$\sum_{k = 0}^n (-1)^k\binom{n+1}{k+1}(1/f^k(x))^{(n)}f^{1+k}(x) = (f(x))^{(n)}$

Question

$\newcommand{\on}[1]{\operatorname{#1}}$

Prove that $$ \sum_{k = 0}^{n}\left(-1\right)^{k} \binom{n + 1}{k + 1} \left[\dfrac{1}{\on{f}^{k}\left(x\right)}\right] ^{\left(n\right)} \on{f}^{1 + k}\left(x\right) = \left[\on{f}\left(x\right)\right]^{\left(n\right)}. $$ Note that $f^k(x) = (f(x))^k$ and $f^{(n)}(x) = \dfrac d{dx^n}f(x)$.

Since $\binom{n+1}{k+1} = \binom{n}{k}+\binom{n}{k+1}$ it follows that the sum is $$\sum_{k = 0}^n (-1)^k\binom{n}{k}\left(\dfrac1{f^k(x)}\right)^{(n)}f^{1+k}(x)+\sum_{k = 0}^n (-1)^k\binom{n}{k+1}\left(\dfrac1{f^k(x)}\right)^{(n)}f^{1+k}(x).$$ Is this the same as $(f(x))^{(n)}$?

Note that $\sum_{k = 0}^n (-1)^k\binom{n+1}{k+1}(k+n+1)^m = n^m$ and $\sum_{k = 0}^n (-1)^k\binom{n+1}{k+1} = 1$.

Answer 1

Suppose for the moment that $x$ is a complex number. We will assume $f(x)$ and $1/f(x)$ are analytic at $x$ so they both have power series there with some non-zero radius of convergence. Take $\rho\ll$ the lesser of the two and take $\rho=1$ if both converge everywhere. We then get for the sum

$$\sum_{k=0}^n (-1)^k {n+1\choose k+1} f^{k+1}(x) \frac{n!}{2\pi i} \int_{|z-x|=\rho} \frac{1}{(z-x)^{n+1}} \frac{1}{f^k(z)} \; dz.$$

This is

$$\frac{n!}{2\pi i} \int_{|z-x|=\rho} \frac{f(z)}{(z-x)^{n+1}} \sum_{k=0}^n (-1)^k {n+1\choose k+1} f^{k+1}(x) \frac{1}{f^{k+1}(z)}\; dz \\ = \frac{n!}{2\pi i} \int_{|z-x|=\rho} \frac{f(z)}{(z-x)^{n+1}} \; dz \\ + \frac{n!}{2\pi i} \int_{|z-x|=\rho} \frac{f(z)}{(z-x)^{n+1}} \sum_{k=-1}^n (-1)^k {n+1\choose k+1} f^{k+1}(x) \frac{1}{f^{k+1}(z)}\; dz \\ = f^{(n)}(x) - \frac{n!}{2\pi i} \int_{|z-x|=\rho} \frac{f(z)}{(z-x)^{n+1}} \left[1-\frac{f(x)}{f(z)}\right]^{n+1} \; dz \\ = f^{(n)}(x) - \frac{n!}{2\pi i} \int_{|z-x|=\rho} \frac{1}{f^n(z)} \left[\frac{f(z)-f(x)}{z-x}\right]^{n+1} \; dz = f^{(n)}(x).$$

Expanding the bracketed term into a power series about $x$ we find

$$g(z) = f'(x)+f''(x)(z-x)/2+f'''(x)(z-x)^2/6+\cdots$$

so it goes to a constant $f'(x)$ with no more pole at $z=x.$ More importantly it is convergent in the neighborhood of $x$ so it represents an analytic function there. Integrating $g^{n+1}(z)/f^n(z)$ which is the product of two analytic terms we obtain a zero contribution from the integral which concludes the argument.

Answer 2

We show the following formula is valid for $n\geq 1$: \begin{align*} \color{blue}{D_x^n\left(\frac{1}{f}\right)=\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}\frac{1}{f^{k+1}}D_x^nf^k}\tag{1} \end{align*} Here we have the $n$-th derivative of a composition $g\circ f$ of functions $f$ and $g$ with $g(x)=\frac{1}{x}$. A formula to calculate the $n$-th derivative of composite functions is stated as Theorem 8.1 in H.W. Gould's Combinatorial Identities called:

Hoppe Form of Generalized Chain Rule.

Let $D_f$ represent differentiation with respect to $f$ and $f=f(x)$. Hence $D^n_x g(f)$ is the $n$-th derivative of $g$ with respect to $x$. The following is valid for $n\geq 1$: \begin{align*} D_x^n g(f)=\sum_{q=0}^n\left(D_f^qg(f)\right)\frac{(-1)^q}{q!}\sum_{k=0}^q(-1)^k\binom{q}{k}f^{q-k}D_x^nf^k\tag{2} \end{align*}

We obtain from (2) with $g(f)=1/f$: \begin{align*} \color{blue}{D_x^n\left(\frac{1}{f}\right)} &\color{blue}{=\sum_{q=0}^n\left(D_f^q\left(\frac{1}{f}\right)\right) \frac{(-1)^q}{q!}\sum_{k=0}^q(-1)^k\binom{q}{k}f^{q-k}D_x^nf^k}\tag{3}\\ \end{align*} We calculate the $q$-th derivative of $\frac{1}{x}$ and get \begin{align*} D_x^q\left(\frac{1}{x}\right)=(-1)^qq!\frac{1}{x^{q+1}} \end{align*} Using this identity with $x$ replaced by $f$ we can put it into (3) and obtain \begin{align*} \color{blue}{D_x^n\left(\frac{1}{f}\right)} &=\sum_{q=0}^n\left((-1)^qq!\frac{1}{f^{q+1}}\right) \frac{(-1)^q}{q!}\sum_{k=0}^q(-1)^k\binom{q}{k}f^{q-k}D_x^nf^k\\ &=\sum_{q=0}^n\sum_{k=0}^q(-1)^k\binom{q}{k}\frac{1}{f^{k+1}}D_x^nf^k\\ &=\sum_{k=0}^n(-1)^k\frac{1}{f^{k+1}}\left(D_x^nf^k\right)\sum_{q=k}^n\binom{q}{k}\tag{4}\\ &\,\,\color{blue}{=\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}\frac{1}{f^{k+1}}D_x^nf^k}\tag{5} \end{align*} and the claim (1) follows.

Comment:

• In (4) we exchange the order of the sums.

• In (5) we apply the Hockey stick identity.