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using pade approx

asked 2017-12-01 04:13:14 -0600

florin gravatar image

Hi 1) From the only example I found s = PowerSeriesRing(QQ,'s').gen() a=exp(s);a.pade(4, 0) type(a)

it seems this works for type 'sage.rings.power_series_poly.PowerSeries_poly'

But I have a type 'sage.symbolic.expression.Expression' b, and was unable to convert b to the type required to apply pade

2) There's an alternative rational.reconstruct , but that seems to be geared to computations modulo (n) 3) In conclusion, it seems at current stage the simplest is to write one's own Pade? Thanks, Florin

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  • Please give us a, or the b from (1), so that potential helpers have a clear idea what kind of expressions are involved, and give the solution working in that particular case in any case.
    • There is a function rational_reconstruction, indeed. But i do not see it as an alternative, it is designed only for the special use described in its doc string, offered e.g. by typing rational_reconstruction? in the sage interpreter.
    • No, i would say, simplest for me is to use the pade method inside the class where it is genuinely defined. So just start with a b which is some function in s - say, convert is somehow to a power series, get the result, and possibly come back. But why do we need such a complication, why is the input b not directly in the PowerSeriesRing of QQ ?!
dan_fulea gravatar imagedan_fulea ( 2017-12-01 14:51:57 -0600 )edit

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answered 2017-12-01 14:46:54 -0600

FrédéricC gravatar image

You need to convert your symbolic expression to a symbolic Taylor series, then to a formal power series.

Typical example:

sage: f=sin(x)+cos(x)
sage: g=f.series(x,28)
sage: h=QQ[['x']](g)
sage: h.pade(4,4)
(127441/6601*x^4 - 481140/6601*x^3 - 2632500/6601*x^2 + 709560/943*x + 749040/943)/(x^4 - 10940/6601*x^3 + 265500/6601*x^2 - 39480/943*x + 749040/943)
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it works now! Thanks both of you and excuses for my silly novice questions. Both series and taylor which I had tried before work fine (I guess both must have advantages sometimes if they are kept ?)

adding the QQ magic changed the donna Here's the example var('s') L_F=(exp(-s)-1+s)/(s^2)
t=L_F.series(s,6) #t = taylor(L_F,s,0,6) h=QQ['s'] h.pade(2,2)

florin gravatar imageflorin ( 2017-12-02 03:08:41 -0600 )edit

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Asked: 2017-12-01 04:13:14 -0600

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Last updated: Dec 02