Prove that <munderover> &#x2211;<!-- ∑ --> <mrow class="MJX-TeXAtom-ORD"> k =

Jordon Haley

Jordon Haley

Answered question

2022-05-14

Prove that k = 0 n ( 1 ) k k ! ( n k ) = e 0 t n e t n ! J 0 ( 2 t ) d t using only real analysis.

Answer & Explanation

Gillian Kelly

Gillian Kelly

Beginner2022-05-15Added 11 answers

Calculating the generating function for the LHS has already been done following your link. It suffices to calculate the generating function for the RHS i.e.
n = 0 J n x n = e n = 0 x n k = 0 ( 1 ) k k ! ( n + k k ) = e k = 0 ( 1 ) k k ! n = 0 ( n + k k ) x n = e 1 x k = 0 ( 1 ) k k ! 1 ( 1 x ) k = e 1 x e 1 1 x .
Here we used the fact that
n = 0 ( n + k k ) x n = 1 ( 1 x ) k + 1 .
Interchanging summation and integration is justified, since
| k = 0 n ( t ) k k ! 2 | k = 0 n t k k ! 2 k = 0 t k k ! 2 ( k = 0 t k / 2 k ! ) 2 = e 2 t
for any integer n. You can then make use of the DCT.

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