out what I think might stumble, while replies against the math Ipresent here should show clearly any biases of posters. Yup, the mathargument is THAT good.My position is that mathematicians are trying to run from an explosiveresult, which challenges many of their long held views aboutmathematics and their role in discovery.Let's begin.Concentrating on the expression (v^3+1)x^3 - 3vxy^2 + y^3let v= -1 + mf^2.If you have to have a ring, assume it's algebraic integers, thoughthat leads to a contradiction as there is a problem with the ring. Ifthe terminology ring puzzles you, don't worry about it. Here f is aprime integer other than 3, while m is a nonzero integer coprime to f.The expression chosen was picked because it has certain specialproperties. As given with m the variable it's not a polynomial as apolynomial has constant coefficients. I now like to call it anuber-polynomial. Here uber is from the German word for above.Now then substituting for v gives ((-1 + mf^2)^3 + 1)x^3 - 3(-1 + mf^2)xy^2 + y^3which is (m^3 f^6 - 3m^2 f^4 + 3m f^2 ) x^3 - 3(-1 + mf^2 )xy^2 + y^3and now let y = uf, where u is an algebraic integer coprime to f, andusing that substitution gives(m^3 f^6 - 3m^2 f^4 + 3m f^2 ) x^3 - 3(-1 + mf^2)x u^2 f^2 + (uf)^3which is f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2 )x u^2 + u^3 f).Now let P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)and consider that the constant term is given by setting m=0, and is P(0) = f^2(3x u^2 + u^3 f).Dividing f^2 from both sides gives P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f.Now introducing b_1, b_2, b_3, w_1, w_2, and w_3, I have thefactorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so at m=0, two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)proving that w_1 w_2 must equal 1, as f is coprime to 3 from before,which leaves b_3 = 3.Here notice that it's clear that two of the b's must go to 0 from theu symbols, as that's the only way to get that factor of u^2 inP(0)/f^2.Given that w_1 and w_2 are coprime to f, then the factorization is P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf).Now looking at the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)it's clear that a_1 = f b_1, and a_2 = f b_2, which proves that a_3 iscoprime to f, since a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + m),so b_1 b_2 a_3 = (m^3 f^4 - 3m^2 f^2 + m),which is coprime to f as m is coprime to f, and notice that a_3 = b_3.The methods outlined above are the same methods that have beenattacked by posters trying to cast doubt on the correctness of myshort proof of Fermat's Last Theorem, where a few lines in the entireproof have been argued over for quite some time.What I like about the argument above is that it removes past areaswhere posters have looked to cast confusion.Obviously, such a straightforward argument doesn't leave much room fordoubt, but then again I don't want to challenge doubts asunreasonable, as I can understand how hard it could be to believe thatmathematicians would try to cover their butts so diligently.However those who wish to continue doubting should consider that themathematical logic is straightforward. Pay attention to the fact thatwhile the a's and b's given vary with m, the w's do not, as they arenot dependent on m in any way.Finally, remember that mathematicians have a lot at stake here, and ifthey didn't they could answer my math with rational and objectivearguments, so please pay attention to things like personal attacks,and assertions made without presentation of logic or math.James Harris === Subject: > An unending argument over math depends on hidden issues,> misunderstandings, and logical errors.http://b5.sdvc.uwyo.edu/bab5/snds/argcstpd.wavhttp:// w0rli.home.att.net/youare.swfhttp://www.apa.org/journals/psp/ psp7761121.htmlThe problem is discretized. You are either clinically insane fordenying empirical reality with perseveration, or are intrinsicallygenetically defective and cannot self-consistently think. Either waygo away. Infest religious newsgroups where loud jackasses are reveredrather than scorned. Maybe they will set you up with a little boy forto reward your endeavor. After all, the commandment prohibitsshtupping your neighbors's wife not his kids. [snip]-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net! === Subject: >> [.snip.]>(ii) take each root and divide out all the factors (in the>>algebraic integers) that it shares with 5. In other words,>>divide each root r by GCD(r, 5), where GCD means>>greatest common divisor in the algebraic integers.>> [.snip.]>Case 2: Assuming you mean (ii): Asking us to >>divide out all the factors of 5 in the sense of>>(ii) simply doesn't make any sense. Someone>>correct me if I'm wrong: There simply _is_>>no GCD in the algebraic integers!>>(If there _were_ a unique prime factorization in the>>algebraic integers this would make sense.)>>There is a natural way of defining a gcd, but it is not uniquely>determined; it is only determined up to units. >>Since the algebraic integers are a Bezout domain (so every finitely>generated ideal is principal), then given any two algebraic integers a>and b, the ideal (a,b) is principal. Let c be any generator of>(a,b). Then c satisfies the following two properties:>>(1) c|a and c|b;>(2) if d is any algebraic integer such that d|a and d|b, then d|c.>>So it makes sense to call c ->a<- greatest common divisor of a and b,>and it is unique up to units.Oh. Ok, thanks (and thanks to the other people who pointed thisout).Of course nobody seemed to notice my amazing proof that none of the roots are divisible by 5. I'm kinda proud of that -I actually showed that something James said was wrong..> === Subject: === Subject: === Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:= === Subject: === Subject: === Subject:=== Subject:=>Why do you take so much trouble to expose such a reasoner as> Mr. Smith? I answer as a deceased friend of mine used to answer> on like occasions - A man's capacity is no measure of his power> to do mischief. Mr. Smith has untiring energy, which does > something; self-evident honesty of conviction, which does more;> and a long purse, which does most of all. He has made at least> ten publications, full of figures few readers can critize. A great> many people are staggered to this extend, that they imagine there> must be the indefinite something in the mysterious all this.> They are brought to the point of suspicion that the mathematicians> ought not to treat all this with such undisguised contempt,> at least.> -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan> === Subject: === Subject: === Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject: === Subject: === Subject: === Subject:=== Subject:>>Arturo Magidin>magidin@math.berkeley.edu************************David C. Ullrich === Subject: > Sorry, my previous replies to this post were just plain>wrong due to an algebraic error. Note however that >David Ullrich has given a simple valid disproof of >Harris's statement that two of the roots of >> y^3 + 12*y - 65>>are divisible by sqrt(5).Thought nobody noticed, since it was buried after mysilliness about GCD's.(We should also note that if _I_ can come up with adisproof without hardly even thinking about it it followsthat it's not just wrong, it's _obviously_ wrong...)> Nora B.> Well finally there's resolution, but it probably won't be satisfying>> to many of you though it's greatly satisfying to me, as the problem>> I've highlighted is rather easy to demonstrate, and is a problem with>> *actual* decomposition of algebraic integers.>>> Many of you may have considered decomposition of algebraic integers in>> the abstract, but if you'd dug into it, you might have made math>> history.>>> Consider the polynomial>>> y^3 + 12y - 65>>> which of course has 3 roots and is irreducible over Q. >>> Now imagine dividing off all factors of 5 from its roots and getting>> the polynomial for which the results are roots.>>> Give that polynomial.>>> It turns out that you can't as there's a problem with decomposing>> algebraic integers in actuality.>>> What I can do is start you out on the polynomial as you have>>> x^3 + sx^2 + tx - 13 >>> and what neither you, nor any other mortal in all of the Universe can>> do is give s, and t, to *see* the polynomial.>>> Not even I can even though I know that only two of the roots of >> y^3 + 12y - 65>>> have a factor that is sqrt(5), but you see, you can't pick them out!>>> You can prove that only two have that factor but because of the>> cuberoot operator, which must be used to display them, you have a>> fundamental ambiguity, like how sqrt(1) = +/- 1, which prevents you>> from picking out which ones.>>> So you can't calculate, and neither can you display s or t.>>> They are fundamentally unknowable, which is a fascinating result.>>> Don't believe me? Well then, go find s and t.>>>> James Harris************************David C. Ullrich === Subject: Mike:I mean to ask if their is a hierarchy between logic disciplinesthat could be expressed in a tree structure what would thatstructure look like?For example propositional logic is certainly a subset of Predicate logic.David-- David>> If we were to create a tree structure with predicate logic as thehighest> node on the tree what other disciplines would be under that tree? Alsoare> their any disciplines that are of a higher priority then Predicatelogic?>> Explain.> _You_ should explain. The question seems incomprehensible.> Is the tree structure supposed to indicate which courses are> prerequisites to which in a graduate program in mathematical> logic? Or is it about something else? If so, what?>> Mike Hardy> === Subject: > Mike:> I mean to ask if their is a hierarchy between logic disciplines> that could be expressed in a tree structure what would that> structure look like?> For example propositional logic is certainly a subset of Predicate logic. DavidConfusion might result. For exampleIntuitionistic propositional calculus (IPC) is a subset of classicalpropositional calculus (CPC)in the sense that every theorem of IPC considered as an uninterpretedformula is a theorem of CPC. But an intuitionist would not say, becauseof that, that IPC was subordinate in any way to CPC; because theintuitionistic connective are interpreted in a different way to theclassical one. So in another sense, _no_ theorem of IPC is a theorem ofCPC. For this reason one might question a claim that X logic was asubset of Y logic as being a reason to have a branch like Y | Xin such a tree unless one knew just what set X logic was. Note thatpropositional logic isn't a set and therefore can't be a subset ofanything. Though the set of theorems of propositional logic is a set,just _what_ set depends on what you mean by theorem as my remarksabove hoped to show.GC-- === Subject: Mathematica 5 for Students is now out.http://www.wolfram.com/products/student/mathforstudents/ index.html-- Dana= = = = = = = = = = = = = = = = => Is there any decent math software for people who> simply want to brush up on basic algebra, geometry, trig and basic> calculas?>> I've looked at Mathematica 5, but the standard version is nearly> $2000. A bit steep for me.> === Subject: > Let f be a one-one function, and let g be a> one-one and onto function.>> If a function h can be decomposed into f and g, and is> given by h = f o g , then the inverse of h is the composite> function inv_h given by>> inv_h = inv_g o inv_f , (*)>> where inv_g and inv_f are the inverse functions of> g and f, respectively.>> Does anyone know of a proof for (*) ?>> You've got a difficulty: unless f is also onto, it doesn't _have_ an> inverse.>Same as for group theory.h = fg; e = identitye = hh^-1 = fgh^-1f^-1 = f^-1 e = ... = gh^-1g^-1 f^-1 = ... = h^-1 === Subject: >You've got a difficulty: unless f is also onto, it doesn't _have_ an>inverse.Is it really necessary for a function f to be onto(as well as one-one) for it to have an inverse?Suppose we a function ff : A --> Bf(x) = y ,where A and B are the domain and codomain of f,respectively.Then we could define the inverse function of f,f^-1, as follows:f^-1 : I --> Af^-1(y) = x ,where I is the image set of f (which we useas the domain for f^-1).Example:Here, the function g is one-one, but not ontog : [1,5] --> R (where R is the set of all real numbers)x |--> x+1It is continuous and increasing, and so its image setis equal to the interval [f(1),f(5)]I.e. the image set is{ x+1 : x belongs to [1,5] } (by definition)= [f(1),f(5)]= [2,6]We could then describe the inverse of g, g^-1,as follows:g^-1 : [2,6] --> [1,5]x |--> x-1 === Subject: >Same as for group theory.>h = fg; e = identity>e = hh^-1 = fgh^-1>f^-1 = f^-1 e = ... = gh^-1>g^-1 f^-1 = ... = h^-1Also for matrices.(A*B)^-1 = B^-1 * A^-1 === Subject: >You've got a difficulty: unless f is also onto, it doesn't _have_ an>inverse.> Is it really necessary for a function f to be onto> (as well as one-one) for it to have an inverse?Yes. A function f : A -> B, is invertible with inverse h if there is afunction h : B -> A such that, for every a in A, h(f(a)) = a and forevery b in B, f(h(b)) = b. Or, if you prefer, h o f = I_A and f o h = I_B where I_a and I_B are the identity maps. Note f(h(b)) = bforces f to be onto B.It _is_ possible to talk about left and right hand inverses. A functionwhich is one to one but not onto has a left hand inverse but no righthand one. [...]> Example:> Here, the function g is one-one, but not onto> g : [1,5] --> R (where R is the set of all real numbers)> x |--> x+1> It is continuous and increasing, and so its image set> is equal to the interval [f(1),f(5)][1, 5] -> R : x -> x + 1 and[1, 5] -> [2, 6] : x -> x + 1are two different functions. The second has an inverse, the firstdoesn't.How about this:R -> R : x -> | x | andR- -> R+ : x -> | x | where R- is all reals < 0 and R+ is all reals > 0.The second has an inverse. Would that mean the first had an inverse?After all, if it is fair to restrict the codomain to make the functiononto surely it is fair to restrict the domain to make it one to one.> I.e. the image set is> { x+1 : x belongs to [1,5] } (by definition)> = [f(1),f(5)]> = [2,6]> We could then describe the inverse of g, g^-1,> as follows:> g^-1 : [2,6] --> [1,5]> x |--> x-1>-- Paul SperryColumbia, SC (USA) === Subject: I've made a new type of the Paper Challeran puzzle game for all ages. http://www5.ocn.ne.jp/~pachalle/pachalleEnglishpage.htmlPlease challenge it.Time limit of 24th contest registration is the end of AugustPlease tell your family, friends, and students of your school thisinteresting contest puzzle as many as possible and enjoy to solve iteach other.Ryosuke Ito === Subject: For continuous data f(x) on [a,b], we can calculate the mean oraverage value asA = 1/(a-b) int{a,b} f(x) dxMy question is can we calculate the variance of a continuousfunction as1/(a-b) int{a-b} (f(x) - A)^2 dxFor N discrete data points, we can divide by either N or N-1 toget biased or unbiased estimator, but is there this distinction incontinuous case? === Subject: > For continuous data f(x) on [a,b], we can calculate the mean or> average value as> A = 1/(a-b) int{a,b} f(x) dxThis gives the mean value of any function f(x) over the interval [a,b].> My question is can we calculate the variance of a continuous> function as> 1/(a-b) int{a-b} (f(x) - A)^2 dx> For N discrete data points, we can divide by either N or N-1 to> get biased or unbiased estimator, but is there this distinction in> continuous case?> I think what you are trying to do here is calculate the variance of a distribution whose *probability density function* is f(x).This suggests that you were actually trying to do something quite different with your first equation. Perhaps you wanted an expression for the mean of a distribution whose probability density function is f(x), rather than the mean value of the function f(x).If so the expression you wanted was mu = int{-inf,+inf} x.f(x) dxand the equation for the variance is V = int{-inf,+inf} (x-mu)^2.f(x) dxHope that helps.Mark Atherton === Subject: >> For continuous data f(x) on [a,b], we can calculate the mean or>> average value as>> A = 1/(a-b) int{a,b} f(x) dx> This gives the mean value of any function f(x) over the interval [a,b].>> My question is can we calculate the variance of a continuous>> function as>> 1/(a-b) int{a-b} (f(x) - A)^2 dx>> For N discrete data points, we can divide by either N or N-1 to>> get biased or unbiased estimator, but is there this distinction in>> continuous case?> I think what you are trying to do here is calculate the variance of a > distribution whose *probability density function* is f(x).> This suggests that you were actually trying to do something quite > different with your first equation. Perhaps you wanted an expression for > the mean of a distribution whose probability density function is f(x), > rather than the mean value of the function f(x).> If so the expression you wanted was> mu = int{-inf,+inf} x.f(x) dx> and the equation for the variance is> V = int{-inf,+inf} (x-mu)^2.f(x) dx> Hope that helps.> Mark Atherton> I've just noticed that perhaps you mean something else by the variance of a continuous function, like how wavy a function is on an interval. If so, then I suppose that would be the expression. Since there are infinitely many points on this interval then presumably it doesn't matter whether you use N or N-1.Mark Atherton === Subject: > I've just noticed that perhaps you mean something else by the variance> of a continuous function, like how wavy a function is on an interval.> If so, then I suppose that would be the expression. Since there are> infinitely many points on this interval then presumably it doesn't> matter whether you use N or N-1.This is exactly what I mean: how wavy the function is. f(x) is not the PDFfor the random variable X. Better notation would be f(t) where t is time.And f(t) would be continuous data observation as time progresses. The thingis I do not see the continuous variance formula in any book I've seen.r. === Subject: K. Kunen has shown( www.cs.wisc.edu/~kunen/moufang.psJ. Algebra 83 (1996) 231-234 )that any quasigroup satisfying((xy)z)x = x(y(zx)) all x,y,zmust have an identity element. But the proof is a couple of pageslong, and I wonder if any quicker proof has been found yet.For comparison, if this identity holds in a quasigroup Q:(x(yz))x = (xy)(zx)then we can quite quickly see that Q has an identity element.TIA,Larry === Subject: I have compiled a list of hundreds of mathematics problem books.It can be found athttp://www.mathpropress.com/mathBooks/bookFrames.htmlOf particular interest to this group is the sublist ofabout two dozen books of unsolved problems.(Click on special to see them.) === Subject: Good work! Many people will appreciate this.Now wouldn't be nice if someone also compiled a page of links to various lists of open mathematical problems on the Internet.> I have compiled a list of hundreds of mathematics problem books.> It can be found at> http://www.mathpropress.com/mathBooks/bookFrames.html > Of particular interest to this group is the sublist of> about two dozen books of unsolved problems.> (Click on special to see them.) === Subject: > Now wouldn't be nice if someone also compiled a page of links to various> lists of open mathematical problems on the Internet.You may like to visithttp://dmoz.org/Science/Math/Research/Open_Problems/-- Richard Pinch === Subject: There is a lot of work in this area. In particular,look for papersby Bill Symes at Rice University and his formerstudent Christiaan Stolk. Papers should be availableon the arXiv, and some of Stolk's work has beenpublished in Communications in PDE.Rafe> I may be applying Fourier Integral Operators to seismic> ray theory (and probably the transport equation portion)> for my Ph.D. thesis and would like to know of any> good online or printed recent or standard references> in the subject and I guess also in its subset> pseudodifferential operators.> David === Subject: > I may be applying Fourier Integral Operators to seismic> ray theory (and probably the transport equation portion)> for my Ph.D. thesis and would like to know of any> good online or printed recent or standard references> in the subject and I guess also in its subset> pseudodifferential operators.J. Chazarain, A. Piriou, Introduction to the Theory of Linear Partial Differential Equations, Elsevier Science, September 1982, 0444864520.-- David Marcus === Subject: 1. What about the role of p-adic and/or rigid cohomology? Are theseany better than motivic or etale cohomology? (especially those ofBerthelot and his co-workers)2. Why is there a need to prove the Weil Conjectures using p-adictechniques? (Is there a manageable reference for these topics?) === Subject: > 1. What about the role of p-adic and/or rigid cohomology? Are these > any better than motivic or etale cohomology? (especially those of > Berthelot and his co-workers) > 2. Why is there a need to prove the Weil Conjectures using p-adic > techniques? (Is there a manageable reference for these topics?) >theories, their uses, and their interrelatioships?Also, an explanation of their relationships to homology theories wouldbe nice. It's often more subtle than the prefix co- would indicate.Nemo === Subject: http://www.geocities.com/dharwadker/Has this proof been peer reviewed? Is it correct or is it just garbage?Charles === Subject: > http://www.geocities.com/dharwadker/> Has this proof been peer reviewed? Is it correct or is it just garbage?> Charles As far as I am aware this proof has not been peer reviewed, nor hasin appeared in any journal. I have read it before and don't think itis a correct proof, but I didn't check it that hard. At least one person (not me) has critiqued this proof, this critiqueis available on-line (I can't recall the web address).HTH,Paul. === Subject: The TeaBag problem below seeks to find maximum volume enclosable in aflexible square bag made by gluing together edges of two unit squareinextensional sheets.http://www.ics.uci.edu/~eppstein/junkyard/ teabag.htmlhttp://mathforum.org/discuss/sci.math/a/m/509288/ 509622More fundamental than the TeaBag appears to be the alliedFlexibleCubeBox Problem proposed herein. Six inextensional butflexible unit square sheets are glued together at edges to form a unitCubeBox. When it is (somehow) made to stand on flat sheets, its volumeis 1 cubic unit. When filled with water or when air at a smallpressure is let in, the bag puffs up to a higher volume, approximately1.6 cubic units. This volume figure is to be found more accurately bytheoretical means. I pose this as a separate problem not only because it is simpler thanTeaBag but due a reason following :When TeaBag is fully blown, one notices that the three girthdimensions are not same, there is no cuboidal or spheroidal three axissymmetry. Girth dimensions in yz and zx midplanes is 2 units each, butin xy midplane it is 4 units, being the perimeter of the TeaBag! The FlexibleCubeBox has 4 units girth in all the three midplanes. It isexpected that hydrostatic pressure loading yields a simple symmetricgeometric form to evaluate its volume.The TeaBag enthusiasts should perhaps have attended to this easierproblem at first, and I request them to address it now as a separatenew problem.Narasimham G.L. Ex Head Product Design and Advisor, CompositesIndian Space Research Orgn === Subject: You notice that the infamous JSH hasn't posted in a day or two?? Hmmmmm~Bhuvan === Subject: >> I don't know how I managed to miss this business of onto functions >> and codomains, but I did. I apologize if my questions are too >> elementary.>>Not at all. Keep 'em coming if you like. I hope there are some>interested lurkers.I guess what bothers me about your examples is that there doesn't seem to be any motivation to 'em. I don't mean that as a criticism of _you_, just as an explanation of why they don't seem to be clicking in my brain.Compare the problem of integrating sec x. The step of multiplying by (sec x + tan x)/(sec x + tan x) seems to come out of the blue. However, at least one can appreciate the reason for it when a little manipulation resolves the integrand to ln|sec x + tan x|. Without that resolution, multiplying by that rather odd form of 1 would seem, well, pointless.That's what bothers me about the idea of defining f(x) = x to have a domain of (0,1) and range (image) of [0,1]: it seems artificial, but I'm not seeing the payoff. What purpose is there to defining a function other than by its inputs and outputs?Maybe I need to let these ideas fester a bit. Can you recommend any treatment of codomain for me to read that's not too long but goes into more detail than Mathworld does?-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comFortunately, I live in the United States of America, where we aregradually coming to understand that nothing we do is ever ourfault, especially if it is really stupid. --Dave Barry === Subject: >> I don't know how I managed to miss this business of onto functions >> and codomains, but I did. I apologize if my questions are too >> elementary.>>Not at all. Keep 'em coming if you like. I hope there are some>interested lurkers.> I guess what bothers me about your examples is that there doesn't > seem to be any motivation to 'em. I don't mean that as a criticism > of _you_, just as an explanation of why they don't seem to be > clicking in my brain.The deal is that definitions are important - they are a distinguishingfeature of math. The function I gave, f : (0, 1) -> [0, 1] : x -> x,was shorthand for the _real_ definition of the function which I won'tgo into, but note that it has three parts _all_ of which werearbitrarily chosen by me. If I say let h(x) = x^2, you are perfectly within your rights to sayOh yeah? What are the domain and codomain? I _must_ be able toanswer.just that: _my_ choice. I could just as well used, (-5, 5) or (0, 1] or(0, 1). Only in the last case, however, would I have given aninvertible function. For the inverse, if there is one, I have nodiscretion whatsoever; I _must go by the definition of inverse or riskhaving no one know what I am talking about.Having said all that, as best I recall at the moment, there are veryfew places in the undergrad math curriculum where the codomain is ofany particular importance. One of them is the definition of theinverse.If you ask why the inverse is defined that way, I have two answers: Ithas to be defined _some_ way and, it turns out (consider theendomorphism ring of a group - if that doesn't mean anything to you,don't worry about it) that the definition is just what is needed.> Compare the problem of integrating sec x. The step of multiplying by > (sec x + tan x)/(sec x + tan x) seems to come out of the blue. > However, at least one can appreciate the reason for it when a little > manipulation resolves the integrand to ln|sec x + tan x|. Without > that resolution, multiplying by that rather odd form of 1 would > seem, well, pointless.> That's what bothers me about the idea of defining f(x) = x to have a > domain of (0,1) and range (image) of [0,1]: it seems artificial, > but I'm not seeing the payoff. What purpose is there to defining a > function other than by its inputs and outputs?You might findinteresting. (Sorry about the long URL - if command click doesn't work,cut and paste.) > Maybe I need to let these ideas fester a bit. Can you recommend any > treatment of codomain for me to read that's not too long but goes > into more detail than Mathworld does?Not about codomain per se but maybe you will get an idea why anyonecares:, Stan Brown>> I don't know how I managed to miss this business of onto functions >> and codomains, but I did. I apologize if my questions are too >> elementary.>>Not at all. Keep 'em coming if you like. I hope there are someinterested lurkers.> I guess what bothers me about your examples is that there doesn't > seem to be any motivation to 'em. I don't mean that as a criticism > of _you_, just as an explanation of why they don't seem to be > clicking in my brain.The deal is that definitions are important - they are a distinguishingfeature of math. The function I gave, f : (0, 1) -> [0, 1] : x -> x,was shorthand for the _real_ definition of the function which I won'tgo into, but note that it has three parts _all_ of which werearbitrarily chosen by me. If I say let h(x) = x^2, you are perfectly within your rights to sayOh yeah? What are the domain and codomain? I _must_ be able toanswer.just that: _my_ choice. I could just as well used, (-5, 5) or (0, 1] or(0, 1). Only in the last case, however, would I have given aninvertible function. For the inverse, if there is one, I have nodiscretion whatsoever; I _must go by the definition of inverse or riskhaving no one know what I am talking about.Having said all that, as best I recall at the moment, there are veryfew places in the undergrad math curriculum where the codomain is ofany particular importance. One of them is the definition of theinverse.If you ask why the inverse is defined that way, I have two answers: Ithas to be defined _some_ way and, it turns out (consider theendomorphism ring of a group - if that doesn't mean anything to you,don't worry about it) that the definition is just what is needed.> Compare the problem of integrating sec x. The step of multiplying by > (sec x + tan x)/(sec x + tan x) seems to come out of the blue. > However, at least one can appreciate the reason for it when a little > manipulation resolves the integrand to ln|sec x + tan x|. Without > that resolution, multiplying by that rather odd form of 1 would > seem, well, pointless.> That's what bothers me about the idea of defining f(x) = x to have a > domain of (0,1) and range (image) of [0,1]: it seems artificial, > but I'm not seeing the payoff. What purpose is there to defining a > function other than by its inputs and outputs?You might findinteresting. (Sorry about the long URL - if command click doesn't work,cut and paste.) > Maybe I need to let these ideas fester a bit. Can you recommend any > treatment of codomain for me to read that's not too long but goes > into more detail than Mathworld does?Not about codomain per se but maybe you will get an idea why anyonecares:-- Paul SperryColumbia, SC (USA) === Subject: >Compare the problem of integrating sec x. The step of multiplying by >(sec x + tan x)/(sec x + tan x) seems to come out of the blue. >However, at least one can appreciate the reason for it when a little >manipulation resolves the integrand to ln|sec x + tan x|. Oops -- sorry, in editing I removed too many words. I meant manipulation resolves the integrand to du/u where u = sec x + tan x, and the integral is therefore ln|sec x + tan x|.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comFortunately, I live in the United States of America, where we aregradually coming to understand that nothing we do is ever ourfault, especially if it is really stupid. --Dave Barry === Subject: >> I guess what bothers me about your examples is that there doesn't >> seem to be any motivation to 'em. I don't mean that as a criticism >> of you , just as an explanation of why they don't seem to be >> clicking in my brain.>>The deal is that definitions are important - they are a distinguishing>feature of math. The function I gave, f : (0, 1) -> [0, 1] : x -> x,>was shorthand for the real definition of the function which I won't>go into, but note that it has three parts all of which were>arbitrarily chosen by me. And the point of making the range arbitrary is what escapes me. But is it arbitrary, really? You could not for instance define the range as (.2,.5) because that would contradict your domain and your function rule. >just that: my choice.Yes, I understand that. What I don't understand is what possible benefit could come from such a choice. Alternatively, what I don't understand is what difference there is between the behavior of such a function and one defined in the conventional way, as f(x)=x with domain (0,1) and range (0,1). Sure, one can create a definition of function that says those are different functions, but what is the use of such a definition?A quotation from may be on point here:Poincar.8e remarked with regard to the proliferation of pathological functions, 'Formerly, when one invented a new function, it was to further some practical purpose; today one invents them in order to make incorrect the reasoning of our fathers, and nothing more will ever be accomplished by these inventions.' :-)>> That's what bothers me about the idea of defining f(x) = x to have a >> domain of (0,1) and range (image) of [0,1]: it seems artificial, >> but I'm not seeing the payoff. What purpose is there to defining a >> function other than by its inputs and outputs?>You might find>n&ie=UTF-8>>interesting. (Sorry about the long URL - if command click doesn't work,>cut and paste.)http://makeashorterlink.com is our friend :-)thttp://makeashorterlink.com/?N2D952795is (now) equivalent to the above link.It is interesting, I agree, but I'm not sure why you posted it. One range of a function is the actual set of output values. Consider for instance:Let A and B be sets and let there be a given rule which assigns exactly one member to[sic] B to each member of A. The rule, together with the set A, is said to be a function and the set A is said to be its domain. The set of all members of B actually assigned to members of A by the rule is said to be the range of the function.>Not about codomain per se but maybe you will get an idea why anyone>cares:>>Is this actually trustworthy? I've heard it said that wikipedia entries tend to be full of wrong information because it's just a matter of whoever made the last updates, whether they truly know or just think they know.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comFortunately, I live in the United States of America, where we aregradually coming to understand that nothing we do is ever ourfault, especially if it is really stupid. --Dave Barry === Subject: Firstly, I believe their was a misunderstanding of notation, or if not, wecertainly have one now...definition. I may be completely off base, but it seems to me that we aredefining a function named f that is a mapping from X to Y, where X is(0,1), Y is [0,1], and x->x gives the rule of the image, ie the image of funder x is x in this case. IOW, f(x)=x.When we say y=f(x)=x in contexts such as Larson's, which is a very broadcontext as we say this sort of thing all the time, we are only explicitlydenoting one thing about f: The range. Yeah, we know the dummy variableto, but what I means is of the domain, codomain, and range, we only knowone--the range. That's it. That's what f(x) *is* by definition--therange-- at least by this definition.If [0,1] denoted the *range* and not the codomain, then such notation asabove would not even make sense, much less be of any use.That said, I empathize somewhat with the confusion regarding inversefunctions.What purpose is there defining a function other than by inputs and outputs(domains/ranges?) Well,f: X -> ? : x -> xWhat is ? if we are not given it? We must assume, in similar fashion thatwe assume the domain. If we don't, then we don't have a function, just likewe don't have a function without assuming the domain is R (or whateversubset of R we happen to be talking about).Again, the notation f(x)=x denotes only the range of f and the dummyvariable, however a function is arguably more than that. Larson definesfunction as a correspondence, and goes on to define what correspondencesexists, and between what sets. This means, a statement such as f(x)=x islacking in properly defining a function, even if we assume the domain, evenby Larson's standard.We usually assume domains in such contexts. No big deal, right? We assume,unless told otherwise, the domain to be the largest set, x E R, such thatf(x) E R unless told otherwise. This is just a convention, though. Paul'sright, it's kind of a trick question to ask for the domain of a functionsuch as f(x)=x when that's all you know about f. The same can be said forthe codomain. I was always tempted to answer you're supposed to tell MEwhat the domain is but instead I answered in the manner expected by theinstructor, like everyone else.The question remains, *what* do we assume the codomain (the set Y) to be?Larson does not speak of this, really. All we know is its R or someappropriate subset of R. Well, which is it? R itself or some appropriatesubset of R? If always R, why? If some appropriate subset, then *what*subset and why? Does it even matter in contexts such as elementarycalculus?Consider:f: (0,1) -> R : x -> xHere is where I empathize wrt inverse function. Also in Larson we find thedefinition of inverse function:A function g is the inverse of the function f if f(g(x))=x for each x inthe domain of g, and g(f(x))=x for each x in the domain of f.You may also notice him stating just below that certainobservations implied by the definition:1. If g is the inverse of f, then f is the inverse of g.2. The domain of g is equal to the range of f, and the range of f is equalto the domain of g.3. A function need not posses an inverse, but if it does, it is unique.Pay particular attention to observation #2. First realize it's justthat--an observation--and is not part of the actual definition. Thisparticular observation seems sufficiently different from the one Paul made.Paul said something to the effect it's the domain and CODOMAIN that areinterchanged between inverse functions, desirably.Strictly at face value (disregarding his observations), Larson seems to betelling us just from the definition that the following canbe considered an inverse of f:g: (0,1) -> R : x -> xAfter all, for each x in (0,1) we have f(g(x))=g(f(x))=x and x is in R.Also, at the same face value, Larson seems to be telling us that:h: (0,1) -> [0,1] : x -> x...is also an inverse of f as are any number of other functions, withvarying appropriate codomains.BUT, this would mean that the inverse of f is not unique, according toobservation #3. I take this to mean, what Larson really is saying is thatf^-1 is unique given appropriate codomains for f and g.But under Paul's reasoning, we must have:f(g(x)) = g(f(x)) = x...for all x in *R*, not just all x in (0,1), if I understand him correctly.This is clearly not the case, since x is always in (0,1). Hence, in Paul'sview, f can have no inverse.But, and this is a *big* but, if we define an appropriate codomain for f:f: (0,1) -> (0,1) : x -> x...then we can safely say, even under Paul's reasoning, this has *the*unique inverse function:g: (0,1) -> (0,1) : x -> xThe point is, I infer from Paul's remarks that this definition of inversefunction (from Larson) is a special case, so to speak, a case wheredistinction of codomain is beyond the scope of the text. What he's nottelling us, is that in other contexts such as what Paul alludes to, f and gnot only have their domains and ranges interchanged if they are inverses,but need to have their domains and CODOMAINS interchanged as well.Why does Larson not mention this? Presumably, because nothing whatsoever issaid about codomains in any part of the text save for that statementregarding the set Y in the definition of function. He is telling what weneed to know for now, nothing more, nothing less.IOW, I assume Paul is speaking of, wrt inverse functions, is someextension of the definition of inverse function that we may moreaccustomed to (eg the one from Larson) where such additional restriction isforeign, at least for the moment.-- Darrell === Subject: > I was wondering...If someone gave you their birth dates and all the info you> required (apart from their name of course), could you come up with their> name by reversing the process? Just curious, I don't want my name done or> anything.> MarcNo, of course not, that is not what I do, I have never claimed to,where is your head? -Daryl === Subject: >>About probability variable X,Y>>Bivariate probability density function f(x,y) is given :>>f(x,y)=(1/2)xy if 0=>Question : Find probability density function of Z=Y-X.>>If anyone can help me, please post reply...>I admit up front that I don't know the answer. But since no one else >has answered, maybe my thinking out loud will help get you >started.>Visualize f(x,y). The original density is a curved surface rising >above a triangular base. The edges of the base are the three lines >x=0, y=x, y=2; and the height above any (x,y) is xy/2. The total >probability is the double integral> INT[0, 2, INT[0, y, xy/2 dx] dy]>which equals 1 (as it should).>To answer your question, I think you need to begin by looking at >what z = y-x looks like in the region where the probability function >is nonzero. Since z=y-x, y=x+z. The defined region must have>0 <= z <= 2 as well as 0 <= x <= y <= 2.> z = 2 represents point (x,y) = (0,2)> z = 1 represents line segment y=x+1 between (0,1) and (1,2);> z = 0 represents line segment y=x between (0,0) and (2,2);>and so forth. More generally, any z is represented y the line >segment y = x+z between the points (0,z) and (2-z,2).>So far I'm pretty confident. >Now I _think_ that the probability density for each z value is the >area of the vertical slice that cuts the base in the indicated line >segment (or point, in the case of z=2).Can't be: when you integrate those areas with respect to z,you're essentially integrating wrt y, since dz = dy, andtherefore not integrating in the direction normal to the slices.If you take one of those slices and give it thickness dy in the ydirection, its real thickness normal to itself is dy/sqrt(2).Thus, you're going to get an integral that's sqrt(2) times thevolume of the solid -- as in fact you did.If you had the joint pdf of X and Z, say g(x, z), you could getthe desired pdf of Z by integrating g(x, z) wrt x, so the problemboils down to finding this g.Define a function h from R^2 to R^2 by h(x, y) = (x, x+y); thismap is area-preserving since its Jacobian is 1. (You can alsosee this geometrically: a little square in the domain istransformed to a rhombus of the same area in the range.) Let Tbe the triangular region bounded by the coordinate axes and theline x + y = 2, and let R be the original triangular region onwhich f is non-zero, 0 <= x <= y <= 2; it's pretty easy to seethat h maps T onto R bijectively.Note that the event (X, Z) = (x, z) is identical to the event (X, Y) = (x, x+z) = h(x, z). Now set g(x, z) = f(x, x+z) =f(h(x, z)) for (x, z) in T, g(x, z) = 0 otherwise. Because h isarea-preserving, integrating g over any region A will yield thesame value as integrating f over h[A], and it follows that g isthe joint pdf of X and Z. The pdf of Z can then be found byintegrating g(x, z) wrt x. But this is exactly your integrationwithout the extra factor of sqrt(2).If Z had been defined as (Y - X)/2, the region corresponding to Tabove would have been the triangle with vertices at the origin,(0, 1), and (2, 0), and the map corresponding to h (and thereforesending (x, z) to (x, x+2z)) would have been an area-doublingmap. To get the joint pdf of X and Z in this case you'd have toset g(x, z) = 2f(h(x, z)).[...]Brian === Subject: >>Since z=y-x, y=x+z. The defined region must have>>0 <= z <= 2 as well as 0 <= x <= y <= 2.>>... any z is represented y the line >>segment y = x+z between the points (0,z) and (2-z,2).>Now I _think_ that the probability density for each z value is the >>area of the vertical slice that cuts the base in the indicated line >>segment (or point, in the case of z=2).>>Can't be: when you integrate those areas with respect to z,>you're essentially integrating wrt y, since dz = dy, and>therefore not integrating in the direction normal to the slices.>If you take one of those slices and give it thickness dy in the y>direction, its real thickness normal to itself is dy/sqrt(2).>Thus, you're going to get an integral that's sqrt(2) times the>volume of the solid -- as in fact you did.Here's what I'm not understanding, I guess. I wasn't actually integrating with respect to z, but with respect to t. I thought I had constructed a straightforward line integral: f(z) = INT[0, 2-z, f(x,y) * sqrt([dx/dt]^2 + [dy/dt]^2), dt] f(z) = (z-2)^2 (z+4) sqrt(2)/12Are you saying _that_ integral is wrong, or were you talking about the integral that came further down to find the total probability: INT[0, 2, f(z), dz]I _think_ you're talking about the later one, but I'm not sure.Either way, I see that dy = dz as you say, but I don't see how that means integrating over z and integrating over y are the same thing. After all, dz = -dx also, yet the y and x directions are perpendicular so an integral can't be over both. Can you explain a bit more please?-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/Walrus meat as a diet is less repulsive than seal. === Subject: >Since z=y-x, y=x+z. The defined region must have>0 <= z <= 2 as well as 0 <= x <= y <= 2.>... any z is represented y the line >segment y = x+z between the points (0,z) and (2-z,2).>Now I _think_ that the probability density for each z value is the >area of the vertical slice that cuts the base in the indicated line >segment (or point, in the case of z=2).>>Can't be: when you integrate those areas with respect to z,>>you're essentially integrating wrt y, since dz = dy, and>>therefore not integrating in the direction normal to the slices.>>If you take one of those slices and give it thickness dy in the y>>direction, its real thickness normal to itself is dy/sqrt(2).>>Thus, you're going to get an integral that's sqrt(2) times the>>volume of the solid -- as in fact you did.>Here's what I'm not understanding, I guess. I wasn't actually >integrating with respect to z, but with respect to t. No, as you suspected, I mean after you've found the areas and gotwhat you thought was going to be the desired pdf for Z. Call itp(z). When you then integrate p(z) wrt z, you're 'adding up' allof the elements p(z) dz. Now p(z) is the area of a sliceparallel to the line y = x through the original solid defined byf(x, y) = xy/2, and that solid has unit volume. 'Adding up'these areas will give you the total volume, 1, *if* you add themup with the right infinitesimal thicknesses. Going from theslice defined by z to the one defined by z + dz shifts you up dyor over -dx, but in the direction normal to the slice it onlyshifts you dz/sqrt(2). Thus, the appropriate volume element isp(z)/sqrt(2) dz, not p(z) dz.[...]>Are you saying _that_ integral is wrong, or were you talking about >the integral that came further down to find the total probability:> INT[0, 2, f(z), dz]This one.>I _think_ you're talking about the later one, but I'm not sure.>Either way, I see that dy = dz as you say, but I don't see how that >means integrating over z and integrating over y are the same thing. >After all, dz = -dx also, yet the y and x directions are >perpendicular so an integral can't be over both. Can you explain a >bit more please?It's not the direction of integration so much as the direction inwhich you have to measure the thickness. (It's also not reallyall that important: it's just how I was able to see that p(z)couldn't possibly be a pdf without actually doing theintegrations to find and it and check that it would yield a totalintegral of 1.)Brian === Subject: [big snip]I think now I understand what you mean. To put what (I think) you're saying in different words, p(z) = (z+4) (z-2)^2 * sqrt(2) / 12is the correct probability density function for random variable Z. But the z axis isn't parallel to either the x or the y axis, so it's not valid to integrate over z.What I did, integrating over z, was essentially integrating over x or y and taking the _slant_ lines as the integrand. Since the slant lines are all at a 45 degree angle, this overstates the integrand by a factor of sqrt(2), which is why my total integral gave sqrt(2) instead of 1.So I did actually manage to answer the OP's question correctly, but I blundered in doing the check. :-)-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comFortunately, I live in the United States of America, where we aregradually coming to understand that nothing we do is ever ourfault, especially if it is really stupid. --Dave Barry === Subject: i've been trying to find out in other textbooks about those threestatements...one other says that each one shows/proves the otherstatement...thanks...-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: > .... > note that CD is given as the vertical diameter of a circle...and AB is> the chord it intersects... > (i) given CD is a diameter and given CD is perpendicular to AB, then> CD bisects AB. > (ii) given CD is a diameter and given CD bisects AB, then CD is> perpendicular to AB. > (iii) given CD bisects AB and given CD is perpendicular to AB, then CD> is a diameter....> .... > i've been trying to find out in other textbooks about those three> statements...one other says that each one shows/proves the other> statement... Well, if that means that one of them follows from the other two as amatter of pure logic, then it's wrong. Conceivably CD could be theperpendicular bisector of AB without being a diameter, which would make(i) and (ii) true but (iii) false. This shows that you can't infer (iii)from (i) and (ii) purely by logic. Proving (iii) requires some freshgeometry. You've given me a good chance to recommend reading Euclid'sElements. Your (iii) is the porism (i.e. corollary) to Euclid III.1:....if in a circle a straight line cut a straight line into two equalparts and at right angles, the centre of the circle is on the cuttingstraight line. Your (i) and (ii) are the second and first halves of Euclid III.3:If in a circle a straight line through the centre bisect a straight linenot through the centre, it also cuts it at right angles; and if it cut itat right angles, it also bisects it.(Notice the words not through the centre which Paul Tanner pointed outwere needed.) Now, if you like, I can tell you how to prove all three. But why notgo to the fountainhead? English translations of Euclid are in manylibraries, and also at the following three web sites.http://aleph0.clarku.edu/~djoyce/java/elements/ elements.htmlhas Java applets for manipulating diagrams.http://sunsite.ubc.ca/DigitalMathArchive/Euclid/ byrne.htmlreproduces the edition of Oliver Byrne (1847) which used colours on thediagrams to make the proofs clearer. It includes only Books I-VI.http://thales.vismath.org/euclid/also has only Books I-VI, but gives a succession of diagrams to show howeach proof builds up. You may find this the most helpful version, even ifnot the prettiest. :-) Ken Pledger.-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: I tried to use the distance formula trying to use a common ratio amoung twodistances, and this yielded a mess; finally, I saw how some conic equationswere developed in an intermediate algebra textbook, and my attempt wasdifficult because I made the situation too complicated. The book used fociwhihc were aligned with the coordinate axes, but my attempt did not specifysimilarly. Now, your suggestion seems more workable.G C>: Can the distance formula be applied to determine equational forms for>: anything in ADDITION TO conic sections, or is it only possible to create>: the conic section type equations and nothing more?>>Certainly distance is involved in lots of other curves.>Take for example, the locus of points whose distances to two fixed points>have a constant product. Or the locus of points whose total distance>from THREE fixed points has a constant sum. Or the spiral whose distance>from the original equals the angle (in radians) through which it has>spiraled. None of these are conics.>>The last is a fairly well studied curve (R=theta in polar coordinates).>not long ago about the locus with a constant total distance from>three or more points.>>I've seen other (simplier?) examples involving the locus of points>a fixed distance along the tangent (or normal) from a given curve.>Or perhaps the curves you get from Spirograph?>Anyone have a favorite example?> |)|/| || Burnaby South Secondary School> || |orewood@olc.ubc.ca || Beautiful British Columbia>Mathematics & Computer Science || (Canada)-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: A new version 0.9.7 of GNU Dr. Geo is ready to be download athttp://www.ofset.org/drgeo. GNU Dr Geo is a free interactive geometrysoftware.Follow is the change log:* Add the getAngle function in the DGS API. This method expects oneparameter, a reference to an angle object, it returns one measure inDEGREE. To get a measure in RADIAN, use the getValue function.* First implementation to turn Dr. Geo as a TeXmacs plug-in. It isexperimental and still very buggy. However when installing Dr.Geo, thein Text->Session->Dr. Geo* Implement a minimal unit grid. It can be display per figure, also itis saved a long the figure.* Implement an automatic upgrade system for the user preferences file.The system is mostly entirely implemented in Scheme. This will avoidproblem seen when upgrading Dr. Geo from one version to a subsequentone.* Updated translations of the user interface in Czech, Dutch, French,German, Italian, Marathi, Polish, Portuguese, Spanish and Swedish.More information concerning Dr. Geo can be found athttp://www.ofset.org/drgeoA debian package of the 0.9.7 release will come out in the followingdays.For bugs reports, documentations, screenshots, see the specific pages atthe Dr. Geo web page.Friendly,http://www.ofset.org/projects/edusoft.html-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: I have 5 numbers out of 39 what would the combinations be?-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: > I have 5 numbers out of 39 what would the combinations be?Does this mean How many ways may one choose 5 items out of 39? If itdoes then the answer is 39! 39! 39*38*37*36*35 39 choose 5 = ------------ = -------- = -------------- = 13*19*37*9*7 =575757 (39 - 5)! 5! 34! 5! 5*4*3*2Here's betting that the op does not clarify her question.-- G.C.-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: i really need help on this problem.A withdraws $1500 from the bank. He wants 10x as many $5 bills as $1dollar bills and 2x as many $50 bills than $10 bills. It all equals upto $1500. How many of each bill does he have.I have been trying to figure this out but cant. Any help would beappreciated-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: >i really need help on this problem.>>A withdraws $1500 from the bank. He wants 10x as many $5 bills as $1>dollar bills and 2x as many $50 bills than $10 bills. It all equals up>to $1500. How many of each bill does he have.>>I have been trying to figure this out but cant. Any help would be>appreciatedMake a chart to help clarify things: Denomination # of bills Value of bills $1 x $ x $5 10x 10x times $5 = $ 50x $10 y $ 10y $50 2y 2y times $50 = $ 100yYou have x+50x+10y+100y = 1500, or 51x+110y = 1500, where x and y are whole numbers.Note there are two variables because you are not given any connection between the number of $10 bills and the number of $1 bills. Are you sure you told us everything? Did the problem perhaps tell you the total number of bills?Assuming the problem did not tell you that, you then have to use some logic. Since the grand total of money is a multiple of $100, you know that there can't have been 1, 2, 3, 4, 6, 7, 8, 9, 11, etc. $1 bills.In fact, you know there can't be 5, 15, etc. $1 bills either. Why? because if you divide the equation by 10 you have x/10 + 5x + 1y + 10y = 150Since 5x, 1y, 10y, and 150 are whole numbers, x/10 must be a whole number. So try x = 10: 51*10 + 110y = 1500 ==> 110y = 1500-510 = 990 ==>y = 9. Answer: Denomination # of bills Value of bills $1 10 $ 10 $5 100 $ 500 $10 9 $ 90 $50 18 $ 900 TOTAL $ 1500What if x = 20? Then 51*20 + 110y = 1500 ==> y = 480/110 is not a whole number. And if x = any higher multiple of 10, y will be negative.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comFortunately, I live in the United States of America, where we aregradually coming to understand that nothing we do is ever ourfault, especially if it is really stupid. --Dave Barry-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: > i really need help on this problem.> A withdraws $1500 from the bank. He wants 10x as many $5 bills as $1> dollar bills and 2x as many $50 bills than $10 bills. It all equals up> to $1500. How many of each bill does he have.> I have been trying to figure this out but cant. Any help would be> appreciatedLet the number of one dollar bills be D and the number of ten dollarbills be T. Then 1500 = 10*D*5 + D*1 + 2*T*50 + T*10And you need to solve this with positive integers D and T. CollectingDs and Ts: 1500 = 51*D + 110*T . . . . (1)Since 1500 is divisible by 10 and 110*T is divisible by 10 ,51*D must be divisible by 10 as well. Trial and error suggestsitself: D = 10 = 20 = ... etc.D = 10 gives 1500 = 510 + 110*T and T = 9. _If_ we are told that thereis only one solution we can stop there, otherwiseD = 20 gives 1500 = 1020 + 110*T and T = 48/11 which isn't integral. Trying D = 30 gives a negative T and clearly all bigger D give negativeT, so the answer is T = 9.Putting T = 9, D = 10 in equation (1) gives 1500 = 51*10 + 110*9just to check.-- G.C.-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: > 'the weight of a new statue will be 18 tonnes. what would the weight> be of a similar statue, made of the same material but only half the> height?'> i know that the volume scale factor is the (linear scale factor)^3...Also, if the material is homogenous, weight is proportional to volume. So if the half sized one isn't hollow (or _is_ hollow and the big oneisn't, etc (the phrase is ceteris paribus))weight scale factor = volume scale factor = (1/2)^3 = 1/8, and(18 tonnes)/8 = 9/4 tonnes = 9000/4 kg = 2250 kg.-- G.C.-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: as a basis for Multidimensional Scaling I use different matrices: a distance matrix and a correlation matrix (Spearman). The results are different, too. Why? I use STATISTICA.Torsten === Subject: > Do you expect the same result for different input? !Definitely not, although the underlying data set is the same. My question was: why are the results different? Because the matrices are different? Is the answer so simple? And which conclusions I have to draw for the interpretation of the MDS?Torsten === Subject: > Do you expect the same result for different input? !> Definitely not, although the underlying data set is the same. My > question was: why are the results different? Because the matrices are > different? Is the answer so simple? And which conclusions I have to draw > for the interpretation of the MDS?> I tried to do MDS on correlations, 15 years ago when I was trying to figure out if MDS was worth anything, andI never figured out how I was supposed to reverse thecorrelations in order to have a 'distance' metric that wouldwork. Did you have some concrete advice from somewhere?Since the results I could get on correlations did not looknearly as meaningful to me as what Factor Analysisgave me, I decided to forget about MDS. I googled for < Multidimensional scaling FAQ > and two of the first 4 hitswere to my own FAQ -- that is not encouraging.< Multidimensional scaling tutorial > gave a better lookingset of references.Has somebody cared enough about MDS to update the computer programs? It's long been my impression that like MDS sometimes is included in the tools of data mining.-- Rich Ulrich, wpilib@pitt.eduhttp://www.pitt.edu/~wpilib/index.htmlTaxes are the price we pay for civilization. Justice Holmes. === Subject: >...> I never figured out how I was supposed to reverse the> correlations in order to have a 'distance' metric that would> work. Did you have some concrete advice from somewhere?>> Since the results I could get on correlations did not look> nearly as meaningful to me as what Factor Analysis> gave me, I decided to forget about MDS.>MDS is not intended as a modeling method that helps you to establishanalytical relationships. It is more a visualization tool for preliminarydata analysis. It can help you, for instance, to answerquestions like: Are there any clear clusters in a dataset?Does a data point have many similar data points in a dataset?Are there many isolated outliers in a dataset?You can probably answer these questions with statististics, but with MDSit is more direct and visual. The value and strength of MDS is its mappingfeature which visualizes abstract high dimensional data.> Has somebody cared enough about MDS to update the> computer programs? It's long been my impression that> like MDS sometimes is included in the tools of data mining.That's right, MDS is more appropriated for areas whereconventional analytical modeling have failed.For self-promotion purpose please see our software VisuMap athttp://www.visumap.net which combines traditional MDS, PCA,clustering method with modern navigation user interface. VisuMapbasically turns any numerical table ( or any distance matrix) into map.cheers,James X. Li === Subject: > I was asking, partly from technical curiosity, What algorithm/ or > whatever/ did you use to convert correlations to distances? > The other part of my reason for asking was my suspicion > that you hadn't noticed that conversion is needed. Or, Did > the computer program offer options for default treatment? > You still haven't settled that suspicion.As far as I understand my software (STATISTICA) generats the matrices directly from the raw data. I do not understand why I've to convert correlations to distances. But I do not look behind the software, so maybe I've used procedures which are not assumable. > Hybrid distances? nominal with Interval? Seems awkward, > at best, to do it or to describe it, whatever MDS is offering.At the end I used only ordinal variables: I've ranked the nominal variables after their frequencies of values.Torsten === Subject: me >> I was asking, partly from technical curiosity, What algorithm/ or> whatever/ did you use to convert correlations to distances?> The other part of my reason for asking was my suspicion> that you hadn't noticed that conversion is needed. Or, Did> the computer program offer options for default treatment?> You still haven't settled that suspicion.TF > As far as I understand my software (STATISTICA) generats the matrices > directly from the raw data. I do not understand why I've to convert > correlations to distances. But I do not look behind the software, so > maybe I've used procedures which are not assumable.Now *that* seems difficult to justify, unless thedimension is 'how conventional is the response' --I have seen one scale where that is intended. I suppose you could do that when referring to attitudes,but I have only seen it once.-- Rich Ulrich, wpilib@pitt.eduhttp://www.pitt.edu/~wpilib/index.htmlTaxes are the price we pay for civilization. Justice Holmes. === Subject: Do you expect the same result for different input? !> Definitely not, although the underlying data set is the same. My > question was: why are the results different? Because the matrices are > different? Is the answer so simple? And which conclusions I have to draw > for the interpretation of the MDS?> I tried to do MDS on correlations, 15 years ago when I > was trying to figure out if MDS was worth anything, and> I never figured out how I was supposed to reverse the> correlations in order to have a 'distance' metric that would> work. Did you have some concrete advice from somewhere?Try adding the coauthor name Young to your Google search.If you interpret a correlation coefficient as the cosineof an angle, a, you can represent the data in terms ofpoints on a unit 3-dim sphere. When the angle between two position vectors emanating from the sphere center is a,the distance between the points is 2*sin(a/2).Hope this helps.Greg> Since the results I could get on correlations did not look> nearly as meaningful to me as what Factor Analysis> gave me, I decided to forget about MDS. > I googled for > < Multidimensional scaling FAQ > and two of the first 4 hits> were to my own FAQ -- that is not encouraging.> < Multidimensional scaling tutorial > gave a better looking> set of references.> Has somebody cared enough about MDS to update the > computer programs? It's long been my impression that > like MDS sometimes is included in the tools of data mining. === Subject: Here is a Math Trick I invented.http://w4u.sytes.net/MagicMathLet me know how you find it. gonzalep@.84scientist.com no .84 === Subject: I'm trying to follow a proof that: sum(1...n^2) = n^3/3 + n^2/2 + n/6The author starts with the equality: (k + 1)^3 = k^3 + 3k^2 + 3k + 1I can follow each of his steps up until we get to:3[1^2 + 2^2 + ... + (n-1)^2] + 3[1 + 2 + ... + (n-1)] + (n-1) = n^3 - 1^3I understand the line above (I'll call it E1), however the authorcontinues in his words: The second sum on the left side is the sum of terms in an arithmetic progression and it simplifies to 1/2n(n-1). Therefore this last equation gives us 1^2 + 2^2 + ... + (n-1)^2 = n^3/3 - n^2/2 + n/6 (I'll call this E2)And it's here that I get lost. When I try to derive the same resultas the author I wind up with something like1^2 + 2^2 + ... + (n-1)^2 = (n^3 - 1^3)/3 - (n(n-1))/2 - (n-1)/3Could someone please show me the baby steps required to get from E1to E2. I'd appreciate it. === Subject: > I'm trying to follow a proof that: sum(1...n^2) = n^3/3 + n^2/2 + n/6>Bah, sum(1...n^2) = 1 + 2 + 3 + ... + n^2 = n^2 (n^2 + 1)/2.Do you mean sum(j=1..n) j^2 ? No wonder you're confused.> The author starts with the equality: (k + 1)^3 = k^3 + 3k^2 + 3k + 1>sum [(k+1)^3 - k^3] = sum 3k^2 + sum 3k + sum 1where sum is from 0 to n(n+1)^3 = 3 sum k^2 + 3 sum k + sum 13 sum k^2 = (n+1)^3 - 3n(n+1)/2 - n+1 = n^3 + 3n^2 + 3n + 1 - 3n^2 / 2 - 3n/2 - n+1 = n^3 + 3n^2 / 2 + n/2---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: || I'm trying to follow a proof that: sum(1...n^2) = n^3/3 + n^2/2 + n/6|| The author starts with the equality: (k + 1)^3 = k^3 + 3k^2 + 3k + 1|| I can follow each of his steps up until we get to:|| 3[1^2 + 2^2 + ... + (n-1)^2] + 3[1 + 2 + ... + (n-1)] + (n-1) = n^3 - 1^3|| I understand the line above (I'll call it E1), however the author| continues in his words:|| The second sum on the left side is the sum of terms in an arithmetic| progression and it simplifies to 1/2n(n-1). Therefore this lastequation| gives us|| 1^2 + 2^2 + ... + (n-1)^2 = n^3/3 - n^2/2 + n/6 (I'll call this E2)|| And it's here that I get lost. When I try to derive the same result| as the author I wind up with something like|| 1^2 + 2^2 + ... + (n-1)^2 = (n^3 - 1^3)/3 - (n(n-1))/2 - (n-1)/3|| Could someone please show me the baby steps required to get from E1| to E2. I'd appreciate it.||(k+1)^3 = k^3 + 3*k^2 + 3*k + 1(k+1)^3 - k^3 = k^3 + 3*k^2 + 3*k + 1 - k^3(k+1)^3 - k^3 = 3*k^2 + 3*k + 1Take the sums of both sides of the equation.NB All SIGMA parameters are (For k = 1 to n)SIGMA[(k+1)^3 - k^3] = SIGMA[3*k^2 + 3*k + 1]Itemize the elements of the series of the left hand sideand note that it simply collapses to [(n+1)^3 - 1] (n+1)^3 - 1 = SIGMA[3*k^2] + SIGMA[3*k] + SIGMA[1]n^3 + 3*n^2 + 3*n = 3*SIGMA[k^2] + [3*(1/2)*n*(n+1)] + nSIGMA[k^2] = (1/3)*[n^3 + 3*n^2 + 3*n - (3/2)*n^2 - (3/2)*n - n] = (1/3)*[n^3 + (3/2)*n^2 + (1/2)*n] = (1/3)*[(1/2)*n*(n + 1)*(2*n + 1)]SIGMA[k^2] = (1/6)*n*(n + 1)*(2*n + 1) QEDAn example of the LH series ....... let n = 4[(1+1)^3 - 1^3] + [(2+1)^3 - 2^3] + [(3+1)^3 - 3^3] + [(4+1)^3 - 4^3] +....... [(n+1)^3 - n^3] [2^3 - 1] + [3^3 - 2^3] + [4^3 - 3^3] + [5^3 - 4^3] -1 + 5^3 -1 + (4+1)^3 -1 + (n+1)^3-- ---David Snook ............ dsnook1@telus.net === Subject: >> I'm trying to follow a proof that: sum(1...n^2) = n^3/3 + n^2/2 + n/6>> The author starts with the equality: (k + 1)^3 = k^3 + 3k^2 + 3k + 1>> I can follow each of his steps up until we get to:>> 3[1^2 + 2^2 + ... + (n-1)^2] + 3[1 + 2 + ... + (n-1)] + (n-1) = n^3 - 1^3>> I understand the line above (I'll call it E1), however the author> continues in his words:>> The second sum on the left side is the sum of terms in an arithmetic> progression and it simplifies to 1/2n(n-1). Therefore this lastequation> gives us>> 1^2 + 2^2 + ... + (n-1)^2 = n^3/3 - n^2/2 + n/6 (I'll call this E2)>> And it's here that I get lost. When I try to derive the same result> as the author I wind up with something like>> 1^2 + 2^2 + ... + (n-1)^2 = (n^3 - 1^3)/3 - (n(n-1))/2 - (n-1)/3> = n^3/3 - 1/3 - n^2/2 + n/2 - n/3 + 1/3 = n^3/3 - n^2/2 + n(1/2 - 1/3) = n^3/3 - n^2/2 + n/6which matches E2.So you were doing fine! You just needed the confidence to carry on andsimplify what you got a bit more, expanding some product terms andcollecting together the different powers of n. :)Mike.(sorry if this appears twice, I had PC troubles posting it!)> Could someone please show me the baby steps required to get from E1> to E2. I'd appreciate it.> === Subject: > I've got trouble with a question reviewing for an upcoming test. My> brain is fried at the end of the quarter, so forgive me my blindness.> Any hints at all would be much appreciated.> Assume that f(x) is a continuous function on [0, oo) and that f(x) ->> 0 as x -> oo. Also, assume that f has a bounded second derivative on> [0, oo). Prove that f'(x) -> 0 as x -> 0.> My argument assumes a bit of stuff that we haven't really proved with> proofs a bit beyond my current ability. I'd love any pointers in terms> bbscottWhat about f(x) = 1/(1 + x)?Did you mean to show lin(f'(x), x -> oo) = 0?If so, I'd look at the Mean Value Theorem - no guarantees.-- Paul SperryColumbia, SC (USA) === Subject: > Assume that f(x) is a continuous function on [0, oo) and that f(x) ->> 0 as x -> oo. Also, assume that f has a bounded second derivative on> [0, oo). Prove that f'(x) -> 0 as x -> 0.I assume you mean x -> oo at the end. (Note the continuity assumption is redundant.) Just to get the feel of the problem, assume to reach a contradiction that f'(x_n) > 1 for some sequence x_n -> oo. The boundedness of f'' will then show there exists a t > 0 such that f' > 1/2 on [x_n,x_n + t] for all n. Now look at f(x_n + t) - f(x_n). By the mean value theorem, that equals f'(c_n)*t for some c_n in [x_n,x_n + t]. This implies f(x_n + t) - f(x_n) > (1/2)*t for all n. But f(x_n + t) - f(x_n) -> 0 as n -> oo, contradiction. === Subject: >I've got trouble with a question reviewing for an upcoming test. My>brain is fried at the end of the quarter, so forgive me my blindness.>Any hints at all would be much appreciated.>Assume that f(x) is a continuous function on [0, oo) and that f(x) ->>0 as x -> oo. Also, assume that f has a bounded second derivative on>[0, oo). Prove that f'(x) -> 0 as x -> 0.As x --> oo, I presume.>My argument assumes a bit of stuff that we haven't really proved with>proofs a bit beyond my current ability. I'd love any pointers in termsSince f(x) is bounded, fix M such that |f(x)| < M for x >= 0.Suppose that f'(x) doesn't approach 0 as x --> 0; then there'ssome e > 0 such that |f'(x)| > e for arbitrarily large values ofx. Let d = e/M. Here are some useful steps; you may find ithelpful to make a sketch of f', picturing it as having a localmaximum at y.Suppose that y is such that f'(y) > e. (The other possibility isthat f'(y) < -e, but this can be reduced to the first case byconsidering -f instead of f.) Show that if 0 < t < d, thenf'(y+t) > e - Mt > 0. [Apply the Mean Value Theorem to f'.] Usethis to show that f(y + d/2) - f(y) > (d/2)(e - Md/2) = e^2 / (4M). [Apply the MVT to f itself.] Conclude from thisthat |f(y + d/2)| > e^2 / (8M), a positive constant that isindependent of y. With just a little more work you get acontradiction with one of your hypotheses.Brian === Subject: >> Further, I think this indicates that *in practice* it's not an essential> idea. Not that I'm defending using that as a gold standard (*in> practice* all that is essential is fulfilling biological urges).>> Perhaps I misunderstand. Are you saying that substitution is not an> essential idea? If so I disagree strongly.>> The reason that calculators/computers were introduced into the> curriculum was to counteract an emphasis on calculation over> understanding. How does knowing tricks of extremely limited> applicability aid understanding?>> There was no emphasis on calculation in any of the mathematics courses I> ever took. There is now, in the wake of calculators and the onslaught of> the push-button world. I watched this disaster as it occurred. Its main> function, in my experience, has been to give teachers with weak insight> into mathematics something to do. If their students are typing and looking> at LED's and blinking lights all day, at least they're doing something.You are correct. === Subject: > I am having some trouble with my trig homework if anybody would like to> help?> 1. Find all values of x or y such that the distance is 5> (x,7) (2,3)> 2. Find all values of y such that the distance is 12> (3,y) (-2,9)> All the rest I could get, but these two are giving me trouble, any help> would be greatly appriciated.> Use the distance formula, solve for the only variable.-- Will Twentyman === Subject: >> I am having some trouble with my trig homework if anybody would like to> help?>> 1. Find all values of x or y such that the distance is 5> (x,7) (2,3)>> 2. Find all values of y such that the distance is 12> (3,y) (-2,9)>> All the rest I could get, but these two are giving me trouble, any help> would be greatly appriciated.>>> Go to www.mathtutor1.com/trig.htm to see the solution in video format.>> === Subject: >2. Find all values of y such that the distance is 12>(3,y) (-2,9)What is the formula for the distance between two points, in rectangular coordinates? Apply that to the given pair of points, set it equal to 12, and solve for y.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comFortunately, I live in the United States of America, where we aregradually coming to understand that nothing we do is ever ourfault, especially if it is really stupid. --Dave Barry === Subject: I am using GMRES to iteratively solve a non-symmetric, non-Hermitiansystem. All implementations of GMRESI have found stop the iterationon the preconditioned residual rather than unpreconditioned. We usea fairly large tolerance (10e-2, which is tied to the relative accuracyof the matrix elements), and experience has shown that thepreconditioned residual can sometimes be considerably larger than the unpreconditioned residual when we get a particularly bad problem.Obviously stopping on the preconditioned residual could be disasterous.Is it possible to stop GMRES on the unprecondtioned residual withoutresorting to an extra matrix-vector multiply?-- James C. West (Jim) jwest@okstate.eduProfessorElectrical and Computer EngineeringOklahoma State University === Subject: > Is it possible to stop GMRES on the unprecondtioned residual without> resorting to an extra matrix-vector multiply?I don't believe so. The whole iterative method is based on a Krylovspace for the preconditioned operator, so there is no need or place forthe unpreconditioned residual to be computed.Btw, I hope you're not using a stopping test like |r_i|> Is it possible to stop GMRES on the unprecondtioned residual without>> resorting to an extra matrix-vector multiply?> I don't believe so. The whole iterative method is based on a Krylov> space for the preconditioned operator, so there is no need or place for> the unpreconditioned residual to be computed.I figured that was the case, but was hoping there might be a trickinvolved somewhere I was missing. I will continue following up withan extra mat-vec to make sure it has sufficiently converged prior tostopping. > Btw, I hope you're not using a stopping test like |r_i| http://www.netlib.org/linalg/html_templates/node83.html for better> stopping tests; since these involve the norm of the operator (which> includes the preconditioner) this may actually solve your problem.We actually are using the |r_i| < eps |b| criterion (#2 in the Templatesbook, which is my standard reference on these type things). It is a fast multipole expansion technique so we do not have the entries of Adirectly available for estimating the norm. I'm sure that there areprobably other ways to estimate it, but since we are stopping at eps = 10e-2 I've always been more comfortable using the more stringent stopping condition of #2 anyway. A few extra iterations aren't really thatbig of deal provided we know that it is not stopping to soon. === Subject: > I don't believe so. The whole iterative method is based on a Krylov> space for the preconditioned operator, so there is no need or place for> the unpreconditioned residual to be computed.> I figured that was the caseActually, what I said above is pure nonsense.But I still don't know if what you want is possible.GMRES generates intermediate error estimates (q_{1n} or something). Arethose preconditioned or un? I don't have time right now to look into it.V.-- === Subject: if you use right-preconditioning in GMRES, you can get the un-preconditionedresidual.(see Saad's book)Jim West escribi.97 en el mensaje>>> Is it possible to stop GMRES on the unprecondtioned residual without>> resorting to an extra matrix-vector multiply?> I don't believe so. The whole iterative method is based on a Krylov> space for the preconditioned operator, so there is no need or place for> the unpreconditioned residual to be computed.>> I figured that was the case, but was hoping there might be a trick> involved somewhere I was missing. I will continue following up with> an extra mat-vec to make sure it has sufficiently converged prior to> stopping.>> Btw, I hope you're not using a stopping test like |r_i| http://www.netlib.org/linalg/html_templates/node83.html for better> stopping tests; since these involve the norm of the operator (which> includes the preconditioner) this may actually solve your problem.>> We actually are using the |r_i| < eps |b| criterion (#2 in the Templates> book, which is my standard reference on these type things). It is a> fast multipole expansion technique so we do not have the entries of A> directly available for estimating the norm. I'm sure that there are> probably other ways to estimate it, but since we are stopping at eps =10e-2> I've always been more comfortable using the more stringent stopping> condition of #2 anyway. A few extra iterations aren't really that> big of deal provided we know that it is not stopping to soon.> === Subject: Big Bertha Thing pasturesCosmic Ray SeriesPossible Real World System Constructshttp://web.onetel.com/~tonylance/cricket.html Access page to 138K ZIP fileAstrophysics net ring Access siteNewsgroup Reviews including alt.support.attn-deficitcomplete with programs, source code in listing format and documentation.Pastures Software Package.Sub-atomic Mesons, Baryons and Leptons Classification System.(C) Copyright Tony Lance 1997Distribute complete and free of charge to comply.Big Bertha Thing cricket The Rules of Cricketby Tony Lance 1. 1st side goes in till all out. (10 out of 11) 2nd side goes in till all out. 1st side wins.Exceptions;-i Rain stops play is draw.ii 2nd side scores more, game stops they win.iii Equal final scores are a draw. 2. Every run counts one.Exceptions;-i A no runs boundary hit counts 4.ii A no runs clean over boundary hit counts 6.iii A no hit too wide counts 1.iv A no hit boundary counts 4.v A no hit over boundary counts 6. (missing rule)vi A no hit can still be run. 3. Caught out or bowled out. (owzat!)Exceptions;-i Stumped out.ii Run out.iii Thrown out.iv Trod on wicket.v Retired injured. Exception to all of above.Runner for slightly injured. Street Cricket Forget all the exceptions and rules 1 and 2.Equipment;-i Cut out bat.ii Rubber ball.iii Piece of chalk. Owner of bat goes first.Owner of ball goes first.Fielder bowls next.Bowled out, bowler bats.Caught out, catcher bats and swaps bat for ball.The end(C) Copyright Tony Lance 2000Distribute complete and free of charge to comply.Tony Lancepobox47@big-bertha-thing.comBig Bertha Thing NASA IIhttp://groups.google.com/groups?hl=en&safe=off&group= comp.std.c&th=683a5cac33f8ee2f,69Further to your above link, you have missed one or two sub-plots:- 1. Case documentation has been published on my web site. 2. My nbci web site has been closed. 3. You did not do this. 4. The opposition did it. (3rd Battle of Cyberspace) 5. You do not believe the opposition exists. 6. All Big Bertha Thing postings on that Saturday had the same prefix. 7. NASA would tell you to shut up. 9. There is no problem my end.10. A rocket scientist counts double.11. John Prescott, Minister of the Crown counts double.12. Ken Livingstone, Mayor of Greater London counts double.13. Both have appointed directors to LPFA. (see mayor) http://web.onetel.com/~tonylance/mayor.html14. Pensions Ombudsman's judgements cannot be enforced on these directors.15. Magna Carta does not apply; they are above the law.16. It is feudal personal political patronage; all opticians to a man.Tony Lancepobox47@big-bertha-thing.comhttp://users.mweb.co.za/d/da/ dalen/tclockex.htmKeep site owners from using too much of your computersystems resources, with free software. Systems info:-dd/MM/yy h':'m'Sy'S'CP'C'MM'LWhen Sys is equal to zero, then windows crashes due to shortage ofresources. Use Ctrl Alt Del to restart. === Subject: I'm trying to solve an integral on the interval t=0..infinity with Gauss-Laguerre quadrature. I've been testing my subroutine usingdifferent functions whose integral I know, and I find that the accuracy of the result depends on how I choose the scaling for t. Relatively small changes in the choice of s when integrating f(s*t) can make a difference of several orders of magnitude. Does any one know of an optimal way (or at least sensible way) of rescaling t? The function I am integrated in integrating is only available numerically, so I don't know its properties except that it is very smooth, is positive with a single mode and has an exponentially decaying tail.P.S. Some preliminary tests tell me that neither the decay rate of the tail, nor the mean of the distribution are particulary good choices. === Subject: > I'm trying to solve an integral on the interval t=0..infinity with> Gauss-Laguerre quadrature. I've been testing my subroutine using> different functions whose integral I know, and I find that the accuracy> of the result depends on how I choose the scaling for t. Relatively> small changes in the choice of s when integrating f(s*t) can make a> difference of several orders of magnitude. Does any one know of an> optimal way (or at least sensible way) of rescaling t? The function I am> integrated in integrating is only available numerically, so I don't know> its properties except that it is very smooth, is positive with a single> mode and has an exponentially decaying tail.> P.S. Some preliminary tests tell me that neither the decay rate of the> tail, nor the mean of the distribution are particulary good choices.If you wnat to use Gauss-Laguerre, and the integrand has the behavior f(x) ~ g(x)*exp(-ax)where g(x) is either a polynomial or -> 0 as x -> infty, then your_only_ choice is to scale with the asymptotic behavior of theintegrand, i.e. let x -> y/a .In this kind of problem a good method is to pick a reasonable cutoffX, and write int_0^infty {dx f(x) } approx int_0^X {dx f(x)} + int_X^infty {dx g(x) exp (-ax)}The first term can be done adaptively. The tail integral is of order const. X^n exp(-aX)where n is the highest power in the polynomial g(x). If you transformthe tail integral via x = X +y/a , and pick a Gauss-Laguerre formulawith at least (n+1)/2 points, the tail integral will be exact.-- Julian V. NobleProfessor Emeritus of Physicsjvn@lessspamformother.virginia.edu ^^^^^^^^^^^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ Science knows only one commandment: contribute to science. -- Bertolt Brecht, Galileo. === Subject: > I'm trying to solve an integral on the interval t=0..infinity with> Gauss-Laguerre quadrature. I've been testing my subroutine using> different functions whose integral I know, and I find that the accuracy> of the result depends on how I choose the scaling for t. Relatively> small changes in the choice of s when integrating f(s*t) can make a> difference of several orders of magnitude. Does any one know of an> optimal way (or at least sensible way) of rescaling t? The function I am> integrated in integrating is only available numerically, so I don't know> its properties except that it is very smooth, is positive with a single> mode and has an exponentially decaying tail.> P.S. Some preliminary tests tell me that neither the decay rate of the> tail, nor the mean of the distribution are particulary good choices.If you are set on using Gauss-Laguerre integration, you will probably haveto consider an adaptive algorithm, such as is included in QUADPACK. A analternative you might also consider double exponential integration. This hasbecome a method of choice with commercial packages, such as Matlab andMathematica. Professor Ooura has a website with very good source code for anumber of integrators.(I don't have a URL handy, but a Google search willfind his site.)--There are two things you must never attempt to prove: the unprovable -- andthe obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Does anyone have an equivilent to the matlab/octave/sci.py 'fftshift'routine that is written in C? I am using FFTW directly to computerank 1 and 2 FFTs but need to 'shift' the zero frequency component.--FFTSHIFT Shift zero-frequency component to center of spectrum. For vectors, FFTSHIFT(X) swaps the left and right halves of X. For matrices, FFTSHIFT(X) swaps the first and third quadrants and the second and fourth quadrants. For N-D arrays, FFTSHIFT(X) swaps half-spaces of X along each dimension.--Graham === Subject: > int n; // FFT output vector length> int n2;> int i;> complex x[n]; // you might typedef your own complex type> complex tmp;> n2 = n / 2; // half of vector length> for (i = 0; i < n2; i++)> {> tmp = x[i];> x[i] = x[i+n2];> x[i+n2] = tmp;> }This is close to what I came up with but it only works for even n.Matlab gives:>> fftshift([1,2,3,4,5,6,7,8])ans = 5 6 7 8 1 2 3 4>> fftshift([1,2,3,4,5,6,7,8,9])ans = 6 7 8 9 1 2 3 4 5Above gives:n = 8n2 = 4in = 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 out = 5.00 6.00 7.00 8.00 1.00 2.00 3.00 4.00n = 9n2 = 4in = 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 out = 5.00 6.00 7.00 8.00 1.00 2.00 3.00 4.00 9.00Any ideas on how to get the same answer as matlab for both even and odd n?Graham === Subject: Graham,It can be done out of place rather simply, but note a new array, z[n]was introduced. It also works for even n:for (i = 0; i < n; i++){// find circulant index k = (i+n2) % n;// or, without modulo call k = (i+n2); if (k >= n) k = k - n;// move z[k] = x[i];}For this particular case, an odd n and shift by [n/2], it might bedone in place, but more complicated. As it is seen, the only problemis the overlap of i-th and (i+n2)-th entries:0 1 2 3 4 5 6 7 85 6 7 8 0 1 2 3 4 ^0-th entry is moved into 4-th location, but 4-th location instead ofbeen moved to 0-th position, as in an even length case, is moved intothe last location, etc. This happens due to curcular addressing mode:e.g. n = 9, n2 = 4 and 0+4 mod 9 = 4, 4+4 mod 9 = 8, but 5+4 mod 9 = 0and the 5th entry eventually goes into the first location. In an even length case, e.g. n = 8, the 4-th location is moved into4+4 mod 8 = 0th location, thus the circulant shift might be accomplishedby swap operations.The code works for even n as well (again, n2 = n/2):for (i = 0; i < n2; i++) // swap what can be swapped{ tmp = x[i]; x[i] = x[i+n2]; x[i+n2] = tmp;}if (n & 1) // odd n, shift the rest{ tmp = x[n-1]; x[n-1] = x[0]; for (i = 1; i < n2; i++) { x[i-1] = x[i]; } x[n2-1] = tmp;}It might be done in place more simply, with 4 load/store ops less,but works only for odd n. For even n, (ip1+n2) is n when i becomesn2-1, thus load/store is being done out of array bound on the lastiteration. For odd n, (ip1+n2) is n-1 on the last iteration:for (i = 0; i < n2; i++){ ip1 = i + 1 tmp = x[i+n2]; x[i+n2] = x[i]; x[i] = x[ip1+n2]; x[ip1+n2] = tmp;}Andrew> int n; // FFT output vector length> int n2;> int i;> complex x[n]; // you might typedef your own complex type> complex tmp;> n2 = n / 2; // half of vector length for (i = 0; i < n2; i++)> {> tmp = x[i];> x[i] = x[i+n2];> x[i+n2] = tmp;> }> This is close to what I came up with but it only works for even n.> Matlab gives:> fftshift([1,2,3,4,5,6,7,8])> ans = 5 6 7 8 1 2 3 4>> fftshift([1,2,3,4,5,6,7,8,9])> ans = 6 7 8 9 1 2 3 4 5> Above gives:> n = 8> n2 = 4> in = 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 > out = 5.00 6.00 7.00 8.00 1.00 2.00 3.00 4.00> n = 9> n2 = 4> in = 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 > out = 5.00 6.00 7.00 8.00 1.00 2.00 3.00 4.00 9.00 Any ideas on how to get the same answer as matlab for both even and odd n?> Graham === Subject: > Does anyone have an equivilent to the matlab/octave/sci.py 'fftshift'> routine that is written in C? I am using FFTW directly to compute> rank 1 and 2 FFTs but need to 'shift' the zero frequency component.> --> FFTSHIFT > Shift zero-frequency component to center of spectrum.> For vectors, FFTSHIFT(X) swaps the left and right halves of> X. For matrices, FFTSHIFT(X) swaps the first and third> quadrants and the second and fourth quadrants. For N-D> arrays, FFTSHIFT(X) swaps half-spaces of X along each> dimension.> -->> GrahamGraham,to reorder in 1-D:int n; // FFT output vector lengthint n2;int i;complex x[n]; // you might typedef your own complex typecomplex tmp;n2 = n / 2; // half of vector lengthfor (i = 0; i < n2; i++){ tmp = x[i]; x[i] = x[i+n2]; x[i+n2] = tmp;}to reorder in 2-D:int m, n; // FFT row and column dimensions might be differentint m2, n2;int i, k;complex x[m][n];complex tmp13, tmp24;m2 = m / 2; // half of row dimensionn2 = n / 2; // half of column dimension// interchange entries in 4 quadrants, 1 <--> 3 and 2 <--> 4for (i = 0; i < m2; i++){ for (k = 0; k < n2; k++) { tmp13 = x[i][k]; x[i][k] = x[i+m2][k+n2]; x[i+m2][k+n2] = tmp13; tmp24 = x[i+m2][k]; x[i+m2][k] = x[i][k+n2]; x[i][k+n2] = tmp24; }}Hope this helps,Andrew === Subject: > I just create a small C program to reproduce the newton method of root> finding on equation.> 1.- What will be a good method to guess the initial value x0?> 2.- how can i know if there its an imaginary root ? (complex number).My book A. Neumaier Introduction to Numerical Analysis Cambridge Univ. Press, Cambridge 2001 contains a chapter on solving nonlinear equations, with a globallyconvergent method for complex zeros that does not need a special starting point.Arnold Neumaier === Subject: Apologises if this is the wrong area for this problem i am working on.But formulas isn't my forte and I hoping someone can help me out. I amconvinced there should be a way to resolved this in a math lanuage.This is my first post ever so forgive me if this dosnt come accrosswell its a complex problem for me..........Problem.A source gnerates a random number in a stream one after the other thatis feed into a another source that receives the numbers one by one.Source Ouput----> 2534,35566,36567,3567,3445,7767----->input sourceFor every 50 numbers generated from the source they belong to aspecfic group for example group 'A' will contain a list of 50 numbersA = (7767,3445,3567,36567,35566,2534.....). Groups will coniune to becreated each group holding 50 number and so on. The problem ariseswhen an intermidiate source that takes a number before the numbersarrive at the input source and wants to find out what group the numberbelongs to. I like to call this a neighbourhood theroy or somethinglike that because if the input source knows the neighbour numbers thenit can figure out the group it belongs to. The problem gets worst ifthe input source dosnt receive its immediate neighbour numbers i.e.sometimes numbers don't make it therefore it has too look to its nextneighbour number i.e. two neighbours away from the orginal number. Inthe above list for example if the intermidate source grabs number 3567and asks ether the output source or input source what group it belowstoo the awnser should be group 'A' because the output source will knowits neighbours of 36567 is its left neighbour and 3445 is its rightneighbour.The output source will never loose its neighbour numbers. But if the intermidate sources were to ask the input source what group'3567' belows to too it will also know its neighbours of '36567' fromthe left side and '3445' from its right side but input source might bemissing its first rightside neighbour of 3445 but it has its next onewhich is 7767. Therefor the input source should return group 'A'. Itis possible that the input souce could be missing serveral of itsneighbours therefore the formula would have to go to its nextavailable number which could be N number from the orginal number thatintermidate souce is asking for what group it belongs to. It ispossible that the input souce could return more than one groupresponse like group A & B & C because the input source was missingallot of its neighbours. Therefore the input put must return more thanone group to ensure that the number was in one of the groups of A,B,C.The output source never looses its neighbours numbers so it would bebetter if the intermidate source where to ask the output source forthe group identify but this group identity from the ouptut source maynot be correct because some of the numbers may never make it too theinput souce. If every number were to make it to the input souce thenthis would be fine. 90% of the time all numbers do make it to theinput source but I need to guarentee a 100% group identity. So i couldask output source but I would have to verfiy this with the input soucebefore i return the group identity to ensure that group identity iscorrect.Sorry for long description.....confussing I know but the problem isdoing my head in and I am trying to attempt a formula for it that willcover all possble situations...but i can't even get started..... :(Any help would be much appricated....thanks,Jason. === Subject: |> ...|> I'm trying to get the Template Numerical Toolkit (TNT) working on my|> Powerbook with g++v3.1. However, even though I link to the libraries|> g++ still gives me problems. For example, if I run something like|> |> #include tnt.h|> |> main()|> {|> Array2D A(3,3) = 0.0;|> }|> |> the error tells me Array2D is undeclared. I'm guessing my problem|> ...In C and C++, declaration, initialization and assignmentare different concepts of starting a variable. In your case,you are trying to assign to a variable that has not beendeclared before. You could simply call the default constructorby the declaration (which initializes to zero for complex variables and leaves them initialized) Array2D A(3,3);The assignment is superfluous in this case. === Subject: > Multiple Form Logic is relevant to... Zen because> (like Laws of Form) it derives all conventional> Propositional Logic from three axioms which are> based on Experience: (1) The Law of Oneness,> (2) The Law of Reflection and> (3) The Law of Perception.What a sad cross-posted troll.-- Mind Control: TT&P ==> http://www.datafilter.com/mcHome page: http://www.datafilter.com/albAllen Barker === Subject: > Multiple Form Logic is relevant to... Zen because> (like Laws of Form) it derives all conventional> Propositional Logic from three axioms which are> based on Experience: (1) The Law of Oneness,> (2) The Law of Reflection and> (3) The Law of Perception.> What a sad cross-posted troll.What a boring predictable internet-culture reaction.Instead of moaning, _you_ especially should read theTheory of Multiple Forms. It might make you understandmore clearly some of the things that interest _you_,personally (after reading some of your own web-pages,about mind control, and so on).Here is the URL again:http://multiforms.netfirms.comRead especially the saga and the agony section. ;)Besides, I don't consider it as irrelevant to alt.zen,or to the other newsgroups. I posted it to selected few,in order to attract serious scientific interest, as wellas people who'd like to hire me, for serious work.Revolutionising logic has two benefits: A freer stateof mind, and a more efficient production of hardwareand software.The Theory of Multiple Forms deserves serious attention,yes. By logicians, scientists, gurus, programmers, even...comedians. ;)The internet should be used for serious exchanges ofuseful information, not for slander and intimidation(of the kind we often see in Usenet). If some idea ortheory deserves attention, it _should get it_. No labellingis justified. A troll is a person who seeks attention forpsychological reasons, only; and offering nothing in return.I'm offering you (for free) a theory that is worth millions,since it can be used to make lightning-fast computer hardwareand better software. I'm also offering you free therapy, sincethe same theory can... liberate your mind, by showing yousome common fallacies of logic, and how they still go onwithin traditional logic.What else do you want, maan? Fish? Hooks? Jokes?OK, I'll see what I can do, and I'll let you know. ;)Cheer up and be happyGeorgehttp://www.geocities.com/omadeon === Subject: > I just bought Maple 9 the other day and have noticed that it returns> answers in radians instead of degrees. So, evalf(sin(90)); returns> the answer 0.89399666 etc. instead of 1. I've search the Learning> Guide and the Help file on Maple but cannot figure out how to change> from radians to degrees. If someone knows how, would you please> --> Cindy Smith Unless the LORD build the house,> cms@dragon.com they labor in vain who build.> cms@5sc.net Unless the LORD guard the city,> cms@romancatholic.org in vain does the guard keep watch.> Me transmitte sursum, -- Psalm 127:1> Caledoni! All your base are belong to us.> A Real Live Catholic You are on the way to destruction.> in Georgia! What you say.>->> <<-< Go against the flow! You have no chance to survive make your time.Conversions are fairly easy:> convert(3.14159,units,radians, degrees); 179.9998479Look under Help, units.-- Julian V. NobleProfessor Emeritus of Physicsjvn@lessspamformother.virginia.edu ^^^^^^^^^^^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ Science knows only one commandment: contribute to science. -- Bertolt Brecht, Galileo. === Subject: >> I just bought Maple 9 the other day and have noticed that it returns>> answers in radians instead of degrees. So, evalf(sin(90)); returns>> the answer 0.89399666 etc. instead of 1. I've search the Learning>> Guide and the Help file on Maple but cannot figure out how to change>> from radians to degrees. If someone knows how, would you please>> -->> Cindy Smith Unless the LORD build the house,>> cms@dragon.com they labor in vain who build.>> cms@5sc.net Unless the LORD guard the city,>> cms@romancatholic.org in vain does the guard keep watch.>> Me transmitte sursum, -- Psalm 127:1>> Caledoni! All your base are belong to us.>> A Real Live Catholic You are on the way to destruction.>> in Georgia! What you say.>>->> <<-< Go against the flow! You have no chance to survive make your time.> Conversions are fairly easy:>> convert(3.14159,units,radians, degrees);> 179.9998479> Look under Help, units.know what command to look for in help. convert (Pi/2, units, radians, degrees); gives 90. > -- > Julian V. Noble> Professor Emeritus of Physics> jvn@lessspamformother.virginia.edu> ^^^^^^^^^^^^^^^^^^> http://galileo.phys.virginia.edu/~jvn/> Science knows only one commandment: contribute to science.> -- Bertolt Brecht, Galileo. === Subject: Good day to one and all,I had a small question. Could you please tell me if first order logicand quantification theory are essentially the same. If they are not,can some kind soul enumerate the differences between the two.kindly yoursManik === Subject: > Are math professors willing to teach their students information they> know is wrong?> Consider that I found that by using> f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f)> I can show a flaw with the ring of algebraic integers.> The mathematics is rather basic and mostly involves algebraic> manipulations and the simple lemma that for any factor g of a> polynomial you have g=r+c, where c is a factor of the constant term,> and g varies with your key variable, and equals 0, when that variable> equals 0.> While posters have managed to make VERY involved posts as if it's all> so terribly complicated that doesn't protect mathematicians who teach> information that the math community now knows is flawed.> No, the posts are not VERY involved. Here is a counter-argument to one of your claims, directly related to what yousay above, which you have never answered in any way: === Subject: === Subject: === Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:== === Subject: === Subject: You claim that if you have a polynomial of the form P(x) = (v^3 + 1)*x^3 - 3*v*x*(u*f)^2 + (u*f)^3,where v = -1 + m*f^2, and m, u, and f are integers,with f prime and m coprime to f, then P(x)/f^2 can be factored in the form P(x)/f^2 = (b1*x + u)*(b2*x + u)*(b3*x + u*f) [1]where b1, b2, and b3 are algebraic integers.I say not. Let m = 1, f = 5, and u = 1. Thenv = 24, and v^3 + 1 = 13825 = 25*553. It is easily verified that P(x)/f^2 = 553*x^3 - 72*x + 5.If this is factored in the form [1] as you claim,then -u/b1 = -1/b1 is a root of Q(x) = P(x)/f^2. That is, Q(-1/b1) = 553*(-1/b1)^3 - 72*(-1/b1) + 5 = 0.Multiply through by b1^3: 5*b1^3 + 72*b1^2 - 553 = 0.The expression on the left is a *non-monic*polynomial in b1 with integer coefficients,and it is *irreducible* over the rationals.Therefore b1 cannot be an algebraic integer.Therefore your claim is false. === Subject: === Subject: === Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject: === Subject: [social commentary deleted]Nora B. === Subject: > Are math professors willing to teach their students information they> know is wrong?> Consider that I found that by using f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f)> I can show a flaw with the ring of algebraic integers.> The mathematics is rather basic and mostly involves algebraic> manipulations and the simple lemma that for any factor g of a> polynomial you have g=r+c, where c is a factor of the constant term,> and g varies with your key variable, and equals 0, when that variable> equals 0.> While posters have managed to make VERY involved posts as if it's all> so terribly complicated that doesn't protect mathematicians who teach> information that the math community now knows is flawed.> No, the posts are not VERY involved. Here is a counter-> argument to one of your claims, directly related to what you> say above, which you have never answered in any way:Let's see what you have. > === Subject: === Subject: === Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:== === Subject: > You claim that if you have a polynomial of the form P(x) = (v^3 + 1)*x^3 - 3*v*x*(u*f)^2 + (u*f)^3,> where v = -1 + m*f^2, and m, u, and f are integers,> with f prime and m coprime to f, then P(x)/f^2 > can be factored in the form> P(x)/f^2 = (b1*x + u)*(b2*x + u)*(b3*x + u*f) [1]> where b1, b2, and b3 are algebraic integers.No I don't.The problem is that the ring of algebraic integers doesn't includethem.More comments below... > I say not. Let m = 1, f = 5, and u = 1. Then> v = 24, and v^3 + 1 = 13825 = 25*553. It is > easily verified that > P(x)/f^2 = 553*x^3 - 72*x + 5.> If this is factored in the form [1] as you claim,> then -u/b1 = -1/b1 is a root of Q(x) = P(x)/f^2. > That is, Q(-1/b1) = 553*(-1/b1)^3 - 72*(-1/b1) + 5 = 0.> Multiply through by b1^3:> 5*b1^3 + 72*b1^2 - 553 = 0.> The expression on the left is a *non-monic*> polynomial in b1 with integer coefficients,> and it is *irreducible* over the rationals.> Therefore b1 cannot be an algebraic integer.> Therefore your claim is false.> === Subject: === Subject: === Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:> [social commentary deleted]> Nora B.The problem is that the b's aren't included in the ring of algebraicintegers, which I've pointed out more than once.So how is such an error a moral test?Mathematicians might believe they can sit back with my prime countingfunction or my proof of Fermat's Last Theorem, but ignoring an errorrequires allowing it to be taught to students.My assessment is that mathematicians lack the morals to make the rightdecision, but more interesting to me at this point is whether or notsociety will allow mathematicians to proceed with teaching flawedinformation.That is, will humanity protect its children in this area?James Harris === Subject: [snip]> So how is such an error a moral test?The presence of an error is not a moral issue. The defense of an error, in the face of conclusive proof that itis an error, certainly *is* a moral issue -- it relates to honesty, integrity, and malice.> My assessment is that mathematicians lack the morals to make the right> decision, but more interesting to me at this point is whether or not> society will allow mathematicians to proceed with teaching flawed> information.>> That is, will humanity protect its children in this area?I strongly urge you to stay away from children.--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: > Are math professors willing to teach their students information they> know is wrong?> Consider that I found that by using f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f)> I can show a flaw with the ring of algebraic integers.> The mathematics is rather basic and mostly involves algebraic> manipulations and the simple lemma that for any factor g of a> polynomial you have g=r+c, where c is a factor of the constant term,> and g varies with your key variable, and equals 0, when that variable> equals 0.> While posters have managed to make VERY involved posts as if it's all> so terribly complicated that doesn't protect mathematicians who teach> information that the math community now knows is flawed. No, the posts are not VERY involved. Here is a counter-> argument to one of your claims, directly related to what you> say above, which you have never answered in any way:> Let's see what you have.> === Subject: === Subject: === Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:== === Subject: > You claim that if you have a polynomial of the form> P(x) = (v^3 + 1)*x^3 - 3*v*x*(u*f)^2 + (u*f)^3,> where v = -1 + m*f^2, and m, u, and f are integers,> with f prime and m coprime to f, then P(x)/f^2 can be factored in the form> P(x)/f^2 = (b1*x + u)*(b2*x + u)*(b3*x + u*f) [1]> where b1, b2, and b3 are algebraic integers.> No I don't.> I downloaded Advanced Polynomial Factorization from your website last night. You said, assuming P(x) = (a1*x + uf)*(a2*x + uf)*(a3*x + uf),where a1, a2, and a3 are algebraic integers, thenletting g1 = a1*x + uf, g2 and g3 defined similarly, two of the g's should have a factor of f which would force two of the a's to have a factor that is f.Assume without loss of generality that a1 and a2 have a factor that is f.To equate this to what I said above, let b1 = a1/f, and b2 = a2/f.Since you say essentially that a1 and a2have a factor that is f, this means thatb1 and b2 are algebraic integers. The question here is: why are you denying theresults in your own paper? > The problem is that the ring of algebraic integers doesn't include> them.> More comments below...> I say not. Let m = 1, f = 5, and u = 1. Then> v = 24, and v^3 + 1 = 13825 = 25*553. It is > easily verified that > P(x)/f^2 = 553*x^3 - 72*x + 5.> If this is factored in the form [1] as you claim,> then -u/b1 = -1/b1 is a root of Q(x) = P(x)/f^2. > That is,> Q(-1/b1) = 553*(-1/b1)^3 - 72*(-1/b1) + 5 = 0.> Multiply through by b1^3:> 5*b1^3 + 72*b1^2 - 553 = 0. The expression on the left is a *non-monic*> polynomial in b1 with integer coefficients,> and it is *irreducible* over the rationals.> Therefore b1 cannot be an algebraic integer.> Therefore your claim is false.> === Subject: === Subject: === Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject:=== Subject: [social commentary deleted]> Nora B.> The problem is that the b's aren't included in the ring of algebraic> integers, which I've pointed out more than once.> See above. When you say two of the a's have a factor that is f, and you are working in the ringof algebraic integers, that can mean only one thing:b1 = a1/f and b2 = a2/f are algebraic integers. Or are you now saying that your results in APFare incorrect ???> So how is such an error a moral test?> Yes, perhaps it is. Nora B.> Mathematicians might believe they can sit back with my prime counting> function or my proof of Fermat's Last Theorem, but ignoring an error> requires allowing it to be taught to students.> My assessment is that mathematicians lack the morals to make the right> decision, but more interesting to me at this point is whether or not> society will allow mathematicians to proceed with teaching flawed> information.> That is, will humanity protect its children in this area?> James Harris === Subject: >> [...]>> The question here is: why are you denying the>results in your own paper? Well that one's easy - if you'd written a paper with resultslike that would _you_ admit it?************************David C. Ullrich === Subject: James Harris> It is a test for my own entertainment as the issue has been settled by> humanity's representatives--mathematicians.! Who voted for mathematicians to be the representatives ofhumanity. If anything mathematicians represent the abscense of what itmeans to be human.Mike Helland === Subject: > James Harris>> It is a test for my own entertainment as the issue has been settled by>> humanity's representatives--mathematicians.> ! Who voted for mathematicians to be the representatives of> humanity. If anything mathematicians represent the abscense of what it> means to be human.Absence.And that's a particularly verminous comment by the way.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen === Subject: James Harris>> It is a test for my own entertainment as the issue has been settled by>> humanity's representatives--mathematicians.> ! Who voted for mathematicians to be the representatives of> humanity. If anything mathematicians represent the abscense of what it> means to be human.> Absence.> And that's a particularly verminous comment by the way.He's probably using a non-standard definition of absence.'cid 'ooh === Subject: > Looking for a comprehensive book on the history of Mathematics.is now IMO the best general introduction. Older books can still bevaluable, but I strongly advise reading Katz first. He is a responsibleand up-to-date historian of mathematics who gives sound overviews backedup by references to original material in the excellent bibliographies atthe ends of chapters. Also, his book is good to read. :-) Ken Pledger. === Subject: > I'm supposed to show that for some n, postage of n cents or more can be> achieved by using only p-cent and q-cent stamps provided that gcd(p,q) = 1.> I decided to do an example. Pick p = 17 and q = 3. gcd(17,3) = 1 seeing as> they are both primes. I started by creating 17-cent postage, then 18-cent> postage, but when trying to create 19-cent postage I ran into a problem.> Since I can't create 19-cent postage, I decided to start from 20-cent> postage. I can then create 21-cent postage, but can't with 22-cent postage.> So then I decided to start from 23-cent postage and so on and so forth.> It seems that whenever n = 19 + 3a for some nonnegative integer a, there> exists no amount of p- and q-cent stamps that add up to n. So supppose 19 +> 3a = 3x + 17y for some nonnegative integers x and y.> 19 + 3a = 3x + 17y> 2 = 3x - 3a + 17y - 17 = (x - a)3 + (y - 1)17> This is a contradiction since you can't make 2-cent postage from any> combination of 3- and 17-cent stamps. This must mean the original statement> is false. ??Well, let's check out induction.Suppose n = a*17 + b*3.If a >= 4, n + 1 = (a - 4)*17 +4*17 + b*3 + 1 = (a - 4)*17 + (b + 23)*3.What if a < 4? If b >= 11, thenn + 1 = a*17 + (b - 11)*3 + 3*11 + 1 = (a + 2)*17 + (b - 11)*3. If a < 4, how big must n be to assure b >= 11? Ans: 84.Moreover 84 = 3*17 + 11*3.We conclude that any amount of postage >= 84 cents can be made up of 17cent and 3 cent stamps.Can you spot where gcd(3, 17) = 1 was important? Is 84 minimum? (No)However, (I'm dredging this up from memory - don't bet the ranch on it):Theorem. Let a and b be relatively prime positive integers. Let L(a,b)be the set of all integers of the form a*x + b*y for nonnegativeintegers x and y.Then, for any integer n, exactly one of n or a*b - a - b - n belongs to L(a,b). If I recall correctly, the proof of the Theorem, if it is even true,does not involve induction.Since 0 is in L(a, b), it follows that a*b - a - b is not in L(a,b),and that L(a,b) contains all integers greater than a*b - a - b becausethey are a*b - a - b - n for negative n and L(a, b) contains nonegative integers. Note, as Brian said,a*b - a - b + 1 = (a - 1)*(b - 1).-- Paul SperryColumbia, SC (USA) === Subject: >I'm supposed to show that for some n, postage of n cents or more can be>achieved by using only p-cent and q-cent stamps provided that gcd(p,q) = 1.It might help to know that the minimum n for which this is trueis (p - 1)(q - 1).>I decided to do an example. Pick p = 17 and q = 3. gcd(17,3) = 1 seeing as>they are both primes. I started by creating 17-cent postage, then 18-cent>postage, but when trying to create 19-cent postage I ran into a problem.>Since I can't create 19-cent postage, I decided to start from 20-cent>postage. I can then create 21-cent postage, but can't with 22-cent postage.>So then I decided to start from 23-cent postage and so on and so forth.>It seems that whenever n = 19 + 3a for some nonnegative integer a, there>exists no amount of p- and q-cent stamps that add up to n. Counterexample: 34 = 19 + 3*5 can be made with two 17-centstamps.[...]Brian === Subject: > It might help to know that the minimum n for which this is true> is (p - 1)(q - 1).How is it known to be a minimum? How can on even write this in the form aq +bp?> Counterexample: 34 = 19 + 3*5 can be made with two 17-cent> stamps.Very good. There goes my reasoning...Bernd === Subject: >> It might help to know that the minimum n for which this is true>> is (p - 1)(q - 1).>How is it known to be a minimum? It's a standard result. The proof is just a little tricky, as Irecall, but I remember producing one many years ago. (The fullstatement of the result also includes the fact that exactly halfof the non-negative integers less than (p - 1)(q - 1) can berepresented as a sum of non-negative multiples of p and q.)>How can on even write this in the form aq + bp?Don't recall off the top of my head, but I suspect that you'llneed to look at the general solution to the Diophantine equationthe equation px + qy = n, and with a bit of fiddling around youcan probably show that one of these solutions is non-negative ifn >= (p - 1)(q - 1). I don't guarantee that this approach works,but it's where I'd start.[...]Brian === Subject: > In an equilateral triangle (a triangle were lenght of all sides are equal)> there are inscribed three circles which touch (tangent) each other.> The radii of circles is 3.0 cm.>> How long are the sides of the triangle?>The center of the circles is an equilateral triangle.Find the center of that triangle, the point equidistant from the centers of the circles.That point is equidistant from the sides of the circumscribing triangle. <9d88b76e.0308161722.14696a35@posting.google.com> <262cb2f9.0308280816.498ea550@posting.google.com>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft === Subject: at 09:16 AM, mathphyer@yahoo.com (mathphyer) said:>I am always a little confused when people say that a real number>corresponds to a point on the real axis or straight line. Is the>straight line or the point a mathematical object (it seems closer to>a physical object to me)? No, but that's how you motivate it.>How is it defined mathematically? You have two options: analytic and synthetic. The first approachrequires teaching the properties of the Reals first, then definingGeometry in terms of numbers. The second approach requires giving theaxioms for Geometry and proving things from them.Note that it is possible to derive the properties of the real line,including addition and multiplication, strictly from Geometry, butthat is probably not pedagogically sound.Probably using axiomatic set theory?That's also possible. It would require teachers more qualified thanwhat we currently have in the public schools.-- Shmuel (Seymour J.) Metz, SysProg and JOATReply to domain Patriot dot net user shmuel+news to contact me. Do not replyto spamtrap@library.lspace.org === Subject: > at 09:16 AM, mathphyer@yahoo.com (mathphyer) said:>>I am always a little confused when people say that a real number>>corresponds to a point on the real axis or straight line. Is the>>straight line or the point a mathematical object (it seems closer to>>a physical object to me)? >No, but that's how you motivate it.Even Euclid's students considered it to be a mathematicalobject. Diagrams, etc., were to help understand the formal aspects, and not important in themselves. Thisdoes not mean that diagrams cannot be used both to findresults and for clarification.>>How is it defined mathematically? >You have two options: analytic and synthetic. The first approach>requires teaching the properties of the Reals first, then defining>Geometry in terms of numbers. The second approach requires giving the>axioms for Geometry and proving things from them.>Note that it is possible to derive the properties of the real line,>including addition and multiplication, strictly from Geometry, but>that is probably not pedagogically sound.updating of the Euclidean axioms involves essentiallyputting in what can only be called properties of thereal numbers.>Probably using axiomatic set theory?>That's also possible. It would require teachers more qualified than>what we currently have in the public schools.If they cannot understand axiomatic set theory, they cannotteach Euclid or rigorous arithmetic. But there is littledifficulty in teaching formal logic and axiomatic settheory to a substantial proportion of elementary schoolchildren, with minor modifications in the texts.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University <262cb2f9.0308280816.498ea550@posting.google.com> <3f4eeb51$2$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft === Subject: at 12:16 PM, hrubin@odds.stat.purdue.edu (Herman Rubin) said:>But there is little>difficulty in teaching formal logic and axiomatic set>theory to a substantial proportion of elementary school>children,The difficulty is in teaching it to their teachers. :-(-- Shmuel (Seymour J.) Metz, SysProg and JOATReply to domain Patriot dot net user shmuel+news to contact me. Do not replyto spamtrap@library.lspace.org === Subject: > at 05:59 PM, amorgan@Xenon.Stanford.EDU (Alan Morgan) said:>>At what level? >Any?>>Would you suggest teaching Cauchy sequences to>>junior high school students?>Would you suggest sprinkling them with pixie dust? That has as much>wouldn't suggest introducing real numbers in terms of rational numbers>at all; I'd suggest introducing real numbers geometrically at an early>age and only teaching constructions much later.Introducing them geometrically is only going to lead toconfusion later. The earlier you teach conceptually, theeasier it is for them to understand. >When it was time to introduce them to more rigorous methods, I'd start>them on an axiomatic approach well before I did a construction from>the integers.This belongs in first grade, if not earlier. Children canunderstand concepts, but the emphasis on memorization androutine computation is what kills this.One does not have to be complete, but everything needs tobe rigorous. This does not mean that it cannot also beintuitive; the Peano Postulates are quite intuitive ifproperly presented. And if I did get around to teaching a construction from>Peano's Postulates, it most certainly wouldn't be rdecimals[1]>expansions, because that is by far the *most* complex approach under>discussion.If you are doing a postulational approach, the properties areneeded BEFORE expansions can be done. That way, it is clearthat the use of decimal expansions is only one way, which ischosen for certain purposes of convenience. Look at all the asinine threads on the claim that .9 ...>is not equal to 1.>>but most classes don't do that>Then why introduce the complications inheren't in using repeating>decimals as a definition? Do it geometrically, and teach decimals as>simle nomenclature.They are now taught that recurring decimals are rationalnumbers. This is a fine place to introduce limits quiterigorously. But do not use the usual approach withsymmetric intervals; that one can do it that way should bea corollary or a comment.>>Nice. It's almost impossible to find qualified math teachers for>>public school now. You'll only make it harder.>No. I'd make it easier to find *qualified* teachers. It doesn't bother>me that I'd make it harder to find unqualified teachers.Most are unqualified now; they may be certified.>[1] Why base 10, anyway?-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University <3f4c096b$11$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft === Subject: at 12:20 PM, hrubin@odds.stat.purdue.edu (Herman Rubin) said:>Introducing them geometrically is only going to lead to>confusion later. The earlier you teach conceptually,How is that not teaching conceptually? Whether I start from Peano'sPostulates and construct everything, give axioms for Geometry or giveaxioms for the real, I will be teaching concepts.>This belongs in first grade, if not earlier. Halevai (it should only happen!) Bimheirah byameinu (Speedily, in ourday!)>Children can understand concepts, but the emphasis on memorization >and routine computation is what kills this.-- Shmuel (Seymour J.) Metz, SysProg and JOATReply to domain Patriot dot net user shmuel+news to contact me. Do not replyto spamtrap@library.lspace.org === Subject: > at 12:20 PM, hrubin@odds.stat.purdue.edu (Herman Rubin) said:>>Introducing them geometrically is only going to lead to>>confusion later. The earlier you teach conceptually,>How is that not teaching conceptually? Whether I start from Peano's>Postulates and construct everything, give axioms for Geometry or give>axioms for the real, I will be teaching concepts.There are at least two distinct concepts of integers, andeven if the axioms for geometry are given, these will needto be added. >>This belongs in first grade, if not earlier. >Halevai (it should only happen!) Bimheirah byameinu (Speedily, in our>day!)>>Children can understand concepts, but the emphasis on memorization >>and routine computation is what kills this.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === Subject: I have trouble with understanding following exercise solution.Q> T:V->V, T(v)=a x v - c is a linear transformation, where V is a vector space with orthonormal basis {a,b,c}, c=a x b. Find eigenvalue of T and eigenspace of T.sol) <-in my book answer. Let lamda is a eigenvalue of T then, T(a)=a x a - c=0-0=0=lamda*a T(b)=a x b - c=a x b - a x b=0=lamda*b T(c)=a x c - c=-b=lamda*c_____________________________________________ ______I don't understand above,,,why T(a)=lamda*a ,T(b)=lamda*b, T(c)=lamda*c ?I agree with T(v)=lamda*v if v is a eigenvector of T and lamda is aeigenvalue.but, my point of view, a,b,c are only basis member of T's domain...that is, I wonder how can we see a,b,c as eigenvector of T.And at T(c)=a x c - c=-b=lamda*c, why a x c = -b ?I don't know why this is -b not b.If anyone can help me, please post reply. === Subject: >I have trouble with understanding following exercise solution.>>Q> T:V->V, T(v)=a x v - c is a linear transformation,> where V is a vector space with orthonormal basis {a,b,c}, c=a x b.> Find eigenvalue of T and eigenspace of T.>>sol) <-in my book answer.>These look like lucky guesses to me. As presented -- there's actually a good reason for making these guesses.> Let lamda is a eigenvalue of T then,>Skip this for now.> T(a)=a x a - c=0-0=0=lamda*a>T(a) = 0 (as above) = 0*a (how convenient), so one of the eigenvalues is 0 and an eigenvector is a.> T(b)=a x b - c=a x b - a x b=0=lamda*b>T(b) = 0 = 0*b (how convenient again), so 0 is an eigenvalue (again) with eigenvector b.> T(c)=a x c - c=-b=lamda*c>T(c) = -b is never lambda*c. So this statement is just wrong.>__________________________________________________>>I don't understand above,,,>why T(a)=lamda*a ,T(b)=lamda*b, T(c)=lamda*c ?>I agree with T(v)=lamda*v if v is a eigenvector of T and lamda is a>eigenvalue.>but, my point of view, a,b,c are only basis member of T's domain...>that is, I wonder how can we see a,b,c as eigenvector of T.>>And at T(c)=a x c - c=-b=lamda*c, why a x c = -b ?>Recall that a x b = c. So a x c = -b.>I don't know why this is -b not b.>Well, you can either say right hand rule (holding your hand up to see it better) or you can chase through the calculations.>If anyone can help me, please post reply.>Now, why did we check T(a), T(b), T(c). Because (in spite of our professor's warning against it) we like to write linear transformations as matrices. The matrix is A= (row1, row2, row3) = (0 0 0, 0 0 0, 0 -1 0). It's easy to calculate the eigenvalues now. They're the solutions to det(A - lambda*I)=0. Then we have to solve for vectors v that satisfy Av=lambda*v. But we (fortunately!) already found 2 solutions above, and we know that there are only 2 (T(c) is not 0, so dim(image)>=1). . . .Jon Miller === Subject: ---------------------------------------------------- -----------------In the coming months a black spot will pop up everywhere . . . on store windows and newspaper boxes, on gas pumps and supermarket shelves. Open a magazine or newspaper - it's there. It's on TV. It stains the logos and smears the nerve centers of the world's biggest, dirtiest corporations.This is the mark of the people who don't approve of Bush's plan tocontrol the world, who don't want countries liberated without UNbacking, who can't stand anymore neo-con bravado shoved down theirthroats.This is the mark of the people who want the Kyoto Protocol for theenvironment, who want the International Criminal Court for greaterjustice, who want a world where all nations, including the U.S.A., are free of weapons of mass destruction, and who pledge to take theircountry back.--They are receiving inside the store now, won't grasp cards later.-------------------------------------------------------- -------------Unbrand America
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In the coming months a black spot will pop up everywhere . . . on store windows and newspaper boxes, on gas pumps and supermarket shelves. Open a magazine or newspaper - it's there. It's on TV. It stains the logos and smears the nerve centers of the world's biggest, dirtiest corporations.

This is the mark of the people who don't approve of Bush's plan to control the world, who don't want countries "liberated" without UN backing, who can't stand anymore neo-con bravado shoved down their throats.

This is the mark of the people who want the Kyoto Protocol for the environment, who want the International Criminal Court for greater justice, who want a world where all nations, including the U.S.A., are free of weapons of mass destruction, and who pledge to take their country back.
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=== Subject: -------------------------------------------- -------------------------In the coming months a black spot will pop up everywhere . . . on store windows and newspaper boxes, on gas pumps and supermarket shelves. Open a magazine or newspaper - it's there. It's on TV. It stains the logos and smears the nerve centers of the world's biggest, dirtiest corporations.This is the mark of the people who don't approve of Bush's plan tocontrol the world, who don't want countries liberated without UNbacking, who can't stand anymore neo-con bravado shoved down theirthroats.This is the mark of the people who want the Kyoto Protocol for theenvironment, who want the International Criminal Court for greaterjustice, who want a world where all nations, including the U.S.A., are free of weapons of mass destruction, and who pledge to take theircountry back.--One more bizarre grocers live Rashid, and they weekly kick Fahd too.---------------------------------------------------------- -----------Unbrand America
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height=51
In the coming months a black spot will pop up everywhere . . . on store windows and newspaper boxes, on gas pumps and supermarket shelves. Open a magazine or newspaper - it's there. It's on TV. It stains the logos and smears the nerve centers of the world's biggest, dirtiest corporations.

This is the mark of the people who don't approve of Bush's plan to control the world, who don't want countries "liberated" without UN backing, who can't stand anymore neo-con bravado shoved down their throats.

This is the mark of the people who want the Kyoto Protocol for the environment, who want the International Criminal Court for greater justice, who want a world where all nations, including the U.S.A., are free of weapons of mass destruction, and who pledge to take their country back.
height=100
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height=50 height=50 height=50 height=50 height=50 height=50 height=50 height=50 height=50 height=50 height=50 height=50
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=== Subject: > Has somebody cared enough about MDS to update the > computer programs? It's long been my impression that > like MDS sometimes is included in the tools of data mining.Try this sitehttp://www.newmdsx.com/It offers a very good MDS windows program developed by Prof. T. Coxonand co. I have used it (as a beta tester) and it is rather friendly. Ican run almost any possible MDS model available today.BestNiko Tiliopoulos === Subject: > Try this site> http://www.newmdsx.com/> It offers a very good MDS windows program developed by Prof. T. Coxon> and co. I have used it (as a beta tester) and it is rather friendly. I> can run almost any possible MDS model available today.> Best> Niko TiliopoulosERRATUM!I beg pardon. In the previous message the sentence I can runalmost... should have said IT can run almost... . I cannot I amafraid.BestNiko Tiliopoulos === Subject: > Has somebody cared enough about MDS to update the> computer programs? It's long been my impression that> like MDS sometimes is included in the tools of data mining.MDS has become slightly outdated, but there has been some good work intothe same direction recently, primarily in the direction of locally linearembedding. The problem with MDS is globality: in most stress functionsshort dissimilarities are approximated with a similar precision as largedissimilarities. In reality, what a human analyst expects from MDS is morea structure alike clustering: one that would identify groups of similarobjects.The 'old' solution was Shepard's non-metric MDS. NMDS attempts to modifythe dissimilarity matrix so that the distance rankings are maintained,rather than metric deviations. For example, we are not interested in theexact distances between A, B and C, but we do want distance A-B to begreater than A-C if C is more dissimilar from A than is B. It does not workwell in practice, unfortunately. Today, new methods have emerged. Forexample, locally linear embedding instead only evaluates the distances froman object to K of its nearest neighbors. This yields nice results.Good starting points for further exploration arehttp://www.cs.toronto.edu/~roweis/lle/http:// basis.stanford.edu/carrie-web/Aleks === Subject: >>Has somebody cared enough about MDS to update the>>computer programs? It's long been my impression that>>like MDS sometimes is included in the tools of data mining.> MDS has become slightly outdated, but there has been some good work into> the same direction recently, primarily in the direction of locally linear> embedding. The problem with MDS is globality: in most stress functions> short dissimilarities are approximated with a similar precision as large> dissimilarities. In reality, what a human analyst expects from MDS is more> a structure alike clustering: one that would identify groups of similar> objects.> The 'old' solution was Shepard's non-metric MDS. NMDS attempts to modify> the dissimilarity matrix so that the distance rankings are maintained,> rather than metric deviations. For example, we are not interested in the> exact distances between A, B and C, but we do want distance A-B to be> greater than A-C if C is more dissimilar from A than is B. It does not work> well in practice, unfortunately. Today, new methods have emerged. For> example, locally linear embedding instead only evaluates the distances from> an object to K of its nearest neighbors. This yields nice results.> Good starting points for further exploration are> http://www.cs.toronto.edu/~roweis/lle/> http://basis.stanford.edu/carrie-web/> Aleks> Forrest Young still distributes the Fortran code and Dos executable of Alscal, the descendent of KYST, the Bell Labs NMDS program. MDS, for some reason, is often used as an acronym for NMDS. I follow those who restrict MDS to Shepard's metric scaling, which is identical to Gower's Principal coordinates analysis.http://forrest.psych.unc.edu/MDS still has huge advantages, if you have access to the case by variables matrix. There are a variety of transformations that can be used to transform the data matrix prior to doing the metric scaling. these transformations. Pierre provides code on his web page for mac & pc for doing the transformations and PCA's, and I provide the Matlab 4 & Matlab 6 code for doing the same:http://www.es.umb.edu/edgwebp.htm#LegGallMat6 Strictly speaking the MDS model (not the MDS model) can have problems with non-metric distances, producing negative eigenvalues. Legendre & Legendre (1998) Numerical Ecology, 2nd ed. review 3 solutions to this problem, and their algorithms are programmed in Pierre & my programs. Often it is not the ordination of the cases that is important but explaining why the cases take the positions they do in low-dimension space. The Gabriel Euclidean distance biplot and correlation biplot are important tools for interpreting the low-dimension ordination. Gower's book 'biplots' is the best overall description of the process. MDS has also been revived by the use of constrained ordination techinques. Canonical correspondence analysis can impose conditions that either the distances among cases in low-dimension space must be linear functions of a set of external explanatory variables, or uncorrelated with a set of covariates (partial canonical correspondence analysis). When preserving Euclidean distances, as in the MDS model, the technique is called redundancy analysis. Both redundancy analysis and canonical correspondence analysis (not to be confused with canonical correlation analysis) are available in the CANOCO package. Using algorithms presented by ter Braak or by Legendre & Legendre, the basic CANOCO or redundancy analysis models can be programmed in languages such as Matlab. BTW, you can perform an MDS, aka Gower Principal coordinates analysis, on a correlation matrix after converting it to a distance matrix. This distance matrix will not be Euclidean, so anticipate negative eigenvalues. The standard NMDS programs have an option to specify whether the matrix entered is a distance or similarity matrix.Gene Gallagher === Subject: For my simulations, I have to simulate paths of Normal(mu*, sigma*).random variables. It so happens, that each pathhas few [500] observations, AND sigma is large [compared to mu]. Thusthe realized values of mu are often very different from mu*. Idiscovered that in my particular application, it makes sense to takeeach path, 1) estimate mu and sigma 2) if mu is not equal to m*, then massage the data - for each path,convert the data into N(0,1) [by subtracting the estimated mu [forthis path] from each observation, etc], and then convert the data intoN(M*, Sigma*)I [by adding m* to each value, etc].I was wondering whether there is a formal Statistics term describingthis massaging the data procedure. I really need to know this, tobe able to put my research into context of existing work.Stan === Subject: >Fairly good analysis; but in a real world, a small difference in thenumber ofstudents among different classes may be acceptable.>Sorry, this is not a real world problem that school administrators areexpected to solve and where having one extra student in one class maystill entitle them to claim that all classes are of equal size. Itis a math problem where it is clearly stated that ALL classes are ofequal size. Nothing could be clearer than this (this is not like theclassic bus problem where it is acceptable--indeed necessary-- toeither have one extra bus or to cram a few more passengers into somebuses buses so that they are not left stranded). I'm inclined to think that there was a numerical error in the problemas posed. That would not be the first time. Matt used to come homewith math worksheets riddled with them. The teacher had been too busyto check them before handing them out. There's nothing more calculated(sorry for the pun) to undermine a student's self confidence thanassigning problems that cannot be solved because of such errors. mattsmom -- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: >I'm inclined to think that there was a numerical error in the problem>as posed. That would not be the first time. That was probably the difficulty. Still, we can avoid that typical assumptionand find a practical way to treat the given problem, in case what wasoriginally presented were not a mistake. A teacher is often better able to serve the students if he does notcreateinvent too many problems; teacher should use published sources orpreviously created and tested problems. It keeps everybodies' duties neater(less troublesome and less unnecessary frustration). How To Really Make An Instructional Mess:Create an exercise worksheet based almost entirely on a combination of your ownexperience, imagination, and modifications of book-published problems. Someteacher may at times have particular reasons for doing that, but if he/she isnot careful, that could confuse the hell out of some beginning students.G C-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: >I'm inclined to think that there was a numerical error in the problem>as posed. That would not be the first time.>> That was probably the difficulty. Still, we can avoid that typicalassumption> and find a practical way to treat the given problem, in case what was> originally presented were not a mistake....but the original question *is* a mistake, since there is no solution(more specifically, the hypotheses are impossible).>> A teacher is often better able to serve the students if he does not> createinvent too many problems; teacher should use published sources or> previously created and tested problems.? Mistakes still exist even in 'published' sources. Always proofread.> It keeps everybodies' duties neater> (less troublesome and less unnecessary frustration).>> How To Really Make An Instructional Mess:> Create an exercise worksheet based almost entirely on a combination ofyour own> experience, imagination, and modifications of book-published problems.Some> teacher may at times have particular reasons for doing that, but if he/sheis> not careful, that could confuse the hell out of some beginning students.?Many a worksheet/test exists authored soley by the instructor giving it, orfrom some combination of sources as you describe. This does not constitutean 'instructional mess' unless said instructor hasn't a clue what he isdoing.And that's the _real_ problem. Some instructors haven't a clue what theyare doing.-- Darrell-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: >A teacher is often better able to serve the students if he does notcreateinvent too many problems; teacher should use published sourcesorpreviously created and tested problems. It keeps everybodies' dutiesneater(less troublesome and less unnecessary frustration).>Actually, the error-riddled worksheets Matt came home with werecommercially produced; this induced a false sense of security in theteachers, who did not see any need to check their accuracy.Matt's Dad bought an ARCO AP-Calculus review book because he liked theway the problems were presented. Only after Matt did some problemsdid we realize the book was also full of mistakes. Luckliy, Matt wasself-confident enough not to let them faze him. But a student justlearning arithmetics (and the student's parents) will feel dumb fornot being able to solve an unsolvable problem. The proof is thatKate's mom found it necessary to appeal for help in this newsgroupinstead of concluding that the problem was insoluble because itcontained an error. Sometimes, what's needed from the real world iscommon sense.-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: > The fifth grades has 335 students, and the sixth grades has 264> students. All classes are the same size.gcd(336,264) = 24gcd(335,264) = 1Are you sure it wasn't 336 students, instead of 335?-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: I cannot figure out this problem:I am given three points (0,1); (2,0); and (4,0). I am supposed to makean equation about it but I don't know where to start.I have tried to graph this but it only comes out in a curvy graph andI don't know how to write the equation of a curvy graph except a sinegraph (and this is not a sine graph because it has integral xnumbers).-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: : I cannot figure out this problem:: : I am given three points (0,1); (2,0); and (4,0). I am supposed to make: an equation about it but I don't know where to start.: : I have tried to graph this but it only comes out in a curvy graph and: I don't know how to write the equation of a curvy graph except a sine: graph (and this is not a sine graph because it has integral x: numbers).Integral x-values are not a problem: y = 1/6 + Sin(x*Pi/3 + Pi/2) / 3All you need is any two values A and B for which Sin(A) = Sin(A+B) but Sin(A) is NOT equal Sin(A-B)Then do a horizontal translation by the amount (B-A)and scale the horizontal axis by a factor of B/2.Also do a vertical translation by -Sin(A)and scale the vertical axis by Sin(A-B)+Sin(A). y = {Sin[x*B/2 - (B-A)] - Sin(A)} / {Sin(A-B)-Sin(A)}Done.Of course, with only 3 points to match, mostpeople would try a parabola (quadratic function).There is only ONE quadratic function passing throughthe three given points, while there are infinitelymany Sine curves which do so. Also, many peoplethink parabolas are simplier than Sine waves.Robert |)|/| || Burnaby South Secondary School || |orewood@olc.ubc.ca || Beautiful British ColumbiaMathematics & Computer Science || (Canada)-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: My guess is that you are trying to fit a quadratic curve. In that case, itwould have to be of the form,y = a(x-2)(x-4) since when x = 2 or 4, y = 0 and a is a constant to bedetermined.Since it passes through (0,1), then1 = a (-2)(-4) = 8a, which implies that a =1/8So the (quadratic) curve would be y = (x-2)(x-4)/8.> I cannot figure out this problem:>> I am given three points (0,1); (2,0); and (4,0). I am supposed to make> an equation about it but I don't know where to start.>> I have tried to graph this but it only comes out in a curvy graph and> I don't know how to write the equation of a curvy graph except a sine> graph (and this is not a sine graph because it has integral x> numbers).-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Now that everyone has agreed on the solution, I like to proposeanother one which at least deserves points for being interesting.We would just draw a circle through the three points.So we need prependicular bisecetors.The bisector for the line (2, 0) to (4, 0) is easy, the midpoint is(3, 0) and the bisector is x = 3The midpoint of line (0, 1) to (2, 0) is (1, 1/2), the slope is -1/2,so the bisector slope is 2, the bisector is y = 2 * x + c, and sinceit pass through (1, 1/2) the bisector is y = 2 * x - 3/2The two bisector intersects at (3, 2 * 3 - 3/2), i.e. (3, 9/2) andthat is the center of the circle.The square of the radius of the circle is (3 - 0)^2 + (9/2 - 1)^2So the equation of the circle is (x - 3)^2 + (y - 9/2)^2 = 85/4If you don't like fraction, you can write the equation as4 * (x - 3)^2 + (2 * y - 9)^2 = 85> My guess is that you are trying to fit a quadratic curve. In that case, it> would have to be of the form,> y = a(x-2)(x-4) since when x = 2 or 4, y = 0 and a is a constant to be> determined.> Since it passes through (0,1), then> 1 = a (-2)(-4) = 8a, which implies that a =1/8> So the (quadratic) curve would be y = (x-2)(x-4)/8.> I cannot figure out this problem:>> I am given three points (0,1); (2,0); and (4,0). I am supposed to make> an equation about it but I don't know where to start.>> I have tried to graph this but it only comes out in a curvy graph and> I don't know how to write the equation of a curvy graph except a sine> graph (and this is not a sine graph because it has integral x> numbers).-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: > I cannot figure out this problem:> I am given three points (0,1); (2,0); and (4,0). I am supposed to make> an equation about it but I don't know where to start.Let me make a guess.Let the roots of the polynomial equation (y = ax^n + bx^(n-1) + cx^(n-2) + ....+ jx + k) be (2,0) and (4,0) as given.The simplest such equation is the quadratic. So we have y = (x-a)(x-b) or forour values here. y = (x-2)*(x-4)multiply out y = x^2 -6x + 8I think you can see that this will work okay for x=2 and x= 4What happens when you put in the value x = 0 ? y = 0^2 -6*0 + 8OR y = 8.... but we want 1 not 8So the equation cannot be quadratic.Then lets try a cubic equation y = (x-a)(x-b)(x-c)We know the values for a,b = 2,4 since we know the roots.plug in the third value (0,1) with 1 = y and 0 = x to obtain 1 = (0-2)*(0-4)*(0-c) 1 = -8c c = -1/8This works then y = (x-2)(x-4)(x+1/8) or y = x^3 - 47/8*x^2 + 29/4*x + 1This is the simplest polynomial equation that will work with the 3 values givenabove.I hope the reason to pick a third degree polynomial equation, not a seconddegree quadratic makes sense.Randall-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: > I cannot figure out this problem:>> I am given three points (0,1); (2,0); and (4,0). I am supposed to make> an equation about it but I don't know where to start.>> Let me make a guess.>> Let the roots of the polynomial equation (y = ax^n + bx^(n-1) + cx^(n-2) +....> + jx + k) be (2,0) and (4,0) as given.>> The simplest such equation is the quadratic.> So we have y = (x-a)(x-b) or for> our values here.> y = (x-2)*(x-4)Of course we can't be totally sure without further clarification, but I tendto agree that chances are we are tasked to find a polynomial. And if so,why not one of the least degree.If r_1, r_2,...,r_n are the zeros of some real polynomial P with leadingcoefficient 'a', then P factors as:P(x) = a(x-r_1)(x-r_2)(x-...)(x-r_n)You failed to consider the necessary factor of 'a' in your analysis.>> multiply out> y = x^2 -6x + 8I don't know why, but since you want to:y = ax^2 -6ax + 8a>> I think you can see that this will work okay for x=2 and x= 4You can see clearer from the factored form...y = a(x-2)(x-4) = 0 when x=2 or x=4.>> What happens when you put in the value x = 0 ?when x=0, y=1a(0-2)(0-4) = 1a(8) = 1a = 1/8>> y = 0^2 -6*0 + 8>> OR y = 8.... but we want 1 not 8>> So the equation cannot be quadratic.We have y = (1/8)x^2 + bx + c for some constants b and c. Solve for b andc, which is easily done from plugging in the other points.-- Darrell-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: >>I cannot figure out this problem:>>I am given three points (0,1); (2,0); and (4,0). I am supposed to make>>an equation about it but I don't know where to start.> Let me make a guess.> Let the roots of the polynomial equation (y = ax^n + bx^(n-1) + cx^(n-2) + ....> + jx + k) be (2,0) and (4,0) as given.> The simplest such equation is the quadratic. So we have y = (x-a)(x-b) or for> our values here.> y = (x-2)*(x-4)> multiply out> y = x^2 -6x + 8> I think you can see that this will work okay for x=2 and x= 4> What happens when you put in the value x = 0 ?> y = 0^2 -6*0 + 8> OR y = 8.... but we want 1 not 8So just divide the polynomial by 8 so y(0) = 1 instead of 8. The values of y(2) and y(4) will still be 0 since 0/8 = 0y = (x^2)/8 -3x/4 + 1 goes through the three given points> So the equation cannot be quadratic..... > I hope the reason to pick a third degree polynomial equation, not a > second degree quadratic makes sense.Nope.-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: >I am given three points (0,1); (2,0); and (4,0). I am supposed to make>an equation about it but I don't know where to start.>>I have tried to graph this but it only comes out in a curvy graph and>I don't know how to write the equation of a curvy graph except a sine>graph (and this is not a sine graph because it has integral x>numbers).>What KIND of equation are you asked to make? Some exercises from PreCalculusCollege Algebra give points for which a student should find an equation but acertain type of equation is expected. G C-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: have been preparing:> My six year old child seems to have a natural math ability. I compare> it to a musician with perfect pitch. For example, he learned most> arithmetical skills when he was three and four. I haven't formally> assessed his skills, but he can consistently do 4th and 5th grade> level problems and he's not too shabby on more advanced topics not> necessarily part of the typical curriculum.>> He understands how some infinities are bigger than others. He can make> up and do Fibonacci sequences in his head. He can estimate square> roots.I take it you have some substantial math background yourself? My FAQ iswritten more for people like me, liberal arts majors with math-interestedchildren, but it should help. Many people who are good at math are new tothinking about math pedagogy, which my oldest son's strong interest in mathhas driven me to think about for a long time.> My question: what to do beyond simply accelerating through school> curriculum?>> Are there any books, web resurces, etc. that can point us in the right> direction? Or any other suggestions?Here's my FAQ of suggestions for parents whose children who appear to bebright at a young age in math.Suggestion 1: Run, don't walk, to get and read a copy of Liping Ma's bookKnowing and Teaching Elementary Mathematics. You can request it byinterlibrary loan if it is not in your local library. Ma's book makesapparent what kind of foundation is necessary at the beginning for a childto go as high as he can go in math. Try solving the teaching problems inthat book for yourself, and you'll see what I mean.Suggestion 2: LOOK UP some of the research Professor Tony Gardiner inBritain is doing comparing acceleration to enrichment as a strategy forpreparing bright children for advanced study of mathematics. See, forexample,http://www.m-a.org.uk/association/presidents_page/ able_students/no longer in print.) Gardiner has shown what he means in a series of bookscalled Maths Challenge published in Britainhttp://www.singaporemath.com/supplement_others.htmwhich I have bought and which are full of problems to develop a STRONGunderstanding of math. The Maths Challenge books are designed by Gardinerfor seventh-graders and older students who are in the top 10 percent of theBritish population--younger students with good preparation could start usingthem at a younger age.Suggestion 3: Surf over to Professor Hung-hsi Wu's Web sitehttp://math.berkeley.edu/~wu/and make sure to download and read the draft chapters Whole Numbers(Draft) and Fractions (Draft) to get more than 100 pages each on thoseeasy subjects from a thoughtful mathematician with a deep interest in matheducation. Then read his How to Prepare Students for Algebra for moreinsights.Suggestion 4: Get and read the book Concepts of Modern Mathematics by IanStewart. If your child is an advanced reader that book might be readable byyour child by late elementary age. (My son was a delayed reader, because hehas strabismus, so he hasn't read this book yet. I have read it.) This willshow you what your child will be thinking about if he or she takesuniversity-level math courses.Suggestion 5: Get and read How to Teach Mathematics (2nd edition, 1999) byStephen G. Krantz, which is a book pertaining mostly to university-levelmath study, but with some interesting comments by Krantz on the Saxon mathprogram and on other topics. The book includes essays by other professors ofmathematics. Think about what kind of primary and secondary mathematicseducation (an issue Krantz hardly addresses in his book) would be fitpreparation for university study of mathematics.Suggestion 6: Having done the above, ponder what materials you are using forprimary instruction in mathematics. My top recommendation for a firstmathematics program is Miquon Math,http://www.sonlight.com/miquon.htmlhttp:// www.keypress.com/catalog/products/supplementals/Prod_ Miquon.htmla nominally three-year program that covers almost all of elementary schoolmathematics from a higher math perspective. For people who have already gonethrough early elementary math, my number-one recommendation is the SingaporePrimary Mathematics serieshttp://www.sonlight.com/singapore.htmlhttp:// www.singaporemath.com/primary_math_US_ED.htm#primary% 20mathematics%20Orderwhich is described by many mathematicians as the best mathematics textbookseries available in English, an accurate description. The Singapore PrimaryMathematics series is followed by other series from Singapore that take alearner up to all the mathematics needed for A level examinations in theBritish university entrance system.Sometimes these two programs, Miquon and Singapore, contain problems thatare confusing to American parents who had more conventional mathinstruction. My friendly suggestion is to take the confusing parts of thosebooks as learning opportunities. Mathematicians linger and ask why? and analternative representation of a mathematical operation (and you can count onthe Primary Mathematics series to have ACCURATE representations ofmathematical operations in visual, verbal, and other forms) is anopportunity to THINK about why the content was presented that way. Sure, notevery child gets the content first from the same kind of presentation, butknowing why all the presentations relate to the same idea is part ofunderstanding mathematics thoroughly.Many of the illustrations in the Singapore books show examples ofmanipulatives that could be used in the classroom or at home as a firstintroduction to a topic. In my house, we USUALLY just went straight to thebook, figuring our real life and earlier use of Miquon Math had alreadyprovided the concrete examples that fit into the Singapore concrete -->pictorial --> abstract model of instruction. But the concrete examples arelatent in the coursebook (for example, baking cookies, inviting guests toparties, etc.) and may be helpful for many learners.The Education Program for Gifted Youth (EPGY)http://epgy.stanford.edumathematics program is probably wholly unnecessary at the very earliest agelevel, but it is a great way to move ahead for young people who like thatkind of computer-based instruction, especially beginning at about thefourth- or fifth-grade level. EPGY can take learners all the way up touniversity-level math at their own pace. The ALEKS online programhttp://www.aleks.com/is a useful supplement to any of the other recommended programs, beingalmost as good as and a lot less expensive than EPGY, but lacking a highschool geometry course and not continuing to advanced undergraduatemathematics.Many parents have tried out many other kinds of math programs to help theirprecocious math learners move ahead with a good foundation. Math programsthat I personally do NOT recommend, based on the desirability of a) trulychallenging word problems, b) multiple representations of mathematicalideas, and c) clear, CORRECT explanations of mathematical concepts include1) Saxon Math, 2) Math-U-See, 3) any old, traditional program (e.g., ABeka), 4) exclusive use of gifted learner worksheet books or other booksthat consist mostly of exercise sets, or 5) any reform math program usedin United States public schools, although the best of these are better thansome of the other programs I don't recommend. These negative recommendationsare not intended to offend any parent who has used these programs in agood-faith belief that they are useful math programs, but are mentioned tosuggest trying out the truly superior programs if you haven't done soalready.The very best mathematics textbook series in the world, as best I canascertain, is the Hua Loo-keng School Mathematics Textbook series publishedin China. I may have to turn that series into English to give Americanstudents an opportunity to learn from the very best. The Hua Loo-keng seriesmakes the curriculum expectations of the EPGY series look like slow learnerexpectations. The Hua Loo-keng School series doesn't just go faster but alsodeeper.Suggestion 7: Get involved in competition culture for a reality check onhow your child is doing in math. There is a great variety of mathematicscompetition programs these days, unlike the days when I went to school, withmany programs of differing characteristics. See what local math competitionsthere are in your area, and what the requirements are for forming a team orjoining as an individual. The American Mathematics Competition programshttp://www.unl.edu/amc/are readily available worldwide and start at the prealgebra level and go upto qualifying tests for the International Mathematics Olympiad. The bettermath competitions (MATHCOUNTShttp://www.mathcounts.org/is also in this category) are an excellent reality check on how much mathyour child knows as second nature.Suggestion 8: The talent search testshttp://www.ditd.org/floater.php?location=240http:// www.hoagiesgifted.org/talent_search.htmare another way to get a reality check on a child's math level. The talentsearch tests are typically a standardized achievement test normed for oneage group and given to a younger age group. Most of the regional talentsearch centers will give you DETAILED information about where your childstands in test performance ranking compared to other children who show up totake the test. The tests vary in format, and thus check whether yourcurriculum is developing a well-rounded approach to solving (simple)mathematical problems.Hope this helps! Best wishes to your child.-- Karl M. Bunday Christ has set us free. Galatians 5:1Learn in Freedom (TM) http://learninfreedom.org/-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: > have been preparing...students. Do you have it up on the web, so that I can point to iteasily?Can you explain in more detail why you prefer Miquon to the SingaporeMath series as a first series? (I've been working with my son on theSingapore series, but I did not look very hard at Miquon.)-- Kevin Karplus karplus@soe.ucsc.edu http://www.soe.ucsc.edu/~karpluslife member (LAB, Adventure Cycling, American Youth Hostels)Effective Cycling Instructor #218-ck (lapsed)Professor of Computer Engineering, University of California, Santa CruzUndergraduate and Graduate Director, BioinformaticsAffiliations for identification only.-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: > Don't feel dumb. Many of my students haven't understood fractions because> their teachers didn't. So, its not your fault.That makes sense.> First, a fraction is a way of describing a quantity like part of a pie.> Think about cutting the pie into 2,3,4,... equal sized slices. This is the> number we call the denominator. So, 1/4 is one of the pieces from a pie> cut into 4 equal sized slices and 3/4 would be 3 of them.>> The basic rule is that we need to think about just one size slice at atime.> So, if we are working with 7ths for example, all of the pieces of pie have> to be described as 7ths, ie, a number of slices selected from the 7 equal> slices of a whole pie.>> When someone asks you to add or subtract fractions with different> denominators, you remember that you can'tYes you can. Any two rational numbers (fractions) can be added, no matterwhat denominators they have, with a simple algorithma/b + c/d = (ad+cb)/bdthat is fifth-grade knowledge in several parts of the world, but which seemsto be unknown to most school mathematics textbook authors in the UnitedStates (and thus to most schoolteachers in the United States).Seehttp://math.berkeley.edu/~wu/looking for links on that page referring to fractions, for much more on thissubject.> so you look for a common> denominator.Looking for a common denominator is intuitive, and maybe the first thingto teach to a beginner in most cases, but it is not strictly necessary.[snip]> I don't understand fractions at all.adding,subtracting,none of it.> can you help me...i feel so dumb..thanks christyThe way that fractions are explained in most United States schools couldmake anyone feel dumb. Again, seehttp://math.berkeley.edu/~wu/for the path to enlightenment.Hope this helps!-- Karl M. Bunday Christ has set us free. Galatians 5:1Learn in Freedom (TM) http://learninfreedom.org/-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html