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

Apr 2005
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?
Last edited:
  • Like
Reactions: 2 people