I would like to prove the following result:

Let $f:R\to R$ be such that $f(x)=\sum\limits_{k=0}^\infty a_k(x-c)^k$ for all $x$ in some open set $O\subset R$.

Suppose that $f'(c)\ne 0$.

Then there is a function $g:T\to R$ such that $g(f(x))=x$ and $g(x)=\sum\limits_{k=0}^\infty b_k(x-f(c))^k$, where $T$ is an open set of $R$ containing $f(c)$.

I know how to prove that $f$ is locally invertible but not how to prove that its inverse can be written as a power series on some open set.

There are many proofs for the inverse function theorem which use complex analysis but I would like to write a proof which uses only real analysis. I have been writing a book of proofs of many of the important theorems in calculus and I don't want to introduce complex analysis for the proof of one theorem.

Thanks, Andrew Murdza.

  • $\begingroup$ I suggest consulting section 1.4 of A Primer of Real Analytic Functions by Krantz and Parks. $\endgroup$
    – Umberto P.
    Commented Mar 4, 2019 at 20:38
  • $\begingroup$ This is perfect! This answers my question. If you convert your comment into an answer (with the same text) I will accept it. $\endgroup$ Commented Mar 5, 2019 at 0:26

1 Answer 1


This proof works over any Cauchy-complete field $\mathbb F$ with a non-trivial absolute value. The basic examples are $\mathbb R$ (real numbers), $\mathbb C$ (complex numbers), $\mathbb Q_p$ ($p$-adic numbers), and $\mathbb F_p((X))$ (formal Laurent series over a finite field). In fact $\mathbb F$ necessarily contains one of these four as a sub-field, but we don't need to know that.

After shifting/translating (which is invertible and preserves analyticity), we can assume that $c=f(c)=0$. Likewise, after scaling, we can assume that $f'(c)=1$, but let's not do that yet.

$$f(x)=\sum_{n\geq1}a_nx^n,\quad|x|\leq R$$ $$\sum_{n\geq1}|a_n|R^n<\infty$$ $$g(y)=\sum_{n\geq1}b_ny^n,\quad|y|\leq S$$ $$\sum_{n\geq1}|b_n|S^n<\infty$$

We're given $a_n$ and $R>0$, and we want to find $b_n$ such that $g(f(x))=x$ for all $x$ in some neighbourhood of $0$ and $f(g(y))=y$ for all $y$ in some neighbourhood of $0$.

Expanding the formal power series $g(f(X))$ (as shown at the bottom of this answer) and equating to the identity series $e(X)=X$, we can see that the required coefficients $b_n$ exist and are unique. Each one is a polynomial in $a_k$ for $k\leq n$, with integer coefficients, divided by some power of $a_1\neq0$. Thus $f$ has a unique formal left-inverse: $g\circ f=e$. By similar reasoning, $f$ has a unique formal right-inverse: $f\circ h=e$. Since composition of formal power series is associative, we have $g=g\circ e=g\circ(f\circ h)=(g\circ f)\circ h=e\circ h=h$, so the two inverses are in fact the same.

We only need to show that $g(y)$ converges for some $y\neq0$. That is, we need to find $S>0$ such that $\sum_n|b_n|S^n<\infty$.

Wolfram MathWorld gives this formula for the coefficients:


I've reformulated it somewhat. I haven't verified it, but we can look for a proof of this formula elsewhere.

That factorial stuff looks like a fraction, which might be undefined if $\mathbb F$ has positive characteristic. But we've already seen that the coefficients are integers, so that's not a problem. We can evaluate the fraction in $\mathbb Q$, getting a result in $\mathbb N$, and then map that to $\mathbb F$ in the natural way.

Now let's scale $f$ so that $a_1=1$. (This simplifies the formula, but may actually make it harder to interpret and to verify, which is why I haven't done it yet.)

Applying the triangle inequality:

$$|b_n|\leq\sum_{k_2+2k_3+3k_4+\cdots=n-1}\frac{(2k_2+3k_3+4k_4+\cdots)!}{n!\;k_2!\;k_3!\;k_4!\;\cdots}|a_2|^{k_2}|a_3|^{k_3}|a_4|^{k_4}\cdots$$ $$\sum_{n\geq1}|b_n||y|^n\leq\sum_{n\geq1}\;\sum_{k_2+2k_3+3k_4+\cdots=n-1}\frac{(2k_2+3k_3+4k_4+\cdots)!}{n!\;k_2!\;k_3!\;k_4!\;\cdots}|y|^n|a_2|^{k_2}|a_3|^{k_3}|a_4|^{k_4}\cdots$$

Rearranging the series, and eliminating the variable $n$:


Introducing a new variable $m$:


Distributing factors of $|y|$:


Breaking off the $m=0$ term, where the inner sum reduces to a single term with $k_2=k_3=k_4=\cdots=0$:


Rewriting the factorial stuff with a binomial coefficient, and using $\binom nk\leq\sum_k\binom nk=2^n$:

$$=|y|+\sum_{m\geq1}|y|^{1-m}\sum_{k_2+k_3+k_4+\cdots=m}\binom{2k_2+3k_3+4k_4+\cdots}{m-1}\frac{(m-1)!}{k_2!\;k_3!\;k_4!\;\cdots}|a_2y^2|^{k_2}|a_3y^3|^{k_3}|a_4y^4|^{k_4}\cdots$$ $$\leq|y|+\sum_{m\geq1}|y|^{1-m}\sum_{k_2+k_3+k_4+\cdots=m}2^{2k_2+3k_3+4k_4+\cdots}\frac{(m-1)!}{k_2!\;k_3!\;k_4!\;\cdots}|a_2y^2|^{k_2}|a_3y^3|^{k_3}|a_4y^4|^{k_4}\cdots$$

Distributing factors of $2$:

$$=|y|+\sum_{m\geq1}|y|^{1-m}\sum_{k_2+k_3+k_4+\cdots=m}\frac{(m-1)!}{k_2!\;k_3!\;k_4!\;\cdots}\big(2^2|a_2y^2|\big)^{k_2}\big(2^3|a_3y^3|\big)^{k_3}\big(2^4|a_4y^4|\big)^{k_4}\cdots$$ $$=|y|+\sum_{m\geq1}\frac{|y|^{1-m}}{m}\sum_{k_2+k_3+k_4+\cdots=m}\frac{m!}{k_2!\;k_3!\;k_4!\;\cdots}\big(2^2|a_2y^2|\big)^{k_2}\big(2^3|a_3y^3|\big)^{k_3}\big(2^4|a_4y^4|\big)^{k_4}\cdots$$

Applying the multinomial formula:

$$=|y|+\sum_{m\geq1}\frac{|y|^{1-m}}{m}\Big(2^2|a_2y^2|+2^3|a_3y^3|+2^4|a_4y^4|+\cdots\Big)^m$$ $$=|y|+\sum_{m\geq1}\frac{2^2|y|^{1+m}}{m}\Big(2^0|a_2y^0|+2^1|a_3y^1|+2^2|a_4y^2|+\cdots\Big)^m$$

The parenthesized series will converge if $|y|\leq R/2$, since we assumed $\sum_n|a_n|R^n$ converges:

$$=|y|+\sum_{m\geq1}\frac{2^2|y|^{1+m}}{m}\left(\sum_{n\geq2}|a_n|(2|y|)^{n-2}\right)^m$$ $$\leq|y|+\sum_{m\geq1}\frac{2^2|y|^{1+m}}{m}\left(\sum_{n\geq2}|a_n|R^{n-2}\right)^m$$ $$=|y|+4|y|\sum_{m\geq1}\frac{1}{m}\left(|y|\sum_{n\geq2}|a_n|R^{n-2}\right)^m$$

The outer series will converge if $|y|<1/\sum_{n\geq2}|a_n|R^{n-2}$, since the Taylor series of the real logarithm is $-\ln(1-u)=\sum_{m\geq1}u^m/m$ for $-1<u<1$:

$$=|y|-4|y|\ln\left(1-|y|\sum_{n\geq2}|a_n|R^{n-2}\right)$$ $$<\infty$$

Thus, we've shown that $g(y)$ converges absolutely, with a radius at least


  • $\begingroup$ What did I do after "Applying the multinomial formula"? I factored out $2^2$ instead of $2^{2m}$... That changes the last few lines of the answer, but not too badly. The radius should be $$S=\min\left\{\frac R2,\;\frac{0.99R^2}{\mathbf4\sum_{n\geq2}|a_n|R^n}\right\}.$$ $\endgroup$
    – mr_e_man
    Commented Jan 11, 2023 at 19:59

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