Taylor expands into a series a0 + a1*x + a2*x^2 + ...., where the exponents
of x are constant integers.
It can only expand into series like a0 + a1*x^m + a2*x^(2*m) + ..., where m
remains symbolic, under rather restricted circumstances.
There are two approaches you could take here. One is to expand
1/(q+1)^(1/m) then substitute q=x^m; the other is to use the powerseries
routine.
powerseries can expand some series with closed-form coefficience. In this
case, for example:
powerseries(1/(x^m+1)^(1/m),x,0) =>
('sum(x^(i1*m)/beta(-1/m-i1+1,i1+1),i1,0,inf))/(1-1/m)
two-dimensional display:
inf
==== i1 m
\ x
> --------------------------
/ 1
==== beta(- - - i1 + 1, i1 + 1)
i1 = 0 m
---------------------------------
1
1 - -
m
Using the substitution approach:
subst(x^m,q,taylor(1/(1+q)^(1/m),q,0,6)) =>
(120*m^5+274*m^4+225*m^3+85*m^2+15*m+1)*x^(6*m)/(720*m^6)-(24*m^4+50*m^3+35*m^2+10*m+1)*x^(5*m)\
/(120*m^5)+(6*m^3+11*m^2+6*m+1)*x^(4*m)/(24*m^4)-(2*m^2+3*m+1)*x^(3*m)/(6*m^3)+(m+1)*x^(2*m)/(2*m^2)-x\
^m/m+1
Hope this helps,
-s
On 6/1/07, Gottfried Helms <helms at uni-kassel.de> wrote:
>
> Hi,
>
> possibly my question can be solved differently, but
> I didn't get it this way.
>
> With TAYLOR() I get the first six series coefficients for
>
> TAYLOR( (1/(1+x) , x, 0 ,6)
>
> This works also for m'th powers of the expression:
>
> TAYLOR( (1/(1+x)^m , x, 0 ,6)
>
> What I wanted to get is
>
> TAYLOR( (1/(1+x^m)^(1/m) , x, 0 ,6)
>
> but this gives only one coefficients and three ellipses.
> The final exponent doesn't matter, also for
>
> TAYLOR( (1/(1+x^m) , x, 0 ,6)
>
> I get only one coefficient.
>