Finding a Taylor series expansion.

**C.E** · May 20th 2009, 10:30 AM

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?

**Gamma** · May 20th 2009, 01:19 PM

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.

**C.E** · May 20th 2009, 01:25 PM

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.

**Gamma** · May 20th 2009, 01:26 PM

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.

**awkward** · May 20th 2009, 02:38 PM

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.

Thread: Finding a Taylor series expansion.

LinkBack

Thread Tools

Display

Finding a Taylor series expansion.

Similar Math Help Forum Discussions

Taylor series expansion

Taylor Series expansion

taylor series expansion for log(1-x/1+x)

Taylor Series Expansion of an eqn

Taylor Series Expansion

Search Tags