Thread: Representing functions as Power Series

1. Representing functions as Power Series

Hi, can someone please help me with this problem? I started it but couldn't finish. Thanks a lot!!

f(x) = 3x/1-5x^3
= 3x * 1/1-5x^3
= 3x sigma (5x^3)^n...
Now I think I need to integrate both sides. but I am not sure if what I have done is correct so far?

2. Originally Posted by jsu03
Hi, can someone please help me with this problem? I started it but couldn't finish. Thanks a lot!!

f(x) = 3x/1-5x^3
= 3x * 1/1-5x^3
= 3x sigma (5x^3)^n...
Now I think I need to integrate both sides. but I am not sure if what I have done is correct so far?
The singular $\displaystyle 3x$ in the numerator is meaningless(why?). So let us consider $\displaystyle \frac{1}{1-5x^3}$ well if we rewrite this as $\displaystyle \frac{1}{1-z}\quad z=5x^3$ then it is clear from the geometric sum $\displaystyle \frac{1}{1-u}=\sum_{n=0}^{\infty}u^n$ that $\displaystyle \frac{1}{1-5x^3}=\sum_{n=0}^{\infty}\left(5x^3\right)^n=\sum_ {n=0}^{\infty}5^nx^{3n}$. Which is precisely what you got. So whats the problem?

3. Now to finish the problem I need to integrate both sides
Here is what I did:

integral 1/1- 5x^3 = integral 5^n x^3n
sigma 5x^3n + sigma 5^n x^5n+1/5^n+1
= sigma 5^n x^5n+1/ 5n+1 + C

so radius of convergence = 5
interval of convergence = (-5,5)

So my question is are these calculations correct cause I am still confuse. Thanks again.

4. Originally Posted by jsu03
Now to finish the problem I need to integrate both sides
Here is what I did:

integral 1/1- 5x^3 = integral 5^n x^3n
sigma 5x^3n + sigma 5^n x^5n+1/5^n+1
= sigma 5^n x^5n+1/ 5n+1 + C

so radius of convergence = 5
interval of convergence = (-5,5)

So my question is are these calculations correct cause I am still confuse. Thanks again.
I am completely lost. The power series representation of $\displaystyle f(x)=\frac{3x}{1-5x^3}$ is $\displaystyle 3x\cdot\sum_{n=0}^{\infty}5^nx^{3n}$...why are you integrating?!?!