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In my work, I have come across the following equation

$$ \sum_{k = 0}^m \sum_{l = 0}^{n-m} {m \choose k} {n - m \choose l} (-1)^k r_{k + l} $$

where $r_n = 1$ if $n$ is congruent to 0 or 1 modulo 4, and $r_n = -1$ if $n$ is congruent to 2 or 3 modulo 4. I've seen some identities involving the products ${m \choose k} {n - m \choose l}$, but I can't seem to find an algebraic trick to reduce this form to a standard identity. Any tips?

1 Answers 1

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This is

$$\sum_{k=0}^m {m\choose k} (-1)^k \sum_{l=0}^{n-m} {n-m\choose l} r_{k+l}.$$

We have that

$$r_{k+l} = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{k+l+1}} (1+w - w^2-w^3 + \cdots) \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{k+l+1}} \frac{1+w - w^2-w^3}{1-w^4} \; dw.$$

We get for the sum

$$\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w} \frac{1+w - w^2-w^3}{1-w^4} \sum_{k=0}^m {m\choose k} \frac{(-1)^k}{w^k} \sum_{l=0}^{n-m} {n-m\choose l} \frac{1}{w^l} \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w} \frac{1+w - w^2-w^3}{1-w^4} \sum_{k=0}^m {m\choose k} \frac{(-1)^k}{w^k} \left(1+\frac{1}{w}\right)^{n-m} \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{n-m}}{w^{n-m+1}} \frac{1+w - w^2-w^3}{1-w^4} \sum_{k=0}^m {m\choose k} \frac{(-1)^k}{w^k} \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{n-m}}{w^{n-m+1}} \frac{1+w - w^2-w^3}{1-w^4} \left(1-\frac{1}{w}\right)^m \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{n-m} (w-1)^m}{w^{n+1}} \frac{1+w - w^2-w^3}{1-w^4} \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{n-m} (w-1)^m}{w^{n+1}} \frac{1+w}{1+w^2} \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{n-m+1} (w-1)^m}{w^{n+1}} \frac{1}{1+w^2} \; dw.$$

Now checking on the residue at infinity by a circular contour of radius $R$ we obtain a term of order $2\pi R \times R^{n+1}/R^{n+1}/R^2$ which is $2\pi/R$ and vanishes so that residue is zero. This leaves for the sum the residues at $w=\pm i$ and we get (remember to flip the sign at the end)

$$\exp(-(n+1)\pi i/2) \sqrt{2}^{n+1} \frac{1}{2i} \exp(+(n-m+1)\pi i/4)\exp(+3m\pi i/4) \\ - \exp(+(n+1)\pi i/2) \sqrt{2}^{n+1} \frac{1}{2i} \exp(-(n-m+1)\pi i/4)\exp(-3m\pi i/4) \\ = \sqrt{2}^{n+1} \sin(-(2n+2)\pi/4 +(n-m+1)\pi/4 + 3m\pi/4).$$

This yields $$- \sqrt{2}^{n+1} \sin((2m-n-1)\pi/4)$$ or

$$\bbox[5px,border:2px solid #00A000]{ \sqrt{2}^{n+1} \sin((n+1-2m)\pi/4).}$$

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    I don't completely understand how you get the integral formula for $r_n$. Is this a general technique? – 2017-02-27
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    Alternatively we may use formal power series notation if you find it easier to process. We obtain $$r_{k,l} = [w^{k+l}] (1+w-w^2-w^3+\cdots)$$ which is seen to hold by inspection. We sum the geometric series to get the formula. Also, there are several users that solve these types of sums on MSE and you may yet see a completely different approach. – 2017-02-27
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    I see what you mean now - you're using Cauchy's integral formula with a generating function - clever :) – 2017-02-27
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    This is the so-called Egorychev method which has been in use in various guises for decades by many mathematicians, not always being identified by that name. – 2017-02-27
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    Nice uniform cascade of integrals. :) – 2017-02-27