How can I get f(x) from its Maclaurin series? I know how to get a Maclaurin series when $f(x)$ is given. I have to find $\sum_{n=0}^{\infty}\frac{f^{(k)}(0)}{k!}x^k$.
But how can I get $f(x)$ from its Taylor series?
The problem is $$f(x) = \sum_{n=0}^{\infty} C_n x^n,$$ where $C_n$ is a Catalan number defined by $C_n = \frac{1}{n+1}\binom{2n}{n}$.
How can I get $f(x)$?
 A: *

*The Catalan numbers $$C_n=\frac{1}{n+1}\binom{2n}{n}$$ are strongly related with the Central binomial coefficients $\binom{2n}{n}$.

*The binomial identity 
\begin{align*}
\frac{1}{2n-1}\binom{2n}{n}=(-1)^{n+1}\binom{\frac{1}{2}}{n}4^n\tag{1}
\end{align*}
provides a relationship with the binomial series expansion
\begin{align*}
\sqrt{1+x}=\sum_{n= 0}^\infty\binom{\frac{1}{2}}{n}x^n\tag{2}
\end{align*}

From (1) we obtain a relationship with the Catalan numbers.
  \begin{align*}
\frac{1}{2}\cdot\frac{1}{2n-1}\binom{2n}{n}&=\frac{1}{2}\cdot\frac{1}{2n-1}\cdot\frac{2n}{n}\binom{2n-1}{n-1}\\
&=\frac{1}{2n-1}\binom{2n-1}{n-1}\\
&=\frac{1}{2n-1}\binom{2n-1}{n}\\
&=\frac{1}{n}\binom{2n-2}{n-1}\\
&=C_{n-1}\tag{3}
\end{align*}
We conclude
  \begin{align*}
\sum_{n=0}^\infty C_nx^n&=\sum_{n=1}^\infty C_{n-1}x^{n-1}\tag{4}\\
&=\frac{1}{2}\sum_{n=1}^\infty\frac{1}{2n-1}\binom{2n}{n}x^{n-1}\tag{5}\\
&=\frac{1}{2}\sum_{n=1}^\infty(-1)^{n+1}\binom{\frac{1}{2}}{n}4^nx^{n-1}\tag{6}\\
&=-\frac{1}{2x}\sum_{n=1}^\infty\binom{\frac{1}{2}}{n}(-4x)^{n}\tag{7}\\
&=-\frac{1}{2x}\left(\sqrt{1-4x}-1\right)\tag{8}\\
&=\frac{1}{2x}\left(1-\sqrt{1-4x}\right)
\end{align*}

Comment:


*

*In (3) we use twice the identity $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$ and the symmetry $\binom{2n-1}{n-1}=\binom{2n-1}{n}$

*In (4) we shift the index by one

*In (5) we use the relationship (3)

*In (6) we use the identity (1)

*In (7) we do a small rearrangement

*In (8) we use the identity (2). Since the series starts with $n=1$ we have to subtract $1$.
A: Here is a different approach based upon the Lagrange inversion theorem. We follow the paper Lagrange Inversion: when and how by R. Sprugnoli etal. It is convenient to apply formula G6 from the paper, which states that if there are functions $F(u)$ and $\phi(u)$, so that
\begin{align*}
C_n=[u^n]F(u)\phi(u)^n\tag{1}
\end{align*}
the following is valid
\begin{align*}
f(x)&=\sum_{n=0}^{\infty}C_nx^n\\
&=\sum_{n=0}^{\infty}[u^n]F(u)\phi(u)^nx^n\\
&=\left.\frac{F(u)}{1-x\phi^{\prime}(u)}\right|_{u=x\phi(u)}
\end{align*}

We obtain
  \begin{align*}
f(x)&=\sum_{n=0}^{\infty}C_nx^n\\
&=\sum_{n=0}^{\infty}\frac{1}{n+1}\binom{2n}{n}x^n\\
&=\sum_{n=0}^{\infty}\left(\binom{2n}{n}-\binom{2n}{n-1}\right)x^n\\
&=\sum_{n=0}^{\infty}\left([t^n](1+t)^{2n}-[t^{n-1}](1+t)^{2n}\right)x^n\tag{2}\\
&=\sum_{n=0}^{\infty}\left([t^n](1+t)^{2n}-[t^{n}]t(1+t)^{2n}\right)x^n\tag{3}\\
&=\left.\frac{1}{1-2x(1+u)}\right|_{u=x(1+u)^2}-\left.\frac{u}{1-2x(1+u)}\right|_{u=x(1+u)^2}\\
&=\frac{1}{1-\frac{2u}{1+u}}-\frac{u}{1-\frac{2u}{1+u}}\\
&=1+u
\end{align*}
  Since $u=x(1+u)^2$, we get
  \begin{align*}
u(x)&=\frac{1}{2x}\left(1-2x\pm\sqrt{1-4x}\right)\\
&=\frac{1}{2x}\left(1\pm\sqrt{1-4x}\right)-1
\end{align*}

Comment:


*

*In (2) we use the coefficient of operator $[t^n]$ to denote the coefficient of $t^n$ in a series. This way we can write
\begin{align*}
[t^k](1+t)^{n}=\binom{n}{k}
\end{align*}

*In (3) we use the rule $[t^{n-k}]A(t)=[t^n]t^kA(t)$. We can set according to (1)
\begin{align*}
\phi(t)=(1+t)^2
\end{align*}
and $F(t)=1$ resp. $F(t)=t$ for the terms.

We select the solution $u(x)$ which can be expanded as power series and obtain
  \begin{align*}
f(x)&=\sum_{n=0}^{\infty}C_nx^n
=1+u\\
&=\frac{1}{2x}\left(1-\sqrt{1-4x}\right)
\end{align*}

Note: Another variant of the Lagrange inversion theorem which derives the generating function of Catalan numbers from a different point of view  is provided in this answer.
