Series representation of a function

[iFargle]

Homework Statement

Find a power series representation for the function and determine the radius of convergence.

Homework Equations

$$f(x)=x^{2}tan^{-1}(x^{3})$$

The Attempt at a Solution

I don't have any idea on how to even start this. First I differentiated $$\frac{d}{dx} arctan(x)=\frac{1}{x^2+1}$$, thinking I could try and manipulate it to fit a geometric series, but that quickly came to a dead end. I've tried thinking of any relate-able series representations, but all I know so far is the geometric series and natural logarithms.

 

[ckutlu]

Google for Taylor Series Expansion formula, and try to apply it to your problem. If you have any questions about it, you can ask :)

 

[iFargle]

I think I have it. I am overjoyed! Tell me if this is correct:

$$f(x)=x^2tan^{-1}(x^3)$$
First,
$$\frac{d}{dx}tan^{-1}(x) = \frac{1}{x^2+1}$$
$$\int(tan^{-1}(x))dx=\int\frac{1}{x^2+1}dx$$
Rearranging it so it resembles the geometric series:
$$\int\frac{1}{1-(-x^2)}dx=\int\sum((-x^2)^n)dx$$
Integrated and re-written:
$$\int\sum((-x^2)^n)dx=\sum\frac{(-1)^nx^{2n+1}}{n+1}$$
So now we put x^3 where x is.
$$\sum\frac{(-1)^n(x^3)^{2n+1}}{n+1}=\sum\frac{(-1)^n(x)^{6n+3}}{n+1}$$
So we finally have:
$$f(x)=x^2\sum\frac{(-1)^n(x)^{6n+3}}{n+1}=\sum\frac{(-1)^n(x)^{6n+5}}{n+1}$$
For the radius of convergence, I just use the ratio test. I get that part pretty easily.

 

[vela]

Almost right. Check your integration.

 

[ckutlu]

$$\int{(tan^{-1}(x))dx} = \int{\frac{1}{x^2+1}dx}$$

I didn't understand how did you make this equality. Am i missing a point?

 

[iFargle]

Sorry, it was supposed to read:
$$tan^{-1}(x)=\int\frac{1}{x^2+1}dx$$
Got carried away with formatting, haha

 

[Char. Limit]

$$\int\sum((-x^2)^n)dx=\sum\frac{(-1)^nx^{2n+1}}{n+1}$$

You have a problem here. The denominator is n+1, but the integral of x²ⁿ isn't x²ⁿ⁺¹/(n+1).

 

[iFargle]

$$\int\frac{1}{1-(-x^2)}dx=\int\sum(-x^{2})^{n}dx$$
$$=\sum(-1)^{n}\frac{x^{2n+1}}{2n+1}dx$$
Correct? I still feel like my answer is wrong, though.

 

[SammyS]

$$\int\frac{1}{1-(-x^2)}dx=\int\sum(-x^{2})^{n}dx$$
$$=\sum(-1)^{n}\frac{x^{2n+1}}{2n+1}dx$$
Correct? I still feel like my answer is wrong, though.

Drop the dx. Where's the integration constant?
What you have so far is:$$\tan^{-1}(x)=C+\sum^{\infty}_{n=0}(-1)^{n}\frac{x^{2n+1}}{2n+1}\,.$$

Evaluate the integration constant. Modify the above to get a series representation for x² tan^-1(x³). Find the radius of convergence.

 

[iFargle]

Drop the dx. Where's the integration constant?
What you have so far is:$$\tan^{-1}(x)=C+\sum^{\infty}_{n=0}(-1)^{n}\frac{x^{2n+1}}{2n+1}\,.$$

Evaluate the integration constant. Modify the above to get a series representation for x² tan^-1(x³). Find the radius of convergence.

So then that would be...
$$C+x^{2}\sum^{\infty}_{n=0}(-1)^{n}\frac{x^{6n+3}}{2n+1}=C+\sum^{\infty}_{n=0}(-1)^{n}\frac{x^{6n+5}}{2n+1}$$
or am I missing something huge? I don't get what you mean by evaluate the integration constant.

 

[vela]

The equation

$$\tan^{-1}(x)=C+\sum^{\infty}_{n=0}(-1)^{n}\frac{x^{2n+1}}{2n+1}$$

holds for only one value of C. You need to figure out what that is.