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I was exploring the Riemann Zetta function, and I observed that $\zeta (s)$ is not normally defined for $s $ such that $\Re (s) \leq 1$, but analytically continued to the whole complex plane. And the famous Riemann Hypothesis is all about the behavior of the function in the region $0 < \Re(s) < 1$. This means we have accepted the validity of analytic continuation.

Here is my question: why is it valid to assume that a function behaves in the analytically continued domain as it behaves where it is normally defined ? I mean analytic continuation results in counter-intuitive conclusions like "the sum of all natural numbers is a negative fraction", yet we still use it in our mathematics. Why is that a valid assumption ?

I hope my question is clear. If any of my statements sound absurd or incorrect, I apologize; that's because I just discovered about analytic continuation.

Thanks.

Edit: My question in short is why is $\zeta (-1) = \frac{-1}{12}$ when it should be infinity ?

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    $\begingroup$ Your question is too vague. Analytic continuation is just what it says. For example for $|x|\lt 1$, we have $\frac{1}{1-x}=\sum_{k=0}^\infty x^k$, so if start with the series (which has that limited domain) the fraction is the definition of the function for all $x\ne 1$. $\endgroup$ – herb steinberg May 17 '20 at 2:11
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    $\begingroup$ It's not correct that $\zeta(s)$ is not normally defined for $s$ such that ${\frak R}(s)\le1$. The analytic continuation of $\sum n^{-s}$ is the normal definition of $\zeta(s)$. To continue herb's comment, while the series $1+2+2^2+\cdots$ may not mean much to you, the fraction $1/(1-2)$ is perfectly sensible. Perhaps we shouldn't treat the analytic continuation as mysterious, but the restricted domain as insufficient. $\endgroup$ – runway44 May 17 '20 at 2:14
  • $\begingroup$ @runway44, May be "normally defined" might not be the right word. What I meant is $\zeta(s) $ blows up to infinity for real part of s less than one by the p-series test right. $\endgroup$ – user669545 May 17 '20 at 2:54
  • $\begingroup$ @herbsteinberg, if it makes it a little clearer, look at the Edit. $\endgroup$ – user669545 May 17 '20 at 3:03
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    $\begingroup$ The usual series for $\zeta(s)$ blows up there but $\zeta$ itself is conventionally defined on its maximal domain after doing as much analytic continuation as can be done. $\endgroup$ – Ian May 17 '20 at 3:05
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It doesn't have to be zeta function to realize the joy of analytic continuation.

You will need to know some background knowledges on power series.

Consider the old geometric series $$ F(x) = 1 + x + x^{2} + \cdots $$ as a function of $x$. Just as $\zeta(s)$ sounds nonsense for $s = -1$, so does, say, $F(2)$.

But, we know that $F(x)$ is written as $$ F(x) = 1/(1 - x) $$ for $|x| < 1$.

Now, analytic continuation. The series $$ 1 + x + x^{2} + \cdots $$ makes sense for $|x| < 1$ (and, differentiable term by term) and so does the expression 1/(1 - x) for all complex numbers $x$ except for $x = 1$.

Therefore, F(x) can be uniquely defined (yes, we are defining) as F(x) = 1/(1 - x) for all complex numbers $x \not = 1$.

As you recall that F(x) was nonsense for x = 2 at first, this sounds a big advance, doesn't it?

Roughly speaking, this depends on the uniqueness of power series representation; that is, suppose you have two functions $f(x) = \sum a_{n}x^{n}$ defined on $x \in X$ and $g(x) = \sum b_{n}x^{n}$ on $x \in Y$, $X \subset Y$, that are equal over some intersection of $X$ and $Y$. Well, power series are unique, so f is g for the larger domain $Y$ as well.

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