find the taylor series expansion of

f(x) = ___1____

.........(3-2$\displaystyle x^3$)

at the point x = 0

Printable View

- Aug 19th 2009, 07:00 AMmanalive04taylor series
find the taylor series expansion of

f(x) = _____1______

.........(3-2$\displaystyle x^3$)

at the point x = 0 - Aug 19th 2009, 07:19 AMJG89
Not sure if this is correct (I hope more knowledgeable users correct me if it isn’t!).

$\displaystyle \frac{1}{3 - 2x^3} = \frac{1}{3(1 - \frac{2x^3}{3})} $. Let $\displaystyle u = \frac{2x^3}{3} $. Then $\displaystyle = \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. - Aug 19th 2009, 07:42 AMmanalive04
can you explain further please

- Aug 19th 2009, 07:43 AMJG89
$\displaystyle

\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} + ...)

$