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I've been looking with a friend at the values of zeta at the odd integers. WolframAlpha can give us over 100 digits in a second or two, but it seems that if you take the sum out to n, say, then you are off by an order of 1/n^2, so getting 100 digits of accuracy would take about 10^50 operations. How does Alpha do it so fast?

user387394
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  • I've never looked into how WA works, but I would guess it's among the following: 1) the response was cached (computed beforehand) and thus returned immediately, 2) they have a large amount of computing power, which can be used to parallelize the computation, and 3) there is some cleverer way of computing the answer than by computing the partial sum directly. – theyaoster May 05 '19 at 03:43
  • Key search phrase: acceleration of convergence. But for $\zeta$, there are integral representations, and fast accurate methods for estimating integrals.See also https://math.stackexchange.com/questions/1459709/can-this-approximate-closed-form-of-aperys-constant-zeta3-be-improved and https://math.stackexchange.com/questions/378912/interesting-phenomenon-with-the-zeta3-series and one of the answers to https://math.stackexchange.com/questions/62878/proving-irrationality and http://math.stackexchange.com/questions/622762/formula-for-zeta3-verification and probably several others. – Gerry Myerson May 05 '19 at 03:57
  • Have you had a look at those links, Greg? – Gerry Myerson May 06 '19 at 13:05
  • Ah yes, I see. OK, cool, that will give us some things to digest. No doubt it is one of these methods. Many thanks! – user387394 May 07 '19 at 20:00

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I am not sure if this is what Mathematica uses, but I do know that "EulerMaclaurin" is one of the summation acceleration methods it uses.

The Euler-Maclaurin Sum Formula gives asymptotic expansions such as $$ \sum_{k=1}^n\frac1{k^3}=\zeta(3)-\frac1{2n^2}+\frac1{2n^3}-\frac1{4n^4}+\frac1{12n^6}-\frac1{12n^8}+\frac3{20n^{10}}-\frac5{12n^{12}}+O\!\left(\frac1{n^{14}}\right)\tag1 $$ Which leads to the accelerated $$ \zeta(3)=\sum_{k=1}^n\frac1{k^3}+\frac1{2n^2}-\frac1{2n^3}+\frac1{4n^4}-\frac1{12n^6}+\frac1{12n^8}-\frac3{20n^{10}}+\frac5{12n^{12}}+O\!\left(\frac1{n^{14}}\right)\tag2 $$ Applying $(2)$ with $n=100000$, we get $70$ decimals places of $\zeta(3)$: $$ \small1.202056903159594285399738161511449990764986292340498881792271555341838\tag3 $$ Taking more terms of the Euler-Maclaurin expansion or more terms of the sum yieds more precision.

robjohn
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  • And the O constant isn't small, the larger we take $m $ (here $m= 14$) the larger we need $n$. This way for any $m$ we get $O(n^{-m})$ series for $\zeta(3)$. When arbitrary precision is needed we should prefer this series – reuns May 05 '19 at 18:43
  • The big-O constant here is less than $2$, so the error with $n=100000$ is less than $2\times10^{-70}$. The actual error is $1.6452\times10^{-70}$. – robjohn May 06 '19 at 02:20
  • Cool, thanks for the answers! There's a lot to learn in these links. – user387394 May 07 '19 at 20:02