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What are some approximate values for $\zeta(3)$ and $\zeta (1.5)$ (upto three decimal places)? Looked at the paper "Chebyshev approximations for Riemann zeta functions" but I don't think those approximations are easily computable (at least for me).

baharampuri
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  • You can work it out: $\zeta(\sigma)=1+1/2^\sigma+1/3^\sigma+\cdots$ until you find the accuracy you need, where $\sigma=3$ and $\sigma=3/2$. To get you started $\zeta(3)=1.202$ to 3 decimal places. You'd need a CAS as you might need 94 terms for $\sigma=3$, for example. – pshmath0 Sep 14 '22 at 19:31
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    I thought the convergence is rather slow. Is that accurate to three decimal places? – baharampuri Sep 14 '22 at 19:31
  • This question linked claims 1.202 is accurate for zeta 3. https://math.stackexchange.com/questions/378912/interesting-phenomenon-with-the-zeta3-series?rq=1 – baharampuri Sep 14 '22 at 20:08
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    Wolfram alpha is your friend ; 2.61238 and 1.20205 – Conrad Sep 14 '22 at 20:14
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    For $ s>-1$, $s\neq 1$ and $N\ge 1$, you can use $$ \zeta (s) = \sum\limits_{n = 1}^N {\frac{1}{{n^s }} + \frac{{N^{1 - s} }}{{s - 1}}} - \frac{{N^{ - s} }}{2} + \theta (s,N)\frac{s}{{12}}N^{ - s - 1} $$ where $0 < \theta (s,N) < 1$ is a suitable number. – Gary Sep 14 '22 at 21:16

1 Answers1

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Let $Z(x, M) = \sum_{n=1}^M n^{-x}$ be a truncated approximation of $\zeta(x)$. Simply using the naïve summation

def Z(x, M):
    return sum(n ** (-x) for n in range(1, M + 1))

The approximation of $\zeta(3) \approx 1.202056903150321$ converges fairly quickly, with maximum accuracy achieved after $10^6$ iterations.

  • $Z(3, 10) \approx 1.197531985674193$
  • $Z(3, 10^2) \approx 1.2020074006596781$
  • $Z(3, 10^3) \approx 1.2020564036593433$
  • $Z(3, 10^4) \approx 1.202056898160098$
  • $Z(3, 10^5) \approx 1.2020569031097323$
  • $Z(3, 10^6) \approx 1.202056903150321$
  • $Z(3, 10^7) \approx 1.202056903150321$
  • $Z(3, 10^8) \approx 1.202056903150321$
  • $Z(3, 10^9) \approx 1.202056903150321$

OTOH, the approximation of $\zeta(1.5)$ converges more slowly, with the result still not being settled after $10^9$ iterations.

  • $Z(1.5, 10) \approx 1.9953364933456017$
  • $Z(1.5, 10^2) \approx 2.412874098703719$
  • $Z(1.5, 10^3) \approx 2.5491456029175756$
  • $Z(1.5, 10^4) \approx 2.5923758486729866$
  • $Z(1.5, 10^5) \approx 2.6060508091764736$
  • $Z(1.5, 10^6) \approx 2.6103753491852295$
  • $Z(1.5, 10^7) \approx 2.611742893169012$
  • $Z(1.5, 10^8) \approx 2.6121753486854478$
  • $Z(1.5, 10^9) \approx 2.6123121030481857$

But if 3 decimal places is all you need, just say $\zeta(1.5) \approx 2.612$.

Or, you could simply follow @Conrad's advice and consult the great oracle of mathematical knowledge, WolframAlpha, which gives:

  • Zeta(3) = 1.2020569031595942853997381615114499907649862923404988817922715553...
  • Zeta(1.5) = 2.6123753486854883433485675679240716305708006524000634075733282488...
Dan
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