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taylor expansion with arbritary precision numbers

asked 2014-07-11 08:58:26 +0200

Marco Caliari gravatar image

updated 2014-07-11 19:48:20 +0200

slelievre gravatar image


if I define a function with arbritary precision numbers (e.g., f=0.123456789123456789*log(1+x)) and then compute its Taylor expansion (f.taylor(x,0,5)) it seems to me that the coefficients are given in double precision, whereas if I compute them (e.g., by derivative(f,x,5)(x=0)/factorial(5)) they are in original precision. First of all, am I right or is it only a visualisation difference? If I'm right, is it possible to compute the Taylor expansion with the original precision?



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answered 2014-07-11 14:35:20 +0200

kcrisman gravatar image

f.taylor() sends things to Maxima, and does not keep precision. I believe this does.

sage: f.series(x,5)
0.12345678912345679*x + (-0.061728394561728395)*x^2 + 0.041152263041152263*x^3 + (-0.030864197280864197)*x^4 + Order(x^5)
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Thanks, it works, And in order to get the coefficients? f.series(x,5).coefficients() seems to me in double precision.

Marco Caliari gravatar imageMarco Caliari ( 2014-07-14 10:34:12 +0200 )edit

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Asked: 2014-07-11 08:58:26 +0200

Seen: 236 times

Last updated: Jul 11 '14