Determining a function by its Taylor series coeficients

Question

We know that if a function $f(z)$ is "nice enough", it can be written as a Taylor series: $$\sum_{n=0}^\infty a_n(z-z_0)^n,$$ where $a_n=\dfrac{f^{(n)}(z_0)}{n!}$. Is there a systematic, explicit or general way of determining this function $f(z)$ purely by knowing only the corresponding $a_n$'s? Any help or reference will be appreciated!

Answer 1

So, the quick and short answer is no. there is not always a nice quick way to identify what a power series looks like in common functions that we like to talk about (exponentials, trig functions, polynomials, rational functions, logarithms, etc).

In fact, we can't even define most power series. simply, consider that a definition of a specific power series needs to be a finite sentence that describes that power series and no other. this defines a 'definition function' from english sentences to functions. the set of finite english sentences is countable (the set of finite strings of a finite set) and the set of power series is uncountable (as $\mathbb{R}$ is uncountable, so would be the set of possibly infinite sequences of real numbers). So necessarily, the 'definition function' cannot be surjective and there are uncountably many power series that we just can't point out explicitly

Answer 2

In some cases, when we have a closed form for the coefficients $a_n$, it is possible to find an ODE verified by the power series, and if this ODE is solvable then you can find back the original function.

But keep in mind, that we also often do the opposite, when we are not able to solve an ODE by conventional means, we try to find a power series verifying it, which is kind of ironical...

Anyway suppose you have $f(x)=\sum\limits_{n=0}^{\infty}\dfrac{(-1)^nx^{2n}}{(2n)!}$

We can calculate the derivatives:

$f'(x)=\sum\limits_{n=1}^{\infty}\dfrac{(-1)^nx^{2n-1}}{(2n-1)!}$

$f''(x)=\sum\limits_{n=2}^{\infty}\dfrac{(-1)^nx^{2n-2}}{(2n-2)!}=\sum\limits_{n=0}^{\infty}\dfrac{(-1)^{n+1}x^{2n}}{(2n)!}=-f(x)$

And $f$ is solution of $y''+y=0$ thus $f(x)=a\cos(x)+b\sin(x)$.

Since the power series gives $f(0)=1$ and $f'(0)=0$ then $f(x)=\cos(x)$.

However, I suspect it would not be so straightforward in general to find the ODE given the coefficients, but who knows, it is a possible method.

Answer 3

No. A function which can be represented by a power series is known as an analytic function, and there are uncountably many analytic functions, most of which obviously cannot be simply expressed in terms of elementary functions. So there is no general method, given a sequence $(a_n)$, to find the closed form of $\sum a_nx^n$ since such an expression might not even exist.

However, for most useful circumstances it could be possible to recognize certain similarities between the given sequence $a_n$ and other known ones. This can be a good way to recognize a closed form when it is possible. For example, if $f(x)=\sum a_nx^n$, then $(xD)f(x)=\sum na_nx^n$, where $D$ is the differential operator (with respect to $x$). Inductively this implies $p(xD)=\sum p(n)a_nx^n$ for any polynomial $p$, so if a sequence is of this form and $f(x)$ is known, we have a nice way to determine $\sum p(n)a_nx^n$.

To illustrate, how can we find $\sum_{n\geq0}(n^2+1)\binom{a}{n}x^n$? Easy: note that $(x+1)^a=\sum_{n\geq0}\binom{a}{n}x^n$, so $$\begin{split}\sum_{n\geq0}(n^2+1)\binom{a}{n}x^n&=((xD)^2+1)(x+1)^a\\&=(xD)(ax(x+1)^{a-1})+(x+1)^a\\&=x(a(x+1)^{a-1}+a(a-1)x(x+1)^{a-2})+(x+1)^a\\&=(x+1)^{a-2}\left((a^2+1)x^2+(a+2)x+1\right)\end{split}.$$

Another method is by recognising some form of recurrence in the sequence $(a_n)$, which can be converted to an ODE which $f(x)$ satisfies. Then solve that ODE to get your closed form. For example, if you somehow didn't recognize immediately that $\sum(1/n!)x^n=\exp(x)$, you can observe that $na_n=a_{n-1}$, which after multiplying throughout by $x^{n-1}$ and summing for all $n$ gives $f'(x)=f(x)$, and then using $f(0)=1$ will tell you $f(x)=\exp(x)$.