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Concrete mathematics : a foundation for computer science

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Kummer was a<br />

summer.<br />

5.5 HYPERGEOMETRIC FUNCTIONS 213<br />

The sum in Problem 2 and Problem 4 likewise yields F( 2,1 - n; 2 - m; 1).<br />

(We need to replace k by k + 1 first.) And the “menacing” sum in Problem 6<br />

turns out to be just F(n + 1, -n; 2; 1). Is there nothing more to sum, besides<br />

disguised versions of Vandermonde’s powerful convolution?<br />

Well, yes, Problem 3 is a bit different. It deals with a special case of the<br />

general sum tk (“kk) zk considered in (5.74), and this leads to a closed-<strong>for</strong>m<br />

expression <strong>for</strong><br />

We also proved something new in (5.55), when we looked at the coefficients<br />

of (1 - z)~( 1 + z)~:<br />

F l-c-2n, - 2 n<br />

(2n)! (c - 1 )!<br />

-1 = (-l)n-<br />

( C 1 ><br />

n! (c+n-l)!’<br />

integer n 3 0.<br />

This is called Kummer’s <strong>for</strong>mula when it’s generalized to complex numbers:<br />

(5.94)<br />

The summer of ‘36. (Ernst Kummer [187] proved this in 1836.)<br />

It’s interesting to compare these two <strong>for</strong>mulas. Replacing c by l -2n- a,<br />

we find that the results are consistent if and only if<br />

(5.95)<br />

when n is a positive integer. Suppose, <strong>for</strong> example, that n = 3; then we<br />

should have -6!/3! = limX+ 3x!/(2x)!. We know that (-3)! and (-6)! are<br />

both infinite; but we might choose to ignore that difficulty and to imagine<br />

that (-3)! = (-3)(-4)(-5)(-6)!,so that the two occurrences of (-6)! will<br />

cancel. Such temptations must, however, be resisted, because they lead to<br />

the wrong answer! The limit of x!/(2x)! as x + -3 is not (-3) (-4) (-5) but<br />

rather -6!/3! = (-4)(-5)(-6), according to (5.95).<br />

The right way to evaluate the limit in (5.95) is to use equation (5.87),<br />

which relates negative-argument factorials to positive-argument Gamma functions.<br />

If we replace x by -n + e and let e + 0, two applications of (5.87)<br />

give<br />

(-n-e)! F(n+e) sin(2n + 2e)rt<br />

(-2n - 2e)! F(2n + 2e) = sin(n + e)rc

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