Finding closed form for the next series:

[MathematicalPhysicist]

i need to find the closed form expressiosn for the next sums:
$$g(x)=\sum_{n=0}^{\infty}\frac {x^{2n}}{(2n)!}$$
and $$f(x)=\sum_{n=0}^{\infty} \frac {x^{4n+1}}{4n+1}$$
well for the second one i thought differentiating, i.e $$f'(x)= \sum_{n=0}^{\infty} x^{4n}=1/(1-x^4)=\frac {1}{(1+x^2)*(1-x^2)}$$, now i could integrate it by parts but in my course i haven't yet got to integrals, so i cannot use integration by parts, do you have any other way to compute this sum.
for the first one, it looks like the power series of e^x, i tried to recursively get to the expression but so far to no success.
any pointers would be helpful.

 

[HallsofIvy]

The first one is, in fact,
[tex]cosh(x)= \frac{e^x+ e^{-x}}{2}[/itex]<br /> <br /> For the second one, your idea of differentiating to get a geometric series and then integrating works nicely (after you learn to integrate, of course!) using partial fractions, not "by parts". But it gives a function involving arctan and logarithms so I doubt you will find any simple way to do that.[/tex]

 

[Gib Z]

Just in case you wanted it, the second ones

$$\frac{1}{4} (2\arctan x -\ln (x^2-1))$$ I think.

 

[Hurkyl]

The summand for g(x) looks similar to the summand for e^x -- this suggests you might be able to express the power series for g(x) in terms of the power series for e^x.

The summand for f(x) looks similar to the summands for log(1+x) and for arctan x -- this suggests you might be able to express the power series for g(x) in terms of the power series for log(1+x) and arctan x.

 

[MathematicalPhysicist]

after one hour i got it by myself.
but i appreciate your help.
btw, iv'e got something like this:
i need to find a power series for (1+x^2)*arctg(x).
i know the power series of arctg(x), but when I am multiplying this sum with (1+x^2), i get x+sum, i.e where the sum starts from n=1, but i can't put the x into the sum. perhaps it should be this way?

 

[Gib Z]

What is arctg(x)?

 

[MathematicalPhysicist]

the same as arctan(x).
it's one of those shorcuts,like for sinh, there's sh, etc.

 

[Gib Z]

So are you saying arctg(x) is the same as arctan(x)? Or that its a similar pattern, its the inverse of g(x)? Because that's argcosh(x).

 

[MathematicalPhysicist]

yes arctg(x) is the same as arctan(x), now back on topic.

 

[Gib Z]

Well then yes, it can be the way you stated, it doesn't matter and isn't wrong.