# What is $\zeta(\varepsilon)$ for Riemann's zeta function $\zeta$ and Dual Numbers' $\varepsilon$?

This is inspired by a previous question of mine: What is $x^\bot$? Is $\zeta(\bot)=\bot$ for Riemann's zeta function $\zeta$ and wheel theory's $\bot$?

## The Question:

What is $$\zeta(\varepsilon)$$ for Riemann's zeta function $$\zeta$$ and the dual number $$\varepsilon$$?

## The Details:

Define $$\varepsilon$$, by fiat, to be a nonzero number such that $$\varepsilon^2=0$$. This is known as a dual number.

Define Riemann's zeta function as

$$\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}.\tag{\Delta}$$

## Thoughts:

We can write

$$\zeta(s)=\sum_{n=1}^\infty n^{-s}$$

and so

$$\zeta(\varepsilon)=\sum_{n=1}^\infty n^{-\varepsilon}$$

requires us to make sense, first of all, of $$n^{-\varepsilon}$$.

Consider

\begin{align} n^{-\varepsilon}&=e^{\log\left(n^{-\varepsilon}\right)}\\ &=e^{-\varepsilon\log(n)}, \end{align}

assuming logarithms make sense for dual numbers. Then, with a further assumption that the following makes sense, we have

\begin{align} e^{-\varepsilon\log(n)}&=\sum_{k=0}^\infty \frac{(-\varepsilon\log(n))^k}{k!}\\ &=1-\varepsilon \log(n), \end{align}

so that

\begin{align} \zeta(\varepsilon)&=\sum_{n=1}^\infty n^{-\varepsilon}\\ &=\sum_{n=1}^\infty (1-\varepsilon\log(n))\\ &=\sum_{n=1}^\infty 1-\varepsilon\sum_{n=1}^\infty \log(n), \end{align}

which I don't know how to evaluate.

## Doubts:

I made a bunch of assumptions in the above. There's nothing to say that $$\Delta$$ makes sense as a definition for $$\zeta$$ for $$\varepsilon$$; indeed, the above would not work for $$s=-1$$, as we all know.

• Really, defining $\zeta(\varepsilon)$ isn’t much different as defining $\zeta(0)$: if you believe hard enough in Taylor (since $\zeta$ is analytic), one expects $\zeta(\varepsilon)=\zeta(0)+\varepsilon \zeta’(0)$. The $\zeta$ series may not converge at $0$, but the $\zeta$ function still has analytic continuation, giving meaning to the expression above. Commented Jun 18, 2023 at 16:53
• The (dual-numbers) tag seems more appropriate than (clifford-algebras) IMO. Commented Jun 18, 2023 at 17:04
• Thank you, @elemelons. I typed in "dual", hoping to find the tag, but it didn't show; you have to include "-numbers". By the way, type [tag:dual-numbers] for dual-numbers.
– Shaun
Commented Jun 18, 2023 at 17:06
• @GEdgar True in general, but it does prove divergence if the terms are elements of an inner product space and you split summands into components and then get divergent series for multiple components. (Here, $\Bbb C$ being a vector space over $\Bbb R$.) Commented Jun 18, 2023 at 17:11
• Just a few months ago, this complex, dual, etc plotter was operational. Try seeing if it works by typing inzeta(z) and choosing “dual numbers” Commented Jun 18, 2023 at 17:38

Your doubt is correct; the usual series definition does not work outside its abscissa of convergence.

If a function $$f(z)$$ is analytic around a value $$z=a$$, it has a valid power series there, and then we can plug in $$z=a+b\varepsilon$$ to get $$f(a+b\varepsilon)=f(a)+bf'(a)\varepsilon$$. For the zeta function, this yields

$$\zeta(\varepsilon)=\zeta(0)+1\zeta'(0)\varepsilon=\frac{-1-\ln(2\pi)\varepsilon}{2}.$$

The actual terms of the power series do not matter in order to say the above, only its existence. But if you're curious about that power series, you would use the series for $$\zeta(s)$$ around $$s=1$$ (involving the stieltjes constants), rewrite it as $$\zeta(1+s)$$ around $$s=0$$, negate $$s$$, then use $$\zeta$$'s famous functional equation to start grinding out terms for $$\zeta(s)$$ around $$s=0$$.

• Joke: If we compare coefficients, we'd have $$\sum_{n=1}^\infty 1=-\frac{1}{2}$$ and $$\sum_{n=1}^\infty \ln(n)=\frac{\ln(2\pi)}{2}.$$
– Shaun
Commented Jun 18, 2023 at 17:16
• Thank you for this answer. I am likely to accept it. Usually, I give things a while to allow other users to contribute . . .
– Shaun
Commented Jun 18, 2023 at 17:19
• Isn't there something wrong with the formula $f(a+b\varepsilon)=f(a)+f'(b)\varepsilon$? According to Wikipedia it's $f(a+b\varepsilon)=f(a)+a\cdot f'(b)\varepsilon$... Wikipedia's derivation of the formula would be the one you suggested as well. Commented Jun 24, 2023 at 21:59
• In fact, it says $$f(a+b\varepsilon)=f(a)+bf'(a)\varepsilon,$$ @KevinDietrich; but thank you. I will revoke my acceptance of this answer until the matter with $\zeta(\varepsilon)$ is resolved.
– Shaun
Commented Jun 24, 2023 at 22:20
• But $b=1$ here, @KevinDietrich.
– Shaun
Commented Jun 24, 2023 at 22:22

Assuming that $$\zeta\left( x \right)$$ is real analytic around a point $$a$$, then we can simply apply the formula for the Taylor series around $$x = a$$ ($$f\left( a + x \right) = \sum_{k = 0}^{\infty}\left[ \frac{1}{k!} \cdot f^{k}\left( a \right) \cdot x^{k} \right]$$): (you can find a derivation of this fundamental formula at the end of my answer)

\fbox{ \begin{align*} f\left( a + x \cdot \varepsilon \right) &= f\left( a \right) + f'\left( a \right) \cdot x \cdot \varepsilon\\ \end{align*} }

The keyword is automatic differentiation (for reference see wikipedia > dual number > differentiation). You'll find the equation more often that in the form $$f\left( a + b \cdot \varepsilon \right) = f\left( a \right) + b \cdot f'\left( a \right) \cdot \varepsilon \wedge \left\{ a,\, b \right\} \in \mathbb{R}$$.

With this we would get: \fbox{ \begin{align*} \zeta\left( a + b \cdot \varepsilon \right) &= \zeta\left( a \right) + b \cdot \zeta'\left( a \right) \cdot \varepsilon\\ \zeta\left( \varepsilon \right) &= \zeta\left( 0 \right) + 1 \cdot \zeta'\left( 0 \right) \cdot \varepsilon\\ \zeta\left( \varepsilon \right) &= -\frac{1}{2} - \frac{1}{2} \cdot \ln\left( 2 \cdot \pi \right) \cdot \varepsilon\\ \zeta\left( \varepsilon \right) &= -0.5 - 0.918... \cdot \varepsilon\\ \end{align*} }

The derivation of $$f\left( a + x \cdot \varepsilon \right) = f\left( a \right) + f'\left( a \right) \cdot x \cdot \varepsilon$$:

We can simply apply the formula for the Taylor series around $$x = a$$ ($$f\left( a + x \right) = \sum_{k = 0}^{\infty}\left[ \frac{1}{k!} \cdot f^{k}\left( a \right) \cdot x^{k} \right]$$) and get: \begin{align*} f\left( a + x \right) &= \sum_{k = 0}^{\infty}\left[ \frac{1}{k!} \cdot f^{k}\left( a \right) \cdot x^{k} \right]\\ f\left( a + x \cdot \varepsilon \right) &= \sum_{k = 0}^{\infty}\left[ \frac{1}{k!} \cdot f^{k}\left( a \right) \cdot \left( x \cdot \varepsilon \right)^{k} \right]\\ f\left( a + x \cdot \varepsilon \right) &= \sum_{k = 0}^{\infty}\left[ \frac{1}{k!} \cdot f^{k}\left( a \right) \cdot x^{k} \cdot \varepsilon^{k} \right]\\ f\left( a + x \cdot \varepsilon \right) &= \sum_{k = 0}^{\infty}\left[ \frac{1}{k!} \cdot f^{k}\left( a \right) \cdot x^{k} \cdot \varepsilon^{k} \right]\\ f\left( a + x \cdot \varepsilon \right) &= \frac{1}{0!} \cdot f^{0}\left( a \right) \cdot x^{0} \cdot \varepsilon^{0} + \sum_{k = 1}^{\infty}\left[ \frac{1}{k!} \cdot f^{k}\left( a \right) \cdot x^{k} \cdot \varepsilon^{k} \right]\\ f\left( a + x \cdot \varepsilon \right) &= f\left( a \right) + \sum_{k = 1}^{\infty}\left[ \frac{1}{k!} \cdot f^{k}\left( a \right) \cdot x^{k} \cdot \varepsilon^{k} \right]\\ f\left( a + x \cdot \varepsilon \right) &= f\left( a \right) + \frac{1}{1!} \cdot f^{1}\left( a \right) \cdot x^{1} \cdot \varepsilon^{1} + \sum_{k = 2}^{\infty}\left[ \frac{1}{k!} \cdot f^{k}\left( a \right) \cdot x^{k} \cdot \varepsilon^{k} \right]\\ f\left( a + x \cdot \varepsilon \right) &= f\left( a \right) + f'\left( a \right) \cdot x \cdot \varepsilon + \sum_{k = 2}^{\infty}\left[ \frac{1}{k!} \cdot f^{k}\left( a \right) \cdot x^{k} \cdot \varepsilon^{k} \right]\\ f\left( a + x \cdot \varepsilon \right) &= f\left( a \right) + f'\left( a \right) \cdot x \cdot \varepsilon + \sum_{k = 2}^{\infty}\left[ \frac{1}{k!} \cdot f^{k}\left( a \right) \cdot x^{k} \cdot \varepsilon^{k} \right]\\ f\left( a + x \cdot \varepsilon \right) &= f\left( a \right) + f'\left( a \right) \cdot x \cdot \varepsilon + \sum_{k = 0}^{\infty}\left[ \frac{1}{\left( k + 2 \right)!} \cdot f^{k + 2}\left( a \right) \cdot x^{k + 2} \cdot \varepsilon^{k + 2} \right]\\ f\left( a + x \cdot \varepsilon \right) &= f\left( a \right) + f'\left( a \right) \cdot x \cdot \varepsilon + \sum_{k = 0}^{\infty}\left[ \frac{1}{\left( k + 2 \right)!} \cdot f^{k + 2}\left( a \right) \cdot x^{k + 2} \cdot \varepsilon^{k} \cdot \varepsilon^{2} \right]\\ f\left( a + x \cdot \varepsilon \right) &= f\left( a \right) + f'\left( a \right) \cdot x \cdot \varepsilon + \sum_{k = 0}^{\infty}\left[ \frac{1}{\left( k + 2 \right)!} \cdot f^{k + 2}\left( a \right) \cdot x^{k + 2} \cdot \varepsilon^{k} \cdot 0 \right]\\ f\left( a + x \cdot \varepsilon \right) &= f\left( a \right) + f'\left( a \right) \cdot x \cdot \varepsilon + \sum_{k = 0}^{\infty}\left[ 0 \right]\\ f\left( a + x \cdot \varepsilon \right) &= f\left( a \right) + f'\left( a \right) \cdot x \cdot \varepsilon + 0\\ f\left( a + x \cdot \varepsilon \right) &= f\left( a \right) + f'\left( a \right) \cdot x \cdot \varepsilon\\ \end{align*}