hi all,

I fear this will be long, but i'll try to keep it short.

the main question I am having trouble with is:

Evalutate f(0), f'(0), f''(0), f'''(0) for the function

$\displaystyle f(x) = (1+x)^{1/3}$

and thus find the Taylor Polynomials of degree 3 about the center c = 0 for this function.

I've been having difficulty with the Taylor Polynomials - here is what I have:

$\displaystyle f(x) = (1+x)^{1/3}$

$\displaystyle f'(x) = \frac{1}{3}(1+x)^{-2/3}$

$\displaystyle f''(x) = \frac{-2}{9}(1+x)^{-5/3}$

$\displaystyle f'''(x) = \frac{10}{27}(1+x)^{-8/3}$

$\displaystyle f''''(x) = \frac{-80}{81}(1+x)^{-11/3}$

Therefore in the

Taylor Polynomial we get:

$\displaystyle (1+0)^{1/3}+\frac{1}{3}(1+0)^{-2/3}(x-0)+\frac{-2}{9}(1+0)^{-5/3} \frac{(x-0)^2}{2!}+$ $\displaystyle \frac{10}{27}(1+0)^{-8/3}\frac{(x-0)^3}{3!}+\frac{-80}{81}(1+0)^{-11/3}\frac{(x-0)^4}{4!}$

which gives:

$\displaystyle 1 + \frac{1}{3}x + (\frac{-2}{9})(\frac{x^2}{2!}) + (\frac{10}{27})(\frac{x^3}{3!}) + (\frac{-80}{81})(\frac{x^4}{4!})$

from here I start trying to find the $\displaystyle \sum$ notation to express this. I can get as far as:

$\displaystyle 1 + \sum \frac{(-1)^{n+1}...}{3^n} * \frac{x^n}{n!}$

As you can probably tell from the '...' that is the part I am having difficulty with. Hopefully my conundrum makes sense.