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Let $f$ be a polynomial of degree $m$ in $t$. The following curious identity holds for $n \geq m$,

\begin{align}

\binom{t}{n+1} \sum_{j = 0}^{n} (-1)^{j} \binom{n}{j} \frac{f(j)}{t – j} = (-1)^{n} \frac{f(t)}{n + 1}.

\end{align}

The proof follows by transforming it into the identity

\begin{align}

\sum_{j = 0}^{n} \sum_{k = j}^{n} (-1)^{k-j} \binom{k}{j} \binom{t}{k} f(j) = \sum_{k = 0}^{n} \binom{t}{k} (\Delta^{k} f)(0) = f(t),

\end{align}

where $\Delta^{k}$ is the $k^{\text{th}}$ forward difference operator. However, I’d like to prove the aforementioned identity directly, without recourse to the calculus of finite differences. Any hints are appreciated!

Thanks.

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This is just Lagrange interpolation for the values $0, 1, \dots, n$.

This means that after cancelling the denominators on the left you can easily check that the equality holds for $t=0, \dots, n$.

I just ran across this question after working on this answer, and realized that the same method could be used here.

Notice that your equation is equivalent to

$$

\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}\frac{f(j)}{t-j}=\frac{n!f(t)}{t(t-1)(t-2)\dots(t-n)}

$$

As long as $f$ is a polynomial of degree $n$ or less, apply the Heaviside Method for Partial Fractions to the right hand side to get the left hand side.

That is, to compute the coefficient of $\frac1{t-j}$ on the left hand side, multiply both sides by $t-j$ and set $t=j$. The right hand side becomes

$$

\frac{n!f(j)}{j(j-1)(j-2)\dots1(-1)(-2)\dots(j-n)}=(-1)^{n-j}\binom{n}{j}f(j)

$$

This can also be done with complex variables. Observe that we have by

inspection that

$$\sum_{j=0}^n (-1)^j {n\choose j} \frac{f(j)}{t-j}

= (-1)^n \times n! \times

\sum_{k=0}^n \mathrm{Res}_{z=k}

\frac{f(z)}{t-z}\prod_{q=0}^n \frac{1}{z-q}.$$

This holds even if $f(t)$ vanishes at some positive integer in the

range.

Recall that the residues sum to zero, so the right is equal to

$$-(-1)^n \times n! \times

\sum_{k\in\{\infty,t\}} \mathrm{Res}_{z=k}

\frac{f(z)}{t-z}\prod_{q=0}^n \frac{1}{z-q}.$$

The residue at infinity of a function $h(z)$ is given by the formula

$$\mathrm{Res}_{z=\infty} h(z)

= \mathrm{Res}_{z=0}

\left[-\frac{1}{z^2} h\left(\frac{1}{z}\right)\right]$$

which in the present case yields

$$-\mathrm{Res}_{z=0} \frac{1}{z^2}

\frac{f(1/z)}{t-1/z}\prod_{q=0}^n \frac{1}{1/z-q}

\\ = -\mathrm{Res}_{z=0} \frac{z^{n+1}}{z^2}

\frac{f(1/z)}{t-1/z}\prod_{q=0}^n \frac{1}{1-qz}

\\ = -\mathrm{Res}_{z=0} z^n

\frac{f(1/z)}{zt-1}\prod_{q=0}^n \frac{1}{1-qz}.$$

Note however that $f(z)$ has degree $m$ and we require that $n\ge m$

which means $f(1/z) z^n$ has no pole at zero and hence the residue at

infinity is zero as well.

That leaves the residue at $z=t$ for a total contribution of

$$-(-1)^n \times n! \times (-1)\times f(t)\times

\prod_{q=0}^n \frac{1}{t-q}

= (-1)^n f(t) \times n! \prod_{q=0}^n \frac{1}{t-q}

\\ = (-1)^n f(t) \times n! \times

\frac{1}{(n+1)!} {t\choose n+1}^{-1}

\\ = (-1)^n \frac{f(t)}{n+1} {t\choose n+1}^{-1}$$

which was to be shown, QED.

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