# Tag Info

5

(This is a math question, not a Mathematica question.) To add to Artes's answer, there is the well-known(!) identity $$\zeta(-n)=\frac{(-1)^n}{n+1}B_{n+1}$$ so you might as well ask why \begin{align*} -\frac12\times B_2&=-\frac1{14}\times B_{14}\\ -\frac12\times \frac16&=-\frac1{14}\times \frac76 \end{align*} A justification for the ...

5

In order to understand the issue, we should provide the underlying definitions. Mathematica helps in verifying appropriate relations and definitions. The main functional equation relating Riemann's zeta function $\zeta\;$, to Euler's $\Gamma\;$, established in Riemann's famous paper Über die Anzahl der Primzahlen unter einer gegebener Grösse (1859, English ...

3

A function from the article that cormullion linked is shorter and faster than what I proposed below. Transcribed in terse style: uf[m_, 1] := {{}} uf[1, n_] := {{}} uf[m_, n_?PrimeQ] := If[m < n, {}, {{n}}] uf[m_, n_] := uf[m, n] = Join @@ Table[Prepend[#, d] & /@ uf[d, n/d], {d, Select[Rest@Divisors@n, # <= m &]}] uf[n_] := uf[n, n] ...

1

In fact there are lots of Zetas which produce the same value, e.g. try these: $\{\zeta(-2), \zeta(-4), \zeta(-6),\ldots ,\zeta(-2n)\}$ {Zeta[-2], Zeta[-4], Zeta[-6]} (* {0, 0, 0} *) Using Plot may shed some light on this: Plot[Zeta[x], {x, -15, 0.5}]

Only top voted, non community-wiki answers of a minimum length are eligible