By Wilf, Zeilberger.
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Extra resources for A=B (symbolic summation algorithms)
2 Identities 21 identity are equal, let’s prove that their ratio is 1 (that will be a frequent tactic in this book). Not only that, we’ll prove that the ratio is 1 by differentiating it and getting 0 (another common tactic here). So define the function F (x, y) = e(x + y)e(−x)e(−y). By direct differentiation we find that Dx F = Dy F = 0. Thus F is constant. Set x = y = 0 to find that the constant is 1. Thus e(x + y)e(−x)e(−y) = 1 for all x, y. Now let y = 0 to find that e(−x) = 1/e(x). Thus e(x + y) = e(x)e(y) for all x, y, as claimed.
To begin, let’s try to simplify some expressions that contain factorials. /n! then what we get back is (1 + n)! , n! which doesn’t help too much. ] then we also get back (1 + n)! , n! so we must be doing something wrong. Well, it turns out that if you would really like to simplify ratios of factorials then the thing to do is to read in the package RSolve, because in that package there lives a command FactorialSimplify, which does the simplification that you would like to see. Out = 28 Tightening the Target So let’s start over, this time with In :=<< DiscreteMath‘RSolve‘.
3F2 d e a! ( a2 − b)! ( a2 − c)! (a − b − c)! (V) Clausen’s 4 F3 identity. If d is a nonpositive integer and a + b + c − d = 12 , and e = a + b + 12 , and a + f = d + 1 = b + g, then 4F3 (2a)|d| (a + b)|d| (2b)|d| a b c d ;1 = . e f g (2a + 2b)|d| a|d| b|d| (VI) Dougall’s 7 F6 identity. If n + 2a1 + 1 = a2 + a3 + a4 + a5 and a1 a6 = 1 + ; a7 = −n; and bi = 1 + a1 − ai+1 (i = 1, . . , 6), 2 then 7 F6 a1 a2 a3 a4 a5 a6 a7 ;1 b1 b2 b3 b4 b5 b6 (a1 + 1)n (a1 − a2 − a3 + 1)n (a1 − a2 − a4 + 1)n (a1 − a3 − a4 + 1)n = .