Get the function from Taylor series on maple

Question

Can I get the reverse process for determining the function if I have its Taylor series, with maple?

Thank you.

Answer 1

If you are lucky and as Pauly says Maple recognises the series, then yes.

If Maple does not recognise it, then you have to do a manual search: Extract each coefficient $a_n$ and search backwards, knowing that if there was such an $f$ which was expanded around $z=z_0$, it would have to satisfy for some domain with $x_0\in\mathfrak{D}$:

$$f(z)=\sum\limits_{n=0}^\infty a_n\cdot (z-z_0)^n$$

therefore by Taylor's Theorem:

$$a_n=\frac{f^{(n)}(z_0)}{n!}$$

At this stage you can try a couple of $f$'s manually and see if you get agreement, using Maple's "series" command.

Note that even the above search is not guaranteed to end successfully. There are many[*] analytic functions which have no closed form and can be expanded into Taylor series around a given point $z=z_0$, so you may stumble into one where although the $a_n$ are well-known, there is no closed form for it, so the process resembles a little like looking for a pin in a haystack.

[*]: It would be a fairly simple exercise to show that the number of functions which admit a Taylor series (aka analytic at $z_0$) and do not have a known closed form far exceeds the number of known analytic functions (those admitting a Taylor series at $z=z_0$).

Answer 2

If you know the taylor series, plugging it in directly (as a series) will usually evaluate it if Maple can simplify it. For example:

If it doesn't evaluate it, then Maple can't do it in its current form :(