Need a hard math problem
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Abel was spot-on here. That paper is shameless.
rian.wrenn
Registered Member
lim
N!1
N Xn=0
an this is what i say to my teacher???
N!1
N Xn=0
an this is what i say to my teacher???
DH said:
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.
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 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.
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:
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.
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.
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.
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.
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.
rian.wrenn
Registered Member
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.
Check out Lubos' explanation of this. He's a pretty bright string theorist.
The story does not end there.DH said:
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.
Do you think you deserve the credit if someone else gives you the equation?
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