Pratice exam confusion Showing the sumation is equal to the other wee fun

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This discussion focuses on solving a practice exam related to differential equations and series solutions. The user seeks clarification on changing indices in summations and demonstrates the application of recurrence relations. Key points include the transformation of indices from n to j and the derivation of conditions for coefficients a2 and a3, which are both determined to be zero. The user successfully applies the recurrence relation a_(n+4) = -a_n/((n+3)(n+4)) to simplify the problem.

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mr_coffee
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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 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
http://img126.imageshack.us/img126/2601/lastscan29ir.jpg

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