# Power series and Frobenius' method

• Aug 1st 2007, 05:30 AM
neven87
Power series and Frobenius' method
I dont work with power seris that much and i have a problem:

(x-1)y''-xy'+y=0

I just need to know what happens to the x-1 when i multiply it into the sumation; since multiplying the x just takes the n on the X to n+1...(basically im having a mathmatical brain fart)

Anywhos it...i could also use help with a power series in which i have to find two linealy indepentdent series about x=0 (using Frobenius' method) on the porblem 2xy''-y'+2y=0

I really appreciate any help, thanks. :D
• Aug 1st 2007, 06:59 AM
ThePerfectHacker
The point x=0 is an ordinary point, there is no need to use Frobenius Method here.

Look for a solution of the form,
$\displaystyle y=\sum_{n=0}^{\infty}a_nx^n$
• Aug 1st 2007, 07:09 AM
neven87
I was told to use the method to get two independant linear sets.
• Aug 1st 2007, 07:22 AM
neven87
never mind, figured it out, but i was wondering if anyone could tel me about the x+1...

im guessing it would basically become (X^(n+1)) +1 ?
• Aug 1st 2007, 12:19 PM
topsquark
Quote:

Originally Posted by neven87
I dont work with power seris that much and i have a problem:

(x-1)y''-xy'+y=0

I just need to know what happens to the x-1 when i multiply it into the sumation; since multiplying the x just takes the n on the X to n+1...(basically im having a mathmatical brain fart)

Say that you are going to guess a solution of the form:
$\displaystyle y = \sum_{n = 0}^{\infty}a_nx^n$

Then
$\displaystyle y^{\prime} = \sum_{n = 1}^{\infty}na_nx^{n-1}$
and
$\displaystyle y^{\prime \prime} = \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2}$

Put this into your differential equation:
$\displaystyle (x - 1) \cdot \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2} - x \cdot \sum_{n = 1}^{\infty}na_nx^{n-1} + \sum_{n = 0}^{\infty}a_nx^n = 0$

The first term becomes:
$\displaystyle (x - 1) \cdot \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2} = x \cdot \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2} - 1 \cdot \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2}$

$\displaystyle = \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 1} - \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2}$

The second term becomes:
$\displaystyle x \cdot \sum_{n = 1}^{\infty}na_nx^{n-1} = \sum_{n = 1}^{\infty}na_nx^n$

$\displaystyle \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 1} - \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2} - \sum_{n = 1}^{\infty}na_nx^n + \sum_{n = 0}^{\infty}a_nx^n = 0$