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Interesting Summation?
I came across a rather interesting identity while messing around with one of my favorite functions (which will be left unidentified, for now). The identity, in text, is
sum{sum{ (-2)^n*x^(2n+2k+1)/[k! (2n+1)!!] }}
= sum{x^(2m+1)/[(2m+1) m!]}
the summations are taken wrt the appropriate variables n, k, and m, and all run from 0 to infinity.
image version:
http://www.ilstu.edu/~bmreini/sumid.bmp
I'd like to see someone give a direct proof of the identity. I want to know if my round-about discovery of this is actually better than trying to show it directly. Bonus points if anyone can deduce how I came up with the thing in the first place.
(Oh, and that is a double factorial in the LHS; that means the product of the odd integers up to 2n+1 in this case, NOT the factorial of the factorial.)
I'll try to put up some helpful hints up here every day or so...
Very nice A Cave!
In the spirit of keeping the question alive, I'm still interested in seeing a proof of the identity by matching coefficients...in essence, prove that the two expressions are equivalent as power series.
Also, how might I have come up with the double summation LHS?...you've already noted that LHS = exp(x^2)*stuff, so perhaps this won't be too hard.
Methinks people have forgotten this question...
Anyway, here's a hint for the coefficient matching: make a change of variables in the double summation. In particular, notice that if we sum over (n+k), the powers of x would match up nicely...in fact, more things than that start to match...
Hmm, more stars...that means people are still around.
The next hint (with 3 days to answer):
Using the last hint, we find that it is equivalent to show
sum{ (-4)^n*n! / [(m-n)! (2n+1)!] } = 1/[(2m+1) m!]
where the sum is wrt n from 0 to m.
http://www.ilstu.edu/~bmreini/sumid2.bmp
This is where I'm out...I have some movement forward from here, but nothing that seems relevant.
1 réponse
- il y a 1 décennieRéponse favorite
I think you were finding the error function series.
Let f(x) be LHS and g(x) be RHS.
Both summings start with x^1, so f(0) = g(0) = 0.
Deriving g(x) formally we get g'(x) = Sum (x^(2m) / m!) = exp(x^2) because of Taylor expansion for exp(x).
Inside the sum of f(x) we can find Sum (x^(2k) / k!) = exp(x^2), so f(x) = exp(x^2) * h(x) where h(x) = Sum( (-2)^n * x^(2n+1) / (2n+1)!!). f'(x) = exp(x^2) * (h'(x) + 2x*h(x)).
It is enough to show that f'(x) = g'(x) or h'(x) + 2x*h(x) = 1.
The latter is true because of the definition of h(x). Again, deriving formally: h'(x) = 1 + Sum ((-2)^n * x^(2n) / (2n-1)!!) (from n = 1 to inf) = 1 + Sum ((-2)^(n+1) * x^(2n+2) / (2n+1)!!) (from n = 0 to inf) = [indeed] = 1 - 2x * h(x), what we needed.
