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

**moonman** · December 8th 2012, 02:03 AM

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

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

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

b) Find a power series of f(x) and it's radius of convergence.

**HallsofIvy** · December 8th 2012, 04:41 AM

Do you understand what the question is asking you to do? You know that $f'= \frac{3x^2}{x^3+ 27}$ (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 $a+ ar+ ar^2+ ar^3+ ...$ and that the sum of such a series is $\frac{a}{1- r}$ as long as |r|< 1 so that the series converges.

Now look at $\frac{3x^2}{x^3+ 27}$ . That is not quite in the form $\frac{a}{1- r}$ but can be put in that form. Divide both numerator and denominator by 27 and you get $\frac{\frac{3}{27}x^2}{\frac{x^3}{27}+ 1}= \frac{\left(\frac{x}{3}\right)^2}{1- (-\left(\frac{x}{3}\right)^3)}$ . Now do you see what "a" and "r" must be?

**moonman** · December 9th 2012, 12:50 PM

$a = \frac{x}{3} , r = \frac{x}{3}$

How do we go about finding the radius of convergence?

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

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