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By Petkovsek M., Wilf H.S., Zeilberger D.

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2(−n − 12 )! (− 12 )! A similar argument shows that (−2n − 12 )! (−1)n (n − 12 )! = , (−n − 12 )! (2n − 12 )! 2 2n . n (2n − 12 )! (− 12 )! But for every positive integer m, 1 1 3 1 1 (m − )! = (m − )(m − ) · · · ( )(− )! 2 2 2 2 2 (2m − 1)(2m − 3) · · · 1 1 = (− )! 2m 2 (2m)! 1 = m (− )!. 4 m! 6 Using the database 47 So we can simplify our answer all the way down to f(n) = labor is that we have found the identity (−1)k k 2n k 2k k 2n 2 . 3) we realize that it is a special case of Dixon’s identity, and we further realize that the “lookup” in the database was not quite a routine matter!

1. Define e(x) to be the famous series n≥0 xn /n!. We will prove that e(x + y) = e(x)e(y) for all x and y. First, the series converges for all x, by the ratio test, so e(x) is well defined for all x, and e (x) = e(x). Next, instead of trying to prove that the two sides of the 1 See page 141. 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).

There are n−k Hence there are nk ways of of the 2n n n k ways to choose k ways to choose n−k letters n n−k = n 2 k ways to make choosing n letters from the such a pair of choices. But every one 2n letters 1, 2, . . , 2n corresponds uniquely to such a pair of choices, for some k. ✷ We must pause to remark that that one is a really nice proof. So as we go through this book whose main theme is that computers can prove all of these identities, please note that we will never7 claim that computerized proofs are better than human ones, in any sense.

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A=B by Petkovsek M., Wilf H.S., Zeilberger D.


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