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

Question

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.

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Please help :)

Answer 1

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 $s=0$.

Answer 2

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

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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*} $$