1. ## power series

The function $\displaystyle f(x) = \frac{5}{(1 - 3 x)^2}$ is represented as a power series $\displaystyle f(x) = \sum_{n=0}^\infty c_n x^n$.

Find the first few coefficients in the power series.
$\displaystyle C_0 =$
$\displaystyle C_1 =$
$\displaystyle C_2 =$
$\displaystyle C_3 =$
$\displaystyle C_4 =$

Find the radius of convergence R of the series.
$\displaystyle R =$

2. Originally Posted by viet
The function $\displaystyle f(x) = \frac{5}{(1 - 3 x)^2}$ is represented as a power series $\displaystyle f(x) = \sum_{n=0}^\infty c_n x^n$.

Find the first few coefficients in the power series.
$\displaystyle C_0 =$
$\displaystyle C_1 =$
$\displaystyle C_2 =$
$\displaystyle C_3 =$
$\displaystyle C_4 =$

Find the radius of convergence R of the series.
$\displaystyle R =$
To the first question: Recall that the power series expansion for $\displaystyle \frac {1}{(1 - x)^2}$ is the series $\displaystyle \sum_{n = 1}^{ \infty}n x^{n - 1} = \sum_{n = 0}^{ \infty} (n + 1) x^n$ (this is derived by taking the derivative of the well know power series for $\displaystyle \frac {1}{1 - x}$)

So we have that:

$\displaystyle \frac {5}{(1 - 3x)^2} = \sum_{n = 0}^{ \infty} 5 \cdot (n + 1)(3x)^n = \sum_{n = 0}^{ \infty} 5 \cdot 3^n (n + 1) x^n$

therefore, we have

$\displaystyle C_n = 5 \cdot 3^n (n + 1)$

use that to find the first few coefficients

For the second question:

use the root test (or if you prefer, the ratio test, either one works nicely in this case) to find the radius of convergence. do you need help with that part?

3. yea looks like i need help on the second question

4. Originally Posted by viet
yea looks like i need help on the second question
Ok, we wish to find the radius of convergence for the series $\displaystyle \sum_{n = 0}^{ \infty} 5 \cdot (n + 1)(3x)^n$

Since you didn't specify what test you wanted me to use, I will pick one myself. I choose the root test, it's less typing than the ration (why do i always do that?! type "ration" when i mean to type "ratio"!) test in this case.

We have $\displaystyle \lim_{n \to \infty} \left| 5 (n + 1) \cdot (3x)^n \right|^{ \frac {1}{n}} = \lim_{n \to \infty} \left| 5^{ \frac {1}{n}} (n + 1)^{ \frac {1}{n}} \cdot 3x \right| = | 1 \cdot 1 \cdot 3x | = 3 |x|$

we have convergence if $\displaystyle 3|x| < 1 \implies |x| < \frac {1}{3}$

Thus the radius of convergence is $\displaystyle \frac {1}{3}$