# Math Help - taylor series

1. ## taylor series

find the taylor series expansion of

f(x) = ___1____
.........(3-2 $x^3$)

at the point x = 0

2. Not sure if this is correct (I hope more knowledgeable users correct me if it isn’t!).
$\frac{1}{3 - 2x^3} = \frac{1}{3(1 - \frac{2x^3}{3})}$. Let $u = \frac{2x^3}{3}$. Then $= \frac{1}{3(1 - \frac{2x^3}{3})} = \frac{1}{3} \frac{1}{1 - u}$ which is the sum of a geometric series, so just use the expansion for a geometric series now.

3. can you explain further please

4. $
\frac{1}{3(1 - \frac{2x^3}{3})} = \frac{1}{3} \frac{1}{1 - u} = \frac{1}{3} (1 + u + u^2 + u^3 + ... )= \frac{1}{3} (1 + \frac{2x^3}{3} + \frac{4x^6}{9} + \frac{8x^9}{27} + ...)
$