Power series Construction Help

[Euler2718]

Very basic issue here.

Using:

$$\frac{1}{1-x} = \sum_{i=0}^{\infty} x^{i} , |x|<0$$

Find the power series representation and interval of convergence for:

$$f(x) = \frac{1}{(1-3x)^{2}}$$

We have that:

$$\frac{d}{dx}\left[\frac{1}{1-x}\right] = \frac{1}{(1-x)^{2}} = \sum_{i=0}^{\infty} ix^{i-1} , |x| < 0$$

Of all things, the algebra of manipulating the function in question is stumping me. For some reason whatever I try doesn't seem to work.

 

[Mark44]

Very basic issue here.

Using:

$$\frac{1}{1-x} = \sum_{i=0}^{\infty} x^{i} , |x|<0$$

Find the power series representation and interval of convergence for:

$$f(x) = \frac{1}{(1-3x)^{2}}$$

We have that:

$$\frac{d}{dx}\left[\frac{1}{1-x}\right] = \frac{1}{(1-x)^{2}} = \sum_{i=0}^{\infty} ix^{i-1} , |x| < 0$$

Do you understand what they're doing here? Particularly in getting the summation at the end? This is basically the same as what you need to do with your series.

Also your inequality at the end is wrong. $|x| \ge 0$ for all real x.

Of all things, the algebra of manipulating the function in question is stumping me. For some reason whatever I try doesn't seem to work.

 

[member 587159]

If I am correct the series converges for all x: |x|<1. When you find the derivative of this series, it can be that the series converges in either -1 and or 1, so you might want to check that.

 

[Ssnow]

yes, the initial series converges for $|x|<1$. Observe that in the second case $x$ will be replaced by $3x$, you must consider that ...