Finding the closed form for a double factorial sum

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[BarryRowlingsonBaz]

So the even term bit is pretty trivial and the odd term bit requires an integral, a change of variable, a trigonometric substitution, another couple of integrals and changes of variables, and ends up with an expression in the error function. Nice.

[dimastus]

I think that the less cumbersome method would be differentiation of these series and getting two ODEs: f'-x*f=0 and f'-x*f=1. The solution for the first on can be easily guessed exp(x^2/2), the second one may solved by representation of solution in a form g(x)*exp(x^2/2), then we get equation for g by substituting f=g*exp(x^2/2) into ODE: g'=exp(-x^2/2). From that we get erf related integral

[pwmiles]

You don't need to split out the odd and even series. If
G(z) = sum(n=0 to inf) z^n/n!!
G = 1/1 + z/1 + z^2/2 + z^3/3 + z^4/8 + z^5/15 + z^6/48 + z^7/105...
G' = 1/1 + z/1 + z^2/1 + z^3/2 + z^4/3 + z^5/8 + z^6/15 ...
z(G-1) = z^2/1 + z^3/2 + z^4/3 + z^5/8 + z^6/15 ...
= G' - z - 1

-z cancels
zG = G' - 1
G' - zG = 1

This is a first order linear ODE. The integrating factor is exp(-z^2/2)
G' exp(-z^2/2) - zG exp(-z^2/2) = exp(-z^2/2)
(G exp(-z^2/2))' = exp(-z^2/2)
G = [c + integ 0 to z exp(-t^2/2) dt] exp(z^2/2)

Boundary condition z=0, G=1 => c=1
Sum S = G(1) = [1 + integ(0 to 1) exp(-t^2/2) dt]exp(1/2)
Substitute u = t/sqrt(2)
dt = du sqrt(2)
t=1 => u=1/sqrt(2)
S = [1 + sqrt(2) integ(0 to 1/sqrt(2)) exp(-u^2)) du] exp(1/2)

[scarletevans]

Not sure if it will be much of helpful, but we can express it with Incomplete Gamma Function, with pleasantly looking Gamma(1/2, 1/2) within it :)

sqrt(2) times our final integral can be written as sqrt(pi/2) erf(1/sqrt(2))
Thus, the solution is equal to:

sqrt(e)[1+sqrt(pi/2) erf(1/sqrt(2))]

Now, using the connection with Incomplete Gamma Function
erf(x) = 1 - 1/sqrt(pi) Gamma(1/2, x^2)
We can rewrite the solution as:

sqrt(e)[1+sqrt(pi/2)-sqrt(1/2)Gamma(1/2, 1/2)]

which is approximately:

[dominicellis]

I noticed that the Taylor series for sqrt(x) uses the odd double factorial, no wonder I couldn’t easily find a closed form for it. I ended up simply using the product definition. It was a little clunky but it works. Is there a link between variations of factorials and nth roots? Is the rising factorial relatated to tower functions, is the triple factorial related to cube roots and a cubed root version of the error function?

[johnsalkeld]

If f(x) = sum(x^n/n!!) then if take a derivative of the sum it looks like 1 + sum(x^(n-1)/(n-2)!!) = 1 + xf(x) (we also know that f(0) = 1 from the sum formula). So rearranging (f’ - xf) = 1 implies (exp(-x^2/2)f(x))’ = exp(-x^2/2)

So exp(-x^2/2)f(x) = Int(exp(-x^2/2)) +c
f(x) = (the c allows you really to pick the integral either from -inf or from zero) integrating from zero to x we look at
F(x) = exp(x^2/2)*(c+int(0, x, -u^2/2, du))
Put x =0
1 = 1 * c
C = 1
The root 2 you had was from picking the z^2 form rather than the normal dist form i guess - nice problem anyway u = x then u/root 2 = x/root 2 suggesting a substitution z = u/root 2 - i think your formula emerges

[elgefe]

1/n!! = (n+2)/(n+2)!! implies the generating function satisfies G(x) = (G'(x) - 1)/x, with G(0) = 1.

[panagiotisapostolidis]

we need more factorials in the denominator that would be fun

[lexinwonderland]

ahhh i wish you showed the erf(x) correction directly!!

[kylecow]

why bother with splitting evens and odds, just use that n/n!! = 1/(n-2)!!, from this you derive the ODE f'(x)=1+xf(x) so dy/dx - xy = 1 which by multiplying through by exp(-x^2/2) you can solve and get

[mrminer]

Well, but then, what's the sum of the 1/(n!)! . . . ? Which is what I thought the "double factorial" was? This series should be NICE and convergent, and rather less than e. GO MICHAEL GO!!

[maciuikanikoda]

cool to think that Fijians can now learn for free all sort of things including cool maths like yours.... no more need to pay big money to universities to get at least some background in different fields. this is like a National Geographic for me.

Regards!

MN

[elijahhallbasketball]

Interestingly a close approximation to the final answer is sqrt(2) + sqrt(e). It is about .001 greater than the actual sum.

I wonder if sum(1/n!) is proportional to sum(1/n!!) ??

[enfasenumerica]

Tens o meu respeito doutor... Obrigado pelo seu excelente trabalho!

[lptotheskull]

okay so can someone explain to me why the double factorial is SMALLER THAN THE REGULAR FACTORIAL

[JonathanMandrake]

Oh, I thought this was the sum 1+1+1/2+1/24+1/(24!)+1/(120!)+..., so not the double factorial but the factorial of n!, so the product n!(n!-1)(n!-2)... 2 1

[jkid]

Fun! Neat to see how related it is to these more famous integrals

[jorgesaxon]

you: sum(x^2n/(2n)!!)
the guy she told you not to worry about: sum(x^2n+1/(2n+1)!!)

[georgekh]

That was a very interesting problem but the integral at the end is very sad 😪😪

[draaagoo]

i think there is a fault, the gamma function should have t to the power n+1, m not so sure thoo. i didnt finish the video tho maybe its correct somehow haha!!