# Need a hard math problem

Discussion in 'Physics & Math' started by rian.wrenn, Sep 7, 2007.

1. ### rian.wrennRegistered Member

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I need a hard math problem so i can stump my teaher for extr credit. Pls, needs to be a good one and cant end with a theoracal answer. THX,

Also anyone elts can solve it too
thx

3. ### paulfrRegistered Senior Member

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227
Sum of Coefficients

What is the sum of the coefficients of
( [3x - 3x^2 +1]^744 ) x ( [- 3x + 3x^2 +1]^745 ) ??

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7. ### AbsaneRocket SurgeonValued Senior Member

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How about the one I am working on: Given a k-connected graph, two longest cycles meet at k or more vertices.

8. ### AbsaneRocket SurgeonValued Senior Member

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WTF? 1 + 1/2 + 1/3 + ... = -1/12?

I MUST be missing something...

By the way, n=1... because 1/n^s for n = 0 is a bad thing

9. ### rian.wrennRegistered Member

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just saying, DAMM you people are smart, like really really smart!!!!

O and what calculater do you use when the equasionis come out as a picture

10. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

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absane--it's just an analyitic continuation.

11. ### AbsaneRocket SurgeonValued Senior Member

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So -1/12 has a different meaning?

Sorry, analysis is not my cup of tea.

12. ### iceauraValued Senior Member

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Point. Ben might want to fix that.

Does anyone know what that odd summation is good for? The link describes it as having properties useful for the study of divergent series - like what properties, exactly?

13. ### Tom2Registered Senior Member

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Would you be so kind as to explain how summing infinitely many positive integers can possibly lead to a negative result.

I did check them out. The wiki reference says that the series converges for all s such that Re(s)>1. -1 is not greater than 1. Unless of course you would like to explain how -1 can be analytically continued to be greater than 1.

14. ### CANGASRegistered Senior Member

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The query was focused on Tom2 asking how Tom2 can explain summing an unlimited number of positive integers.

Tom2 has fouled off the query by invoking a third party ( and a notoriously unreliable one) rather than personally providing an opinion and a proof.

There is no way in H(expletive deleted) that an unlimited quantity of positive integers can sum to a negative answer.

In dreams many strange things are seen, so probably Tom2 is speaking of dream hallucinations rather than provable science matters.

15. ### PeteIt's not rocket surgeryRegistered Senior Member

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CANGAS,
Your animosity toward Tom2 seems to have blinded you. You might want to check who made the claim in question.

Last edited: Sep 9, 2007
16. ### CANGASRegistered Senior Member

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I have expressed no animosity.

Your expostulation which tries to form a thing which is not real is alarming.

Do your doctors know of your tendencies to imagine animosities which are are not real?

17. ### CANGASRegistered Senior Member

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I repete:

please explain in specific detail how any sum of positive integers can add up to be a negative answer.

18. ### PeteIt's not rocket surgeryRegistered Senior Member

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:bugeye:
It can't. Just as Tom2 said.

19. ### CANGASRegistered Senior Member

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So, as a kind of a parting shot, you cute little baby head thing, does the sum of an unlimited number of positive integers sum to a positive answer or a negative answer?

Or do you have any clue ?

20. ### PeteIt's not rocket surgeryRegistered Senior Member

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:runaway:
Are you insane?
Obviously it's positive infinity.

21. ### Tom2Registered Senior Member

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Do you really have to ask?

CANGAS, this is not some big mystery. Every one who's ever taken a full course in high school calculus knows that any p-series converges when p is greater than one, and diverges otherwise.

Last edited: Sep 9, 2007
22. ### D HSome other guyValued Senior Member

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Ben the Texan was toying with all of y'all, and it went over almost all of y'all's heads. To summarize, the Reimann zeta function is defined as

$\zeta(s) = \sum_{n=1}^{\infty} \frac 1 {n^s}$

By analytic continuation, $\zeta(-1) = -1/12$ and thus, by analytic continuation,

$\zeta(-1) = \sum_{n=1}^{\infty} n = -1/12$

Nice trick, Ben. So what's wrong with this?

Simple: The analytic continuation of some function f(z) is some function F(z) such that F(z)=f(z) everywhere f(z) is defined. Here, f(z) is the series definition of the zeta function and F(z) is its analytic continuation to the complex plane less the line $\Re z = 1$. The original series diverges for $\Re s <= 1$. The analytic continuation does not change the fact that the series diverges for s=-1.

What Ben did was the analytic equivalent of the various devices using division by zero that "prove" 1=2.

Last edited: Sep 9, 2007
23. ### §outh§taris feeling causticRegistered Senior Member

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Riemann's hypothesis will do