# Find Taylor Series

1. Aug 19, 2009

### JG89

1. The problem statement, all variables and given/known data
Find the Taylor series about the point x = 0 for the function $$\frac{1}{3-2x^3}$$

2. Relevant equations

3. The attempt at a solution

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

I learned that if a function can be expanded in a power series with the point x = 0 in its interval of convergence, then that power series expansion is also the expansion for the Taylor series, so this seems correct to me. Any mistakes?

Last edited: Aug 19, 2009
2. Aug 19, 2009

### foxjwill

Nope!

3. Aug 19, 2009

### benorin

Sure, looks ok... where does it converge?

Your answer to my last question says that the above equality holds for $$|x|<C$$. Try to find a series expansion for the same function that converges for $$|x|>C$$.

4. Aug 19, 2009

### JG89

Well it converges for $$|u| = |\frac{2x^3}{3}| < 1$$. How would I find an expansion that converges for u > 1?

5. Aug 19, 2009

You won't.

6. Sep 4, 2009

### benorin

Note: this implies that the series expansion given in the OP converges for $|x|<\sqrt[3]{{\scriptstyle \frac{3}{2}}}$.

I think you mean, "How do I find an expansion for $|x|> \sqrt[3]{{\scriptstyle \frac{3}{2}}}$?"

Easy, do pretty much the same thing...

$$\frac{1}{3 - 2x^3} = -\frac{1}{2x^3}\cdot\frac{1}{1 - \frac{3}{2x^3}}$$​

Now let $u = \frac{3}{2x^3}$ ... you, finish.

Last edited: Sep 4, 2009