Tag Archives: zeta function

Notes from a talk at Dartmouth on the Fibonacci zeta function

I recently gave a talk “at Dartmouth”1. The focus of the talk was the (odd-indexed) Fibonacci zeta function:
$$ \sum_{n \geq 1} \frac{1}{F(2n-1)^s},$$
where $F(n)$ is the nth Fibonacci number. The theme is that the Fibonacci zeta function can be recognized as coming from an inner product of automorphic forms, and the continuation of the zeta function can be understood in terms of the spectral expansion of the associated automorphic forms.

This is a talk from ongoing research. I do not yet understand “what’s really going on”. But within the talk I describe a few different generalizations; firstly, there is a generalization to other zeta functions that can be viewed as traces of units on quadratic number fields, and secondly there is a generalization to quadratic forms recognizing solutions to Pell’s equation.

I intend to describe additional ideas from this talk in the coming months, as I figure out how pieces fit together. But for now, here are the slides.

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Response to bnelo12’s question on Reddit (or the Internet storm on $1 + 2 + \ldots = -1/12$)

bnelo12 writes (slightly paraphrased)

Can you explain exactly how $latex {1 + 2 + 3 + 4 + \ldots = – \frac{1}{12}}$ in the context of the Riemann $latex {\zeta}$ function?

We are going to approach this problem through a related problem that is easier to understand at first. Many are familiar with summing geometric series

$latex \displaystyle g(r) = 1 + r + r^2 + r^3 + \ldots = \frac{1}{1-r}, $

which makes sense as long as $latex {|r| < 1}$. But if you’re not, let’s see how we do that. Let $latex {S(n)}$ denote the sum of the terms up to $latex {r^n}$, so that

$latex \displaystyle S(n) = 1 + r + r^2 + \ldots + r^n. $

Then for a finite $latex {n}$, $latex {S(n)}$ makes complete sense. It’s just a sum of a few numbers. What if we multiply $latex {S(n)}$ by $latex {r}$? Then we get

$latex \displaystyle rS(n) = r + r^2 + \ldots + r^n + r^{n+1}. $

Notice how similar this is to $latex {S(n)}$. It’s very similar, but missing the first term and containing an extra last term. If we subtract them, we get

$latex \displaystyle S(n) – rS(n) = 1 – r^{n+1}, $

which is a very simple expression. But we can factor out the $latex {S(n)}$ on the left and solve for it. In total, we get

$latex \displaystyle S(n) = \frac{1 – r^{n+1}}{1 – r}. \ \ \ \ \ (1)$

This works for any natural number $latex {n}$. What if we let $latex {n}$ get arbitrarily large? Then if $latex {|r|<1}$, then $latex {|r|^{n+1} \rightarrow 0}$, and so we get that the sum of the geometric series is

$latex \displaystyle g(r) = 1 + r + r^2 + r^3 + \ldots = \frac{1}{1-r}. $

But this looks like it makes sense for almost any $latex {r}$, in that we can plug in any value for $latex {r}$ that we want on the right and get a number, unless $latex {r = 1}$. In this sense, we might say that $latex {\frac{1}{1-r}}$ extends the geometric series $latex {g(r)}$, in that whenever $latex {|r|<1}$, the geometric series $latex {g(r)}$ agrees with this function. But this function makes sense in a larger domain then $latex {g(r)}$.

People find it convenient to abuse notation slightly and call the new function $latex {\frac{1}{1-r} = g(r)}$, (i.e. use the same notation for the extension) because any time you might want to plug in $latex {r}$ when $latex {|r|<1}$, you still get the same value. But really, it’s not true that $latex {\frac{1}{1-r} = g(r)}$, since the domain on the left is bigger than the domain on the right. This can be confusing. It’s things like this that cause people to say that

$latex \displaystyle 1 + 2 + 4 + 8 + 16 + \ldots = \frac{1}{1-2} = -1, $

simply because $latex {g(2) = -1}$. This is conflating two different ideas together. What this means is that the function that extends the geometric series takes the value $latex {-1}$ when $latex {r = 2}$. But this has nothing to do with actually summing up the $latex {2}$ powers at all.

So it is with the $latex {\zeta}$ function. Even though the $latex {\zeta}$ function only makes sense at first when $latex {\text{Re}(s) > 1}$, people have extended it for almost all $latex {s}$ in the complex plane. It just so happens that the great functional equation for the Riemann $latex {\zeta}$ function that relates the right and left half planes (across the line $latex {\text{Re}(s) = \frac{1}{2}}$) is

$latex \displaystyle \pi^{\frac{-s}{2}}\Gamma\left( \frac{s}{2} \right) \zeta(s) = \pi^{\frac{s-1}{2}}\Gamma\left( \frac{1-s}{2} \right) \zeta(1-s), \ \ \ \ \ (2)$

where $latex {\Gamma}$ is the gamma function, a sort of generalization of the factorial function. If we solve for $latex {\zeta(1-s)}$, then we get

$latex \displaystyle \zeta(1-s) = \frac{\pi^{\frac{-s}{2}}\Gamma\left( \frac{s}{2} \right) \zeta(s)}{\pi^{\frac{s-1}{2}}\Gamma\left( \frac{1-s}{2} \right)}. $

If we stick in $latex {s = 2}$, we get

$latex \displaystyle \zeta(-1) = \frac{\pi^{-1}\Gamma(1) \zeta(2)}{\pi^{\frac{-1}{2}}\Gamma\left( \frac{-1}{2} \right)}. $

We happen to know that $latex {\zeta(2) = \frac{\pi^2}{6}}$ (this is called Basel’s problem) and that $latex {\Gamma(\frac{1}{2}) = \sqrt \pi}$. We also happen to know that in general, $latex {\Gamma(t+1) = t\Gamma(t)}$ (it is partially in this sense that the $latex {\Gamma}$ function generalizes the factorial function), so that $latex {\Gamma(\frac{1}{2}) = \frac{-1}{2} \Gamma(\frac{-1}{2})}$, or rather that $latex {\Gamma(\frac{-1}{2}) = -2 \sqrt \pi.}$ Finally, $latex {\Gamma(1) = 1}$ (on integers, it agrees with the one-lower factorial).

Putting these together, we get that

$latex \displaystyle \zeta(-1) = \frac{\pi^2/6}{-2\pi^2} = \frac{-1}{12}, $

which is what we wanted to show. $latex {\diamondsuit}$

The information I quoted about the Gamma function and the zeta function’s functional equation can be found on Wikipedia or any introductory book on analytic number theory. Evaluating $latex {\zeta(2)}$ is a classic problem that has been in many ways, but is most often taught in a first course on complex analysis or as a clever iterated integral problem (you can prove it with Fubini’s theorem). Evaluating $latex {\Gamma(\frac{1}{2})}$ is rarely done and is sort of a trick, usually done with Fourier analysis.

As usual, I have also created a paper version. You can find that here.

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An Application of Mobius Inversion to Certain Asymptotics I

In this note, I consider an application of generalized Mobius Inversion to extract information of arithmetical sums with asymptotics of the form $latex \displaystyle \sum_{nk^j \leq x} f(n) = a_1x + O(x^{1 – \epsilon})$ for a fixed $latex j$ and a constant $latex a_1$, so that the sum is over both $latex n$ and $latex k$. We will see that $latex \displaystyle \sum_{nk^j \leq x} f(n) = a_1x + O(x^{1-\epsilon}) \iff \sum_{n \leq x} f(n) = \frac{a_1x}{\zeta(j)} + O(x^{1 – \epsilon})$.

(more…)

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