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

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

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

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.

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 :p

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

absane--it's just an analyitic continuation.

absane--it's just an analyitic continuation.

So -1/12 has a different meaning?

Sorry, analysis is not my cup of tea.

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?

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. :rolleyes:


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. :p

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.

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

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?

I repete:

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

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

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 ?

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

:runaway:
Are you insane?


Do you really have to ask? :p

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.

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.

Riemann's hypothesis will do :p

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.

Ouch!

http://www.math.luc.edu/~mgb/courses/SummerSeminar2006/DivergentSeries.pdf

http://www.math.luc.edu/~mgb/courses/SummerSeminar2006/DivergentSeries.pdf
Abel was spot-on here. That paper is shameless.

lim
N!1
N Xn=0
an this is what i say to my teacher???

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

Hmmm. I don't think so. This idea of mathematically continuing a function outside it's radius of convergence is a mathematically valid thing to do, unlike dividing by zero to prove 1=2.

We do these things in physics all of the time. For example, Euler's gamma function is

\Gamma(n) = (n-1)!, n\in\mathbb{Z}.

In quantum field theory, we end up with answers like \Gamma(-2). So how do you take a factorial of a negative number??? -3! = ? The answer is to analytically continue the gamma function, and regulate it.

Tha amazing thing is that doing this mathematical trickery gets you answers that are more accurate than any other theory man has ever written down.

someone asked about the zeta functions use---in string theory, this result is necessary for vacuum energy cancellations.

Hmmm. I don't think so. This idea of mathematically continuing a function outside it's radius of convergence is a mathematically valid thing to do, unlike dividing by zero to prove 1=2. Still might want to clarify the indices of summation, back there, to show that you aren't dividing by zero - unless I'm reading everything wrong, somehow ?

yeah you're right.

I'm home in Texas and have been more concerned with fishing and drinking beer than zeta functions:)

Oh well.

I can't edit the post anymore.

The zeta function should be defined as

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

One can show (by analytic continuation), that

\zeta(-1) = \sum_{s=1}^{\infty} n = -\frac{1}{12}.

One can show that the analytic continuation of the zeta function is indeed -1/12 at -1. This does not mean the series evaluates to -1/12 at -1. The sum of a set of elements, all of which are positive, is never negative. The series diverges at -1, end of story.

Given a function f(z) with a limited domain, the analytic continuation of f(z) is some other function F(z) with a domain larger than that of f(z) and such that F(z)=f(z) everywhere f(z) is defined. This does not mean f(z) magically been given a broader domain.

Physicists are just too damn loose when they use math.

D H, that was exactly my point. Here was the original quote:


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


The middle member of the equality shouldn't be there. In truth, I didn't know that \zeta(-1)=\frac{-1}{12} by analytic continuation, but I'm perfectly happy to believe you on that one. But if \zeta(-1) does equal \frac{-1}{12}, then it does not equal the indicated sum. My point is that complex analysis simply does not and can not overturn the results of real analysis.


So how do you take a factorial of a negative number??? -3! = ? The answer is to analytically continue the gamma function, and regulate it.


Right, but when you analytically continue \Gamma(n) for n not in \mathbb{Z}, then obviously the expression for computing \Gamma(n) that is valid in \mathbb{Z} is no longer applicable.

My point is that complex analysis simply does not and can not overturn the results of real analysis.

Physicists are just too damn loose when they use math.

I'll drink to that :)

Oh well.

I can't edit the post anymore.

The zeta function should be defined as

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

One can show (by analytic continuation), that

\zeta(-1) = \sum_{s=1}^{\infty} n = -\frac{1}{12}.

i'm not sure if this has been said before, but i think you can only reduce the second part of the riemann zeta function, i.e. the part that is taken over the gamma function, to that series when the real part of 's' is greater then one.

i think it also says that the riemann zeta function can only be defined that way when the real part of 's' is greater than 1 on that mathworld.wolfram.com article on the riemann zeta function.

actually, you could say that it can be defined that way when the real part of 's' is equal to one, with imaginary part of zero, though, i think, since that series would give you infinity when the real part of 's' equals 1 and the imaginary part of 's' equals 0.

dam you ppl r smart

http://motls.blogspot.com/2007/09/zeta-function-regularization.html

Check out Lubos' explanation of this. He's a pretty bright string theorist.

One can show that the analytic continuation of the zeta function is indeed -1/12 at -1. This does not mean the series evaluates to -1/12 at -1. The sum of a set of elements, all of which are positive, is never negative. The series diverges at -1, end of story. The story does not end there.

The continuation would be: " the series diverges, therefore its sum is undefined". There is a big difference between "undefined" and "impossible". The difference is that the possibility of defining a sum for the series exists. This has been accomplished.

So the statement "the sum of a set of elements, all of which are positive, is never negative" needs attention - it appears to be an illegitimate extrapolation of finite-conditioned intuition to manipulations of infinities, with the key an overlooking of some problems with the word "never".

More hints might be drawn from geometrical language - the "point at infinity", the "circle at infinity", etc.

Or modular arithmetic, in which sums of positive numbers are negative. Such might be better grounds for extrapolation and intuition than a directed number line, in some situations involving infinities.

That's already enough for many physics anti-talents to argue that string theory is not even wrong and it surely can't be tested, and so forth. However, what I haven't told you so far is that the same sum also appears in the calculation of the Casimir effect that has, in fact, been experimentally measured. The measurement - an experiment - confirms that ths sum is equal to "-1/12". Fine, so let's avoid further general clichés and accept the fact that the people who say that theoretical physics is not even wrong are just a waste of time and their writing is spam - one that can't even cure their readers' impotence.

I had forgotten about the Casimir effect, and the fact that the zeta function pops up there, too.

Math is difficult.

think there is no problem since it is not really the sum, but some extension of a function defined by certain sum we are talking about here.

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

Do you think you deserve the credit if someone else gives you the equation?

think there is no problem since it is not really the sum, but some extension of a function defined by certain sum we are talking about here.

Read Lubos's blog entry. The sum pops up in physical problems, and can (in essence) be measured. An example of this is in quantum field theory, where we get expressions like

\Gamma(-2) = \frac{1}{\epsilon} + finite + \mathcal{O}(\epsilon)

for epsilon small. You substract the infinite part, neglect the small part, and use the finite part. Doing so gives you physical results which are more accurate than any physical results ever measured (for example, precision QED processes).

So in some sense, adding all of the positive integers and getting a negative fraction is experimentally verified :) Just like taking a negative integer factorial :)

Mmm... That's funny. I was just reading up on analytic continuation...