# taylor series

• August 19th 2009, 07:00 AM
manalive04
taylor series
find the taylor series expansion of

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

at the point x = 0
• August 19th 2009, 07:19 AM
JG89
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.
• August 19th 2009, 07:42 AM
manalive04
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