Power series of inverse function Let, $f$ be a bijective function on set of Real numbers. Let, $f(x) =\sum_{n=1}^{\infty}a_{n}x^n$ such that $a_{1}=2,a_{2}=4$ let, $f^{-1}(x) =\sum_{n=1}^{\infty} b_nx^n$
Then find value of $b_1$.
My approach:
we know,
$$\frac{1}{1-2x}=1+2x+4x^2+8x^3+\ldots $$
$$\frac{2x}{1-2x} = \sum_{n=1}^{\infty}(2x)^n $$
Hence,
$$f(x) = \frac{2x}{1-2x}$$
Let, $f(x)=y$
then, $$2x =y(1-2x)$$
$$2x= \frac{y}{1+y}$$
$$x= \frac{y}{2(1+y)}$$
Now,
$$\frac{1}{1+y}= 1-y+y^2-y^3+\ldots$$
$$\frac{y}{2(1+y)}= \frac{y}{2} -\frac{y^2}{2}+\ldots $$
Hence, $b_1=\frac{1}{2}$.
Am I correct?
 A: You cannot assume anything about $f$ other than the facts that $a_1=2$ and $a_2=4$.
If $f^{-1}(x)=\sum_{n=1}^\infty b_nx^n$, then $f^{-1}\bigl(f(x)\bigr)=x$ means that$$b_1(a_1x+a_2x^2+a_3x^3+\cdots)+b_2(a_1x+a_2x^2+a_3x^3+\cdots)^2+\cdots=x.\label{a}\tag1$$But the coefficient of $x$ on the LHS of \eqref{a} is $b_1a_1$. So, $b_1a_1=1$. But $a_1=2$. Therefore, $b_1=\frac12$.
A: Here's an approach related to that of José Carlos Santos: if we differentiate the expression $f(f^{-1}(x))=x$ implicitly, we find that
$$f'(f^{-1}(x))(f^{-1})'(x)=1,$$
so that $(f^{-1})'(x)=1/f'(f^{-1}(x))$.  This tells us that $b_1=(f^{-1})'(0)=1/f'(f^{-1}(0))$.  Because $f(0)=0$, we know that $f^{-1}(0)=0$, and can conclude that $b_1=(f^{-1})'(0)=1/f'(0)=1/a_1$.
As noted in the comments, you are only given $a_1$ and $a_2$, and cannot conclude anything else about the function $f$.  The fact that you get the correct answer is because any injective function with $a_0=0$ will have $a_1b_1=1$, so your example had to give the correct answer.  But an example does not prove that the answer is correct.
