Prev: Excel Data Managing
Next: DiscretePlot
From: Luc Roy on 13 Jun 2010 18:53
FullSimplify[JacobiAmplitude [Series[EllipticF[x, m], {x, 0, 10}], m]]

should give x + O[x]^10 because the two functions are the inverse of
one another.

however it gives

x - (m x^3)/3 + 1/30 m (2 + 3 m) x^5 - 1/630 m (4 + m (39 + 16 m)) x^7
+ ( m (8 + 3 m (120 + 13 m (20 + 3 m))) x^9)/22680

Am I missing something
or
Is there an implementation problem with Mathematica

Note : the usage of series is very important.
I know the following works and that is not what I need to resolve or
understand.
PowerExpand[JacobiAmplitude [EllipticF[x, m], m]]
Answer: x

From: Luc Roy on 15 Jun 2010 02:29
On Jun 13, 6:53 pm, Luc Roy <luc.rg....(a)gmail.com> wrote:
> FullSimplify[JacobiAmplitude [Series[EllipticF[x, m], {x, 0, 10}], m]]
>
> should give x + O[x]^10 because the two functions are the inverse of
> one another.
>
> however it gives
>
> x - (m x^3)/3 + 1/30 m (2 + 3 m) x^5 - 1/630 m (4 + m (39 + 16 m)) x^7
> + ( m (8 + 3 m (120 + 13 m (20 + 3 m))) x^9)/22680
>
> Am I missing something
> or
> Is there an implementation problem with Mathematica
>
> Note : the usage of series is very important.
> I know the following works and that is not what I need to resolve or
> understand.
> PowerExpand[JacobiAmplitude [EllipticF[x, m], m]]
> Answer: x

I found the problem.

The following evaluates to the series decribed above.
FullSimplify[InverseSeries[EllipticF [Series[EllipticF[x, m], {x, 0,
10}], m]]]


This implies that JacobiAmplitude[[fx]] is evaluated as
InverseSeries[EllipticF[f[x], m]] where f[x] is evaluated first,
EllipticF second and InverseSeries third.
However, the InverseSeries only applies to EllipticF and not to f[x]

Mathematica has a bad implementation for JacobiAmplitude.

Instead of JacobiAmplitude use ArcSin[JacobiSN[x,m]]


 | 
Pages: 1
Prev: Excel Data Managing
Next: DiscretePlot