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2019-02-04 21:04:03 +0100 commented answer Problem with Padé approximation

Sorry! I read too fast, and misunderstood "f.pade(2,4)(x)" as the Pade approximation!

So the newer version seems to work, I should get a newer one.

2019-02-04 21:01:06 +0100 commented answer Problem with Padé approximation

Merci! I am running version 6.6, which gives error messages already for your "f = x.cos()".

However, the answer which you get is not correct either. By definition the (2,4) Pade approximation is a rational function p/q where p is of degree at most 2 and q of degree at most 4.

2019-02-04 16:09:34 +0100 asked a question Problem with Padé approximation

I let

f(x)=cos(x)

and calculate the (2,4) Padé approximation of f:

f.taylor(x,0,10).power_series(QQ).pade(2,4)

Sage answers with

(-244/3x^2 - 1/1512000x + 200)/(x^4 - 1/18144000x^3 + 56/3x^2 - 1/1512000*x + 200)

which seems incorrect. (The Taylor series of this rational function has a x^3 term which differs from that of f.)

Am I doing something wrong, or is there a problem with sage?

2019-02-04 16:09:33 +0100 asked a question Problem with Padé approximation

I let

f(x)=cos(x)

and calculate the (2,4) Padé approximation of f:

f.taylor(x,0,10).power_series(QQ).pade(2,4)

Sage answers with

(-244/3x^2 - 1/1512000x + 200)/(x^4 - 1/18144000x^3 + 56/3x^2 - 1/1512000*x + 200)

which seems incorrect. (The Taylor series of this rational function has a x^3 term which differs from that of f.)

Am I doing something wrong, or is there a problem with sage?

2019-02-04 16:09:33 +0100 asked a question Problem with Padé-approximation

I let

f(x)=cos(x)

and calculate the (2,4) Padé approximation of f:

f.taylor(x,0,10).power_series(QQ).pade(2,4)

Sage answers with

(-244/3x^2 - 1/1512000x + 200)/(x^4 - 1/18144000x^3 + 56/3x^2 - 1/1512000*x + 200)

which seems incorrect. (The Taylor series of this rational function has a x^3 term which differs from that of f.)

Am I doing something wrong, or is there a problem with sage?