Derivative of the binomial $\binom x n$ with respect to $x$

Question

My background is not mathematics and I need to implement (in C++) the derivative of a binomial, with wxMaxima and wolfram.alpha as a helper. So far, the binomial can be written as: $$\binom x n = \frac 1 {n!}\prod_{k=1}^n (x-k+1)$$ This reduces to a continued convolution. For my specific needs, the binomial needs to be of the form: $$\binom{\frac{n+1}{2}x+\frac{n-1}{2}}{n}$$ But I also need the derivative of it, which wxMaxima solves as $$-\frac{1}{2}(n+1) \,\left( \psi_0\left( \frac{(n+1) x-n+1} 2 \right) -\psi_0 \left( \frac{(n+1) \,(x+1) }{2}\right) \right) \,\begin{pmatrix}\frac{( n+1) x+n-1}{2}\\ n \end{pmatrix}$$

while wolfram goes a bit further and, instead of $\psi_0$ gives $H_n$, which they call harmonic number. (this link). That $\psi$ seems to have quite an involved formula, but $H_n$,as functions.wolfram has it, is a simple $\sum_{k=1}^n 1/k$, which is a lot simpler in terms of C++.

Now, because I have trust issues, I went on to verify the answer given by wolfram, in wxMaxima, for $n=4$. Here's the code:

n:4$
g:diff(binomial((n+1)/2*(x+1)-1,n),x),expand,numer$
h:-(n+1)/2*binomial((n+1)/2*(x+1)-1,n)*(sum(1/k,k,1,(n+1)/2*(x-1))-sum(1/k,k,1,1/2*(n*(x+1)+x-1)));
wxplot2d([g,h],[x,0,1]);

and here's the output of it:

plot

As you can see, they don't match; plotting wxMaxima's derivation is a match, but that involves $\psi$ as an infinite sum. So I'm left wondering what's wrong: is the implementation of the harmonic number? Is the derivation formula? Is it the way I transcribed it?

TL;DR: I need a derivation formula (not the actual code, that's up to me) for the binomial that is (fairly) simple to implement and doesn't take ages to compute, in C++, as the whole function will be called in a bracketed root-finding algorithm. And I'm also using GMP from gmplib dot org (need 10 rep to post more than 2 links).

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Following G Cab's excellent post, and modifying the formulas according to my needs, I managed to come up (with a bit of hammering) to this formula:

$$\frac{d}{dx}\binom{\frac{n+1}{2}x+\frac{n-1}{2}}{n}=(-1)^n \frac{n+1}{2} \binom{\frac{n+1}{2}x + \frac{n-1}{2}}{n} \sum_{k=0}^{n-1} \frac{1}{\frac{n+1}{2}x +\frac{n-1}{2}-k}$$

The $(-1)^n$ takes care of odd $n$. Thank you very much everyone that answered.

Answer 1

For the purpose of computing the derivatives, you can also profitably make use of the expression of the binomials via the Stirling Numbers of 1st kind $$ \binom y n = \frac{y^{\,\underline {\,n\,} }}{n!} = \frac{1}{n!} \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} (-1)^{n - k} \left[ \begin{gathered} n \\ k \end{gathered} \right]y^k $$ Now, to explain about the alternative formulas with $\psi$ and $H$, consider the binomial expressed in terms of the Gamma function $$ \binom y n = \frac{\Gamma (y + 1)}{\Gamma (n + 1)\Gamma (y - n + 1)} = \frac 1 {n!} \frac{\Gamma (y + 1)} {\Gamma (y - n + 1)} $$ then $$ \begin{gathered} \frac d {dy} \binom y n = - \frac 1 {n!} \frac{\Gamma '(y + 1)\Gamma (y - n + 1) - \Gamma (y + 1)\Gamma '(y - n + 1)} {\Gamma (y - n + 1)^2} = \hfill \\ = \frac 1 {n!} \left( {\frac{\Gamma (y + 1)}{{\Gamma (y - n + 1)}} \frac{{\Gamma '(y - n + 1)}} {\Gamma (y - n + 1)} - \frac{{\Gamma (y + 1)}} {\Gamma (y - n + 1)}\frac{\Gamma '(y + 1)} {\Gamma (y + 1)}} \right) = \hfill \\ = \binom y n (\psi _0 (y - n + 1) - \psi _0 (y + 1)) \hfill \\ \end{gathered} $$

Consider instead the binomial expressed in terms of the product,

$$ \binom y n = \frac{y^{\,\underline {\,n\,} }}{n!} = \frac 1 {n!} \prod\limits_{j = 0}^{n - 1} (y - j) $$ then for the derivative you have $$ \begin{gathered} \frac d {dy} \binom y n = \frac 1 {n!} \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\prod\limits_{\begin{array}{*{20}c} {j \ne k} \\ {j = 0} \\ \end{array} }^{n - 1} {(y - j)} } = \frac 1 {n!} \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\frac{1}{y - k} \prod\limits_{j = 0}^{n - 1} {\left( {y - j} \right)} = } \hfill \\ = \binom y n \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\frac{1} {{y - k}}} = \binom y n \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} \frac 1 {y - n + k} \hfill \\ \end{gathered} $$ and for computational purposes, this formula is already quite viable, and I think you do not need to consider the further expansion leading to: $$ \frac d {dy} \binom y n = \binom y n \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} \frac 1 {y - n + k} = \binom y n \sum_{y - n + 1\, \leqslant \,k\, \leqslant \,y} \frac 1 k = \binom y n (H_y - H_{y - n}) $$

Answer 2

We can start with the product representation \begin{align*} \binom{\frac{n+1}{2}x+\frac{n-1}{2}}{n}&=\frac{1}{n!}\prod_{k=1}^n\left(\frac{n+1}{2}x+\frac{n-1}{2}-k+1\right)\\ &=\frac{1}{2^nn!}\prod_{k=1}^n\left((n+1)x+n+1-2k\right) \end{align*} and recall that \begin{align*}\frac{d}{dx}\prod_{k=1}^nf_k(x) =\sum_{j=1}^nf_j^\prime(x)\prod_{{k=1}\atop{k\neq j}}^nf_k(x) \end{align*}

We obtain \begin{align*} \frac{d}{dx}\binom{\frac{n+1}{2}x+\frac{n-1}{2}}{n} &=\frac{d}{dx}\left(\frac{1}{2^nn!}\prod_{k=1}^n\left((n+1)x+n+1-2k\right)\right)\\ &=\frac{1}{2^nn!}\sum_{j=1}^n \left(\frac{d}{dx}\left((n+1)x+n+1-2j\right)\right) \prod_{{k=1}\atop{k\neq j}}^n\left((n+1)x+n+1-2k\right)\\ &=\frac{n+1}{2^nn!}\sum_{j=1}^n\prod_{{k=1}\atop{k\neq j}}^n \left((n+1)x+n+1-2k\right) \end{align*}