1
$\begingroup$

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)

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

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.

Improve this answer
$\endgroup$
9
  • $\begingroup$ Thank you very much! I wouldn't have thought it would take 16.5h to calculate though… In fact, in my answer I will be using $A(B(x))$ up to the power 20. Is there a way to put this cutoff in the code so that I don't have to wait 16+ h for the unnecessary output. (or does simply putting $k=20$ in the last line work?) Thanks again! $\endgroup$
    – GregVoit
    May 13, 2014 at 16:14
  • $\begingroup$ Just tried putting it in, for $k=20$, but can't make sense of the output: it says "a very large output was generated". A piece looks smith like "...+(((-1 + <<1>>^2)^3 <<1>> (-(<<1>>/( 25 <<1>>)) + <<27>> + <<1>>/<<1>>))/r^3)+". $\endgroup$
    – GregVoit
    May 13, 2014 at 16:19
  • $\begingroup$ Going up to $k=20$ should be pretty fast - a few minutes at most. For all except the first few outputs, you will see an output like this one; clicking "Show Full output" will display it in full. As you can see, it's displayed in a summarized form because it is very long. However, you can still use it without problems, unless you need manual access to it (in which case: what made you expect simple expressions in the first place?). It's hard to go beyond this without knowing what you want out of this. $\endgroup$
    – Emilio Pisanty
    May 13, 2014 at 16:55
  • $\begingroup$ I see, thanks. Well, ultimately what I want is to bring the composition in the "$u,v_{k}$" form, like I wrote in the question. In fact, I only need an estimate for the norm $||A(B(x))||$, however, my norm is only defined for the expression brought to that "special" power series form. For example, given that $A(x)$ is given as in the question, then $||A||=|a|+\sum_{k}|b_{k}|$ $\endgroup$
    – GregVoit
    May 13, 2014 at 17:00
  • $\begingroup$ The reason I asked for the coefficients of the usual power series, is that I thought if I have a standard power series expression, then I can find the $u$ and $v_{k}$ by equating coefficients for powers of $x$, by putting two arrays equal. Or is there a better/more efficient way of doing it? $\endgroup$
    – GregVoit
    May 13, 2014 at 17:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.