Challenge Problem - Page 2

jinydu · 2004-12-21, 11:16

Quote:

Originally Posted by wpolly

By Newton's Formula, we have that all the elementary symmetric polynomials of A1, A2, ... An is equal to zero.
Thus these A's are roots to the equation x^n=0 and thus are zero(By Vieta's Theorem).

Could you elaborate on that more please? Evidently, if I knew Newton's Formula (specifically, the formula you have in mind; Newton proved many formulas), I wouldn't consider this much of a challenge.

jinydu · 2004-12-22, 08:23

I did a search on Mathworld, and this is the closest thing I could find: http://mathworld.wolfram.com/Newton-GirardFormulas.html

wpolly · 2004-12-23, 03:53

Quote:

Originally Posted by jinydu

I did a search on Mathworld, and this is the closest thing I could find: http://mathworld.wolfram.com/Newton-GirardFormulas.html

That's exactly what I wanted to say.

jinydu · 2004-12-23, 04:48

But what is that symbol that looks like a big pi? It also resembles the product sign, but its missing text above and below the product sign, so I don't know to multiply from what to what.

Also, I'm pretty sure a (finite) polynomial is supposed to be a function in one variable that looks something like:

c(0) + c(1)*x + c(2)*x^2 + c(3)*x^3 + ... + c(n)*x^n

Orgasmic Troll · 2004-12-23, 08:59

Jinydu:

http://mathworld.wolfram.com/SymmetricPolynomial.html

mfgoode · 2004-12-23, 17:03

Jinydu: I have checked the websites given by yourself and Travis. Yes the eqn.s including large ‘ PI’s’ are rigorous and ultimate solutions but you will agree it is cloaked in high falutin formlae difficult to follow from one step to the other (refer to your recent post on large ‘PI’ in this very thread)
Studying the formulae given I find they are similar to what I’m expressing in simple mathematical language.

We must remember that the evolution of complex functions are based on the axioms applied to those on real numbers.
“God made the integers; the rest is the work of man” Kronecker.

Whatever axioms we derive on real numbers must be true in the complex numbers too.Otherwise the mathematics of it will not be consistent and contrary in its different disciplines..
Hence the set of eqn.s you have given are in contradiction if the other eqn.s are dependent on your 1st premise i.e. A +B =0.
Goven A + B =0 then A^2 +B^2 = 0 does not follow except if A and B are Zero

Similarly for the other eqn.s.
If A + B + C =0 then A^3 +B^3 + C^3 is only true if A = B = C =0.
Test Case:
If A + B = 0 then A^2 and B^2 = 0 cannot also be implied unless A=B=0 .
It will instead be 2*A*B.This is from simple algebra!

But what about complex numbers ?
Let A = (p + iq) and B =(x+iy).
If A + B =0 (given) then B can only have the value (- p –iq ). There is no other value it can be given.
Now substitute the complex values ( p + iq) and (-p - iq) into A^2 +B^ 2 = 0 and you will obtain the same result i.e. A^2 + B^ 2 = 2 AB (*sign dispensed with). If the L.H.S =0 then the R.H.S. gives either A=0 or B=0 and finally A =B =0.

Similarly if A + B + C =0 (given) unless A, B, C, are all zero the result will be valid only if you take the sum of the cubes as Zero.
Otherwise they will be A^3 + B^3 + C^3 = 3ABC (derived from higher algebra) and if this is put equal to Zero then either A, B or, C, = 0 which eventually leads back to A +B =0

If you compare this to the Math World eqns you will see the similarity in the terms
Hope this is clear as I intended it to be.

It will be instructive to read T.Rex’s posts and his views on maths in the thread “Does any know why LL tests work?” :surprised

So your conjecture is true but I sincerely feel that it is of no theoretical or practical value. All you have done is add a lot of Zeros to give Zero in a complicated and confusing way.

Still I have an open mind and consider the possibility that perhaps I may be wrong after all

Mally.

jinydu · 2004-12-23, 23:59

Whoa, thanks for the reply Mally.

I suppose this is the T.Rex quote you had in mind:

"You're right. But, the first time I tried to understand the LLT, I read a proof provided by http://www.utm.edu/research/primes , I think. It was really badly explained. Very short. Boring. But it is a general complain I have against Maths: most of the books provide the shortest possible proof, rather than explaining the long and difficult way used by the first discoverers. (I put group theory inside quotes in order to "group" the 2 words, so that it is more readable.)"

I would have to say that in spite of the vast amount of information available, most of the articles on Mathworld, unfortunately, meet this description. To Weisstein's credit, the proofs do score top points in terms of logical rigor, which is very important. But unfortunately, much of the material is very hard to read unless I've already studied the topic elsewhere.

As for the problem itself, your proof seems to rest on a crucial identity, which for the cases n = 2 and n = 3 are:

If A + B = 0
then A^2 + B^2 = 2AB

If A + B + C = 0
then A^3 + B^3 + C^3 = 3ABC

For the case of n = 2, I can clearly see why it is true. I simply square both sides and move 2AB to the other side.

But I don't really see how I can show that it works for n = 3, other than expanding (A + B + C)^3 and going through some long and tedious algebra. Could you explain that step please?

mfgoode · 2004-12-24, 03:37

Jinydu,

Kindly note due to a typo error the eqn should read A^2 +B^2 =(-2ab)
The cubic formula is correct
I will give you a proof later and definitely will do as its a beautiful piece of Algebra

Mally

jinydu · 2004-12-24, 03:51

Quote:

Originally Posted by mfgoode

Jinydu,

Kindly note due to a typo error the eqn should read A^2 +B^2 =(-2ab)
The cubic formula is correct
I will give you a proof later and definitely will do as its a beautiful piece of Algebra

Mally

Thanks again. And I assume that a similar formula holds for nth powers?

mfgoode · 2004-12-24, 04:04

Quote:

Originally Posted by jinydu

Thanks again. And I assume that a similar formula holds for nth powers?

Jinydu:

Try the proof by Induction.
IF it is true for (3) and we know it is true for (2) then it must be true for n
Mally

xilman · 2004-12-24, 11:51

Quote:

Originally Posted by mfgoode

Jinydu:

Try the proof by Induction.
IF it is true for (3) and we know it is true for (2) then it must be true for n
Mally

Not quite.

Here is a "proof" that all primes are less than 100, using your statement. Let p(n) be the n-th prime. p(2) = 3, which is < 100. p(3) = 5, and 5 < 100. Therefore the nth prime is less than 100.

The resolution, of course, is that for induction to work, you not only have to prove a base case, you also have to show that if the proposition is true for any value of n at all, it must be true for the value n+1.

In the case in question, the proposition was shown by jinydu to be true for n = 1, 2 and 3. He hasn't (as far as I'm aware) shown that the assumption it is true for arbitrary n leads inevitably to the conclusion that it must be true for n+1.

Paul

2004-12-23, 04:48	#15
jinydu Dec 2003 Hopefully Near M48 2×3×293 Posts	But what is that symbol that looks like a big pi? It also resembles the product sign, but its missing text above and below the product sign, so I don't know to multiply from what to what. Also, I'm pretty sure a (finite) polynomial is supposed to be a function in one variable that looks something like: c(0) + c(1)x + c(2)x^2 + c(3)x^3 + ... + c(n)x^n Last fiddled with by jinydu on 2004-12-23 at 04:49

2004-12-23, 17:03	#17
mfgoode Bronze Medalist Jan 2004 Mumbai,India 2²·3³·19 Posts	Challenge Problem Jinydu: I have checked the websites given by yourself and Travis. Yes the eqn.s including large ‘ PI’s’ are rigorous and ultimate solutions but you will agree it is cloaked in high falutin formlae difficult to follow from one step to the other (refer to your recent post on large ‘PI’ in this very thread) Studying the formulae given I find they are similar to what I’m expressing in simple mathematical language. We must remember that the evolution of complex functions are based on the axioms applied to those on real numbers. “God made the integers; the rest is the work of man” Kronecker. Whatever axioms we derive on real numbers must be true in the complex numbers too.Otherwise the mathematics of it will not be consistent and contrary in its different disciplines.. Hence the set of eqn.s you have given are in contradiction if the other eqn.s are dependent on your 1st premise i.e. A +B =0. Goven A + B =0 then A^2 +B^2 = 0 does not follow except if A and B are Zero Similarly for the other eqn.s. If A + B + C =0 then A^3 +B^3 + C^3 is only true if A = B = C =0. Test Case: If A + B = 0 then A^2 and B^2 = 0 cannot also be implied unless A=B=0 . It will instead be 2AB.This is from simple algebra! But what about complex numbers ? Let A = (p + iq) and B =(x+iy). If A + B =0 (given) then B can only have the value (- p –iq ). There is no other value it can be given. Now substitute the complex values ( p + iq) and (-p - iq) into A^2 +B^ 2 = 0 and you will obtain the same result i.e. A^2 + B^ 2 = 2 AB (*sign dispensed with). If the L.H.S =0 then the R.H.S. gives either A=0 or B=0 and finally A =B =0. Similarly if A + B + C =0 (given) unless A, B, C, are all zero the result will be valid only if you take the sum of the cubes as Zero. Otherwise they will be A^3 + B^3 + C^3 = 3ABC (derived from higher algebra) and if this is put equal to Zero then either A, B or, C, = 0 which eventually leads back to A +B =0 If you compare this to the Math World eqns you will see the similarity in the terms Hope this is clear as I intended it to be. It will be instructive to read T.Rex’s posts and his views on maths in the thread “Does any know why LL tests work?” :surprised So your conjecture is true but I sincerely feel that it is of no theoretical or practical value. All you have done is add a lot of Zeros to give Zero in a complicated and confusing way. Still I have an open mind and consider the possibility that perhaps I may be wrong after all Mally.

2004-12-24, 03:37	#19
mfgoode Bronze Medalist Jan 2004 Mumbai,India 2²·3³·19 Posts	Challenge problem Jinydu, Kindly note due to a typo error the eqn should read A^2 +B^2 =(-2ab) The cubic formula is correct I will give you a proof later and definitely will do as its a beautiful piece of Algebra Mally

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2004-12-22, 08:23	#13
jinydu Dec 2003 Hopefully Near M48 2·3·293 Posts	I did a search on Mathworld, and this is the closest thing I could find: http://mathworld.wolfram.com/Newton-GirardFormulas.html

2004-12-23, 08:59	#16
Orgasmic Troll Cranksta Rap Ayatollah Jul 2003 1010000001₂ Posts	Jinydu: http://mathworld.wolfram.com/SymmetricPolynomial.html

2004-12-23, 23:59	#18
jinydu Dec 2003 Hopefully Near M48 2·3·293 Posts	Whoa, thanks for the reply Mally. I suppose this is the T.Rex quote you had in mind: "You're right. But, the first time I tried to understand the LLT, I read a proof provided by http://www.utm.edu/research/primes , I think. It was really badly explained. Very short. Boring. But it is a general complain I have against Maths: most of the books provide the shortest possible proof, rather than explaining the long and difficult way used by the first discoverers. (I put group theory inside quotes in order to "group" the 2 words, so that it is more readable.)" I would have to say that in spite of the vast amount of information available, most of the articles on Mathworld, unfortunately, meet this description. To Weisstein's credit, the proofs do score top points in terms of logical rigor, which is very important. But unfortunately, much of the material is very hard to read unless I've already studied the topic elsewhere. As for the problem itself, your proof seems to rest on a crucial identity, which for the cases n = 2 and n = 3 are: If A + B = 0 then A^2 + B^2 = 2AB If A + B + C = 0 then A^3 + B^3 + C^3 = 3ABC For the case of n = 2, I can clearly see why it is true. I simply square both sides and move 2AB to the other side. But I don't really see how I can show that it works for n = 3, other than expanding (A + B + C)^3 and going through some long and tedious algebra. Could you explain that step please?