Question

Infinite series question:
Does anyone have an idea as to what the following series is equal to?
\[1+\frac{x^2}{2^2}+\frac{x^4}{2^2 \times 4^2}+\frac{x^6}{2^2 \times 4^2 \times 6^2}+ \dots\]

Answer 1

well, it would have to most likely equate to some function since its got an x in it ....

Answer 2

the coefficient function might look scary tho .. why do you need to try to find a function for it?

Answer 3

It's for an integral that I need to evaluate.

Answer 4

can we see the integral to determine that this power series applies to it?

Answer 5

\[\int_0^\infty \left(x-\frac{x^3}{2}+\frac{x^5}{2\times4}-\frac{x^7}{2\times4\times6}+\dots\right)\times\\\left(1+\frac{x^2}{2^2}+\frac{x^4}{2^2 \times 4^2}+\frac{x^6}{2^2 \times 4^2 \times 6^2}+ \dots\right)dx\]
I already evaluated the first term. It's on the second term where I am having trouble.

Answer 6

i assume that "x" means these are multiplied together ...

Answer 7

Yes, that's multiplication.

Answer 8

and your instructions (assuming this is for a class) tell you to remodify each series into ... what, a certain type of function perhaps? or was that your own concept to approach a solution?

Answer 9

It's just my idea for the solution since I was able to simplify the first term into \(xe^{-x^2/2}\). I figured I would be able to do the same to the second term.

Answer 10

hmmm
\[\sum_0 \frac{(-1)^n}{f(n)}x^{2n+1}~~\sum_0 \frac{1}{[f(n)]^2}x^{2k}\]

Answer 11

can you show me how you got the first one modified?

Answer 12

It went like this:
\[\large \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2^n n!}=x\sum_{n=0}^\infty (-1)^n \frac{\frac{x^{2n}}{2^n}}{n!}=x\sum_{n=0}^\infty \frac{\left(\frac{-x^2}{2}\right)^n}{n!}=xe^{-x^2/2}\]

Answer 13

nice .... so the next one should amount to \[\sum\frac{x^{(2n)}}{(2^n~n!)^2}\]

Answer 14

do you know what a bezel function is by chance?

Answer 15

if we can show that one is the derivative of the other, this would amount to undoing a chain rule

Answer 16

if not, then a by parts is an idea that might simplify the structure

Answer 17

No idea what a bezel function is. I'll try the other two ideas for now.

Answer 18

i cant be certain but i believe the chain rule is working on this stuff

Answer 19

I finally solved the integral!
I didn't simplify the series and just multiplied each term by \(xe^{-x^2}\). I then let \[\large I_n=\int_0^\infty \frac{1}{2^{2n}(n!)^2}x^{2n}xe^{-x^2/2}\ dx\]
so that the original integral becomes \(I=\sum_{n=0}^\infty I_n\).
I then got a closed form expression for \(I_n\) by letting \(u=x^2/2\) so that \(du=x\ dx\):
\[\large I_n=\int_0^\infty \frac{1}{2^{2n}(n!)^2}2^nu^ne^{-u}\ du\\
\large=\frac{1}{2^nn!}\int_0^\infty \frac{1}{n!}u^ne^{-u}\ du=\frac{1}{2^nn!}\]
because that last integral is the area of the gamma pdf \(f_Y(y)=\frac{\lambda^r}{\Gamma(r)}y^{r-1}e^{-\lambda y}\) with \(\lambda=1, r=n+1\).
The integral then becomes \[I=\sum_{n=0}^\infty I_n =\sum_{n=0}^\infty \frac{1}{2^nn!}=\sum_{n=0}^\infty \frac{\left(\frac{1}{2}\right)^n}{n!}=e^{1/2}\].