# Thread: Finding a Taylor series expansion.

1. ## Finding a Taylor series expansion.

Hi, can somebody please show me how to find the following Taylor series expansion? $(3+2x)^\frac{1}{3}$ (about x=0).

I get that f(0)= $3^\frac{1}{3}$
f'(0)= $\frac{1}{3} 3^\frac{-2}{3}$
f''(0)= $\frac{-2}{9} 3^\frac{-5}{3}$
etc...

and I know that f(x)=f(0)+f'(0)x+ f''(0)x^2/2! + ...

but I am having a lot of trouble writing it as a sum. Any ideas?

2. Originally Posted by C.E
Hi, can somebody please show me how to find the following Taylor series expansion? $(3+2x)^\frac{1}{3}$ (about x=0).

I get that f(0)= $3^\frac{1}{3}$
f'(0)= $\frac{1}{3} 3^\frac{-2}{3}$
f''(0)= $\frac{-2}{9} 3^\frac{-5}{3}$
etc...

and I know that f(x)=f(0)+f'(0)x+ f''(0)x^2/2! + ...

but I am having a lot of trouble writing it as a sum. Any ideas?
You sure you are doing your derivatives right in the MacLaurin Polynomial?
$f(x)=(3+2x)^{\frac{1}{3}}$
$f'(x)=\frac{1}{3}(3+2x)^{\frac{-2}{3}}2=\frac{2}{3}(3+2x)^{\frac{-2}{3}}$
$f''(x)=\frac{-2^2}{3^2}(3+2x)^{\frac{-5}{3}}2=\frac{-2^3}{3^2}(3+2x)^{\frac{-5}{3}}$
$f'''(x)=\frac{2^35}{3^3}(3+2x)^{\frac{-8}{3}}2=\frac{2^45}{3^3}(3+2x)^{\frac{-8}{3}}$

I think you forgot a few things like chain rule and I am not sure what else went wrong when you did it.

So i get like
$f(x) = 3^{\frac{1}{3}} + \frac{2}{3}3^{\frac{-2}{3}}x+\sum_{n=2}^{\infty}\frac{(-1)^{n+1}2^n2*5*8*11*...*3n-4}{3^nn!}3^{\frac{-(3n-1)}{3}}x^n$

I did it kinda fast, so you should look over it to be sure, but it looks like this works out. I think the pattern really gets going here starting at the second term of the expansion. There might be a slicker way of writing it, but I think this will do.

3. How could you find the radius of convergence of a series like this one? I was expecting it to come out a lot simpler so, the ratio test could be easily applied.

4. ratio test, and carefully, most of the nasty stuff will cancel out with the ratio test and you will just have kind of an ugly coefficient out front. I would guess you could also probably make use of the alternating series test, that (-1) out there probably makes a big difference. It has been quite some time since I have worked with these things, you are probably more fresh on it than me to be honest.

5. Originally Posted by C.E
Hi, can somebody please show me how to find the following Taylor series expansion? $(3+2x)^\frac{1}{3}$ (about x=0).

I get that f(0)= $3^\frac{1}{3}$
f'(0)= $\frac{1}{3} 3^\frac{-2}{3}$
f''(0)= $\frac{-2}{9} 3^\frac{-5}{3}$
etc...

and I know that f(x)=f(0)+f'(0)x+ f''(0)x^2/2! + ...

but I am having a lot of trouble writing it as a sum. Any ideas?
Try expanding $(3+2x)^\frac{1}{3}$by the Binomial Theorem. This should give you the series in a simpler form.