# Find a power series for f'(x) by starting with the convergent geometric series

• Dec 8th 2012, 02:03 AM
moonman
Find a power series for f'(x) by starting with the convergent geometric series
a) Find a power series for f'(x) by starting with the convergent geometric series

$\displaystyle f(x) = ln(x^3 + 27)$

$\displaystyle f'(x) = \frac{2x^2}{x^3 +27}$

b) Find a power series of f(x) and it's radius of convergence.
• Dec 8th 2012, 04:41 AM
HallsofIvy
Re: Find a power series for f'(x) by starting with the convergent geometric series
Do you understand what the question is asking you to do? You know that $\displaystyle f'= \frac{3x^2}{x^3+ 27}$ (note the "3", not "2", in the numerator. I assume that was a typo). You should also know that a "geometric series" is a series of the form $\displaystyle a+ ar+ ar^2+ ar^3+ ...$ and that the sum of such a series is $\displaystyle \frac{a}{1- r}$ as long as |r|< 1 so that the series converges.

Now look at $\displaystyle \frac{3x^2}{x^3+ 27}$. That is not quite in the form $\displaystyle \frac{a}{1- r}$ but can be put in that form. Divide both numerator and denominator by 27 and you get $\displaystyle \frac{\frac{3}{27}x^2}{\frac{x^3}{27}+ 1}= \frac{\left(\frac{x}{3}\right)^2}{1- (-\left(\frac{x}{3}\right)^3)}$. Now do you see what "a" and "r" must be?
• Dec 9th 2012, 12:50 PM
moonman
Re: Find a power series for f'(x) by starting with the convergent geometric series
$\displaystyle a = \frac{x}{3} , r = \frac{x}{3}$