Compose two special power series expansions

Question

I have two functions $A(x), B(x)$, given in a special power series form:

$A(x)=1-x^{2}\left(\frac{a}{10}-\sum_{k=1}^{9}b_{k}\left(\frac{(x^{2}-1)}{r}\right)^{k}\right)$

$B(x)=1-x^{2}\left(\frac{c}{10}-\sum_{k=1}^{9}d_{k}\left(\frac{(x^{2}-1)}{r}\right)^{k}\right)$

where $a,b_{k},c,d_{k}$ are real numbers, and $r\in\mathbb{R}^{+}$.

What I want to do, is to find the composition of these series $A(B(x))$ in a similar form, i.e. find such $u=u(a,b_{k},c,d_{k},r)$ and $v_{k}=v_{k}(a,b_{k},c,d_{k},r)$ so that

$A(B(x))=1-x^{2}\left(\frac{u}{10}-\sum_{k}v_{k}\left(\frac{(x^{2}-1)}{r}\right)^{k}\right)$

I tried defining these series by brute force:

A[x_] := 1 - x^2*((a/10) + Sum[b[k]*((x^2 - 1)/r)^k, {k, 1, 9}]); 


B[x_] := 1 - x^2*((c/10) + Sum[d[k]*((x^2 - 1)/r)^k, {k, 1, 9}]);

And then calling A[B[x]]//Expand but it took forever long to run. Is there an efficient way to get an explicit power series expansion for A[B[x]]?

Then I can just get the desired form by simple equating coefficients, but what my question is: how to get an output for A[B[x]] in the standard power series form?

Thanks! (this is not homework)

Answer 1

Your problem can be simplified a bit before you put it into Mathematica. For starters, because of the shared external form, you need only worry about the 'core' of the $A$ function, $$ \tilde A(x)=\sum_{k=1}^{9}b_{k}\left(\frac{(x^{2}-1)}{r}\right)^{k}, $$ and you will always have $u=a$. Further, because the expansion parameter $y=(x^2-1)/r\geq-1/r$ is the same in $B$ and in your expected expansion for $A\circ B$, you can simply use that: $$ B(y)=1-(r\,y+1)\left(\frac{c}{10}-\sum_{k=1}^{9}d_{k}\,y^{k}\right). $$ With both these objects, your problem reduces to finding the $v_k$ such that $$ \tilde A(B(y))=\sum_{k}v_{k}\,y^{k}. $$ To implement this, use

a[x_] := Sum[b[k] ((x^2 - 1)/r)^k, {k, 1, 9}];
B[y_] := 1 - (r y + 1) ((c/10) + Sum[d[k]*y^k, {k, 1, 9}]);

and you can simply find the kth coefficient $v_k$ by running

SeriesCoefficient[a[B[y]], {y, 0, k}]

This will return an explicit, exact expression whenever k is a nonnegative integer. Low numbers are reasonably fast, though as you go up into the twenties and thirties it starts to show some sluggishness. The highest coefficient is $k=180=2\times9\times10$, and takes about 11 minutes on my machine. Asking for $k=181$ returns zero as expected, also after about 11 minutes. I would therefore expect the whole series to take about $\tfrac12 180\times 11\,\text{min}=16.5\,\text{h}$ to calculate. This is not an amazingly exciting prospect but from the coefficients onwards it depends on exactly what you want to do with all of this.