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Topic: Series 1 + 1/(1 . 3) + 1/(1 . 3. 5) + 1/(1. 3 . 5. 7) .....
Replies: 3   Last Post: Aug 18, 2012 11:13 AM

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 hagman Posts: 1,923 Registered: 1/29/05
Re: Series 1 + 1/(1 . 3) + 1/(1 . 3. 5) + 1/(1. 3 . 5. 7) .....
Posted: Aug 18, 2012 11:13 AM

Am Samstag, 18. August 2012 15:46:56 UTC+2 schrieb steine...@gmail.com:
> That's what I did. I got f'(x) = 1 + x f(x). Then I got stuck. This equation is equivalent to f'(x) - xf(x) = 1. The integrator factor is e^(-x^2/2). Hence, d/dx(e^(-x^2/2) f(x)) = e^(-x^2/2). I can't integrate this.
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> Artur

With
erf(x) = 2/sqrt(pi) * int_{-oo}^x exp(-t^2/2) dt
we have
erf'(x) = 2/sqrt(pi) * exp(-x^2/2)
Hence with
f(x) := sqrt(pi)/2 * exp(x^2/2)*erf(x),
we have
f'(x) = sqrt(pi)/2 * x*exp(x^2/2) * erf(x) + 1,
i.e.
f'(x) = x*f(x) + 1.

We are looking for f(1), i.e.
sqrt(pi)/2 * exp(1/2)*erf(1) = sqrt(e pi)/2 * erf(1)
hence your original guess is off by a factor of
erf(1) ~~ 0.8427

hagman

f(x) =exp(x^2/2) * g
f' = exp(x^2/2) * g' -x f
exp(x^2/2) * g' = 1

f(x)+ln(x) -> f'(x) +1/x
(f g)' = g
y' = 1 + x y
y'/y - 1/y = x
ln(y)'

Date Subject Author
8/18/12 steinerartur@gmail.com
8/18/12 G. A. Edgar
8/18/12 steinerartur@gmail.com
8/18/12 hagman