Need a hard math problem

[rian.wrenn]

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

 

[paulfr]

Sum of Coefficients

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

 

[BenTheMan]

You could ask her ``What is the sum of all the positive integers?''

She says ``Infinite''

You say ``No!''

$$\zeta(s) = \sum_{n=0}^{\infty}\frac{1}{n^s}$$

and

$$\zeta(-1) = \sum_{n=0}^{\infty} n = \frac{-1}{12}$$

Check out

http://mathworld.wolfram.com/RiemannZetaFunction.html
http://en.wikipedia.org/wiki/Riemann_zeta_function#Applications

 

[Absane]

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

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

 

[Absane]

You could ask her ``What is the sum of all the positive integers?''

She says ``Infinite''

You say ``No!''

$$\zeta(s) = \sum_{n=0}^{\infty}\frac{1}{n^s}$$

and

$$\zeta(-1) = \sum_{n=0}^{\infty} n = \frac{-1}{12}$$

Check out

http://mathworld.wolfram.com/RiemannZetaFunction.html
http://en.wikipedia.org/wiki/Riemann_zeta_function#Applications

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

 

[rian.wrenn]

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

 

[BenTheMan]

absane--it's just an analyitic continuation.

 

[Absane]

absane--it's just an analyitic continuation.

So -1/12 has a different meaning?

Sorry, analysis is not my cup of tea.

 

[iceaura]

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

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?

 

[Tom2]

You could ask her ``What is the sum of all the positive integers?''

She says ``Infinite''

You say ``No!''

$$\zeta(s) = \sum_{n=0}^{\infty}\frac{1}{n^s}$$

and

$$\zeta(-1) = \sum_{n=0}^{\infty} n = \frac{-1}{12}$$

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

Check out

http://mathworld.wolfram.com/RiemannZetaFunction.html
http://en.wikipedia.org/wiki/Riemann_zeta_function#Applications

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.

 

[CANGAS]

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.

 

[Pete]

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

 

[CANGAS]

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

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?

 

[CANGAS]

I repete:

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

 

[Pete]

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

 

[CANGAS]

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 ?

 

[Pete]

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

 

[Tom2]

:runaway:
Are you insane?

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.

 

[D H]

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.

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

 

[§outh§tar]

Riemann's hypothesis will do