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The Chudnovsky algorithm based on hypergeometric series seems to appear prominently. Why are approaches based around $\arctan$ slower?

Is there some intuitive, conceptual description of the "redundancy" or inefficiency in $\arctan$ based calculations that explain why they are slower?

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1 Answer 1

Sure, the coefficients in the power series of $\arctan$ are $\sim\frac1n$, hence decrease very slowly. So to compute $n$ digits of $\arctan x$ with $x\approx 10^{-k}$, you need $\frac nk$ summands.

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You don't need to use $\arctan 1$ as your basis. Look at Machin's formula for example, which is based on the fact that $$\frac{\pi}{4} = 4\arctan\frac15 - \arctan\frac{1}{239}$$ –  mrf Mar 6 '13 at 7:55
    
@Hagen But philosophically why "should" the power series of $\arctan$ be converging like $\frac{1}{n}$ and not converge faster. I am asking is there a way of understanding why some series converge faster, other than we tried a bunch randomly and Chudnovsky happens to have very fast convergence and its fast just because that's the way it just happened to end up. –  user782220 Mar 6 '13 at 8:02
    
@mrf Ye3s, sure. So one $x$ is $\frac 1{239}\approx10^{-3}$, thus you need only about 350000 summands for 1000000 digits. The other summand has $x\approx 10^{-1}$, so we need about one summand per digit (in fact more as $5$ is signoificantly smaller than $10$). –  Hagen von Eitzen Mar 6 '13 at 17:26
    
@user782220 Of course the power series of arctan "should" behave exactly the way we prove it does. :) –  Hagen von Eitzen Mar 6 '13 at 17:30

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