I'm doing a little project on the $\zeta$ function, and I am at a complete loss of what it is actually doing. I understand it is way over my head, but when I am plugging say $\zeta(1 + i)$ into WolframAlpha, what is it even calculating? Wikipedia and .edu sites don't seem to have an answer, which is making me think there is no single answer.

Thanks, smart people of MathStack!


Why is everyone voting to close this? I understand it's similar to a different question, but this question might provide someone with some different intuition. Either way - this helped me a ton! Thanks @Mixed_Math.

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    $\begingroup$ See the Wikipedia and Wolfram Math World articles on the subject. $\endgroup$ – Lucian Mar 8 '15 at 6:00
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    $\begingroup$ Oh gee, I didn't think of looking at what Wikipedia had to say!... But seriously, of course I looked at Wikipedia, had it made sense I wouldn't have come here. Thanks for the suggestion, though. $\endgroup$ – Bliebervik Mar 8 '15 at 6:23
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    $\begingroup$ Questions about the analytic continuation of the Riemann zeta function have been asked, and answered, on this site before. Have a look at the questions listed under "related", or do a site search for the keywords. $\endgroup$ – Gerry Myerson Mar 8 '15 at 6:34
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    $\begingroup$ I have done a lot of research thus far @GerryMyerson, but you're right. I'm wasting everybody's time - I should have looked before. Thanks! $\endgroup$ – Bliebervik Mar 8 '15 at 6:39
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    $\begingroup$ @Bliebervik Well what didn't you understand from the Wikipedia article? We're not psychic so, if you don't explain what you didn't understand from Wikipedia, how can we avoid spending hours on an answer and then you saying, "Nah, I don't understand that, either." $\endgroup$ – David Richerby Mar 8 '15 at 12:48

The Riemann zeta function $\zeta(s)$ is a sum of reciprocals of powers of natural numbers,

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

As written, this makes sense for complex numbers $s$ so long as $\text{Re } s > 1$. For these numbers, there is little more to be said.

But you've asked about an interesting number: $\zeta(1 + i)$, and $\text{Re }(1 + i) \not > 1$. What's happening there is a bit subtle, and a bit abusive in terms of notation.

It turns out there is another function (let's call it $Z(s)$) which makes sense for all complex numbers $s$ except for $s = 1$, and which exactly agrees with $\zeta(s)$ when $\text{Re } s > 1$. If you're familiar with some calculus or complex analysis, then you should also know that the function $Z(s)$ is also complex differentiable everywhere except for $s = 1$. This is a very special property that distinguishes $Z(s)$. The theory of complex analysis (in particular, the theory of "analytic continuation") gives that there can be at most one function that extends $\zeta(s)$ to a larger region, like $Z(s)$ does.

In this sense, we could realize that $Z(s)$ is uniquely determined by $\zeta(s)$. As it agrees with $\zeta(s)$ everywhere $\zeta(s)$ (initially) makes sense, it might even be reasonable to just use the name $\zeta(s)$ instead of $Z(s)$. That is, when I write $\zeta(s)$, what I'm really saying is $$\zeta(s) = \begin{cases} \zeta(s) & \text{if Re }s > 1 \\ Z(s) & \text{otherwise } \end{cases}$$ It is this function that W|A computes when you ask it for $\zeta(1 + i)$.

Although what I've written is true (and important), it doesn't answer one aspect of your question

What is it even calculating?

I mentioned there exists this function $Z(s)$, or rather that it is possible to give meaningful values to $\zeta(s)$ for all $s \neq 1$. But how? Stated differently, yo're asking what is the analytic continuation of the Riemann zeta function?

The continuation is unique, but the steps to get there are not. I'll give a very short, incomplete proof that describes one way to calculate $\zeta(1+i)$.

We start by considering $\displaystyle h(s) = \sum_{n \geq 1} \frac{2}{(2n)^s}$. Performing some rearrangements, $$\begin{align} h(s) &= \sum_{n \geq 1} \frac{2}{(2n)^s} \\ &= \frac{1}{2^{s - 1}} \sum_{n \geq 1} \frac{1}{n^s} \\ &= \frac{1}{2^{s - 1}} \zeta(s) \end{align}$$ Let's subtract this from the regular zeta function. On the one hand, $$ \zeta(s) - h(s) = \zeta(s)(1 - \frac{1}{2^{s-1}}).$$

On the other hand, $$ \begin{align}\zeta(s) - h(s) &= \sum_{n \geq 1} \left( \frac{1}{n^s} - \frac{2}{(2n)^s} \right) \\ &= \sum_{n \geq 1} \frac{(-1)^{n+1}}{n^s}, \end{align}$$ and this last series makes sense for $\text{Re } s > 0$. (If you haven't looked at alternating series before, this might not be obvious. But the idea is that the sign changes cancel out a lot of the growth, so much that it converges for a larger region).

In total, this means that $$\zeta(s) = (1 - 2^{s - 1})^{-1} \sum_{n \geq 1} \frac{(-1)^{n+1}}{n^s},$$ and you can just "plug in" $1+i$ here. [Notice that the problem when $s = 1$ is apparent here, as you cannot divide by $0$.] In practice, it's an infinite sum, so you'll take the first very many terms to get the value of $\zeta(1+i)$ to any precision you want.

For completeness, it also turns out that $$\pi^{-s/2} \zeta(s) \Gamma(\tfrac{s}{2}) = \pi^{(s-1)/2} \zeta(1-s) \Gamma(\tfrac{1-s}{2}),$$ which lets us transform values of $\zeta(s)$ for $\text{Re } s > 0$ into values when $\text{Re } s < 1$. The $\Gamma(z)$ function here is called the "Gamma function" (it's an integral, a sort of generalization of a factorial) and this equation is called the symmetric functional equation of the zeta function.

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    $\begingroup$ Very good read, interesting stuff! $\endgroup$ – user142198 Mar 8 '15 at 6:00
  • $\begingroup$ In that last equation, can't you cancel the $\pi^{-s}$? $\endgroup$ – k_g Mar 8 '15 at 6:25
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    $\begingroup$ Very, very helpful. Thank you so much for taking the time to that. You are saying, though, that $\zeta(s) = (1 - 2^{s - 1})^{-1} \sum_{n \geq 1} \frac{(-1)^{n+1}}{n^s}$ is not the "all-encompassing" $\zeta$ function, as it were, then? $\endgroup$ – Bliebervik Mar 8 '15 at 6:31
  • $\begingroup$ You wrote "makes sense for complex numbers as long as $\text{Re } s > 1$". Doesn't it converge for all complex numbers except 1? $\endgroup$ – alancalvitti Mar 8 '15 at 7:16
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    $\begingroup$ Even though I'm a number theorist, no one has ever accused me of getting factors of pi and two right. $\endgroup$ – davidlowryduda Mar 10 '15 at 12:28

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