# [✓] Obtain a proof of this result?

GROUPS:
 I'm seeking a proof that the sum shown below evaluates to shown result. In[1]:= factorialPowerOneHalf[k_] := FactorialPower[1/2, k] In[2]:= f[n_, k_] := Binomial[n, k]*factorialPowerOneHalf[k]*factorialPowerOneHalf[n - k] In[3]:= Sum[f[n, k], {k, 0, n}] Out[3]= FactorialPower[1, n] I'm confident Mathematica is correct, but I'm interested to see the proof. Is there any way Mathematica can help me find the proof?
2 months ago
7 Replies
 Jim Baldwin 1 Vote What you have is a special case of something analogous to the binomial expansion but with factorial powers rather than integer powers. $$(a+b)_n = \sum_{j=0}^n \binom{n}{j} (a)_j (b)_{n-j}$$with $a=b={1\over 2}$ .
2 months ago
 Thank you. I recognize the relation of my problem to the Binomial Expansion: In[18]:= FullSimplify[ Binomial[n, k] FactorialPower[1/2, k] FactorialPower[1/2, n - k] == n!*Binomial[1/2, k]*Binomial[1/2, n - k]] Out[18]= True and that equation I can prove easily.Curiously---but not essentially to my line of inquiry---Mathematica doesn't solve the sum when I express it more like the Binomial Expansion: In[13]:= FullSimplify[n!*Binomial[1/2, k]*Binomial[1/2, n - k]] Out[13]= Binomial[1/2, k] Binomial[1/2, -k + n] n! or more simply: In[13]:= FullSimplify[Binomial[1/2, k]*Binomial[1/2, n - k]] Out[13]= Binomial[1/2, k] Binomial[1/2, -k + n] n! I could solve my problem (finding a proof as described in my initial post) by working through an analysis textbook; but what I'm really after is a proof that doesn't rely on derivatives. A proof modeled after (e.g.) Rudin would make heavy use of derivatives and Taylor's Theorem, which I'm trying to avoid.I was hoping Mathematica could help me find such a proof, but my hope is a long shot, as Mathematica seems to excel at calculating true answers, but less so at providing proofs.
2 months ago
 Jim Baldwin 1 Vote Because your example was a special case of the more general expansion, it wasn't clear that you were aware of that identity. In the link I used the "Sheffer Sequence" is referenced and a search on that term might find some established proofs. Wolfram MathWorld gives a blurb on the Sheffer Sequence.I'm not sure what you would expect from Mathematica. If FullSimplify ended up with FactorialPower[1, n], how would the "black box" function FullSimplify be considered a proof?
2 months ago
 Thanks for the additional references. "Sheffer Sequence" is something I hadn't heard of; I'll research it. I'm not any expert in the Binomial Expansion; I've just seen it asserted, and I've browsed some analysis textbooks for their line of proof.Software could show proofs. I wasn't expecting Mathematica to be that software; I was just asking if it was.
 Updating Name 1 Vote A proof by induction of Jim's identity is probably what you want. Multiply by both sides by {((a - k), (b - (n - k)))} and use the fact that FactorialPower[a, k + 1] == (a - k) FactorialPower[a, k] etc. to get terms like Binomial[n, k] FactorialPower[a, k + 1] FactorialPower[b, -k + n] Binomial[n, k] FactorialPower[a, k] FactorialPower[b, -k + n + 1] Shift the index of the first type to get terms like Binomial[n, k - 1] FactorialPower[a, k] FactorialPower[b, -k + n + 1] Then use the binomial identity for Binomial[n, k - 1] Binomial[n, k] to write the LHS as a sum up to n + 1.