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I need to evaluate EllipticK[m] very close to 1. However, when I get too close to 1 the function defaults to the exact solution for 1 , which is ComplexInfinity. I can enter a comparable number in Wolfram|Alpha and get both the exact and the numerical approximation, but nothing I have tried gets me that approximate number in Mathematica:

Wolfram|Alpha:

EllipticK[0.9999999999999999999999999999999999999999999999]

54.3458...

Mathematica:

EllipticK[0.9999999999999999999999999999999999999999999999]

ComplexInfinity

EllipticK[0.9999999999999999999999999999999999999999999999]//N

ComplexInfinity

I've tried various other combinations of precision manipulation as well, so far nothing has worked. Does anyone know how to override the exact answer?

Edit

I discovered half the solution immediately after posting. When I put a 0 at the end of the string of 9s, it will evaluate the decimal approximation. However, I don't know how to implement this for the purpose of, say, plotting the EllipticK.

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  • $\begingroup$ Try EllipticK[0.9999999999999999999999999999999999999950]` $\endgroup$
    – Somos
    Commented Apr 16, 2019 at 23:43
  • $\begingroup$ Somos means EllipticK[0.99999999999999999999999999999999999999`50]. The editing is wonky with the backticks, and I don't remember how to escape them. $\endgroup$
    – march
    Commented Apr 17, 2019 at 0:05
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    $\begingroup$ @march Use double back-ticks: EllipticK[0.99999999999999999999999999999999999999`50] $\endgroup$
    – Michael E2
    Commented Apr 17, 2019 at 1:12

1 Answer 1

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Your choice to evaluate EllipticK[m] for values of m near the logarithmic singularity at $1$ is doomed from the start. The numerically proper way to go about this is to use the relationship of the complete elliptic integral of the first kind with the arithmetic-geometric mean, so that you are evaluating an argument near $0$:

N[Pi/(2 ArithmeticGeometricMean[1, Sqrt[1*^-46]]), 20]
   54.345751499982941351
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  • $\begingroup$ (I'm not quite back yet, I still don't have my own computer, and I just evaluated this on Wolfram One only to answer this question.) $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Jul 23, 2019 at 2:26
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    $\begingroup$ Glad you're at least partially back. Looking forward to you being fully back. $\endgroup$
    – JimB
    Commented Jul 23, 2019 at 3:45

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