# Homework Help: Pratice exam confusion! Showing the sumation is equal to the other! wee fun!

1. Apr 10, 2006

### mr_coffee

Hello everyone, i'm studying for my exam. He gave us anice little pratice exam. Am i thinking too much about this one? Part (a) seems to easy.
http://img235.imageshack.us/img235/1554/lastscan9dx.jpg [Broken]

I think i'm trying to make x^(n-c) be x^(n)
well to do this, i would add c to (n-c), but if u do this, you will have to subtract c from the indicies (the n under the sumation). So that would change, n = b, to n=b-c.

Is that all i had to do here?

Also in part b, i'm lost now! The directions say: show that a2 = 0, a3 = 0, and that the recurrance relation for this differential equation is: a_(n+4) = -a_n/((n+3)(n+4))
I've been using this as a referance to help me:http://tutorial.math.lamar.edu/AllBrowsers/3401/SeriesSolutions.asp [Broken]
http://img126.imageshack.us/img126/2601/lastscan29ir.jpg [Broken]

THanks!

Last edited by a moderator: May 2, 2017
2. Apr 11, 2006

### HallsofIvy

I think it is simpler to follow if you change indices- don't use "n" for both.
Let j= n-c. Then n= j+ c so an= aj+c, f(n)= f(j+ c) and xn-c= xj. Also, when n= b, j= b-c. Of course, when n= $\infty$, j= $\infty$. That is:
$$\Sigma_{n=b}^\infty a_n f(n) x^{n-c}= \Sigma{j= b-c}^{\infty}a_{j-c}f(j-c}x^j$$.
Now, just change the index in the second sum from j to n- it doesn't change the value.

If $y= \Sum_{n=0}^\infty a_n x^n$ then
$$y"= \Sum_{n=2}^\infty n(n-1)a_n x^{n-2}$$
(the sum starts at n= 2 because if n= 0 or n= 1 the term n(n-1) is 0). Putting that into the equation y"- x2y= 0, we get:
$$\Sigma_{n=2}^\infty n(n-1)a_n x^{n-2}- \Sigma_{n=0}^\infty a_n x^{n+2}$$

In the first sum, let j= n-2 so n= j+ 2. The sum becomes
$$\Sigma_{j=0}^\infty (j+2)(j+1)a_{j+2}x^j$$
In the second sum, let j= n+2 so n= j- 2. The sum becomes
$$\Sigma_{j= 2}^\infty a_{j-2}x^j$$
The point of that is, of course, to have the same exponent on x in both sums so we can "combine like terms". Since the second sum doesn't start until j= 2, we can write j=0 and j= 1 terms in the first sum separately:
$$2a_2+ 6a_3x+ \Sigma_{j=2}^{\infty}((j+2)(j+1)a_{j+2}- a{j-2})x^j= 0$$
From that, it should be obvious that 2a2= 0 and 6a3= 0.

3. Apr 13, 2006

### mr_coffee

Thanks again Ivey!! it makes perfect sense, sorry about the delayed responce!