Showing equivalence of two binomial expressions

Question

I wish to show that $\sum_{k=0}^n {n\choose k}(\alpha + k)^k (\beta + n - k)^{(n-k)} = \sum_{k=0}^n {n\choose k}(\gamma + k)^k (\delta + n - k)^{(n-k)}$ given that $\alpha + \beta = \gamma + \delta$.

Both sides appear to be the n-th coefficient in the product of some exponential generating series, namely $F(y,x) = \sum_{n=0}^{\infty} (y+n)^n / n!$, and that $F(y,x) = \exp(y\times T(x))$ for some series $T(x)$. It seems to be exponential because we could then have $F(a+b,x) = F(a,x) F(b,x)$.

I have thought to use the Lagrange inversion theorem in a manner similar to some proofs of Abel's binomial theorem. Namely, if $T(x)$ is the solution to $T(x) = x \phi (T(x))$ for some invertible $\phi(x)$ in $\mathbb{Z} [[x]]$, then for $f(y,x) = exp(yx)$ we would have:

$[x^n]f(yT(x)) = \frac{1}{n} [x^{n-1}] f'(x) \phi(x)^n = (y+n)^n / n!$.

The issue is then in showing such a $\phi(x)$ exists. Unfortunately, my attempts using this approach have not been successful. Could I have some hints or suggestions on how to proceed?

Regards,

Garnet

Answer 1

We can prove this using a close relative of the labelled tree function that is known from combinatorics.

This will provide a closed form of the exponential generating function of the four terms that are involved.

The species of labelled trees has the specification $$\mathcal{T} = \mathcal{Z} \times \mathfrak{P}(\mathcal{T})$$ which gives the functional equation $$T(z) = z \exp T(z).$$

Now consider the function $$Q_\rho(z) = \frac{\exp(\rho T(z))}{1-T(z)}$$ with $\rho$ a real parameter.

Extracting coefficients via Lagrange inversion we have $$Q_n = n! \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{\exp(\rho T(z))}{1-T(z)} dz.$$

Put $T(z)=w$ so that $z=w/\exp(w) = w\exp(-w)$ and $dz = \exp(-w) - w\exp(-w)$ to get $$n! \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{\exp(w(n+1))}{w^{n+1}} \frac{\exp(\rho w)}{1-w} (\exp(-w) - w\exp(-w)) dw \\ = n! \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{\exp(wn)\exp(w\rho)}{w^{n+1}} \frac{1}{1-w} (1 - w) dw \\ = n! \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{\exp(w(\rho+n))}{w^{n+1}} dw.$$

But we have $$n! [w^n] \exp(w(\rho+n)) = n! \times \frac{(\rho+n)^n}{n!} = (\rho+n)^n$$ which means that $Q_\rho(z)$ is the exponential generating function of $(\rho+n)^n.$

The equality that we seek to prove is a convolution of two exponential generating functions on the left and on the right and to verify it we must show that $$Q_\alpha(z) Q_\beta(z) = Q_\gamma(z) Q_\delta(z).$$

But this is simply $$\frac{\exp(\alpha T(z))}{1-T(z)} \frac{\exp(\beta T(z))}{1-T(z)} = \frac{\exp(\gamma T(z))}{1-T(z)} \frac{\exp(\delta T(z))}{1-T(z)}$$ which is $$\frac{\exp((\alpha+\beta) T(z))}{(1-T(z))^2} = \frac{\exp((\gamma+\delta) T(z))}{(1-T(z))^2}$$ which holds since $\alpha+\beta = \gamma+\delta.$

The labelled tree function recently appeared at this MSE link.