Find Taylor Series of \frac{1/3}{1-2x^3/3}

[JG89]

Homework Statement

Find the Taylor series about the point x = 0 for the function $$\frac{1}{3-2x^3}$$

Homework Equations

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?

 

[foxjwill]

Any mistakes?

Nope!

 

[benorin]

Homework Statement

Find the Taylor series about the point x = 0 for the function $$\frac{1}{3-2x^3}$$

Homework Equations

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?

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 $$|x|>C$$.

 

[JG89]

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

 

[foxjwill]

How would I find an expansion that converges for u > 1?

You won't.

 

[benorin]

Well it converges for $$|u| = |\frac{2x^3}{3}| < 1$$.

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

How would I find an expansion that converges for u > 1?

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.