mm-1679 === Subject: Re: Precalculus Help! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0HDbN029490; i need help. I am doing Trigonometric Functions and i don't understand.Im trying to evalute the TM function using its period === Subject: Re: Precalculus Help! > i need help. I am doing Trigonometric Functions and i don't > understand.Im trying to evalute the TM function using its period What's the TM function? -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: more perfect numbers then thought by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0HDbsP29630; 2^0+2^1+(2^2-2^0)=(2^3-2^1) This here is accepted as a perfect number. The next number is 2^0+2^1+2^2+(2^3-2^0)+(2^4-2^1)=(2^5-2^2)This number is 28.One might say now ,ok the next one is 496,right?USING THE (2^6-2^2)= (2^7-2^3) which is 120. i shall restate here the terms 1+2+4+8+15+30+60=120 .Any comments? === Subject: Re: more perfect numbers then thought > 2^0+2^1+(2^2-2^0)=(2^3-2^1) This here is accepted as a perfect number. > The next number is 2^0+2^1+2^2+(2^3-2^0)+(2^4-2^1)=(2^5-2^2)This > number is 28.One might say now ,ok the next one is 496,right?USING THE > (2^6-2^2)= (2^7-2^3) which is 120. i shall restate here the terms > 1+2+4+8+15+30+60=120 .Any comments? 3 times 40 is 120. -- Larry Lard Replies to group please === Subject: Re: more perfect numbers then thought by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0HG0W910629; >> 2^0+2^1+(2^2-2^0)=(2^3-2^1) This here is accepted as a perfect >number. >> The next number is 2^0+2^1+2^2+(2^3-2^0)+(2^4-2^1)=(2^5-2^2)This >> number is 28.One might say now ,ok the next one is 496,right?USING >THE >> (2^6-2^2)= (2^7-2^3) which is 120. i shall restate here the terms >> 1+2+4+8+15+30+60=120 .Any comments? >3 times 40 is 120. >-- >Larry Lard >Replies to group please Mr Lard you can do better then that . === Subject: Re: more perfect numbers then thought by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0HHeGk19377; > 2^0+2^1+(2^2-2^0)=(2^3-2^1) This here is accepted as a perfect >>number. > The next number is 2^0+2^1+2^2+(2^3-2^0)+(2^4-2^1)=(2^5-2^2)This > number is 28.One might say now ,ok the next one is 496,right?USING >>THE > (2^6-2^2)= (2^7-2^3) which is 120. i shall restate here the terms > 1+2+4+8+15+30+60=120 .Any comments? >>3 times 40 is 120. >>-- >>Larry Lard >>Replies to group please >Mr Lard you can do better then that . So can you: observe that you missed the factors 3 and 40, which is what LL is pointing out to you. Do all the proper factors of 120 add to 120? You're right that all *even* perfect numbers are necessarily of the form 2^(2p-1) - 2^(p-1) = 2^(p-1)(2^p - 1). But not all numbers of this form are perfect; the criterion is that 2^p - 1 be prime. Todd Trimble === Subject: Re: more perfect numbers then thought by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0HIXCs24263; >> 2^0+2^1+(2^2-2^0)=(2^3-2^1) This here is accepted as a perfect >number. >> The next number is 2^0+2^1+2^2+(2^3-2^0)+(2^4-2^1)=(2^5-2^2)This >> number is 28.One might say now ,ok the next one is 496,right?USING >THE >> (2^6-2^2)= (2^7-2^3) which is 120. i shall restate here the >terms >> 1+2+4+8+15+30+60=120 .Any comments? >3 times 40 is 120. >-- >Larry Lard >Replies to group please >>Mr Lard you can do better then that . >So can you: observe that you missed the factors 3 and 40, >which is what LL is pointing out to you. Do all the proper >factors of 120 add to 120? >You're right that all *even* perfect numbers are necessarily >of the form 2^(2p-1) - 2^(p-1) = 2^(p-1)(2^p - 1). But not >all numbers of this form are perfect; the criterion is that >2^p - 1 be prime. >Todd Trimble yes yes mr trimble i see the error of my thinking.Ok just to save sum face i could call these semiperfect. have a nice day === Subject: ARCHIMEDES AND FERMAT LAST THEOREM by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0HDbLA29413; HI MO, HERE is the begining ofARCHIMEDES PROOF OF LAST THEOREM: EQF: X^n+Y^n=Z^n where Z,X,Y are integers and n>3a prime numbe isimpossible. We choose Z = not divisible by n X+Y-Z=B 2*B+Q+P=X+Y B+Q+P=Z EQF implies that 2*B+Q+P=U^n and Z=B+Q+P=U*V ( ifZ is not divisible by n) B=U^n-U*V ThereforeB=b*U and Q+P=U*K NOWwe write the EQF as : EQ1: (Z-Q)^n +(Z-P)^n=Z^n EQ2: {Z^(n-1) -(a1)*Z^(n-2)+ ...........+D*(Z^2)-c*Z+(Q^n+P^n)/(Q+P)=V^ WE seethat (from EQ2) weget that (Q^n+P^n) is didisible by Z c={n*Q^(n-1)+n*P^(n-1) -2*[Q^n+P^n}/(Q+P)}/(Q+P) and D={(1/2)*n*(n-1)*[Q+P)^(n-2)-Q^(n-2)-P^(n-2)]-2*c/(Q+P) NOWwe multiply EQ2 with (Q+P) and then we substract from both sides of equation thevalue [2*Z*V^n +Z^n] andgetEQ3 IF the EQ1 andEQ 2are reproduced in EQ3 then wecan dividethe equation by Z, get a equation whereZ has a lower power than (n-1) ,and get that A=[(Q+P)^(n-2)]-2[Q^(n-2)+P^(n-2)] is divisible by Z Here I stop.Why? Because till now i made the calculations only in mind and I need to take the case n=5 to see where I am heading with the proof.May be using a EUCLID polinomial Algorytm Your input is welcome. george ghiata === Subject: Re: Your baby was damned to death, but who made the choice? >Nothing to do with any of the newsgroups posted to. It does have something to do with the math and physics groups. He said abortion is a sin, which is a trigonometric function. Of course he could be wrong, abortion could be a cos or a tan, who knows. === Subject: Re: Your baby was damned to death, but who made the choice? >>Nothing to do with any of the newsgroups posted to. > It does have something to do with the math and physics groups. He > said abortion is a sin, which is a trigonometric function. Of course > he could be wrong, abortion could be a cos or a tan, who knows. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + But the sin referred to by the original poster does not have an inverse. Aristotle Polonium + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + === Subject: proof? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0HGEK312284; the number 120 is a product of 1*120,2*60,4*30,8*15,4*30 .Now i add 1+2+4+8+15+30+60=120 Does this not qualify the definition for a perfect number? The same line of reasoning can be applied to the number 2016. 1*2016,2*1008,4*504,8*252,16*126,32*63 .now i add 1+2+4+8+16+32+63+126+252+504+1008=2016 .Does this not also qualify? === Subject: Re: proof? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0HHeGG19401; >the number 120 is a product of 1*120,2*60,4*30,8*15,4*30 .Now i add >1+2+4+8+15+30+60=120 Does this not qualify the definition for a >perfect number? The definition is that *all* proper factors add to the given number. You omitted the factors 3 and 40. Oh, also 6 and 20. This has already been explained elsewhere. The same line of reasoning can be applied to the >number 2016. 1*2016,2*1008,4*504,8*252,16*126,32*63 .now i add >1+2+4+8+16+32+63+126+252+504+1008=2016 .Does this not also qualify? Well, let's see. 2016 = 32*9*7. You omitted the factors 3, 7, 9, 21, ... shall I go on? Todd Trimble === Subject: Re: proof? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0HINIp23459; >>the number 120 is a product of 1*120,2*60,4*30,8*15,4*30 >.Now i add >>1+2+4+8+15+30+60=120 Does this not qualify the definition for a >>perfect number? >The definition is that *all* proper factors add to the given >number. You omitted the factors 3 and 40. Oh, also 6 and 20. >This has already been explained elsewhere. >The same line of reasoning can be applied to the >>number 2016. 1*2016,2*1008,4*504,8*252,16*126,32*63 .now i add >>1+2+4+8+16+32+63+126+252+504+1008=2016 .Does this not also qualify? >Well, let's see. 2016 = 32*9*7. You omitted the factors >3, 7, 9, 21, ... shall I go on? >Todd Trimble Sir ,u r correct. I STAND CORRECTED. === Subject: Re: proof? > the number 120 is a product of 1*120,2*60,4*30,8*15,4*30. > Now i add 1+2+4+8+15+30+60=120 Does this not qualify the > definition for a perfect number? The same line of reasoning can > be applied to the number 2016. > 1*2016,2*1008,4*504,8*252,16*126,32*63 .now i add > 1+2+4+8+16+32+63+126+252+504+1008=2016. > Does this not also qualify? A useful result from Euclid and Fermat is: The number (2^(n-1))(2^n - 1) is perfect if and only if (2^n - 1) is prime. Note: 2^n - 1 = 2^(n-1)+2^(n-2)+...+2+1. === Subject: Re: proof? >the number 120 is a product of 1*120,2*60,4*30,8*15,4*30 .Now i add >1+2+4+8+15+30+60=120 Does this not qualify the definition for a >perfect number? The same line of reasoning can be applied to the >number 2016. 1*2016,2*1008,4*504,8*252,16*126,32*63 .now i add >1+2+4+8+16+32+63+126+252+504+1008=2016 .Does this not also qualify? sum of its divisors except itself. The divisors of 120 are 1,2,3,4,5,6,8,10,12,15,20,24,30,40 and 60. These sum to 240, not 120, but, interestingly, 2x120. jps jps === Subject: Re: proof? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0HINIJ23455; >>the number 120 is a product of 1*120,2*60,4*30,8*15,4*30 .Now i add >>1+2+4+8+15+30+60=120 Does this not qualify the definition for a >>perfect number? The same line of reasoning can be applied to the >>number 2016. 1*2016,2*1008,4*504,8*252,16*126,32*63 .now i add >>1+2+4+8+16+32+63+126+252+504+1008=2016 .Does this not also qualify? >> the >sum of its divisors except itself. >The divisors of 120 are 1,2,3,4,5,6,8,10,12,15,20,24,30,40 and 60. >These sum to 240, not 120, but, interestingly, 2x120. others. >jps >jps Sir,you are CORRECT and i stand corrected.thank u === Subject: Re: Ramanujan's Constant (e^pi, Sqrt(163)) And Its Cousins by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0HIg8F24894; >Hello all, >If you are curious about e^(pi*Sqrt(163)) and other numbers like it, >this paper might be of interest: href=http://www.geocities.com/titus_piezas/Ramanujan_a.htm>http://www.ge ocities.com/titus_piezas/Ramanujan_a.htm >-Titus >(tpiezasIII@uap.edu.ph -> remove III for email) e is an irrational number.pi is also an irrationl number.To involve both in an EQUATION will not yield an equal ,but an APPROXIMATION .Im not going to apologize to anyone for my total lack of interest in these APPROXIMATIONS.If its not an exact number its not really a number as i understand numbers.Lets get to the nitty gritty to even write down an infinite number would take infinity to complete the task. === Subject: Re: Ramanujan's Constant (e^pi, Sqrt(163)) And Its Cousins >>Hello all, >>If you are curious about e^(pi*Sqrt(163)) and other numbers like it, >>this paper might be of interest: >> href=http://www.geocities.com/titus_piezas/Ramanujan_a.htm>http://www.geo cities.com/titus_piezas/Ramanujan_a.htm >>-Titus >>(tpiezasIII@uap.edu.ph -> remove III for email) > e is an irrational number.pi is also an irrationl number.To involve > both in an EQUATION will not yield an equal ,but an APPROXIMATION .Im > not going to apologize to anyone for my total lack of interest in > these APPROXIMATIONS.If its not an exact number its not really a > number as i understand numbers.Lets get to the nitty gritty to even > write down an infinite number would take infinity to complete the > task. Mathematica 4.0 gives: 2.625374126407687439999999999992500725971981856888793538563373x10^17 to 50 decimal places -- -- Geo. Michael Henry No! Bad dog! I said sit! anonymous === Subject: Re: Diophantine approximation or something? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0HIsd626052; >Maybe this can be considered a cute bit of recreational math? > Every schoolchild knows that 5, 12, 13 is a Pythagorean >triple. Accordingly, (5/13, 12/13) is a point on the >proverbial unit circle. Call it (cos(t), sin(t)). Applying >double angle formulas for sine and cosine we get that >sin(4t) = 28560/28561 exactly; the numerator and denominator >in this fraction differ by only 1. So 45 falls less than >half-a-degree short of a right angle. This seems like an >unexpectedly good approximation for a starting denominator >of only 13. Is some lesson about Diophantine approximations >or some such thing exemplified here? -- Mike Hardy Mr.Hardy you have an interesting point.Both 12/13 and 5/13 have as their demoninators a prime number,so by attempting to rewrite as a decimal would be foolish.Also,the folling is also true 5^4 + 12^4 + 5*12 + 5*12 = 13^4 === Subject: Re: Order of a recurrence relation > I know how to calculate the order of T(n)=2T(n/2) + 1. This i can do > it by T(n)=4T(n/4) + 2 +1 ie T(n)=2^i*T(n/2^i) + sum(2^i-1 to 1). > Now i have a recurrence relation T(n)=T(an)+T(bn) + cn where a+b=1 > and T(1)=1. > Now how do i solve this recurrence relation. Well, here's my attempt, which yields a solution for a>0, b>0, n>0. The procedure is not very systematic or generalisable I'm afraid. To get an idea of the likely form of the solution, I started by looking at the simple case a=1/2, b = 1/2, c=1. I.e. solve T(n) = 2T(n/2) + n Starting with T(1) = 1, it's fairly clear that T(2) = 2*2^1 T(4) = 3*2^2 T(8) = 4*2^3 ... and, in general T(2^i) = (i+1)*2^i Setting n = 2^i we get the answer T(n) = (log_2(n) + 1)*n = n*log_2(n) + n might be of the form T(n) = p*n*log(n) + q*n with p and q to be found. The condition T(1) = 1 immediately gives that q = 1, so we're left with T(n) = p*n*log(n) + n with p to be found Plugging this into T(n) = T(an) + T(bn) + cn gives p*n*log(n) + n = p*a*n*log(a*n) + a*n + p*b*n*log(b*n) + b*n + c*n = p*n*log(n)*(a + b) + n*(a + b + p*a*log(a) + p*b*log(b) + c) = p*n*log(n) + n*(1 + p*a*log(a) + p*b*log(b) + c) (since a + b = 1) The term p*n*log(n) has come right, so it remains to set 1 + p*a*log(a) + p*b*log(b) + c = 1 which gives p = -c/(a*log(a) + b*log(b)) and the final solution is therefore T(n) = -c/(a*log(a) + b*log(b))*n*log(n) + n Unfortunately this only works for a>0, b>0, n>0. I'm not sure what happens outside this range. === Subject: Re: Order of a recurrence relation > I know how to calculate the order of T(n)=2T(n/2) + 1. This i can do > it by T(n)=4T(n/4) + 2 +1 ie T(n)=2^i*T(n/2^i) + sum(2^i-1 to 1). > Now i have a recurrence relation T(n)=T(an)+T(bn) + cn where a+b=1 > and T(1)=1. > Now how do i solve this recurrence relation. > Well, here's my attempt, which yields a solution for a>0, b>0, n>0. The > procedure is not very systematic or generalisable I'm afraid. > To get an idea of the likely form of the solution, I started by looking > at the simple case a=1/2, b = 1/2, c=1. I.e. solve > T(n) = 2T(n/2) + n > Starting with T(1) = 1, it's fairly clear that > T(2) = 2*2^1 > T(4) = 3*2^2 > T(8) = 4*2^3 > ... > and, in general > T(2^i) = (i+1)*2^i > Setting n = 2^i we get the answer > T(n) = (log_2(n) + 1)*n > = n*log_2(n) + n c > might be of the form > T(n) = p*n*log(n) + q*n > with p and q to be found. > The condition T(1) = 1 immediately gives that q = 1, so we're left with > T(n) = p*n*log(n) + n > with p to be found > Plugging this into T(n) = T(an) + T(bn) + cn gives > p*n*log(n) + n = p*a*n*log(a*n) + a*n + p*b*n*log(b*n) + b*n + c*n > = p*n*log(n)*(a + b) + n*(a + b + p*a*log(a) + p*b*log(b) + c) > = p*n*log(n) + n*(1 + p*a*log(a) + p*b*log(b) + c) > (since a + b = 1) > The term p*n*log(n) has come right, so it remains to set > 1 + p*a*log(a) + p*b*log(b) + c = 1 > which gives > p = -c/(a*log(a) + b*log(b)) > and the final solution is therefore > T(n) = -c/(a*log(a) + b*log(b))*n*log(n) + n > Unfortunately this only works for a>0, b>0, n>0. I'm not sure what > happens outside this range. Actually, looking again I think this formula does extend OK to any non-zero a, b and n. Just needs rewriting as: T(n) = -c/(a*log|a| + b*log|b|)*n*log|n| + n === Subject: Re: Order of a recurrence relation Posting postscripts to my postscripts .... can't be good! Purely for the record, my statement that this formula works for any non-zero a, b and n kind of implies that I've discarded the original condition a+b = 1. I haven't. a+b=1 is still required. === Subject: Re: Order of a recurrence relation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0HDbK229384; I will reply in two parts; -short term : write it T(2n)=2T(n)+1 (1) you can solve T(2n)= 2T(n) T(n)=a*n {a} constant, for 2T(n)+1 add an other constant so T(n)=a*n+b identification LHS and RHS within (1) gives 2an + b=2an+2b+1 ,b=-1 . T(n)=an-1 , a any constant a = 0 yields the constant function T(n)=-1 . -long term* : solve a functional equation: f(2x)=2*f(x)+1 as a process bound to transformation x->2x, when x->2x f->2*f + 1 ;each transformation x->2x is counted by phi(x)=ln(x)/ln(2) to a constant c' (observe phi(2x) ! ) so the L-iterated solution is f(x)=(2L+1)^[phi(x)] (2) computing (2) gives f(x)=(2L+1)^[ln(cx)/ln(2)] = c*x(L+1)-1 ,here we must consider L constant. so f(x)= a*x-1 , a=0 ... *long term means openings towards other mathematical realms... Friendly,Alain. === Subject: No positive integers less than on by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0HKvXY03410; I have been assigned to prove that there are no positive integers less than one. My teacher gave the hint to use the well-ordering principle. Here is what I have so far and where I am stuck: Let x be a pos. int. less than one, so that 0=x. So, 0 I have been assigned to prove that there are no positive integers less > than one. My teacher gave the hint to use the well-ordering principle. HINT 0 < n < 1 => 0 < nn < n < 1 The same type of descent can be used in irrationality proofs, e.g. see my prior post [1]. One is essentially exploiting the fact that rings of (algebraic) integers are DISCRETE, whereas the corresponding rings of (algebraic) fractions are DENSE. For much more see [1]. --Bill Dubuque === Subject: Re: No positive integers less than on days. My association with the Department is that of an alumnus. >I have been assigned to prove that there are no positive integers less >than one. My teacher gave the hint to use the well-ordering >principle. Here is what I have so far and where I am stuck: >Let x be a pos. int. less than one, so that 0By well ordering, any other pos. ints less than one are greater than >or equal to x. This is very poorly phrased. Remember what the well ordering principle ->is<-: the well ordering principle says that Any nonempty subset of the positive integers has a least element. So, say P is a property that positive integers may or may not have. In general, if you want to use the well ordering principle to show that no positive integer has property P, one proceeds by contradiction: you assume that there is at least one positive integer which has property P; that means that the set S = { x : x is a positive integer and x has property P} is nonempty; by the well ordering principle S must have a smallest element. Then you attempt to derive a contradiction from this (the most common way to do so is to show that there would have to be an even smaller element in the set). So: you want to prove that there are no positive integers less than one. So let S = { x: x is a positive integer and x<1} We want to show that S is empty. If S were nonempty, then it would have (by the well ordering principle) a SMALLEST element. Call that element x. That means that if y is any positive integer with y<1, then it follows that y>=x. >Let y be a pos. int. less than one, so that 0=x. >So, 0simplifies to 0Case 1: 2 divides x+y. If this is true, then by rules of >divisibility, 2 divides x and 2 divides y. If 2 divides x, then there >exists a positive integer z less than x, which contradicts the assumed >well-ordering principle. >Case 2: 2 does not divide x+y. The (x+y)/2 can't be an integer >because it can only be written in the form of this fraction..... >... and here is where I am stuck; I don't know what this tells me >about x. I can't get a solid contradiction from this case. Exactly. >Any insight into my proof or hints for another route would be grately >appreciated!!! HINT: For arbitrary positive integers n and m, n-m is a positive integer IF AND ONLY IF m>I have been assigned to prove that there are no positive integers less >>than one. My teacher gave the hint to use the well-ordering >>principle. Here is what I have so far and where I am stuck: >>Let x be a pos. int. less than one, so that 0>By well ordering, any other pos. ints less than one are greater than >>or equal to x. >This is very poorly phrased. Remember what the well ordering principle >->is<-: the well ordering principle says that Any nonempty subset of >the positive integers has a least element. >So, say P is a property that positive integers may or may not have. In >general, if you want to use the well ordering principle to show that >no positive integer has property P, one proceeds by contradiction: >you assume that there is at least one positive integer which has >property P; that means that the set > S = { x : x is a positive integer and x has property P} >is nonempty; by the well ordering principle S must have a smallest >element. Then you attempt to derive a contradiction from this (the >most common way to do so is to show that there would have to be an >even smaller element in the set). >So: you want to prove that there are no positive integers less than >one. So let > S = { x: x is a positive integer and x<1} >We want to show that S is empty. >If S were nonempty, then it would have (by the well ordering >principle) a SMALLEST element. Call that element x. That means that >if y is any positive integer with y<1, then it follows that y>=x. >>Let y be a pos. int. less than one, so that 0=x. >>So, 0>simplifies to 0Bad, bad, bad, bad. You do not know that (x+y)/2 is a positive >integer, because in general you cannot divide in the positive >integers. At this point, you're lost. There is no way to salvage your >argument that I can see. >>Case 1: 2 divides x+y. If this is true, then by rules of >>divisibility, 2 divides x and 2 divides y. If 2 divides x, then there >>exists a positive integer z less than x, which contradicts the assumed >>well-ordering principle. >>Case 2: 2 does not divide x+y. The (x+y)/2 can't be an integer >>because it can only be written in the form of this fraction..... >>... and here is where I am stuck; I don't know what this tells me >>about x. I can't get a solid contradiction from this case. >Exactly. >>Any insight into my proof or hints for another route would be grately >>appreciated!!! >HINT: For arbitrary positive integers n and m, > n-m is a positive integer IF AND ONLY IF mNow, you have x, the smallest positive integer smaller than 1. Can >you come up with an even smaller positive integer which would >necessarily have that property? >-- >It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) >Arturo Magidin >magidin@math.berkeley.edu It would seem according to Peano's Postulates that 1 is the smallest positive integer by definition. Peano's Postulates: 1) 1 is a natural number (positive integer) 2) if a is a natural number then a+ (the successor of a) is a natural number. 3) There is no natural number a such that a+ = 1. (This postulate defines 1 to be the smallest natural number.) 4) If a+ = b+ then a = b. 5) If 1 is a member of a set S, and a in S implies a+ in S then S is the set of natural numbers. - MO === Subject: Re: No positive integers less than on days. My association with the Department is that of an alumnus. >It would seem according to Peano's Postulates that 1 is the smallest >positive integer by definition. >Peano's Postulates: >1) 1 is a natural number (positive integer) >2) if a is a natural number then a+ (the successor of a) is a natural >number. >3) There is no natural number a such that a+ = 1. >(This postulate defines 1 to be the smallest natural number.) No, it does not, because the postulate does not define any ordering among the natural numbers. We can define the integers less than or equal to 1 by saying: (i) 1 is an integer less than or equal to 1; (ii) If a is an integer less than or equal to 1, then a+ (the sucessor of a) is an integer less than or equal to 1. (iii) There is no integer less than or equal to one a such that a+ = 1. (iv) If a+ = b+ then a=b. (v) If S is a subset of the integers less than or equal to 1 such that (1) 1 is in S; and (2) For every integer less than or equal to 1, a, a in S -> a+ in S then S is the set of all integers less than or equal to 1. Does that mean that 1 is the smallest integer less than or equal to 1? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: No positive integers less than on by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0KE4LD10570; >>It would seem according to Peano's Postulates that 1 is the smallest >>positive integer by definition. >>Peano's Postulates: >>1) 1 is a natural number (positive integer) >>2) if a is a natural number then a+ (the successor of a) is a natural >>number. >>3) There is no natural number a such that a+ = 1. >>(This postulate defines 1 to be the smallest natural number.) >No, it does not, because the postulate does not define any ordering >among the natural numbers. >We can define the integers less than or equal to 1 by saying: > (i) 1 is an integer less than or equal to 1; > (ii) If a is an integer less than or equal to 1, then a+ (the > sucessor of a) is an integer less than or equal to 1. >(iii) There is no integer less than or equal to one a such that a+ = > 1. > (iv) If a+ = b+ then a=b. > (v) If S is a subset of the integers less than or equal to 1 such > that > (1) 1 is in S; and > (2) For every integer less than or equal to 1, a, > a in S -> a+ in S > then S is the set of all integers less than or equal to 1. >Does that mean that 1 is the smallest integer less than or equal to 1? >-- >It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) >Arturo Magidin >magidin@math.berkeley.edu You are right. Peano's Postulates do not impose an order on the naturals. So how does one impose an order on the naturals? Does one simply say that a+ > a for all a in N? - MO === Subject: Re: No positive integers less than on days. My association with the Department is that of an alumnus. [.snip.] >You are right. Peano's Postulates do not impose an order on the >naturals. So how does one impose an order on the naturals? Does >one simply say that a+ > a for all a in N? I'm sure there are other ways of doing so; but the way we did it when I learned was after some work. First, define addition recursively by letting (using your notation; I actually had the Naturals starting at 0) x + 1 = x+; x + (y+) = (x+y)+. for each x. Then you define a > b if and only if there exists x such that b+x = a. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: No positive integers less than on by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0KJkMo11190; > [.snip.] >>You are right. Peano's Postulates do not impose an order on the >>naturals. So how does one impose an order on the naturals? Does >>one simply say that a+ > a for all a in N? >I'm sure there are other ways of doing so; but the way we did it when >I learned was after some work. >First, define addition recursively by letting (using your notation; I >actually had the Naturals starting at 0) > x + 1 = x+; > x + (y+) = (x+y)+. >for each x. >Then you define > a > b if and only if there exists x such that b+x = a. >-- >It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) >Arturo Magidin >magidin@math.berkeley.edu I am pretty sure that the definition you give above for the ordering is equivalent to a+ > a. Here are some more thoughts. When I learned Peano's Postulates (from Landau) 0 was not included. Thus Peano's Postulates define the naturals (1, 1+ = 2, 2+ = 3,...). Peano's Postulate 3 (there is no natural a such that a+ = 1) and a+ > a, we should be able to get to 1 is the first number pretty quickly and very nearly by definition. - MO === Subject: Re: No positive integers less than on days. My association with the Department is that of an alumnus. >> [.snip.] >You are right. Peano's Postulates do not impose an order on the >naturals. So how does one impose an order on the naturals? Does >one simply say that a+ > a for all a in N? >>I'm sure there are other ways of doing so; but the way we did it when >>I learned was after some work. >>First, define addition recursively by letting (using your notation; I >>actually had the Naturals starting at 0) >> x + 1 = x+; >> x + (y+) = (x+y)+. >>for each x. >>Then you define >> a > b if and only if there exists x such that b+x = a. >I am pretty sure that the definition you give above for the ordering >is equivalent to a+ > a. Provided you take this relation and extend it transitively, yes; the definition I gave above yields a relation which is already transitive. The definition you gave does not. >Here are some more thoughts. >When I learned Peano's Postulates (from Landau) 0 was not included. Yes; that's up to you. I learned it based on Set Theory, so 0 is a more natural starting point. In addition, the weak inequality is considered more basic for people coming from set theory than the strict inequality, so if you allow 0 and use the same definition I gave, you get the weak inequality instead of the strict one (that is, a binary relation which is reflexive, antisymmetric, and transitive). >Thus Peano's Postulates define the naturals (1, 1+ = 2, 2+ = 3,...). >Peano's Postulate 3 (there is no natural a such that a+ = 1) and >a+ > a, we should be able to get to 1 is the first number pretty >quickly and very nearly by definition. Yes; but the Original Poster was asked to prove it using the Well Ordering principle. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: No positive integers less than on by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0KMr5608396; > [.snip.] >>You are right. Peano's Postulates do not impose an order on the >>naturals. So how does one impose an order on the naturals? Does >>one simply say that a+ > a for all a in N? >I'm sure there are other ways of doing so; but the way we did it when >I learned was after some work. >First, define addition recursively by letting (using your notation; I >actually had the Naturals starting at 0) > x + 1 = x+; > x + (y+) = (x+y)+. >for each x. >Then you define > a > b if and only if there exists x such that b+x = a. >>I am pretty sure that the definition you give above for the ordering >>is equivalent to a+ > a. >Provided you take this relation and extend it transitively, yes; the >definition I gave above yields a relation which is already >transitive. The definition you gave does not. Yes, I now see that my definition also needs transitivity added to it to fully define a well ordering among the naturals. Your definition, which seems to be the standard defintion (at least I was able to find it in a couple of my set theory books) has transitivity built into it. >>Here are some more thoughts. >>When I learned Peano's Postulates (from Landau) 0 was not included. >Yes; that's up to you. I learned it based on Set Theory, so 0 is a >more natural starting point. In addition, the weak inequality is >considered more basic for people coming from set theory than the >strict inequality, so if you allow 0 and use the same definition I >gave, you get the weak inequality instead of the strict one (that is, >a binary relation which is reflexive, antisymmetric, and transitive). >>Thus Peano's Postulates define the naturals (1, 1+ = 2, 2+ = 3,...). >>Peano's Postulate 3 (there is no natural a such that a+ = 1) and >>a+ > a, we should be able to get to 1 is the first number pretty >>quickly and very nearly by definition. >Yes; but the Original Poster was asked to prove it using the Well >Ordering principle. >-- >It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) >Arturo Magidin >magidin@math.berkeley.edu === Subject: Re: No positive integers less than on days. My association with the Department is that of an alumnus. >I have been assigned to prove that there are no positive integers >less >than one. My teacher gave the hint to use the well-ordering >principle. [.snip.] >It would seem according to Peano's Postulates that 1 is the smallest >positive integer by definition. >Peano's Postulates: >1) 1 is a natural number (positive integer) >2) if a is a natural number then a+ (the successor of a) is a natural >number. >3) There is no natural number a such that a+ = 1. >(This postulate defines 1 to be the smallest natural number.) >4) If a+ = b+ then a = b. >5) If 1 is a member of a set S, and a in S implies a+ in S then S is >the set of natural numbers. Could you please point out to me where in the Peano axioms the ordering of the positive integers is defined? Because if 1 is the smallest by definition, then surely the Peano axioms must include a definition of ordering. You know, something that says the natural number n is smaller than the natural number m if and only if XXXX. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: No positive integers less than on > I have been assigned to prove that there are no positive integers less > than one. My teacher gave the hint to use the well-ordering > principle. Here is what I have so far and where I am stuck: > Let x be a pos. int. less than one, so that 0 By well ordering, any other pos. ints less than one are greater than > or equal to x. > Let y be a pos. int. less than one, so that 0=x. > So, 0 simplifies to 0 Case 1: 2 divides x+y. If this is true, then by rules of > divisibility, 2 divides x and 2 divides y. If 2 divides x, then there > exists a positive integer z less than x, which contradicts the assumed > well-ordering principle. 3+3 is divisible by 2, yet 3 is not divisible by 2. > Case 2: 2 does not divide x+y. The (x+y)/2 can't be an integer > because it can only be written in the form of this fraction..... > ... and here is where I am stuck; I don't know what this tells me > about x. I can't get a solid contradiction from this case. > Any insight into my proof or hints for another route would be grately > appreciated!!! > Alej -- Will Twentyman email: wtwentyman at copper dot net === Subject: Orthogonality of bessel functions by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0HLA7E04767; How to prove the orthogonality for bessel function J(x) of order p?? when J(k)=0 I want to find integral from 0 to k for xJ(x)J(x) where the order is p can anyone provide solution using recurrence relation of bessel functions? === Subject: Re: Orthogonality of bessel functions > How to prove the orthogonality for bessel function J(x) of order p?? > when J(k)=0 > I want to find > integral from 0 to k for > xJ(x)J(x) where the order is p > can anyone provide solution using recurrence relation of bessel > functions? Must the proof be by using the recurrence relation? If not, how about this? Let a and b be zeroes of J_p. Write u for J_p(ax), and v for J_p(bx) then x(xu')' + (a^2 x^2 - p^2)u = 0 . . . . . . . . . . . . . (1) x(xv')' + (b^2 x^2 - p^2)v = 0 . . . . . . . . . . . . . (2) get v(xu')' - u(xv')' + (a^2 - b^2)xuv = 0 . . . . . . . . . (3) The first two terms of (3) equals d(vxu' - uxv')/dx Use this to integrate (3) to get (vxu' - uxv')|^1_0 + (a^2 - b^2)int^1_0 xuv dx = 0 . . (4) At the lower limit (a^2 - b^2)int^1_0 xuv dx = 0 because x = 0 and u, v, u' and v' are finite. For the upper limit note that at x = 1, u = J_p(a) = 0, v = J_p(b) = 0 so the integral is 0 at the upper limit. So (4) is (a^2 - b^2)int^1_0 xuv dx = 0 or (a^2 - b^2)int^1_0 xJ_p(ax)J_p(bx) dx = 0 . . . . . . . (5) If a and b are distinct zeroes of J_p the integral = 0. If a = b the integral is not 0 and can be evaluated as follows. Let a be a root and b any root then from (4) get int^1_0 xuv dx = (J_p(b)*a*J'_p(a))/(b^2 - a^2) Find the limit as b --> a by the much-reviled l'Hospital's rule to get int^1_0 xuv dx = (1/2)(J'_p(a))^2 for a = b which is non-zero. === Subject: Re: Orthogonality of bessel functions > How to prove the orthogonality for bessel function J(x) of order p?? > when J(k)=0 > I want to find > integral from 0 to k for > xJ(x)J(x) where the order is p > can anyone provide solution using recurrence relation of bessel > functions? > Must the proof be by using the recurrence relation? If not, how about > this? > Let a and b be zeroes of J_p. > Write u for J_p(ax), and v for J_p(bx) then > x(xu')' + (a^2 x^2 - p^2)u = 0 . . . . . . . . . . . . . (1) > x(xv')' + (b^2 x^2 - p^2)v = 0 . . . . . . . . . . . . . (2) > get > v(xu')' - u(xv')' + (a^2 - b^2)xuv = 0 . . . . . . . . . (3) > The first two terms of (3) equals > d(vxu' - uxv')/dx > Use this to integrate (3) to get > (vxu' - uxv')|^1_0 + (a^2 - b^2)int^1_0 xuv dx = 0 . . (4) > At the lower limit (a^2 - b^2)int^1_0 xuv dx = 0 because x = 0 and u, > v, u' and v' are finite. > For the upper limit note that at x = 1, u = J_p(a) = 0, v = J_p(b) = 0 > so the integral is 0 at the upper limit. > So (4) is (a^2 - b^2)int^1_0 xuv dx = 0 or > (a^2 - b^2)int^1_0 xJ_p(ax)J_p(bx) dx = 0 . . . . . . . (5) > If a and b are distinct zeroes of J_p the integral = 0. If a = b the > integral is not 0 and can be evaluated as follows. > Let a be a root and b any root then from (4) get Sorry, this should read: Let a be a root and b any number then from (4) get > int^1_0 xuv dx = (J_p(b)*a*J'_p(a))/(b^2 - a^2) > Find the limit as b --> a by the much-reviled l'Hospital's rule to get > int^1_0 xuv dx = (1/2)(J'_p(a))^2 for a = b > which is non-zero. === Subject: Re: Algebra, lines, equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0HMVmL10865; No, the slope is different. The first is 1/3, the second is 1/2. >The lines y= 1/3x + 2 >and >2y= x+6 ARE >----- >I say the lines are Parallel. >Correct? >-Ulie === Subject: What if the Nazis had kindness? Kind Nazis could have got the Jews to help them with the bomb. Kind Nazis would have got far more Russian collaborators and perhaps not have inspired the stubborn resistance of the Russians and the British in WWII. Kindness might have made the difference. And while they would not have won they would have concluded a beneficial peace treaty. === Subject: Re: What if the Nazis had kindness? > Kind Nazis could have got the Jews to help them with the bomb. Kind > Nazis would have got far more Russian collaborators and perhaps not > have inspired the stubborn resistance of the Russians and the British > in WWII. Kindness might have made the difference. And while they would > not have won they would have concluded a beneficial peace treaty. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + Were you a writer for Seinfeld? Aristotle Polonium + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + === Subject: Re: What if the Nazis had kindness? If the Nazis had kindness, they would've invented the KareBear. === Subject: Re: What if the Nazis had kindness? > Kind Nazis could have got the Jews to help them with the bomb. Kind > Nazis would have got far more Russian collaborators and perhaps not > have inspired the stubborn resistance of the Russians and the British > in WWII. Kindness might have made the difference. And while they would > not have won they would have concluded a beneficial peace treaty. You want to know? Look at Joey Stalin. Hitler was a totalitarian Stalin (NKVD) in persecuting other ethnic minorities and rounding up 10's of millions to slave and die in the gulag. But other Jews were persecuted by other communists. When Stalin was just about to purge Jews in the 50's, he got sick and died. No doubt his Jewish doctor was overjoyed, if not implicated =) The moral of the story is, it ain't about labels, its about gang warfare for the political power to dominate, degrade, tax and enslave others. Nevermind what letters are assigned to the variable (gang) names, examine the relationship between them when analyzing a social plasma. === Subject: Re: What if the Nazis had kindness? >Kind Nazis could have got the Jews to help them with the bomb. Kind >Nazis would have got far more Russian collaborators and perhaps not >have inspired the stubborn resistance of the Russians and the British >in WWII. Kindness might have made the difference. And while they would >not have won they would have concluded a beneficial peace treaty. Mhmm - Sorry, I don't think that equation can be solved analytically. Best approach would be to use some sort of numerical algorithm. Thomas === Subject: Re: What if the Nazis had kindness? >>Kind Nazis could have got the Jews to help them with the bomb. Kind >>Nazis would have got far more Russian collaborators and perhaps not >>have inspired the stubborn resistance of the Russians and the British >>in WWII. Kindness might have made the difference. And while they would >>not have won they would have concluded a beneficial peace treaty. > Mhmm - Sorry, I don't think that equation can be solved analytically. > Best approach would be to use some sort of numerical algorithm. > Thomas either that or (T)roll another joint... Bob === Subject: Re: What if the Nazis had kindness? >Kind Nazis could have got the Jews to help them with the bomb. Kind >Nazis would have got far more Russian collaborators and perhaps not >have inspired the stubborn resistance of the Russians and the British >in WWII. Kindness might have made the difference. And while they would >not have won they would have concluded a beneficial peace treaty. >> Mhmm - Sorry, I don't think that equation can be solved analytically. >> Best approach would be to use some sort of numerical algorithm. >> Thomas >either that or (T)roll another joint... Hey, come on - why so hostile? This was posted to a math-NG, so a mathematical answer should be in order (Question is what the question was - but that's another problem). Thomas >Bob === Subject: Re: What if the Nazis had kindness? > Kind Nazis could have got the Jews to help them with the bomb. No, smart Nazis would have had two classes of Jews - privileged Reichjuden and furnace feed. After the war, economy of consolidation could have proceeded. Letting Houdry (and Houdry catalyst) out of France into America was a terrible strategic error. Ditto Fermi, Szilard, von Neumann... most of the Manhattan Project slipped through their fingers. If Japan had kept it in its pants for anther year, Europe would have fallen and been united. England would have sued for peace. Russia would have been corked. Joseph P. Kennedy Sr. would have brokered an American/German peace agreement. We would have had Homeland Severity 50 years earlier and a whole lot fewer niggers bottom line. [snip crap] -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: What if the Nazis had kindness? >>Kind Nazis could have got the Jews to help them with the bomb. > No, smart Nazis would have had two classes of Jews - privileged > Reichjuden and furnace feed. After the war, economy of consolidation > could have proceeded. Letting Houdry (and Houdry catalyst) out of > France into America was a terrible strategic error. Ditto Fermi, > Szilard, von Neumann... most of the Manhattan Project slipped through > their fingers. Hitler was too arrogant (or stupid) to follow the Roman model's details. Specifically, in according levels of citizenship with concomittant levels of privileges. Top-down government works if you keep the brutality down. > If Japan had kept it in its pants for anther year, Europe would have > fallen and been united. England would have sued for peace. Russia > would have been corked. Joseph P. Kennedy Sr. would have brokered an > American/German peace agreement. We would have had Homeland Severity > 50 years earlier and a whole lot fewer niggers bottom line. Hitler should have detailed Japan to pacify Asia and Italy to pacifying the Mediterranean, while pacifying us with cheap iron, oil, etc. But then, while the Cold War wouldn't have happened, WWIII would have about thirty years ago. I strongly suspect we'd have lost. Whew! Mark L. Fergerson === Subject: Re: What if the Nazis had kindness? Aha...and if my grandma had two wheels she'd probably be a bicycle...but she hasn't, Japan attacked, Germany declared war on USA, the good ones won, the bad ones got screwed and THAT was happened...*sigh*...so simple. Tonico === Subject: Re: What if the Nazis had kindness? > Aha...and if my grandma had two wheels she'd probably be a > bicycle...but she hasn't, Japan attacked, Germany declared war on USA, > the good ones won, the bad ones got screwed and THAT was > happened...*sigh*...so simple. > Tonico Except that the bad ones reincarnated as the neocons. In answer to your question, if they had had kindness, they wouldn't have been nazis, simple as that. Rich === Subject: Re: What if the Nazis had kindness? > Aha...and if my grandma had two wheels she'd probably be a > bicycle...but she hasn't, Japan attacked, Germany declared war on USA, > the good ones won, the bad ones got screwed and THAT was > happened...*sigh*...so simple. > Tonico Suppose Germany had declared war on Japan and the USA became Germany's ally? -- Paul Townsend Pair them off into threes Interchange the alphabetic letter groups to reply === Subject: Re: What if the Nazis had kindness? > Kind Nazis could have got the Jews to help them with the bomb. Kind > Nazis would have got far more Russian collaborators and perhaps not What if noone replied to trolling crossposts? Michael -- Still an attentive ear he lent Her speech hath caused this pain But could not fathom what she meant Easier I count it to explain She was not deep, nor eloquent. The jargon of the howling main -- from Lewis Carroll: The Three Usenet Trolls === Subject: Re: What if the Nazis had kindness? >Kind Nazis could have got the Jews to help them with the bomb. Kind >Nazis would have got far more Russian collaborators and perhaps not >have inspired the stubborn resistance of the Russians and the British >in WWII. Kindness might have made the difference. And while they would >not have won they would have concluded a beneficial peace treaty. --- Kind Nazis would never have started their in the first place. -- John Fields === Subject: Re: What if the Nazis had kindness? >>Kind Nazis could have got the Jews to help them with the bomb. Kind >>Nazis would have got far more Russian collaborators and perhaps not >>have inspired the stubborn resistance of the Russians and the >>British in WWII. Kindness might have made the difference. And >>while they would not have won they would have concluded a beneficial >>peace treaty. > --- > Kind Nazis would never have started their in the first place. > Oh, they didn't start fascism or antisemitism or racism or > warmongering or nationalism. Europe was full of all of that, > everywhere (remember the Dreifuss affair in France?). But they > definitely took them to unprecedented heights. > -- > David Kastrup, Kriemhildstr. 15, 44793 Bochum What tbe Nazis started was anti-Bolshevism (as the Germans called it) or anti-Communism. To that purpose they enlisted anti-Semitism (or racism, by defining not only the Slavs but also the Jews as a race) and nationalism by asserting their racial and national interests against other nations' interests. I believe fascism is properly defined in economic terms, but if you want to define it in political terms, that is fine with me too. Kind family Nazis did not start this political rumpus. It was unkind political Nazis, at the top of the food chain, who did. David Ames === Subject: Re: What if the Nazis had kindness? >Kind Nazis would never have started their in the first place. Well, the Kaiser went and started a war against the rest of Europe without persecuting Jews. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: What if the Nazis had kindness? <41ed5348.1504015@news.ecn.ab.ca> In <41ed5348.1504015@news.ecn.ab.ca>, on 01/18/2005 >Well, the Kaiser went and started a war against the rest of Europe >without persecuting Jews. I'd hardly call WW I kind, nor is it clear that he started it. Wasn't the root cause a series of mutual-defense treaties that left no room for maneuver? Diplomacy is too important to be left to the diplomats. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: What if the Nazis had kindness? >In <41ed5348.1504015@news.ecn.ab.ca>, on 01/18/2005 >>Well, the Kaiser went and started a war against the rest of Europe >>without persecuting Jews. >I'd hardly call WW I kind, nor is it clear that he started it. Wasn't >the root cause a series of mutual-defense treaties that left no room >for maneuver? Diplomacy is too important to be left to the >diplomats. heck, no. The idea that just because you've signed a treaty stating that in situation X you'll do Y, then, if situation X arises, you actually have to do Y, is a naive belief having little to do with historical evidence. France had a mutual defense treaty with Chechoslovakia, in 1938. So? European histroy of the past few centuries is full of examples of treaties made, then disregarded when not convenient. The root cause of WWI was the fact that nearly everybody wanted it, for various reasons. Germany felt that its standing in European and world affairs is not quite commensurate with its population, economic and military might and was determined to correct the balance (and, it was worried that once the Russian modernization process which was already underway will get into high gear, the combination of France and Russia will be too much for Germany to handle). France wanted Alsace and Lorraigne (sp?) back. Both Russia and the Austro-Hungarian Empire though that war may distract the attention of their populace from internal problems. Italy (which wasn't really obliged to anything) had some territorial accounts to settle with Austria. With the possible exception of britain, the European powers not only didn't wish to avoid a war but were actively courting one, viewing it not as a grim necessity but as a glorious opportunity. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: What if the Nazis had kindness? <41ed5348.1504015@news.ecn.ab.ca> >>Kind Nazis would never have started their in the first place. >Well, the Kaiser went and started a war against the rest of Europe >without persecuting Jews. ...any more than they were being persecuted in other European countries at that time. -- The good news is that nothing is compulsory. The bad news is that everything is prohibited. http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk === Subject: Re: What if the Nazis had kindness? Nazi soldiers were kind, considerate, loving sons, husbands and fathers, just like other nationalities. === Subject: Re: What if the Nazis had kindness? >Nazi soldiers were kind, considerate, loving sons, husbands and fathers, >just like other nationalities. Indeed. It's a pity the term has attained such pejorative connotations over the years. A Nazi was simply a German National Socialist. Being nasty wasn't a requirement to join the party. The Liberal bigots who cast it around as a term of abuse could use some education on the subject! -- What is now proved was once only imagin'd. - William Blake, 1793. === Subject: Sum of series by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0I1IYP24210; I need the sum of the following series. Notation i pow 2 = i*i or i^2 N is a natural number for N=1 it is a G.P. so T(i) = i^N / 4 ^i T(i)= (i pow N)/(4 pow i) I need the sum of T(0)+T(1)+T(2)......inf ie 0+1/4 + 2^N/4^2 + 3^N/4^3+...... Sum of T(i) where i = 0 to inf. How do i go about this. Do we have to use integrals ? === Subject: Re: Sum of series by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0IDiVm20019; >I need the sum of the following series. > Notation i pow 2 = i*i or i^2 > N is a natural number for N=1 it is a G.P. > so T(i) = i^N / 4 ^i > T(i)= (i pow N)/(4 pow i) > I need the sum of T(0)+T(1)+T(2)......inf ie 0+1/4 + 2^N/4^2 + >3^N/4^3+...... > Sum of T(i) where i = 0 to inf. >How do i go about this. Do we have to use integrals ? Hint: Consider sum (i=1,...oo) i^N * x^i/4^i . Deduce sum (i=1,...oo) i^N * x^i/4^i = x* d/dx (sum (i=1,...oo) i^(N-1) * x^i/4^i ) which is a recurrence relation for the sum you want to obtain with the function sum (i=1,...oo) i^0 * x^i/4^i = x/(4-x) to start with. Evaluate at x=1. Best wishes Torsten. === Subject: Expression of plane I wonder how plane can be expressed as X(u,v)=a+b*u+c*v. (in here a,b,c are constant vector) If some one can give me explanation why plane can be expressed as X(u,v)=a+b*u+c*v, then please post reply. === Subject: Re: Expression of plane Herb dixit: >I wonder how plane can be expressed as X(u,v)=a+b*u+c*v. >(in here a,b,c are constant vector) >If some one can give me explanation why plane can be >expressed as X(u,v)=a+b*u+c*v, >then please post reply. http://www.netcomuk.co.uk/~jenolive/vect11.html 'a' is the position vector of a point inthe plane, and b, and c are 2 linearly independent vectors that lie in the plane. And u and v are scalar parameters, you need 2 of them because a plane is a 2 dimensional object. === Subject: x^y=y^x /representations ? Suppose that 0 < x < y are such that x^y = y^x . Prove or disprove that there exists an integrable function K:(0,1)-->[0,infty) such that for every t> 0 x= e - 1/(t+1) - INT{K(v)/(t+v)} , y= e + 1/t + INT{K(v)/(t+1-v} , where INT{G(v)}:=Integral{v=0 to v=1}G(v) dv and e=Napier's constant. K(v)= ?? === Subject: Re: x^y=y^x /representations ? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0JF2a321245; >Suppose that 0 < x < y are such that x^y = y^x . >Prove or disprove that there exists an integrable function >K:(0,1)-->[0,infty) such that for every t> 0 >x= e - 1/(t+1) - INT{K(v)/(t+v)} , >y= e + 1/t + INT{K(v)/(t+1-v} , >where INT{G(v)}:=Integral{v=0 to v=1}G(v) dv and e=Napier's constant. >K(v)= ?? Should the statement read forall (x, y) x < y and x^y = y^x there _exists_ t > 0 such that ...? Otherwise the problem doesn't make much sense. Unless I'm missing something, such K cannot exist. Suppose such K exists. Writing x, y as functions x(t), y(t) of the form given above, it's easy to see x'(t) > 0 and y'(t) < 0. So x(t) is increasing for t in (0, 1), and therefore is bounded above by lim_(t -> 1) x(t) <= e - 1/2. But for each x-value in [e - 1/2, e] there exists y > e such that x^y = y^x (or equivalently, such that log(x)/x = log(y)/y). So not all (x, y) satisfying x < y, x^y = y^x are of the form (x(t), y(t)). Todd Trimble === Subject: Re: x^y=y^x /representations ? > Todd Trimble === Subject: Quotient Ring What do the ideals of F[x]/(p(x)) look like when p(x) is not irreducible in the field F? Because p(x) is not irreducible, F[x]/(p(x)) is not a field nor a domain, so there's something other than (0) and (1) in there. Can the you in advance. === Subject: Re: Quotient Ring > What do the ideals of F[x]/(p(x)) look like when p(x) is not irreducible in > the field F? Because p(x) is not irreducible, F[x]/(p(x)) is not a field > nor a domain, so there's something other than (0) and (1) in there. Can the > you in advance. F[x]/(x^2) = { ax + b | a,b in F } with operations ++ and * ax + b ++ rx + s = (a + r)x + b + s (ax + b)*(rx + s) = arx^2 + brx + asx + bs = bs + (br + as)x For example ax * rx = 0 ax * (rx + s) = axs === Subject: Re: What if the Nazis had kindness? WWII, hitler, nazi, eng, fra, us, blah blah blah.... lets get back to maths & improvement pH === Subject: quadrilateral problem How can it be shown that the sum of the squares of the sides of a quadrilateral equals the sum of the diagonals + 4x^2 where x is the distance between the midpoints of the diagonals? === Subject: Re: quadrilateral problem > How can it be shown that the sum of the squares of the sides of a > quadrilateral equals the sum of the diagonals + 4x^2 where x is the > distance between the midpoints of the diagonals? According to Heath (Euclid's Elements, I, pp.401-2) this theorem is due to Euler. Here's a way to prove it, using (six times!) the theorem about the square on a median of a triangle PQR: if S is the mid-point of PR, then PQ^2 + QR^2 = 2(PS^2 + QS^2). This can be proved from Euclid II.9 & 10 (Heath p.401), or Euclid II.12 & 13, or the cosine law. Now in any quadrilateral ABCD let the mid-points of the diagonals AC, BD be M, N respectively. Apply the above theorem to triangles ABC, BCD, CDA, DAB, add all four equations, and halve, to get AB^2 + BC^2 + CD^2 + DA^2 Now apply the same theorem to triangles ANC, BMD, and add both equations to get Substituting into the previous equation then gives = AC^2 + BD^2 + 4(MN^2) q.e.d. Ken Pledger. === Subject: Re: quadrilateral problem > How can it be shown that the sum of the squares of the sides of a > quadrilateral equals the sum of the diagonals + 4x^2 where x is the > distance between the midpoints of the diagonals? I should add that I solved it first by putting the quadrilateral in a coordinatesystem, and with the Pytagorean theorem, it is easy to show. But I'm looking for other ways to show it, using trigonometry, for example. === Subject: Re: quadrilateral problem > How can it be shown that the sum of the squares of the sides of a > quadrilateral equals the sum of the diagonals + 4x^2 where x is the > distance between the midpoints of the diagonals? > I should add that I solved it first by putting the quadrilateral in a > coordinatesystem, and with the Pytagorean theorem, it is easy to show. But > I'm looking for other ways to show it, using trigonometry, for example. Easy with vectors. Distance between midpoints of the diagonals: x = AB + (BA + AD) / 2 - (AB + BC) / 2 = (AB + AD) / 2 - (AB + BC) / 2 = (AD - BC) / 2 4x^2 = (AD - BC)^2 Difference between sum of squares of sides and sum of squares of diagonals: AB^2 + BC^2 + CD^2 + DA^2 - AC^2 - BD^2 = AB^2 + BC^2 + CD^2 + DA^2 - (AB + BC)^2 - (BA + AD)^2 = AB^2 + BC^2 + CD^2 + DA^2 - AB^2 - BC^2 - AB^2 - AD^2 - 2 AB BC - 2 BA AD = CD^2 - AB^2 - 2 AB BC - 2 BA AD = (CD + AB) (CD - AB) + 2 AB (CB + AD) = (AD + CB) (CD - AB) + 2 AB (CB + AD) = (AD + CB) (CD + AB) = (AD + CB) (AD + CB) = (AD - BC)^2 meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: Re: Why do two negatives equal a positive? > Why do two negatives equal a positive? How is it proven that two negatives > equal a positive when multiplied together? I have always known that a > negative times a negative equals a positive, but it has never made sense to > me. Will someone please make sense of it? Others have posted empirical support for the law of signs, using the distributive law to extend computations from positive to negative integers. In fact it is the desire to preserve the distributive law for signed integers that forces us to adapt the law of signs. In fact once we assume the distributive law holds for signed integers then we can easily employ this to *prove* the law of signs as follows: LAW OF SIGNS a b equals (-a)(-b) PROOF: ----------- a b + a(-b) + (-a)(-b) ---------------- Evaluate above expression in two different ways: FIRST: the overlined expression evaluates to 0, i.e. ab + a(-b) = a(b + -b) = a(0) = 0. So, on one hand, the expression equals (-a)(-b). SECOND: the underlined expression evaluates to 0, i.e. a(-b) + (-a)(-b) = (a + -a)(-b) = 0(-b) = 0. So, on the other hand, the expression equals ab. Therefore, a b equals (-a)(-b), as claimed. QED Notice how the above proof depends crucially upon the DISTRIBUTIVE LAW (among other minor laws). Note that the proof can be summarized concisely as: a(-b) is a common inverse to ab and (-a)(-b) so the latter are equal by uniqueness of additive inverses. -Bill Dubuque === Subject: Re: Why do two negatives equal a positive? Why is minus a minus a plus ? The often quoted -1 x -1 = 1 is more than just a definition or convention. Neg x neg = pos is required in order to have the certain basic properties of arithmetic hold, and for the answers we get to continue to follow the pattern, and thereby make sense. The requirement for consistency is an acceptable explanation. In seeking an explanation that satisfies because it is pictorial and easy, (for students at an earlier stage), we introduce the often used association between sign and direction. Positive multiplication = repeated hops in the direction you're facing Negative multiplication = repeated hops in the other direction. Although it works out nicely, this association is just stated without justification. I thought that, (for those of you that do not know about it already), it might be interesting to give a slightly deeper reason for the association between sign and direction. Suppose, two complex numbers are represented on an Argand diagram in the normal way, so that their 'size' and the angle they make to the x axis, are characterised by [r, a], and [s, b]. According to a discovery first made by Caspar Wesell in 1797, (and expanded on by Argand, Gauss etc.), the product of these two complex numbers will be characterised by [(r * s), (a+b)]. Since real numbers are just a special case of complex numbers where the imaginary component is zero, we can represent them on an Argand diagram in the same way. For example: - -2, (minus two), might look something like <------------ and be represented as (2, pi) -3, (minus two), might look something like <------------------ and be represented as (3, pi) According to the rule discovered by Wesell their product will be (6, 2pi). Well of course when you have rotated 2pi radians, (equivalent to 360 degrees), you are back where you started, so this could equally well be represented as (6, 0). It is clear that the result would point in the real positive direction, and be 6 units long. It would look like ------------------------------------> Thus (-2) * (-3) = +6 and the association between sign and direction is established. Perhaps you prefer to rely on the more familiar expressions for complex numbers. One of Euler's many contributions included e^(i*theta)= cos (theta) + i* sin (theta) from which e^(i*2pi)= cos (2pi) + i* sin (2pi) = +1 and e^(i*pi)= cos (pi) + i* sin (pi) = -1 follow. Since e^(i*pi)*e^(i*pi) = e^(i*2pi) we obtain (-1) * (-1) = +1 Hope you find this interesting - Ian Hutcheson === Subject: Re: Why do two negatives equal a positive? Answer By definition (-2)*(-3) means = you add twice (-3)+(-3) and in the result (-6) you find the opposit one = opposit of (-6) = +6 This is the physical sense of multiply two negatives. ---------------------------------------------------------------------------- --------- ? Bill Dubuque ?????? ??? ?????? > Why do two negatives equal a positive? How is it proven that two negatives > equal a positive when multiplied together? I have always known that a > negative times a negative equals a positive, but it has never made sense to > me. Will someone please make sense of it? > Others have posted empirical support for the law of signs, > using the distributive law to extend computations from > positive to negative integers. In fact it is the desire > to preserve the distributive law for signed integers that > forces us to adapt the law of signs. In fact once we assume > the distributive law holds for signed integers then we can > easily employ this to *prove* the law of signs as follows: > LAW OF SIGNS a b equals (-a)(-b) > PROOF: ----------- > a b + a(-b) + (-a)(-b) > ---------------- > Evaluate above expression in two different ways: > FIRST: the overlined expression evaluates to 0, > i.e. ab + a(-b) = a(b + -b) = a(0) = 0. > So, on one hand, the expression equals (-a)(-b). > SECOND: the underlined expression evaluates to 0, > i.e. a(-b) + (-a)(-b) = (a + -a)(-b) = 0(-b) = 0. > So, on the other hand, the expression equals ab. > Therefore, a b equals (-a)(-b), as claimed. QED > Notice how the above proof depends crucially upon > the DISTRIBUTIVE LAW (among other minor laws). > Note that the proof can be summarized concisely as: > a(-b) is a common inverse to ab and (-a)(-b) > so the latter are equal by uniqueness of additive inverses. > -Bill Dubuque === Subject: Re: Why do two negatives equal a positive? Answer > By definition (-2)*(-3) means = you add twice (-3)+(-3) and in the result > (-6) you find the opposit one = opposit of (-6) = +6 > This is the physical sense of multiply two negatives. No, it's not by DEFINITION but, rather, by CONSEQUENCE of distributivity: Namely (-x)(-y) = -(x(-y)) (which is what you're implying above) <=> (-x)(-y) + x(-y) = 0 <=> (-x + x) (-y) = 0 (by DISTRIBUTIVE LAW) <=> 0 (-y) = 0 <=> 0 = 0 PRESERVING THE DISTRIBUTIVE LAW is what is fundamental. This means LINEARITY of multiplication c(x+y) = cx+cy is preserved also for signed quantities so. E.g. scaling transformations (e.g. doubling, tripling etc) behave in the desired linear manner. E.g. physically, when calculating distance = speed * time, it will behave correctly for negative speed,time; e.g. when run backwards. --Bill Dubuque === Subject: Re: Why do two negatives equal a positive? Answer The proof of this and other results can be found in such textbooks as Birkhoff and MacLane, A Survey of Modern Algebra, 4th ed., ch. 1, and Walter Rudin, Principles of Mathematical Analysis, 3rd ed., ch. 1. Others in this thread have shown how (-a)(-b) = ab is derived; I just wanted to provide references which the original poster or anyone else who isn't familiar with this material can turn to for further background. === Subject: Using standard deviation to make predictions Hi group, I understand and I can calculate mean, median, variance and standard deviation. However, I've read that one can make predictions using standard deviation. How would I do this? snarly === Subject: Re: Using standard deviation to make predictions >Hi group, >I understand and I can calculate mean, median, variance and standard >deviation. However, I've read that one can make predictions using >standard deviation. >How would I do this? >snarly see my response in alt.math --Lynn === Subject: Re: Using standard deviation to make predictions > Hi group, > I understand and I can calculate mean, median, variance and standard > deviation. However, I've read that one can make predictions using > standard deviation. > How would I do this? > snarly You have to know something about the unerlying distribution. Suppose it has a normal distribution then I think you can compute that something like 95% of the trials will fall between the mean and + or - 1.95 sigma) (sigma = standard deviation.) (Double check me, it has been a while, but that is the idea. ) Bill === Subject: Re: Orthogonality of bessel functions by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0JDFNQ11004; Hi Georg, relations if u have if u don't have can u give a good site or a book to look for that. === Subject: Question about Trends and Statistics by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0JGcUY29975; guys I'm new to statistics. I like math. I need to do a lot of statistics for my new job, but did not take stats in school. Will doing trends be a difficult thing to learn? How can i tell if a book teaches trends. Most of the Stats books i've seen say nothing or little able trends. I do see the word trend from time to time but i'm not sure if it is teaching how to use trends in analysis or if it is just a synonym for something else. PLEASE HELP! === Subject: Re: [OT] I hate being American > Furthermore, 'nuking' N Korea and Iran would kill a great number of > people in neighbouring countries. Is that just unfortunate 'collateral'? Since when have USA citizens cared about killing people. They kill many thousands of themselves every year. What has prevented them from nuking since Hiroshima and Nagasaki is fall-out on USA towns, cities and prairies. === Subject: Re: [OT] I hate being American >> Furthermore, 'nuking' N Korea and Iran would kill a great number of >> people in neighbouring countries. Is that just unfortunate 'collateral'? >Since when have USA citizens cared about killing people. They kill many >thousands of themselves every year. >What has prevented them from nuking since Hiroshima and Nagasaki is fall-out >on USA towns, cities and prairies. Same thing that's kept the UK from using its nukes? John === Subject: Re: [OT] I hate being American > Same thing that's kept the UK from using its nukes? Mutual Assuree Destruction is what kept WWW3 from happening while the Soviet Union still stood. Bob Kolker === Subject: Re: [OT] I hate being American >> Same thing that's kept the UK from using its nukes? >Mutual Assuree Destruction is what kept WWW3 from happening while the >Soviet Union still stood. More importantly, the Russians believed that the West would retaliate immediately with lots of nukes if they ever tried it. The West tried to disspell this belief in the late 70s with the philosophy that talking at the Russians would keep them from attacking. This was only 25 years ago and we're back to to bad philosophy again. / BAH Subtract a hundred and four for e-mail. === Subject: Re: [OT] I hate being American > > > Same thing that's kept the UK from using its nukes? >>Mutual Assuree Destruction is what kept WWW3 from happening while the >>Soviet Union still stood. >More importantly, the Russians believed that the West >would retaliate immediately with lots of nukes if they >ever tried it. The West tried to disspell this belief >in the late 70s with the philosophy that talking at >the Russians would keep them from attacking. This was >only 25 years ago and we're back to to bad philosophy >again. Can there really be people in 8 unrelated groups interested in reading this stuff? And do you have any evidence at all that the Russians wanted to nuke anybody? AFAIK, the only time in the last 55 years when this was seriously considered was when MacArthur wanted to nuke the Chinese; and I don't recall that he was a Russian. === Subject: Re: [OT] I hate being American >> >> >> Same thing that's kept the UK from using its nukes? >Mutual Assuree Destruction is what kept WWW3 from happening while the >Soviet Union still stood. >>More importantly, the Russians believed that the West >>would retaliate immediately with lots of nukes if they >>ever tried it. The West tried to disspell this belief >>in the late 70s with the philosophy that talking at >>the Russians would keep them from attacking. This was >>only 25 years ago and we're back to to bad philosophy >>again. >Can there really be people in 8 unrelated groups interested in reading >this stuff? I have no idea. I'd pare down the cross-post if I knew where everybody was. >And do you have any evidence at all that the Russians wanted to nuke >anybody? The Russians have a goal of ruling the world. They believed that their political and economic philosophies were better than and/or group of people have to through the political growing pain that is called colonialism. The Russians were still in this phase during the 70s and 80s. I'm not sure about now. If they thought that they could take over Europe with a token nuke of a small city and threats of larger cities without any retaliation, they would have done it in nanosecond. >AFAIK, the only time in the last 55 years when this was seriously >considered was when MacArthur wanted to nuke the Chinese; and I don't >recall that he was a Russian. Right now I'm reading about the nuclear armaments of the Russians aimed at Europe during the 70s and early 80s. At the same time, Europe was trying to reduce research and armaments because of the mistaken assumption that, if they dropped their weapons, the Russians would drop theirs. They assumed that Russia was a kind and gentle people who had no expansionist goals, even though this country bred and followed viscious people like Lenin and Stalin. The reason atom bombs were getting installed in Cuba was because the Russians believed that Kennedy wouldn't do anything substantial about it, other than bleat. I need to warn you that I've only just started learning about all of this stuff. Beware of people who know just a little bit ;-). /BAH Subtract a hundred and four for e-mail. === Subject: Re: [OT] I hate being American >AFAIK, the only time in the last 55 years when this was seriously >considered was when MacArthur wanted to nuke the Chinese; and I don't >recall that he was a Russian. The French wanted to nuke Dien Bien Phu - but didn't, as I recall, because they couldn't get flares of a colour that would show through the cloud and mist before the (conventional) battle was lost. For want of a nail the world was saved... Martin -- Martin Frey http://www.hadastro.org.uk N 51 02 E 0 47 === Subject: Re: [OT] I hate being American >>AFAIK, the only time in the last 55 years when this was seriously >>considered was when MacArthur wanted to nuke the Chinese; and I don't >>recall that he was a Russian. >The French wanted to nuke Dien Bien Phu - but didn't, as I recall, >because they couldn't get flares of a colour that would show through >the cloud and mist before the (conventional) battle was lost. For want >of a nail the world was saved... Interesting - I hadn't heard this story before. Do you have any references? The French themselves didn't explode their first bomb until 1960, 6 years after Dien Bien Phu fell. A little Googling showed that there were some discussions on the Americans using 3 tactical weapons around DBP, but I couldn't find out how advanced the discussions were. Besides, the idea was crazy; it would probably have wiped out the French defenders. === Subject: Re: [OT] I hate being American >>Same thing that's kept the UK from using its nukes? >Mutual Assuree Destruction is what kept WWW3 from happening while the >Soviet Union still stood. >>More importantly, the Russians believed that the West >>would retaliate immediately with lots of nukes if they >>ever tried it. The West tried to disspell this belief >>in the late 70s with the philosophy that talking at >>the Russians would keep them from attacking. This was >>only 25 years ago and we're back to to bad philosophy >>again. > Can there really be people in 8 unrelated groups interested in reading > this stuff? There are replies generated from all groups, so... > And do you have any evidence at all that the Russians wanted to nuke > anybody? Yeah, sure; the Cuban missile crisis was all a misunderstanding; the Soviets were merely giving some of their missile techs a nice vacation in sunny Cuba, and the techs simply took some of their work with them, right? > AFAIK, the only time in the last 55 years when this was seriously > considered was when MacArthur wanted to nuke the Chinese; and I don't > recall that he was a Russian. You are either ignorant or were educated by revisionists. Krushchev in particular would have been extremely pleased to nuke the US if he thought he could have gotten away with a pre-emptive strike. But MAD won out; he knew in his bones that if he struck first, World Communism would have been permamently killed, and even at his table-pounding worst he couldn't allow that. Mark L. Fergerson === Subject: Re: [OT] I hate being American >> And do you have any evidence at all that the Russians wanted to nuke >> anybody? > Yeah, sure; the Cuban missile crisis was all a misunderstanding; the >Soviets were merely giving some of their missile techs a nice vacation >in sunny Cuba, and the techs simply took some of their work with them, >right? And I suppose US nukes in Turkey were a 'misunderstanding'? Why exactly should the US have had the right to park nuclear missiles next to the USSR, when the Soviets couldn't have the right to park theirs next to the US? If your reasoning is that proximity proves the desire to bomb, as it appears to be, then you would have to accept that the US had already demonstrated a desire to nuke the USSR; the Soviets were merely responding to that. >> AFAIK, the only time in the last 55 years when this was seriously >> considered was when MacArthur wanted to nuke the Chinese; and I don't >> recall that he was a Russian. > You are either ignorant or were educated by revisionists. Krushchev >in particular would have been extremely pleased to nuke the US if he >thought he could have gotten away with a pre-emptive strike. But MAD won >out; he knew in his bones that if he struck first, World Communism would >have been permamently killed, and even at his table-pounding worst he >couldn't allow that. You appear to be posting to a number of groups that should, in principle, have well-educated readers. If you're aware of any evidence that Krushchev 'would have been extremely pleased to nuke the US' then I'm sure that those readers, and myself, would like to see it, rather than being treated to a lecture on ideology. There's clear evidence that MacArthur wanted to bomb the Chinese; is there clear evidence that Krushchev wanted to bomb the US? You tell me. === Subject: Re: [OT] I hate being American >And do you have any evidence at all that the Russians wanted to nuke >anybody? >> Yeah, sure; the Cuban missile crisis was all a misunderstanding; the >>Soviets were merely giving some of their missile techs a nice vacation >>in sunny Cuba, and the techs simply took some of their work with them, >>right? > And I suppose US nukes in Turkey were a 'misunderstanding'? Nope. The significance of their presence was clearly understood by all participants. Think it through; why was their presence known at all? > Why > exactly should the US have had the right to park nuclear missiles next > to the USSR, when the Soviets couldn't have the right to park theirs > next to the US? None at all, or every right, depending on one's viewpoint. Again, think it through; why would the Soviets allow us to know they were there? Hiding them from the satellites of the day wouldn't have been very difficult. > If your reasoning is that proximity proves the desire > to bomb, as it appears to be, then you would have to accept that the > US had already demonstrated a desire to nuke the USSR; the Soviets > were merely responding to that. But that isn't my reasoning. I'll get to that below. >AFAIK, the only time in the last 55 years when this was seriously >considered was when MacArthur wanted to nuke the Chinese; and I don't >recall that he was a Russian. >> You are either ignorant or were educated by revisionists. Krushchev >>in particular would have been extremely pleased to nuke the US if he >>thought he could have gotten away with a pre-emptive strike. But MAD won >>out; he knew in his bones that if he struck first, World Communism would >>have been permamently killed, and even at his table-pounding worst he >>couldn't allow that. > You appear to be posting to a number of groups that should, in > principle, have well-educated readers. If you're aware of any evidence > that Krushchev 'would have been extremely pleased to nuke the US' then > I'm sure that those readers, and myself, would like to see it, rather > than being treated to a lecture on ideology. There's clear evidence > that MacArthur wanted to bomb the Chinese; is there clear evidence > that Krushchev wanted to bomb the US? You tell me. Sigh. You appear to believe I'm a My Country Right Or Wrong jingoist, which I'm not. Nor am I addicted to any particular ideology, but I try to understand the ideologies that drive others' decisions. You ask for clear evidence; the problem with that is that the Soviets only aired their dirty laundry when forced to, but the US makes a point of doing it voluntarily. Any direct documentation of Krushchev's intentions were either destroyed to minimize the collateral guilt of his Politburo or are deeply buried and will show up eventually. Inferring his intentions is relatively easy from reading history. Question my interpretation if you want to, but how else do you explain the Cuban missiles, as a reasonable counter to the Turkish sitings? In a sense they were, but the reasoning was flawed. Yes, MacArthur wanted to bomb the Chinese, because he saw them as a future threat to US security since they'd gotten over their traditional isolationism. Patton wanted to roll his tanks through Berlin all the way to Moscow, for similar reasons. Both were of the Best Defense Is A Strong Offense school of thought, which teaches that a future threat is best nipped in the bud. Both were restrained only because the military was not largely in charge of the US government, which was not the case with China or the USSR. Krushchev wanted to bomb the US for the exact same reason except he'd have couched it differently; he considered the security of the USSR to be dependent on expansion in Europe, and the US was the sole viable announced threat to that. He'd have been happy to convert the US to Communism peacefully, but he was absolutely certain that it couldn't happen so active defense was his only possible posture, which had to hinge on nuclear threat because nothing else would be believable, and Cuba was the only place he could site his missiles. Soviet expansionism was the threat that was being countered by the Turkish deployments. Later, bigger boosters made the futility of that sort of subtle chess move obvious. My general reasoning is that once nuclear weapons were known to work, they were simply irresistible to those that had the capacity to make and field them. However, things got rapidly out of hand once it also became known how easy they were to use over great distances, much greater than any previous WMD and to do so much more damage than any previous weapon. The military minds of the day just couldn't grasp how much bang they were getting for their buck until it was too late. Remember that the bitter logic of MAD was not obvious from the get-go, but only became apparent as the standoffs developed; the presence of US short-range missiles in Turkey, the Soviet missiles in Cuba, and other secondhand deployments were opportunities that the fight the last war military minds on both sides simply couldn't resist. They did not think through the implications of nuclear missiles that were immune to countermeasures until the facts were shoved in their faces. Fortunately we know a little better now, which is why nuke reduction agreements are (slowly) working. Call me an optimist, but I think that the next nuke used (by a desperate Third-worlder with delusions of grandeur) will be the last one; pretty much everyone else dreads the consequences of escalation more than they lust for the glory of conquest. MADness may yet save us from our madness. Mark L. Fergerson === Subject: Re: [OT] I hate being American >>And do you have any evidence at all that the Russians wanted to nuke >>anybody? > Yeah, sure; the Cuban missile crisis was all a misunderstanding; the >Soviets were merely giving some of their missile techs a nice vacation >in sunny Cuba, and the techs simply took some of their work with them, >right? >> And I suppose US nukes in Turkey were a 'misunderstanding'? > Nope. The significance of their presence was clearly understood by > all participants. Think it through; why was their presence known at all? Not to mention that the US missles in Turkey were obsolete Atlas', more dangerous to their crews than the Rooskies. The Cuba settlement was a trade for antiques, which we no longer needed. The pulling of the Turk Atlas' were simply face-saving for Kruschev. He had to have something. >> Why >> exactly should the US have had the right to park nuclear missiles next >> to the USSR, when the Soviets couldn't have the right to park theirs >> next to the US? > None at all, or every right, depending on one's viewpoint. Again, > think it through; why would the Soviets allow us to know they were > there? Hiding them from the satellites of the day wouldn't have been > very difficult. We were going to dump the aAtlas' in Turkey anyway. This argiument is sillier than how many angels dance on pins. The fact is that Kruschev blinked. He knew he'd been called on his bluff. The bum Castro got a promise not to invade. Kruschev got to save another shoe. -- Keith === Subject: Re: [OT] I hate being American >And I suppose US nukes in Turkey were a 'misunderstanding'? >> Nope. The significance of their presence was clearly understood by >>all participants. Think it through; why was their presence known at all? > Not to mention that the US missles in Turkey were obsolete Atlas', more > dangerous to their crews than the Rooskies. The Cuba settlement was a > trade for antiques, which we no longer needed. The pulling of the Turk > Atlas' were simply face-saving for Kruschev. He had to have something. Yup. I'm convinced that the US made that trade to keep Krushchev in power because he was predictable and had a firm grip on the Politburo. >Why >exactly should the US have had the right to park nuclear missiles next >to the USSR, when the Soviets couldn't have the right to park theirs >next to the US? >> None at all, or every right, depending on one's viewpoint. Again, >>think it through; why would the Soviets allow us to know they were >>there? Hiding them from the satellites of the day wouldn't have been >>very difficult. > We were going to dump the aAtlas' in Turkey anyway. This argiument is > sillier than how many angels dance on pins. The fact is that Kruschev > blinked. He knew he'd been called on his bluff. The bum Castro got a > promise not to invade. Kruschev got to save another shoe. And the Politburo didn't have to try to throw him out if he didn't. Saved a lot of grief all around. Of course it's a silly argument, but I was trying to see if Rick was capable of independent reasoning or is merely spouting Soviet apologist's BS. Mark L. Fergerson === Subject: Re: [OT] I hate being American > And do you have any evidence at all that the Russians wanted to nuke > anybody? >> Yeah, sure; the Cuban missile crisis was all a misunderstanding; the >>Soviets were merely giving some of their missile techs a nice vacation >>in sunny Cuba, and the techs simply took some of their work with them, >>right? >And I suppose US nukes in Turkey were a 'misunderstanding'? Why >exactly should the US have had the right to park nuclear missiles next >to the USSR, when the Soviets couldn't have the right to park theirs >next to the US? If your reasoning is that proximity proves the desire >to bomb, as it appears to be, then you would have to accept that the >US had already demonstrated a desire to nuke the USSR; the Soviets >were merely responding to that. Well, the Soviets did invade Hungary, Czechoslovakia, Poland, and East Germany. They did reject the US offer of total abolition of nuclear weapons and free-skies. They did build hundreds of nukes and aim them at Europe. They did pound on desks at the UN with shoes. None of that was confidence-inspiring. And it was Western Europe that those US nukes were protecting from the million-man Soviet army. The French and the British built nukes of their own, for the same reasons, and still have them. The French were the last Western country to give up live nuclear weapons tests. The US was an isolationist, poorly-armed country before WWII, and would have cheerfully gone back to that state except for Soviet and Chinese/Korean expansionism. But then we get all the blame for being militaristic. John === Subject: Re: [OT] I hate being American <10tc5udmmevdnab@corp.supernews.com> <33mmiiF4099avU1@individual.net> <33oba4F41amfjU1@individual.net> <8M$aFIE$4v1BFwgu@jmwa.demon.co.uk> <357sc5F4guah4U2@individual.net> <0vkvu05020soe573e59pdg15kbpp96bqbg@4ax.com> <1NRHd.6384$0B.462@fed1read02> <2o50v0ltafpbl2riae7c7qp6f83n9ths7o@4ax.com> I read in sci.electronics.design that John Larkin Well, the Soviets did invade Hungary, Czechoslovakia, Poland, and East >Germany. During WW2, originally, in all our interests. They did re-invade Hungary and Czechoslovakia again later, and that was very wrong. >They did reject the US offer of total abolition of nuclear >weapons and free-skies. They did build hundreds of nukes and aim them at >Europe. They were frightened of Western technology; they didn't trust their own scientists. The ethos of Communism involved a dilemma over distrust of intellectuals. Intellectuals were not hated capitalists but they were also not 'workers'. Unlike the latter stage of the French Revolution, that executed Lavoisier, they realised they could not simply dispense with intellectuals but they saw a need to keep them closely under control. They were particularly wary of composers. Some of the junta probably knew what Verdi achieved for Italian independence with the 'Chorus of Hebrew Slaves' in 'Nabucco'. (Verdi himself was not only Joe Green, he was also a political statement in his own right - Victor Emmanuel, Rei D'Italia.) >They did pound on desks at the UN with shoes. None of that was >confidence-inspiring. I am frequently involved in international discussions. Many times, I am tempted to borrow a shoe to pound with. (The pictures of K in mid-pound show him wearing two shoes. The implement must have come from an acolyte.) -- The good news is that nothing is compulsory. The bad news is that everything is prohibited. http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk === Subject: Re: [OT] I hate being American <357sc5F4guah4U2@individual.net> <0vkvu05020soe573e59pdg15kbpp96bqbg@4ax.com> <1NRHd.6384$0B.462@fed1read02> <2o50v0ltafpbl2riae7c7qp6f83n9ths7o@4ax.com> >I read in sci.electronics.design that John Larkin >Well, the Soviets did invade Hungary, Czechoslovakia, Poland, and East >>Germany. >During WW2, originally, in all our interests. ermmm....No. The Russians had an agreement with Hitler which gave some of Poland to the Russians. /BAH Subtract a hundred and four for e-mail. === Subject: Re: [OT] I hate being American >I read in sci.electronics.design that John Larkin >Well, the Soviets did invade Hungary, Czechoslovakia, Poland, and East >>Germany. >During WW2, originally, in all our interests. > ermmm....No. The Russians had an agreement with Hitler which > gave some of Poland to the Russians. The Brits also had an agreement with Hitler. By luck that did not involve any transfer of land. the three major members of the alliance, the USA, the USSR and the UK pursued the war jointly. Franz === Subject: Re: [OT] I hate being American >>I read in sci.electronics.design that John Larkin ><2o50v0ltafpbl2riae7c7qp6f83n9ths7o@ >> >Well, the Soviets did invade Hungary, Czechoslovakia, Poland, and >East >Germany. >> >>During WW2, originally, in all our interests. >> ermmm....No. The Russians had an agreement with Hitler which >> gave some of Poland to the Russians. >The Brits also had an agreement with Hitler. By luck that did not >involve any transfer of land. Just Czechoslovakia. John === Subject: Re: [OT] I hate being American >>I read in sci.electronics.design that John Larkin ><2o50v0ltafpbl2riae7c7qp6f83n9ths7o@ 2005: >> >Well, the Soviets did invade Hungary, Czechoslovakia, Poland, and >East >Germany. >> >>During WW2, originally, in all our interests. >> >> ermmm....No. The Russians had an agreement with Hitler which >> gave some of Poland to the Russians. >The Brits also had an agreement with Hitler. By luck that did not >involve any transfer of land. > Just Czechoslovakia. Sudetenland? Franz === Subject: Re: [OT] I hate being American >>The Brits also had an agreement with Hitler. By luck that did not >>involve any transfer of land. >> Just Czechoslovakia. >Sudetenland? and the Rhineland, Poland, Austria, Hungary, Balkans, ..... (any private property belonging to political prisoners, jews...) But it was worth it to establish a 'peace in our time'. >Franz === Subject: Re: [OT] I hate being American > But it was worth it to establish a 'peace in our time'. Aw come on! All Hitler wanted was a little peace. A little peace of Poland, a little peace of France. Bob Kolker === Subject: Re: [OT] I hate being American >>I read in sci.electronics.design that John Larkin >>><2o50v0ltafpbl2riae7c7qp6f83n9ths7o@ >> >> >Well, the Soviets did invade Hungary, Czechoslovakia, Poland, and >> >>East >Germany. >> >>During WW2, originally, in all our interests. >ermmm....No. The Russians had an agreement with Hitler which >gave some of Poland to the Russians. >>The Brits also had an agreement with Hitler. By luck that did not >>involve any transfer of land. > Just Czechoslovakia. > John No- the Sudatenland... === Subject: Re: [OT] I hate being American > You are either ignorant or were educated by revisionists. Krushchev >in particular would have been extremely pleased to nuke the US if he >thought he could have gotten away with a pre-emptive strike. But MAD won >out; he knew in his bones that if he struck first, World Communism would >have been permamently killed, and even at his table-pounding worst he >couldn't allow that. > Mark L. Fergerson The inventor of the machine gun and the inventor of dynamite both thought that their gadget would make war so horrible that nobody would dare start one. Oppenheimer and Teller finally pulled it off. John === Subject: Re: [OT] I hate being American > You are either ignorant or were educated by revisionists. Krushchev >in particular would have been extremely pleased to nuke the US if he >thought he could have gotten away with a pre-emptive strike. But MAD won >out; he knew in his bones that if he struck first, World Communism would >have been permamently killed, and even at his table-pounding worst he >couldn't allow that. > Mark L. Fergerson > The inventor of the machine gun and the inventor of dynamite both > thought that their gadget would make war so horrible that nobody would > dare start one. Oppenheimer and Teller finally pulled it off. > John So now they start them anyway, betting that nobody dares open Pandora's box. === Subject: Re: [OT] I hate being American >> You are either ignorant or were educated by revisionists. Krushchev >>in particular would have been extremely pleased to nuke the US if he >>thought he could have gotten away with a pre-emptive strike. But MAD won >>out; he knew in his bones that if he struck first, World Communism would >>have been permamently killed, and even at his table-pounding worst he >>couldn't allow that. >> >> Mark L. Fergerson >> The inventor of the machine gun and the inventor of dynamite both >> thought that their gadget would make war so horrible that nobody would >> dare start one. Oppenheimer and Teller finally pulled it off. >> John >So now they start them anyway, betting that nobody dares open Pandora's box. Right. But only little ones. Nukes have probably been a stabilizing force in the world. The major powers, the kind that start big, world wars, have mostly skirmished by proxy. The interesting case now is India and Pakistan. John === Subject: Re: [OT] I hate being American by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0NNrDU07061; Is this not supposed to be a forum of mathematical topics? Why this digression, then? Joseph A. in >>message > You are either ignorant or were educated by revisionists. Krushchev >in particular would have been extremely pleased to nuke the US if he >thought he could have gotten away with a pre-emptive strike. But MAD won >out; he knew in his bones that if he struck first, World Communism would >have been permamently killed, and even at his table-pounding worst he >couldn't allow that. > > Mark L. Fergerson > The inventor of the machine gun and the inventor of dynamite both > thought that their gadget would make war so horrible that nobody would > dare start one. Oppenheimer and Teller finally pulled it off. > John >>So now they start them anyway, betting that nobody dares open Pandora's box. >Right. But only little ones. Nukes have probably been a stabilizing >force in the world. The major powers, the kind that start big, world >wars, have mostly skirmished by proxy. The interesting case now is >India and Pakistan. >John === Subject: Re: [OT] I hate being American > Oppenheimer and Teller finally pulled it off. With the help of several thousand scientists and workers, all under the direction of Gen. Groves. Heisenberg couldn't pull it off for Hitler. It was used selectively to end the Pacific bloodbath. The Rosenbergs sold the secrets to the USSR. Gen. Marshall advised President Truman against a war on the Asian mainland. He managed to get in anyway in Korea. He had to enlist McArthur to prevent US defeat. Mac turned it around and chased the N. Koreans to the Yalu River and confrontation with hundreds of thousands of Chinese troops. Mac recommended the use of tactical nuclear weapons. Truman fired him and got thousands of marines and army troops trapped in midwinter at the reservoir. They managed to escape to the sea with many causalities. Subsequently after more war and causalities the US signed a truce with a demilitarized zone. What would the world look like today if Mac had got his weapons which were only available to the US Army at that time. One result was that Truman was never elected to a second term. If we had a quality military leader, no democrat political intervention from Washington, and controlled the gutter traitors of street protests, we could have defeated North Viets quickly. We sometimes forget what a defeat we inflicted during their tet offensive. Democrats seem to want war for an `issue', but are reluctant to win one. === Subject: Re: [OT] I hate being American >> Oppenheimer and Teller finally pulled it off. >With the help of several thousand scientists and workers, all under >the direction of Gen. Groves. Heisenberg couldn't pull it off for >Hitler. >It was used selectively to end the Pacific bloodbath. The Rosenbergs >sold the secrets to the USSR. Gen. Marshall advised President Truman >against a war on the Asian mainland. He managed to get in anyway in >Korea. He had to enlist McArthur to prevent US defeat. Mac turned it >around and chased the N. Koreans to the Yalu River and confrontation >with hundreds of thousands of Chinese troops. Mac recommended the use >of tactical nuclear weapons. Truman fired him and got thousands of >marines and army troops trapped in midwinter at the reservoir. MacArthur was fired because he stopped obeying his commander in chief. Truman knew that MacArthur was going to be trouble when MacArthur didn't salute him. Truman had experience in the military when he fought in WWI. > ..They >managed to escape to the sea with many causalities. Subsequently >after more war and causalities the US signed a truce with a >demilitarized zone. What would the world look like today if Mac had >got his weapons which were only available to the US Army at that time. Atom bombs wouldn't have killed every Chinese soldier. I just read about the Koreas and the guy in North Korea apparently set himself up as head of his own religion; this was from a US foreign studies think tank and I haven't figured out the bias yet. >One result was that Truman was never elected to a second term. Truman served two terms. The US decided that a three-term presidency was not a good idea. It became an amendment. >If we had a quality military leader, no democrat political >intervention from Washington, and controlled the gutter traitors of >street protests, we could have defeated North Viets quickly. We >sometimes forget what a defeat we inflicted during their tet >offensive. Democrats seem to want war for an `issue', but are >reluctant to win one. The Viet Nam conflict was something that my books call a village war. I've got Gen. Giap's book and intend to find out more about this. /BAH === Subject: Re: [OT] I hate being American > The Viet Nam conflict was something that my books call a village > war. I've got Gen. Giap's book and intend to find out more about > this. The U.S. lost that one because Lyndon ing Johnson would not bust the dykes and starve the beggars out. Nor would he use tac-nukes. That is what happens when one has a liberal pinko stinko comsymp president. What the U.S. needs to do now, that it is king of the -heap is to learn from the Romans during their prime and impose the Pax Americana on the world. In those (good old) days anyone giving the Roman government any cheek was visited by several cohorts of legions. The leigions were a mean lean kick-ass fighting machine (HOO-rah!) and ground any opposition into the mud. If thine enemy smite thee on the check rip his head of and down his neck. Then rape his wife and cut the throats of his children. Bob Kolker === Subject: Re: [OT] I hate being American >> The Viet Nam conflict was something that my books call a village >> war. I've got Gen. Giap's book and intend to find out more about >> this. >The U.S. lost that one because Lyndon ing Johnson would not bust the >dykes and starve the beggars out. Nor would he use tac-nukes. That is >what happens when one has a liberal pinko stinko comsymp president. Johnson?!! I figured that one of reasons Johnson's time in Nam was so bad was due to the fact that he and his cabinet played armchair generals. They would have a Tuesday breakfast meeting and decide what the military would bomb this week. Then send the orders. This not how to wage war. Even I, who know nothing about military matters, know this. /BAH Subtract a hundred and four for e-mail. === Subject: Re: [OT] I hate being American >> >> The Viet Nam conflict was something that my books call a village >> war. I've got Gen. Giap's book and intend to find out more about >> this. >The U.S. lost that one because Lyndon ing Johnson would not bust the >dykes and starve the beggars out. Nor would he use tac-nukes. That is >what happens when one has a liberal pinko stinko comsymp president. > Johnson?!! > I figured that one of reasons Johnson's time in Nam was so bad > was due to the fact that he and his cabinet played armchair > generals. They would have a Tuesday breakfast meeting and > decide what the military would bomb this week. Then send > the orders. This not how to wage war. Even I, who know > nothing about military matters, know this. > /BAH > Subtract a hundred and four for e-mail. The current administration is also in the armchair general business: === Subject: Re: [OT] I hate being American >> Oppenheimer and Teller finally pulled it off. > With the help of several thousand scientists and workers, all under > the direction of Gen. Groves. Heisenberg couldn't pull it off for > Hitler. > It was used selectively to end the Pacific bloodbath. The Rosenbergs > sold the secrets to the USSR. Gen. Marshall advised President Truman > against a war on the Asian mainland. He managed to get in anyway in > Korea. He had to enlist McArthur to prevent US defeat. Mac turned it > around and chased the N. Koreans to the Yalu River and confrontation > with hundreds of thousands of Chinese troops. Mac recommended the use > of tactical nuclear weapons. Truman fired him and got thousands of > marines and army troops trapped in midwinter at the reservoir. They > managed to escape to the sea with many causalities. Subsequently > after more war and causalities the US signed a truce with a > demilitarized zone. What would the world look like today if Mac had > got his weapons which were only available to the US Army at that time. > One result was that Truman was never elected to a second term. > If we had a quality military leader, no democrat political > intervention from Washington, and controlled the gutter traitors of > street protests, we could have defeated North Viets quickly. We > sometimes forget what a defeat we inflicted during their tet > offensive. Democrats seem to want war for an `issue', but are > reluctant to win one. Man! You are some politically-incorrect bastard, eh! We don't cotton to the truth 'round here much! -- Keith === Subject: Re: [OT] I hate being American >> You are either ignorant or were educated by revisionists. Krushchev >>in particular would have been extremely pleased to nuke the US if he >>thought he could have gotten away with a pre-emptive strike. But MAD won >>out; he knew in his bones that if he struck first, World Communism would >>have been permamently killed, and even at his table-pounding worst he >>couldn't allow that. > The inventor of the machine gun and the inventor of dynamite both > thought that their gadget would make war so horrible that nobody would > dare start one. Oppenheimer and Teller finally pulled it off. World wars, anyway. Mark L. Fergerson === Subject: Re: [OT] I hate being American > Furthermore, 'nuking' N Korea and Iran would kill a great number of > people in neighbouring countries. Is that just unfortunate 'collateral'? > Since when have USA citizens cared about killing people. They kill many > thousands of themselves every year. > What has prevented them from nuking since Hiroshima and Nagasaki is fall-out > on USA towns, cities and prairies. That and fear of damaging the oil fields. === Subject: Re: [OT] I hate being American > That and fear of damaging the oil fields. We (U.S.) import or acquire over 60 percent of our petroleum from non-middle eastern sources. Push comes to shove we could use coal to generate electricity. Bob Kolker === Subject: Re: [OT] I hate being American >> That and fear of damaging the oil fields. > We (U.S.) import or acquire over 60 percent of our petroleum from > non-middle eastern sources. Push comes to shove we could use coal to > generate electricity. That would certainly piss off the EuroPeon greenies! I say go for it! -- Keith <1104420898.eb8c37161c2b8c47f4a498360246e13e@bubbanews> <10t8dnjav3nvub7@corp.supernews.com> <10t8vamnad16j2d@corp.supernews.com> <33la9rF3urjj3U1@individual.net> <10tb4bhgb9e4a06@corp.supernews.com> <33lidrF40j0fhU1@individual.net> <10tbdpgjp9rcla0@corp.supernews.com> <33lv3dF41m22lU1@individual.net> <10tc5udmmevdnab@corp.supernews.com> <33mmiiF4099avU1@individual.net> <33oba4F41amfjU1@individual.net> <8M$aFIE$4v1BFwgu@jmwa.demon.co.uk> <357s9sF4guah4U1@individual.net> === Subject: Re: [OT] I hate being American >> We (U.S.) import or acquire over 60 percent of our petroleum from >> non-middle eastern sources. Push comes to shove we could use coal to >> generate electricity. >That would certainly piss off the EuroPeon greenies! I say go for it! The US has the largest repository of 'clean burning coal' known to exist on the planet, but our sexual predator president locked it up in a 'monument' in southern Utah. Do we all remember where the second largest repository is, and how much money the owners of that coal illegally gave to the pervert's re-election campaign? === Subject: Re: [OT] I hate being American > We (U.S.) import or acquire over 60 percent of our petroleum from > non-middle eastern sources. Push comes to shove we could use coal to > generate electricity. >>That would certainly piss off the EuroPeon greenies! I say go for it! > The US has the largest repository of 'clean burning coal' known to exist > on the planet, but our sexual predator president locked it up in a > 'monument' in southern Utah. Do we all remember where the second largest > repository is, and how much money the owners of that coal illegally gave > to the pervert's re-election campaign? OooH! OooH! can I guess? Please! Please! Tibet? ...err no... Taiwan? ...no, wait, wait I'm getting it... *Hong Kong*! ...or whoever owns it now. -- Keith === Subject: Re: multiply, divide by zero > 0 means nothing (having no value). > lets assume vacuum as zero. > now if we put in some air in it, it'll get some value say a. > so, vacuum+air=a > i.e. we have added something. > what do we do when we multiply or divide it with something. > please give me a similar example. > (Is the assumption correct i.e.vacuum=0 ?) > thank you in advance. If you mean mathematical zero, then 0 _doesn't_ mean having no value. And the rest of what you have written is silly as well. x*0=0 and x/0 is undefined. === Subject: Join help-for-students@yahoogroups.com = OK! If you need a quality solution of your assignments, i.e. to have them done by experienced professionals, subscribe to the help-for-students@yahoogroups.com The leading scientists who are this group members will help you to do your assignments almost on any subject. They also provide learning by example, i.e. demonstrative working out your problems for educational purposes. High quality is guaranteed, because it is paid service. Rates are flexible and reasonable. To subscribe just send an email to help-for-students-subscribe@yahoogroups.com You may unsubscribe at ant time by sending an email to help-for-students-unsubscribe@yahoogroups.com === Subject: Join help-for-students@yahoogroups.com = OK! If you need a quality solution for your assignments, i.e. to have them done by experienced professionals, join the help-for-students[AT]yahoogroups.com Leading scientists who make up this group will help you with your assignments, whatever the subject. They also provide learning by example, i.e. demonstrative working out your problems for educational purposes. They will help you find the answers and learn how to find them. High quality is guaranteed, because it is a paid service. Rates are flexible and reasonable. To subscribe just send an email to help-for-students-subscribe[AT]yahoogroups.com You may unsubscribe at any time by sending an email to help-for-students-unsubscribe[AT]yahoogroups.com === Subject: Join help-for-students@yahoogroups.com = OK! If you need a quality solution for your assignments, i.e. to have them done by experienced professionals, join the help-for-students@yahoogroups.com Leading scientists who make up this group will help you with your assignments, whatever the subject. They also provide learning by example, i.e. demonstrative working out your problems for educational purposes. They will help you find the answers and learn how to find them. High quality is guaranteed, because it is a paid service. Rates are flexible and reasonable. To subscribe just send an email to help-for-students-subscribe@yahoogroups.com You may unsubscribe at any time by sending an email to help-for-students-unsubscribe@yahoogroups.com === Subject: Re: Join help-for-students@yahoogroups.com = OK! Ahh, the super expert is back with yet another anonymous scam. >If you need a quality solution for your assignments, What sort of quality -- low? >i.e. to have them done by experienced professionals, Ah, so you're a bunch of PROFESSIONAL scammers. That's a relief. >join the help-for-students@yahoogroups.com >Leading scientists who make up this group will >help you with your assignments, whatever the subject. So you professional leading scientists occupy your time with doing students' homework? >They also provide learning by example, i.e. demonstrative working out >your problems for educational purposes. The primary educational purpose is clearly to provide solutions that the students can hand in as their own. Why be so coy rather than just coming out and SAYING that: We do homework for you. Reasonable rates. If not satisfied with our service, that's your hard luck. >They will help you find the answers and learn how to find them. How to find the answers: give us the problems to solve for you. >High quality is guaranteed, because it is a paid service. Oooo, that's right, you're PROFESSIONALS. >Rates are flexible and reasonable. >To subscribe just send an email to >help-for-students-subscribe@yahoogroups.com >You may unsubscribe at any time by sending an email to >help-for-students-unsubscribe@yahoogroups.com you sell your subscriber list to? -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: Join help-for-students@yahoogroups.com = OK! > If you need a quality solution of your assignments, > i.e. to have them done by experienced professionals, > subscribe to the help-for-students@yahoogroups.com > The leading scientists who are this group members > will help you to do your assignments almost on any subject. They also > provide learning by example, i.e. demonstrative working out your > problems for educational purposes. > High quality is guaranteed, because it is paid service. > Rates are flexible and reasonable. > To subscribe just send an email to > help-for-students-subscribe@yahoogroups.com > You may unsubscribe at ant time by sending an email to > help-for-students-unsubscribe@yahoogroups.com I should have thought that leading scientists would be too busy doing science to have time for a scheme like this. But perhaps I misunderstood. Who are they leading, and to where? (Sheep to the slaughter comes to mind.) :-| -- Bruce Weaver bweaver@lakeheadu.ca www.angelfire.com/wv/bwhomedir === Subject: Re: Join help-for-students@yahoogroups.com = OK! > > If you need a quality solution of your assignments, > i.e. [sic, comma missing] to have them done by experienced professionals, > subscribe to the help-for-students@yahoogroups.com > The leading scientists who are this group [sic] members > will help you to do your assignments almost on any subject. [sic -- Almost do the assignments?] They also > provide learning by example, i.e. [sic - idem] demonstrative working out your > problems for educational purposes. > High quality is guaranteed, because it is paid service. > Rates are flexible and reasonable. > > To subscribe just send an email to > help-for-students-subscribe@yahoogroups.com > You may unsubscribe at ant time by sending an email to > help-for-students-unsubscribe@yahoogroups.com > I should have thought that leading scientists would be too busy doing > science to have time for a scheme like this. But perhaps I > misunderstood. Who are they leading, and to where? (Sheep to the > slaughter comes to mind.) :-| - better not ask for help with essays. -- Rich Ulrich, wpilib@pitt.edu http://www.pitt.edu/~wpilib/index.html === Subject: derivatives trigonometri by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0KE3nU10370; I've got a problem with the derivative of this function: 1-(1/cos^2 x) If there is anybody who knows how to do it and would like to explain it, I would be grateful. Thanx! Tobias, Sweden === Subject: Re: derivatives trigonometri > I've got a problem with the derivative of this function: > 1-(1/cos^2 x) D[1 - 1/(cos^2 x)] D[1 - (cos x)^(-2)] 0 - D[ (cos x)^(-2) ] -D[ (cos x)^(-2) ] Apply the Chain Rule - (-2) (cos x)^(-3) * d[cos x] 2 (cos x)^(-3) * -sin x -2 (sin x) / (cos x)^3 CHECK: 1 - 1/(cos^2 x) = [cos^2 x - 1] / cos^2 x = - (sin^2 x)/(cos^2 x) = - tan^2 x D[ - tan^2 ] = -2 (tan x) D[tan x] by the Chain Rule = -2 (tan x) (sec^2 x) = -2 (sin x)/(cos x) * 1/(cos^2 x) -2 (sin x) / (cos x)^3 === Subject: Re: derivatives trigonometri > I've got a problem with the derivative of this function: > 1-(1/cos^2 x) I'd express it as 1 - (cos x)^{-2} = y(x), say, and apply the chain rule. dy/dx = (2)(cos x)^{-3} * (-sin x) = 2 sin x/(cos^3 x). === Subject: Re: derivatives trigonometri > I've got a problem with the derivative of this function: > 1-(1/cos^2 x) > I'd express it as 1 - (cos x)^{-2} = y(x), say, and apply the chain > rule. > dy/dx = (2)(cos x)^{-3} * (-sin x) = 2 sin x/(cos^3 x). Ooops. You lost the negative sign. === Subject: Re: derivatives trigonometri > I've got a problem with the derivative of this function: > 1-(1/cos^2 x) > If there is anybody who knows how to do it and would like to explain > it, I would be grateful. Thanx! Just do it out, step-by-step: d/dx[1 - (1/cos^2 x)] = d/dx[1] - d/dx[1/cos^2 x] (because differentiation is a linear operation) = 0 - d/dx[1/cos^2 x] Now use the chain rule to rewrite the remaining derivative in terms of derivatives you know. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: +1 - 1 + 1 - 1 + ... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0KE3nB10419; >> What is wrong with this proof? >> (i.e. in which step is there a fault?) >> 0 >> = >> (1 - 1) + (1 - 1) + (1 - 1) + ... >> = >> 1 + (- 1 + 1) + (-1 + 1) + ... >> = >> 1 + 0 + 0 + 0 + ... >> = >> 1 >> Clearly, 0 = 1. >> MM. >This is just one of the many bizarre and counterintuitive things that >happen with infinite series that are not absolutely convergent. You can >play around rearranging them and get (almost) whatever the hell you >like. divergent >series are the invention of the devil... By using them, one may draw >any conclusion one pleases... Indeed! In fact, in the same way you can 'prove' that 0=2, etc. very bizarre. This series does not converge, in fact, so it is false to say that it is equivalent to zero. This problem was in my Stuart calc text .... I just recalled it after a long while, and almost forgot how to disprove it (ha). Thx to your replies, MM. === Subject: Re: +1 - 1 + 1 - 1 + ... >you set at first 0 = (1 - 1) + (1 - 1) + (1 - 1) + ... >which is wrong, No, that's right. (1 - 1) + (1 - 1) + (1 - 1) + ... is exactly the same as 0 + 0 + 0 + ..., which certainly equals 0. The error is in the next step, regrouping the terms. You can't regroup the terms like that unless the series 1 - 1 + 1 - 1 ... converges, which it doesn't. (A lot of people are talking about how rearranging terms is not valid unless the series is absolutely convergent. Here we're not rearranging anything, just regrouping, and that doesn't require absolute convergence, just convergence: if the series a = a1 + a2 + a3 + ... converges then it follows that (a1 + a2) + (a3 + a4) + ... = a = a1 + (a2 + a3) + (a4 + a5) + ... .) >It is like you are assuming this series (1 - 1) + (1 - 1) + >(1 - 1) + ... will stop at a (1 -1) so you can get 0, but remember this is >infinite, it does not stop. >> What is wrong with this proof? >> (i.e. in which step is there a fault?) >> 0 >> = >> (1 - 1) + (1 - 1) + (1 - 1) + ... >> = >> 1 + (- 1 + 1) + (-1 + 1) + ... >> = >> 1 + 0 + 0 + 0 + ... >> = >> 1 >> Clearly, 0 = 1. >> MM. ************************ David C. Ullrich === Subject: Re: Equation of a circle by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0KE3mY10337; >My book says: >If a circle has centre (a,b) and radius r >and if then (x,y) is any point on the circle, we have using >Pythagoras' theorem, >(x-a)^2 + (y -b)^2 = r^2 >This is the general equation of a circle >--- >Well working with this is fine. What I am failing to understand is >what Pythagoras theorem on triangles has to do with a circle with a >single line from the centre of the circle to a point on the >circumference. >I would be very grateful if someone could explain it to me. Draw a diagram (didn't your book have one?) and *look* for the right angle triangle. Hint: r is the hypotenuse, |x - a| and |y - b| are the other two sides. === Subject: Re: Equation of a circle by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0KE3nH10423; >My book says: >If a circle has centre (a,b) and radius r >and if then (x,y) is any point on the circle, we have using >Pythagoras' theorem, >(x-a)^2 + (y -b)^2 = r^2 >This is the general equation of a circle >--- >Well working with this is fine. What I am failing to understand is >what Pythagoras theorem on triangles has to do with a circle with a >single line from the centre of the circle to a point on the >circumference. >I would be very grateful if someone could explain it to me. Hi Jo, a good way to think about this problem is that Pythagoras' equation gives you a relationship between two perpendicular distances and the length of their hypotenuse. The radius of your circle is r, a constant, so that the equation x^2 + y^2 = r^2 just says that no matter what I choose for coordinates x and y I must have that the sum of their squares is equal to the square of r (so I am forced to choose the correct x's and y's). Another way: choose a center point in the x-y plane, and also choose some distance r. Take all the points that lie a distance r from that center point ... these points exactly describe a circle (this condition comes from Pythagoras' theorem). Last: draw a picture, or, take a piece of paper, a tack, and string: stick the tack on the paper, and with the string tied to the tack and a pencil, see what figure you make if you always have to be the length of the string away from the tack. Always look at the problem in different ways, never settle on one solution. M.M. === Subject: Re: Is zero even or odd? > ... Halmos in his General > Topology... What's that? Guess (on the basis that they're both GTM): Kelley's General Topology or Halmos's Measure Theory. === Subject: Re: Is zero even or odd? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0KHpA532711; Ryskamp, John Henry, G.9adel's Theorem Disproved (January 19, 2005). http://ssrn.com/abstract=651382 >> See: >> Cohen,P.J., The Independence of the Continuum Hypothesis. Proc. Nat. >> Acad. Sci. U.S.A. 50 1143-148, 1963. > The independence of the continuum hypothesis has no apparent >relation to your statement. Note that the axiom of choice does not >imply the continuum hypothesis. The generalized continuum hypothesis >does imply the axiom of choice. === Subject: Monte Hall Problem... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0L1P2a21486; I know you guys have probably discussed this to death, but I have just come across this tonight, and I have to say that the odds of winning are 1 in 2... No matter how you cut it or what the claims are about a simulation... 1 in 2, PERIOD... but I would like to know if someone has a site they can recommend for me to see what the reasoning is for saying that one has a better chance of winning if they switch doors... B === Subject: Re: Monte Hall Problem... > I know you guys have probably discussed this to death, but I have just > come across this tonight, and I have to say that the odds of winning > are 1 in 2... No matter how you cut it or what the claims are about a > simulation... 1 in 2, PERIOD... but I would like to know if someone > has a site they can recommend for me to see what the reasoning is for > saying that one has a better chance of winning if they switch doors... Despite all your protestations, the fact is, as others have explained, that you DO increase your chances of winning by changing your choice. To get a more intuitive understanding of why this is the case, you might find it helpful to consider the situation where there are many more doors. Say 100 doors, one of which conceals a car, and 99 of which conceal a goat. You choose a door. Monty then opens 98 of the remaining 99 doors that DON'T conceal the car. This leaves two doors unopened: the one that you originally chose, plus the one that Monty left unopened. It should be obvious that there is a vastly greater chance that the car is behind the door that Monty left unopened than behind the door you originally chose. Therefore you greatly increase your chances of winning by changing your choice. It's exactly the same deal with three doors... just not so glaringly obvious. === Subject: Re: Monte Hall Problem... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0LNQ4r03848; and of course, on average, I won about 2/3 of the time when I switched doors... But, I had to finally create a little card game for myself before I fully understood the reasoning behind it... Ultimately what it comes down to is that I am not simply trading my unopened door for Monte's unopened door and thereby making an exact exchange... I am in fact trading my unopened door for BOTH of Monte's doors, and he has done me the favor of eliminating ONE of the losers, thereby leaving responses... B === Subject: Re: Monte Hall Problem... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0N1ADZ30874; Yeah, that is funny... In fact, that is why the concept is hard to grasp if you think of it intuitively instead of mathematically... As for the game, that is EXACTLY the game I came up with the ace... It was an easier way to grasp the idea... For some reason in the Monte Hall gameshow scenario, I was inclined to see Monte as a trivial outside influence that had nothing to do with the odds, however, his foreknowledge of the winning door is key to the odds change... B === Subject: Re: Monte Hall Problem... <30n5wlj0w6gw@legacy> > and of course, on average, I won about 2/3 of the time when I switched > doors... But, I had to finally create a little card game for myself > before I fully understood the reasoning behind it... Ultimately what > it comes down to is that I am not simply trading my unopened door for > Monte's unopened door and thereby making an exact exchange... I am in > fact trading my unopened door for BOTH of Monte's doors, and he has > done me the favor of eliminating ONE of the losers, thereby leaving > responses... Yes, that reminds me actually ... there is a well-known card trick with very similar properties. Quite possibly the one you discovered. Three cards, one of which is an ace, are placed face down. You point at one of the cards. The probability that you've chosen the ace is 1/3. Then someone peeks at the other two cards and turns over one that ISN'T the ace. What is the probability now that you are pointing at the ace? Many people instinctively feel that it has risen to 1/2, but in fact it has remained at 1/3, and the probability that the OTHER face-down card is the ace is now 2/3. === Subject: Re: Monte Hall Problem... Choose car door: - host shows other door -- stay - get car -- switch - get goat Choose goat door: - host shows other door -- stay - get goat -- switch - get car Choose goat door: - host shows other door -- stay - get goat -- switch - get car So: staying - 1/3 chance of car switching - 2/3 chance of car -- Casey === Subject: Re: Monte Hall Problem... > I know you guys have probably discussed this to death, but I have just > come across this tonight, and I have to say that the odds of winning > are 1 in 2... No matter how you cut it or what the claims are about a > simulation... 1 in 2, PERIOD... You're not stupid for getting the incorrect answer of 1 out of 2, but eventually you'll learn how stupid it is to insist you're right no matter what, PERIOD, etc. === Subject: Re: Monte Hall Problem... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0LDBW312454; >> I know you guys have probably discussed this to death, but I have just >> come across this tonight, and I have to say that the odds of winning >> are 1 in 2... No matter how you cut it or what the claims are about a >> simulation... 1 in 2, PERIOD... >You're not stupid for getting the incorrect answer of 1 out of 2, >but eventually you'll learn how stupid it is to insist you're >right no matter what, PERIOD, etc. It may be helpful to consider the monty hall problem in an exaggerated context. Suppose instead of three doors, the game show now consists of 1 million doors. Behind all but one is a goat, and only one door contains a treasure. So, at random you pick, say, door #545. The host then opens up all of the doors except door #545 and door #77348 and behind them are goats. He says now, behind one of these 2 unopened doors remains the prize... do you want to keep your initial selection (545) or switch to the other unopened door (77348). YOU'D SWITCH, RIGHT? Of course you would, because the probability that the prize is behind door 545 is 1 in a million. However, the probability that the prize is behind door 77348 is 999999 in a million. It's been a while since i've seen the exact monty hall problem, but if its anything like i remember it, then this example should help. Joseph A. === Subject: Re: Monte Hall Problem... > I know you guys have probably discussed this to death, but I have just > come across this tonight, and I have to say that the odds of winning > are 1 in 2... No matter how you cut it or what the claims are about a > simulation... 1 in 2, PERIOD... > You're not stupid for getting the incorrect answer of 1 out of 2, > but eventually you'll learn how stupid it is to insist you're > right no matter what, PERIOD, etc. Even if you're right, it does you no harm to switch (1/2 = 1/2). So with all the experts saying switch, why would you stay? === Subject: Re: Monte Hall Problem... > I know you guys have probably discussed this to death, but I have just > come across this tonight, and I have to say that the odds of winning > are 1 in 2... No matter how you cut it or what the claims are about a > simulation... 1 in 2, PERIOD... And that's where you are wrong. Anyways, how about a plain old brute force approach? Label the doors A, B, C. Assume the prize is behind A. Then, if you pick A, Monty will open one of B or C (doesn't matter which). If you stay you win, if you switch you lose. If you pick B, Monty will open C. If you stay you lose, if you switch you win. If you pick C, Monty will open B. If you stay you lose, if you switch you win. By simple tabulation, if you stay you will only win in one of the three scenarios. If you switch, you will win in two of the three scenarios. Since each scenario is equally likely, that means if you stay you will win 1/3 of the time but if you switch you will win 2/3 of the time. Therefore you should switch. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Monte Hall Problem... >I know you guys have probably discussed this to death, > but I have just come across this tonight, and I have to > say that the odds of winning are 1 in 2... No matter how > you cut it or what the claims are about a simulation... 1 in 2, > PERIOD... but I would like to know if someone has a site > they can recommend for me to see what the reasoning is for > saying that one has a better chance of winning if they switch > doors... Simulations all disagree with your conclusion. Some very good mathematicians also got it wrong though. Here's the story. Let's assume Monte always opens a door, which has nothing behind it and invites you to switch. Originally, you had a 1/3 chance of winning. 1/3 of the time you picked the winning door. The other 2/3 of the time you picked a losing door. In this case, when Monte opens another door, if you switch, you win. So, switching wins 2/3 of the time. === Subject: Re: Monte Hall Problem... days. My association with the Department is that of an alumnus. >>I know you guys have probably discussed this to death, >> but I have just come across this tonight, and I have to >> say that the odds of winning are 1 in 2... No matter how >> you cut it or what the claims are about a simulation... 1 in 2, >> PERIOD... but I would like to know if someone has a site >> they can recommend for me to see what the reasoning is for >> saying that one has a better chance of winning if they switch >> doors... >Simulations all disagree with your conclusion. >Some very good mathematicians also got it >wrong though. >Here's the story. >Let's assume Monte always opens a door, >which has nothing behind it and invites you >to switch. And here's the problem: you have to be ->very very careful<- stating the assumptions to the problem. If Monte always opens an unselected door ->at random<- and this time it just happens to have nothing behind it, and he always offers you to switch when he has revealed a door without a prize, then switching does not change your odds of winning. If Monte ->always<- opens a door which has no prize in it (because he ->knows<- where the prize really is), and ->always<- offers you to switch, then switching improves your chances. If Monte sometimes opens doors and sometimes offers a switch, then you go into psychological evaluations and the odds are impossible to calculate. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: How do i set this up as an equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0L2PQo27025; George's first trip was 300 miles. If on his second trip, the time necessary to complete the trip is doubled and his average speed is two-thirds the average speed of his first trip, how many miles is this second trip? === Subject: Re: How do i set this up as an equation > George's first trip was 300 miles. If on his second trip, the time > necessary to complete the trip is doubled and his average speed is > two-thirds the average speed of his first trip, how many miles is this > second trip? Let S1 be the speed of the first trip and S2 be the speed of the second trip. Let D2 be the length of the second trip. Let T1 be the completion time of the first trip and T2 be the completion time of the second trip. What is T1 in terms of 300 miles and S1? What is T2 in terms of D2 and S2? What do we know about how T1 and T2 relate in terms of each other? What do we know about how S1 and S2 relate in terms of each other? How can we use all that to solve for D2 in terms of the given information? -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: How do i set this up as an equation > George's first trip was 300 miles. If on his second trip, the time > necessary to complete the trip is doubled and his average speed is > two-thirds the average speed of his first trip, how many miles is this > second trip? Let the time for the first trip be t, then the average speed of the first trip is (300 miles)/t. The time to complete the second trip is 2t, and the average speed of the second trip is 2(300 miles)/(3t). The distance of the second trip is average speed * time, i.e. (2(300 miles)/(3t)) * 2t = 400 miles. === Subject: Re: How do i set this up as an equation > George's first trip was 300 miles. If on his second trip, the time > necessary to complete the trip is doubled and his average speed is > two-thirds the average speed of his first trip, how many miles is this > second trip? D = r * t 300 = r*t x = (2/3)r*(2t) x = (4/3)rt But rt = 300 x = (4/3)300 = 400 miles === Subject: Logrithmic Functions by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0L2PRi27044; How would you solve the equation, 8+(1/4)e^(x/2)=450 === Subject: Re: Logrithmic Functions > How would you solve the equation, 8+(1/4)e^(x/2)=450 I would first isolate the exponential: (1/4)e^(x/2) = 442 e^(x/2) = 1768 Then I would rewrite the equation in logarithmic form: x/2 = ln(1768) x = 2*ln(1768) === Subject: Re: Logrithmic Functions > How would you solve the equation, 8+(1/4)e^(x/2)=450 8+(1/4)e^(x/2) = 450 (1/4)e^(x/2) = 442 e^(x/2) = 1768 x/2 = log 1768 x = 2(log 1768) Now, _I_ think that that's the answer, but some will tell you that the answer is x = 14.9552 or something similarly false. PS: That log is log base e of course. Some write it as ln. Yuk! === Subject: Re: Logrithmic Functions > 8+(1/4)e^(x/2)=450 (1/4)e^(x/2) = 442 e^(x/2) = 1768 ln(e^(x/2)) = ln(1768) (x/2)ln(e) = ln(1768) x/2 = ln(1768) x = 2ln(1768) === Subject: Identity with Stirling nr.of second kind In the following i) n,m are positive integers, ii) S(n,k) , k=0,1,... , are Stirling numbers if the second kind, i.e. S(n,j)=0 when j>= n+1 and x^n=SUM_{k=0 to k=n}S(n,k)x(x-1)...x(x-k+1) , iii) A = ||a_{i,j}|| is the k x k , (k>=3) matrix having the i-th row i i^2 ...i^{k-2} i^{n} i^{m} . Prove or disprove that det(A)= ( S(k,m)*S(k-1,n) - S(k,n)*S(k-1,m) )*( 1!2!...k! ). === Subject: Distance Functions (Real Analysis Question) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0LDBWl12462; Okay, I'm probably overlooking something simple, but here's my problem: Let u be a point in Euclidean n-space and let r be a positive number. Suppose that the points v and w in n-space are at a distance less than r from the point u. Prove that if 0 <= t <= 1, then the point t*v + (1-t)*w is also at a distance less than r from u. (Note that u, v, and w are vectors and * represents scalar mult.) -- alright so here's how i've been fiddling with it: using reverse triangle inequality, i find that |u| - |v| <= |u - v| < r |u| - |w| <= |u - w| < r so that |u| < r + |v| and so |u| - |w| < r + |v| - |w|... from this point im sort of lost.. im not sure if this is even the right way to look at it, considering i havent even introduced t into the picture. some observations ive made: when t = 0: |u - v| < r when t = 1: |u - w| < r which are both true. its proving what happens it between the endpts of t is what i cant show analytically. any suggestions?? Joseph A. === Subject: Re: Distance Functions (Real Analysis Question) >Okay, I'm probably overlooking something simple, but here's my >problem: >Let u be a point in Euclidean n-space and let r be a positive number. >Suppose that the points v and w in n-space are at a distance less than >r from the point u. Prove that if 0 <= t <= 1, then the point t*v + >(1-t)*w is also at a distance less than r from u. >(Note that u, v, and w are vectors and * represents scalar mult.) In the stuff you did it doesn't look like you started with the obvious thing. You want to show that t*v + (1-t)*w is also at a distance less than r from u. This means that you want to show that |u - (t*v + (1-t)*w)| < r. But we don't see the quantity |u - (t*v + (1-t)*w)| anywhere in what you've done so far! Why not start with |u - (t*v + (1-t)*w)| and try to show that it's less than r? Hint: u = t*u + (1-t)*u. ************************ David C. Ullrich === Subject: Re: Distance Functions (Real Analysis Question) >Okay, I'm probably overlooking something simple, but here's my >problem: >Let u be a point in Euclidean n-space and let r be a positive number. >Suppose that the points v and w in n-space are at a distance less than >r from the point u. Prove that if 0 <= t <= 1, then the point t*v + >(1-t)*w is also at a distance less than r from u. >(Note that u, v, and w are vectors and * represents scalar mult.) When calculating the distance, write u = tu + (1 - t)u and use the triangle inequality. --Lynn === Subject: Re: riddle by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0LDBWr12416; >the person who makes it doesnt use it, the person who uses it doesn't >know that he's using it, but everyone needs it!!!! >wat is the It ? A coffin === Subject: probability by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0LDBVX12374; pls.. discuss the following questions or satatements below. 1. How is probability started? Explain briefly. 2. enumerate and ezplain the succesful attempts to formalize probability === Subject: WHY?! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0LF3Er23483; hello The way we find common points of parabola and linear line ax^2+bx+c=kx+n ax^2+(b-k)x+(c-n)=0 equation doesn't change if you add or substract from both sides.But from quadratic and linear equations point of view this doesn't makes sense to me. Why would combining the two equations give you NEW quadratic equation that just happens to have same x1 and x2 points as ax^2+bx+c and kx+n? thank you === Subject: Re: WHY?! > hello > The way we find common points of parabola and linear line > ax^2+bx+c=kx+n > ax^2+(b-k)x+(c-n)=0 > equation doesn't change if you add or substract from both sides.But > from quadratic and linear equations point of view this doesn't makes > sense to me. > Why would combining the two equations give you NEW quadratic > equation that just happens to have same x1 and x2 points as > ax^2+bx+c and kx+n? It's just how the algebra works out... The point(s) of intersection of the parabola y = ax^2+bx+c and the straight line y = kx+n (if any) are those points (x,y) such that x and y satisfy BOTH equations (i.e. the points lie on the parabola AND the line). To find these point(s), solve the equations simultaneously for x and y, exactly as you have started to write. Solving the resulting equation ax^2+(b-k)x+(c-n)=0 for x gives the a, b, c, k and n). Then plug the x-values back into either equation to get the corresponding y-values. If you get this much, then I'm not sure what's left to understand. Taking a wild guess ... POSSIBLY what your missing is that the resulting quadratic equation, ax^2+(b-k)x+(c-n)=0, is a completely different beast to the original equation for the parabola. In the former case, x is an independent variable which can take on any value to give a corresponding value for y. In the latter case x CANNOT just take on any old value. To satisfy the equation ax^2+(b-k)x+(c-n)=0, the variable x can take only one of two SPECIFIC values, which are the You are therefore NOT combining a parabola and a line to get some sort of new parabola. === Subject: Differential Operators question by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0LFFdV24414; Hi folks, can anyone tell me how to go about working out the following: 1/(D^2+4D+5){x} i.e. that is carry out the operation as defined by the differential operators on x. It is the sort of thing you would encounter when solving second order DEs. Any help much appreciated Vicky === Subject: Re: Differential Operators question by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0LHUpe05325; >Hi folks, >can anyone tell me how to go about working out the following: >1/(D^2+4D+5){x} i.e. that is carry out the operation as defined by >the differential operators on x. It is the sort of thing you would >encounter when solving second order DEs. >Any help much appreciated >Vicky We may observe that D^n o x^(n-1)= 0 (1) n>= 1, so D^2 o x = 0 ,and here terms D^k k>=2 must be ignored. We'd better write Id/(D^2+4D+5Id) ;Id neutral operator , Id=D^0. Ready-to-compute 1/5*Id/(Id+4D/5+D^2/5) develop like 1/(1+x), For reason (1) neglect D^2/5 => 1/5*Id/(Id+4D/5) o x or 1/5*(Id-4D/5) o x = x/5-4/25. Control: (D^2+4D+5Id)o (x/5-4/25)=(4D o x/5)+(5Id O x/5)-(5Id O 4/25) = 4/5 + 5x/5 - 20/25, = x Hope it is clear... Friendly,Alain. === Subject: |f(z)|<|tgz| by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0LGHMv30757; Hi everybody. Could you please help me with solving the following problem: Function f(z) is analytic in the strip |Rez|Hi everybody. >Could you please help me with solving the following problem: >Function f(z) is analytic in the strip |Rez|f(0)=0. Prove that |f(z)|<|tgz|; What is tgz ? >I'll be grateful for everyone who could help me. ************************ David C. Ullrich === Subject: Re: |f(z)|<|tgz| >Hi everybody. >Could you please help me with solving the following problem: >Function f(z) is analytic in the strip |Rez|f(0)=0. Prove that |f(z)|<|tgz|; > What is tgz ? It's tan(z). Why anyone would write f(z) and tgz in the same sentence eludes me. Hint to student: First, the inequality cannot be strict everywhere; f(0) = 0 = tan(0) (and what if f(z) = tan(z) everywhere?). Second, consider the function f(z)/tan(z). It's analytic at 0 and you should be able to use the maximum modulus theorem by looking at what's happening near the boundary of the region. === Subject: Linear Operations of transfer functions I am trying to figure out how to determine if a giving function is linear or not. Currently I am given the following function and I have to determine whether its linear or non-linear: y = sin (2x + 4) The example that I have shows that if you have something like O(x) then to determine if its linear or not you replace x with x1 + x2 so you will have: O(x1 + x2) = O(x1) + O(x2) which is linear So trying to apply the same method to the function I have to work with I replace x with x1 + x2 and get the following: y = sin [2(x1 + x2) + 4] How do you determine if this is linear. Is there another way to do this instead of replacing x with (x1 + x2)? Steve === Subject: Re: Linear Operations of transfer functions > Currently I am given the following function and I have to determine > whether its linear or non-linear: > y = sin (2x + 4) > The example that I have shows that if you have something like O(x) then > to determine if its linear or not you replace x with x1 + x2 so you > will have: > O(x1 + x2) = O(x1) + O(x2) which is linear > So trying to apply the same method to the function I have to work with > I replace x with x1 + x2 and get the following: > y = sin [2(x1 + x2) + 4] So follow your definition of linear. Does sin[2(x1 + x2) + 4] = sin(2x1 + 4) + sin(2x2 + 4) ? Use the sine-of-a-sum formulae to expand both sides and see if they are equal. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Linear Operations of transfer functions >So follow your definition of linear. >Does sin[2(x1 + x2) + 4] = sin(2x1 + 4) + sin(2x2 + 4) ? >Use the sine-of-a-sum formulae to expand both sides and >see if they are equal. Or, simpler, if you suspect they aren't equal try a value or two of the variables, like perhaps 0. --Lynn === Subject: Re: Proving Associativity/Distributivity in Ord Arith Hi Brian, it because I finally understood how all of these concepts work together. As I sarted working though the proof I found a small follow-up question about the following passage: > In your problem the sets A, B, and C are the ordinals alpha, > beta, and gamma, and you have to show that h is not just a > bijection, but also an order-isomorphism. Let a' in A, b' > in B, and c' in C, and suppose that > > < > [...] Here do you mean that the sets A, B, C are the ordinals or that these sets A, B, C have ordinal numbers alpha, beta, gamma. This is a confusing point for me because ordinals are supposedly transitive sets so that the members of the ordinals are also sets. Or perhaps the set is the same as the ordinal? === Subject: Re: Proving Associativity/Distributivity in Ord Arith in in alt.math.undergrad: [...] >> In your problem the sets A, B, and C are the ordinals alpha, >> beta, and gamma, and you have to show that h is not just a >> bijection, but also an order-isomorphism. Let a' in A, b' >> in B, and c' in C, and suppose that >> > < > > [...] > Here do you mean that the sets A, B, C are the ordinals or > that these sets A, B, C have ordinal numbers alpha, beta, > gamma. This is a confusing point for me because ordinals > are supposedly transitive sets so that the members of the > ordinals are also sets. Or perhaps the set is the same as > the ordinal? I meant that the sets A, B, and C are the ordinals. That is, I was just renaming them, since it seemed easier to write 'A' than 'alpha', etc. It appears from your comment about ordinals being transitive sets that you're following the usual modern approach in which each ordinal is the set of smaller ordinals. (When they're defined this way, they're sometimes called von Neumann ordinals.) With this approach you eventually end up having 0 = ø (the empty set); 1 = {0}, 2 = {0, 1} = {0, {0}}; omega = {0, 1, 2, ...}, the set of all finite ordinals; omega_1 being the set of all countable ordinals; and so on. And with this approach the following are equivalent for any alpha and beta that are ordinals: (1) alpha < beta; (2) alpha is a subset of beta; and (3) alpha is an element of beta. Brian === Subject: Theoretical average trials in a biased random walk by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0LLKcA26141; What would be the theoretical average number of trials in a random walk that had a bias of -1.4 % for one of the coins flipped. The walk starts with the walker at 0, with -1 being one boundary, with no upper boundary. JK === Subject: Re: Theoretical average trials in a biased random walk > What would be the theoretical average number of trials in a random > walk that had a bias of -1.4 % for one of the coins flipped. The walk > starts with the walker at 0, with -1 being one boundary, with no upper > boundary. > JK To my mind this problem isn't very clearly stated. 1. I don't understand what a bias of -1.4 % for one of the coins means. Do you have multiple coins, one of which is biased? How do you decide when to flip the biased coin? 2. It's not clear (to me) what a bias of -1.4% means anyway, in terms of probabilities. (This is not critical though, as one would solve the problem generally for probabilities p and 1-p, and then plug in the appropriate p to get a specific solution.) 3. You don't explain how the random walk proceeds or how it ends. I'm guessing ... does a head count as +1 and a tail as -1 (or vice versa)? Does the walk end when the -1 boundary is hit? === Subject: Rational Functions and their series expansions by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0LNQ3L03844; Hi everyone, can anyone tell me if it is possible to obtain the series expnasion for a rational function such as 1/(4x^2-3x+9)just by ordinary division - i.e. not by using Taylor Series. If so how would I go about doing it. Dave === Subject: Re: Rational Functions and their series expansions > can anyone tell me if it is possible to obtain the series expnasion > for a rational function such as 1/(4x^2-3x+9)just by ordinary division > - i.e. not by using Taylor Series. If so how would I go about doing > it. like ordinary long division, except powers of x in the divisor are arranged in order from lowest to highest. 1/9 + (1/27)x - (1/27)x^2 --------------------------------------- 9-3x+4x^2 ) 1 1 -(1/3)x+(4/9)x^2 ---------------------- (1/3)x-(4/9)x^2 (1/3)x-(1/9)x^2+(4/27)x^3 ------------------------------ -(1/3)x^2-(4/27)x^3 -(1/3)x^2+(1/9)x^3 - (4/27)x^4 -------------------------------- etc. === Subject: Re: Rational Functions and their series expansions > can anyone tell me if it is possible to obtain the series expnasion > for a rational function such as 1/(4x^2-3x+9)just by ordinary division > - i.e. not by using Taylor Series. If so how would I go about doing > it. try to factorize the denominator polynomial. use this to write the inverse polynomial a1/(x-b1)+a2/(x-b2). then apply the geometric series to know the Taylor expansion of 1/(x-c). i hope that helps you. j. === Subject: Going Diophrantic would be interested to know a method, of obtaining integer solutions to a^3 + b^3 = 22*c^3 This is not homework; I came across the problem in a popular maths book and realised I do not know how to approach the problem apart from the trial and error methods. Tried a^3 + b^3 = (a +b)(a^2-ab +b^2) and experimented with (a +b) = 22 leading to search for integer values of c = (3a^2 -66a +484)^(1/3) Graphing only indicated c>5, and not easy to see integer a and c from this. Experimenting with (a^2-ab +b^2) = 22 did not deliver integer values either. Would appreciate if someone could lay out the proper way of doing this. === Subject: Re: Going Diophrantic by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0NGOGo04136; Hi Ian, If you really want to findout I sugest to you to send a E-mail to Professor Andrew Granville. Heis a specialist in this field and you can get from him references which you look for.HisE-mail address is: andrew@dms.umontreal.ca george ghiata > would be interested to know a method, of obtaining integer >solutions to >a^3 + b^3 = 22*c^3 >This is not homework; I came across the problem in a popular >maths book and >realised I do not know how to approach the problem apart >from the trial and error methods. >Tried a^3 + b^3 = (a +b)(a^2-ab +b^2) and experimented with >(a +b) = 22 leading to search for integer values of c = >(3a^2 -66a +484)^(1/3) >Graphing only indicated c>5, and not easy to see integer a >and c from this. >Experimenting with (a^2-ab +b^2) = 22 did not deliver >integer values either. >Would appreciate if someone could lay out the proper way of >doing this. === Subject: Re: Going Diophrantic 22 January 2005 12:15 PM Hi William, are you saying that posting to two groups is not the right thing to do, or are you telling me that there is a better way to do it ? I must admit I am a bit vague about how to post to two groups with a single action, if that is what you mean. Sorry if I have done the wrong thing but posting to two different groups seemed like a good idea to me, because there may be some people who have subscribed to one group but not the other. I do not see why that should be a problem, really. Perhaps I misunderstand you. Care to explain, or provide a reference advising best practise ? === Subject: Re: Going Diophrantic > 22 January 2005 12:15 PM > are you saying that posting to two groups is not the right > thing to do, or are you telling me that there is a better > way to do it ? I must admit I am a bit vague about how to > post to two groups with a single action, if that is what you > mean. Post to the desired groups with a single post. That way you only look one place and everybody else know everybody's answer. Otherwise some may get pissed to give answer and then find in the next group, the problem has alread by answered. I'll try it again, didn't remember other newsgroup correctly before. This should appear in both groups with my posting only from one. If you reply, reply will be common to both groups. === Subject: Re: Going Diophrantic It's netequete to crosspost. Responses will not be re-invented or overlap in separate threads. To learn how to do it, simply find a post where it has been done and begin a reply to the newsgroups. === Subject: Re: Going Diophrantic Same question with answers are at alt.math.recreation. Ian, please learn about cross posting as I've done here. > would be interested to know a method, of obtaining integer > solutions to > a^3 + b^3 = 22*c^3 > This is not homework; I came across the problem in a popular > maths book and > realised I do not know how to approach the problem apart > from the trial and error methods. > Tried a^3 + b^3 = (a +b)(a^2-ab +b^2) and experimented with > (a +b) = 22 leading to search for integer values of c = > (3a^2 -66a +484)^(1/3) > Graphing only indicated c>5, and not easy to see integer a > and c from this. > Experimenting with (a^2-ab +b^2) = 22 did not deliver > integer values either. > Would appreciate if someone could lay out the proper way of > doing this. === Subject: Re: Going Diophrantic It's netequete to crosspost. Responses will not be re-invented or overlap in separate threads. To learn how to do it, simply find a post like this one, where it has been done and begin a reply to the newsgroups. === Subject: Re: Going Diophrantic It's netequete to crosspost. Responses will not be re-invented or overlap in separate threads. To learn how to do it, simply find a post like this one, where it has been done and begin a reply to the newsgroups. === Subject: ti-83+ and complex eigen(values/vectors) I'm trying to use my ti-83+ to find complex eigenvalues and vectors. So far I am able to do this graphically. I graph the characteristic polynomial and find it's minimum vertex (for a 2x2 matrix). example: [A]=(3,-2),(2,3) Characteristic polynomial is x^2-6x+13 graphing this and finding the minimum values gives you the point (3,4) Now this does not lok like a complex point unless you say that a+bi = X+sqrt(-1Y) So the point (3,4) becomes 3+2i and 3-2i. having found these eigenvalues I now want to find there eigenvectors. I realize that doing this for a 2X2 matrix is relatively easy but I want to be able to do it for larger matrixes. Finally my real question is if it is possible to tell a ti-83+ calculator to translate that point, the eigenvalue, to its coresponding eigenvectors. === Subject: Vedic math methods by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0MDo7p04798; try this out.. assume: (2x+5)/(2x+9)=(2x+9)/(2x+5) by current method -> by ancient vedic method -> => (2x+5)^2=(2x+9)^2 sum of numerator=sum of denominator => 4(x^2)+25+20x= 4(x^2)+81+36x i.e. 4x+14 => 20x+25=36x+81 equating to zero.. 4x+14=0 => 16x= (-56) 4x=(-14) => x= -56/16 = -14/4 = -7/2 => x= -14/4= -7/2 !! The above method is applicable to any equation whose sum of numerator and denominator is equal, provided there is no quadratic in the equation.(i.e. no term in x^2) calculating squares of 5 by vedic method.. we know that.. 5^2=25 if square of 45 is to be calculated mentally.. 85^2= 7225 it takes less than 5 seconds by this method. the last two digits of your answer are always 2&5 (5^2=25) to calculate the initial digits.. take the no. before 5 (i.e. 8 in this case) and multiply it with the no. succeding it (on the no. line/system) i.e. 9 so we get 8*9=72 7 & 2 are the first two digits of your ans. resp. try 75^2=5625 (7*8) this method is applicable to any no. ending with 5. === Subject: Re: Vedic math methods > try this out.. > assume: > (2x+5)/(2x+9)=(2x+9)/(2x+5) > by current method -> by ancient vedic method -> > => (2x+5)^2=(2x+9)^2 sum of numerator=sum of denominator > => 4(x^2)+25+20x= 4(x^2)+81+36x i.e. 4x+14 > => 20x+25=36x+81 equating to zero.. 4x+14=0 > => 16x= (-56) 4x=(-14) > => x= -56/16 = -14/4 = -7/2 => x= -14/4= -7/2 !! > The above method is applicable to any equation whose sum of numerator > and denominator is equal, provided there is no quadratic in the > equation.(i.e. no term in x^2) > calculating squares of 5 by vedic method.. > we know that.. 5^2=25 > if square of 45 is to be calculated mentally.. > 85^2= 7225 > it takes less than 5 seconds by this method. > the last two digits of your answer are always 2&5 (5^2=25) > to calculate the initial digits.. > take the no. before 5 (i.e. 8 in this case) and multiply it with the > no. succeding it (on the no. line/system) i.e. 9 so we get 8*9=72 > 7 & 2 are the first two digits of your ans. resp. > try 75^2=5625 (7*8) > this method is applicable to any no. ending with 5. === Subject: Re: lim x->a (x^3 - a^3)/x -a? > x^3 - a^3 = (x-a)(x^2+ax+a^2) N.B. ax not 2ax > Look in your book for sum of cubes and difference of cubes in the > section on factoring. > Your limit is the same as LIM[x->a, x^2+ax+a^2] = 3a^2. Ahh, thank you. I actually ended up solving this, but in a much more roundabout way. I multiplied out (x - a)^3 and got: x^3 - 3x^2a + 3xa^2 - a^3 lim x->a (x^3 - a^3 - 3x^2a + 3xa^2 + 3x^2a - 3xa^2) / (x - a) which (I think) factors into lim x->a [(x - a)^3 + 3xa( x - a )] / (x - a) or lim x->a [(x - a)^2 + 3xa] which works out 3a^2, which is the answer I suspected. I'll have to see if I can figure out your factorization, because it's === Subject: Re: lim x->a (x^3 - a^3)/x -a? > Do it the same way, by factoring the numerator. The formula for the > factoring of the difference of two cubes is: > x^3-a^3 = (x-a)(x^2+ax+a^2) > which can be found in a college algebra text, or perhaps on the inside > cover of your calculus book. Yep, I figured it out not too long after I posted this... it was the factorization of (x^3 - a^3) that had me stumped. Actually, my factorization was a little bit more roundabout - yours is Guess I should have stayed awake in class the first time around, eh? === Subject: Re: Is there an answer for this? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0N1ADk30881; Yes, there is an answer, but you will find that most on this site don't like to DO homework for others and if a question smells of 'homework assignment', you won't get very many helpful responses... However, I can give you a start to give you an idea of how to find out information that you might not have seen there... The 'M', for instance, must be 1... 'SEND' is 4 digits and 'MORE' is 4 digits, and the sum, 'MONEY', is 5 digits, so S+M>10 but S+M<20..., therefore the 'M' in 'MONEY' must be 1... B === Subject: Re: Combination lock by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0NGOGw04116; >Imagine a door lock with 10 buttons numbered 0 through 9 and an >unlocking code of four digits. >Each button can only be depressed once and it doesn't matter in what >order they are depressed so 5382 is the same as 2358 or 8532. >How many possible permutations are there? You have 4 selections with 10 choices each (0 through 9). Since order does not matter, we are not looking for permutations, but combinations rather. So there are 10C4 = 10!/(4!*6!) = 210 possible combinations. The probability of randomly guessing the right code to unlock the door is under 5%. >Moving on from that, how many permutations would I be able to create >by using the same lock with a 3 or 5 digit combination? 10C3 or 10C5 combinations, not permutations. Joseph A. === Subject: Re: Combination lock by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0N1ADu30840; >Imagine a door lock with 10 buttons numbered 0 through 9 and an >unlocking code of four digits. >Each button can only be depressed once and it doesn't matter in what >order they are depressed so 5382 is the same as 2358 or 8532. >How many possible permutations are there? >Moving on from that, how many permutations would I be able to create >by using the same lock with a 3 or 5 digit combination? >Philip Green, >Rotterdam - NL. Permutations not the correct term for you applications. Combinations is. For the first number you have 10 choices. For the second number you have 9 choices. For the third number you have 8 choices. For the four number you have 7 choices. Total number of combinations 10.9.8.7 = 10!/(10-4)! In general if you have n buttons and m choices in the combination then the total number of combinations possible is n!/(n-m)!. See http://mathworld.wolfram.com/Combination.html - MO === Subject: Re: Combination lock >>Imagine a door lock with 10 buttons numbered 0 through 9 and an >>unlocking code of four digits. >>Each button can only be depressed once and it doesn't matter in what >>order they are depressed so 5382 is the same as 2358 or 8532. >>How many possible permutations are there? >>Moving on from that, how many permutations would I be able to create >>by using the same lock with a 3 or 5 digit combination? >>Philip Green, >>Rotterdam - NL. >Permutations not the correct term for you applications. >Combinations is. >For the first number you have 10 choices. >For the second number you have 9 choices. >For the third number you have 8 choices. >For the four number you have 7 choices. >Total number of combinations 10.9.8.7 = 10!/(10-4)! Except this is the formula for the number of permutations of m choices out of n, often written as nPm. The difference between combinations and permutations, as you noted above, is that order does not matter fro combinations. In the example, you need to divide again by 4! to eliminate the duplicates such as 1234 and 1243. >In general if you have n buttons and m choices in the combination >then the total number of combinations possible is n!/(n-m)!. The correct formula for m combinations out of n, written nCm is n!/(m!*(n-m)!) >See http://mathworld.wolfram.com/Combination.html <> === Subject: Re: Combination lock by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0NGOGq04140; >>Imagine a door lock with 10 buttons numbered 0 through 9 and an >>unlocking code of four digits. >>Each button can only be depressed once and it doesn't matter in what >>order they are depressed so 5382 is the same as 2358 or 8532. >>How many possible permutations are there? >>Moving on from that, how many permutations would I be able to create >>by using the same lock with a 3 or 5 digit combination? >>Philip Green, >>Rotterdam - NL. >Permutations not the correct term for you applications. >Combinations is. >For the first number you have 10 choices. >For the second number you have 9 choices. >For the third number you have 8 choices. >For the four number you have 7 choices. >Total number of combinations 10.9.8.7 = 10!/(10-4)! That's total number of permutations. >In general if you have n buttons and m choices in the combination >then the total number of combinations possible is n!/(n-m)!. This isn't true. Order is not important here, so we account for that by dividing your n!/(n-m)! by m!. That is, nCm = (nPm)/m! = n!/(m!*(n - m)!). Right? Joseph A. === Subject: Dependent/independent variables? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0NNrAZ06938; Can some please explain to me in the simplest terms possible what the difference between dependent and independent variables are?? THANKS! M === Subject: Mechanics help needed by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0NNrBv06963; (position vector)r=[Acos(bt)]i + [Bsin(bt)]j with A>B, b is a constant and i,j are unit vectors Explain why |(posn vector)r dot| and |r dot| are not the same in general and find the maximum and minimum values of r. How do I find what just the normal r is (not the position vector)? Any help is much apppreciated