# Math Help - zeta function

1. ## zeta function

Can someone explain how setting x = -2 in the zeta function results in the function converging to 0? The first term = 1, the second = 4, the third = 9, etc. How can the sum of the squares add up to 0? Another mystery (at least to me): setting x = -1 results in the value -1/12? The zeta function is the sum of 1/(n^s) for n = 1 to infinity, is it not?

2. Originally Posted by coolmanoh
Can someone explain how setting x = -2 in the zeta function results in the function converging to 0? The first term = 1, the second = 4, the third = 9, etc. How can the sum of the squares add up to 0? Another mystery (at least to me): setting x = -1 results in the value -1/12? The zeta function is the sum of 1/(n^s) for n = 1 to infinity, is it not?
Disclaimer: I am a novice in complex analysis. Don't take what I'm about to say as absolute truth.

That's it's definition on a limited portion of $\mathbb{C}$. To consider the values you mentioned you need to analytically continue it via the usual functional equation $\displaystyle \zeta(s)=2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)$. Plugging in your values gives the desired results.

3. You have gone way over my head. Everything I've read about the zeta function states that the trivial zeros are -2, -4, -6, etc. The equation you produced is never mentioned, only the one I stated. A Professor Hill of University College, London received a letter from the Indian mathematition Ramanujan where he "proved" that 1 + 2 + 3 + .... + infinity = -1/12. Ramanujan got this result by setting x = -1 in the zeta function. If they are not talking about the zeta function, then why don't they say so?

4. Originally Posted by coolmanoh
You have gone way over my head. Everything I've read about the zeta function states that the trivial zeros are -2, -4, -6, etc. The equation you produced is never mentioned, only the one I stated. A Professor Hill of University College, London received a letter from the Indian mathematition Ramanujan where he "proved" that 1 + 2 + 3 + .... + infinity = -1/12. Ramanujan got this result by setting x = -1 in the zeta function. If they are not talking about the zeta function, then why don't they say so?
They're not. This is a common misconception. Let me lay it out for you. Consider the function $\displaystyle f0.1]\to\mathbb{R}:x\mapsto x\sin\left(\frac{1}{x}\right)" alt="\displaystyle f0.1]\to\mathbb{R}:x\mapsto x\sin\left(\frac{1}{x}\right)" />. We can clearly continuously extend $\tilde{f}:[0,1]\to\mathbb{R}$ by defining $\displaystyle \tilde{f}(x)=\begin{cases}x\sin\left(\frac{1}{x}\r ight) & \mbox{if}\quad x\in(0,1]\\ 0 & \mbox{if}\quad x=0\end{cases}$. In which case it's true that $\tilde{f}(0)=0$ from where the seemingly amazing 'result' occurs that $0\sin\left(\frac{1}{0}\right)=0$ even though that's clearly ludicrous. A similar thing is happening here. We have the function $\displaystyle \zeta1,\infty)\to\mathbb{R}:x\mapsto\sum_{n=1}^{\infty} \frac{1}{n^x}" alt="\displaystyle \zeta1,\infty)\to\mathbb{R}:x\mapsto\sum_{n=1}^{\infty} \frac{1}{n^x}" /> for which you are familiar with. Complex analysts found a particular way (in this case analytic continuation ) to take this function $\zeta$ and 'extend' it to a function $\tilde{\zeta}:\mathbb{C}\to\mathbb{C}$ in such a way that, in fact, $\tilde{\zeta}(-2)=0$. If you aren't careful, and realize that what they call $\zeta$ when the plug in $-2$ is actually what I've called $\tilde{\zeta}$ you'd get the amazing result that $\displaystyle 0=\zeta\left(-2\right)=\sum_{n=1}^{\infty}\frac{1}{n^{-2}}=\sum_{n=1}^{\infty}n^2$. But...that's just plain silly :P