Solving an equation using transformation

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December 16th 2010, 02:03 PM
liquidFuzz

Solving an equation using transformation

Solve $zw'' + w' - w = 0$ where $f(0) = 1$ . I'm not exactly sure of how to do it, but I guess it's easier to comment if I show how I'd like to solve it. What I really need help with are the summations, and how to alter the starting integer while rewriting the sums. I tried to get every step on the way here, that's why the post is bloated. (Wait)

I start out with $\displaystyle w = \sum_{n=0}^\infty a_n z^n$

$\displaystyle w' = \sum_{n=1}^\infty n a_n z^{n-1}$ - I seem to remember adding one, n = 1, when differentiate... Please comment!

$\displaystyle zw'' = z \sum_{n=1}^\infty n(n-1) a_n z^{n-2} = \sum_{n=1}^\infty n(n-1) a_n z^{n-1}$ - This is one thing I'm unsure of. should it be n = 2 in the summation? Please comment!

Back to the equation.

$\displaystyle \sum_{n=1}^\infty n(n-1) a_n z^{n-1} + \sum_{n=1}^\infty n a_n z^{n-1} - \sum_{n=0}^\infty a_n z^n = 0$

$\displaystyle \sum_{n=1}^\infty n^2 a_n z^{n-1} - \sum_{n=0}^\infty a_n z^n = 0$

In the first series use n = k + 1

$\displaystyle \sum_{k = 0}^\infty (k+1)^2 a_{k+1} z^{k} - \sum_{n=0}^\infty a_n z^n = 0$ - Once again I'm unsure of the starting integer of the sum. Is it right to subtract one when I use n = k + 1?

$\displaystyle \sum_{n = 0}^\infty (n+1)^2 a_{n+1} z^{n} - \sum_{n=0}^\infty a_n z^n = 0$

$\displaystyle \sum_{n = 0}^\infty ((n+1)^2 a_{n+1} - a_n) z^{n} = 0$

$\displaystyle (n+1)^2 a_{n+1} = a_n$

$f(0) = 1 \implies a_0 = 1$

$a_n = \frac{1}{(n!)^2}$
December 16th 2010, 02:23 PM
snowtea

For the summation part:

$\sum_{n=0}^\infty n a_n z^{n-1} = \sum_{n=1}^\infty n a_n z^{n-1}$

You can shift the index, but you don't have to. The summand is 0 when n=0, so you can shift the index by 1 to get rid of adding 0.
If you are unsure, just check if the summand is 0 for the indices you are throwing away.
December 16th 2010, 02:44 PM
liquidFuzz

Ehh... I not sure I got that. If I got $\sum_{n=0}^\infty etc...$ I can differentiate and leave the n = 0 as it is..? How do I know that don't disregard anything exept zero..?
December 16th 2010, 02:49 PM
snowtea

When you different a summation, the safe thing is to not discard anything. If you do discard make sure the parts you discard are 0.

The reason you want to discard is when you need to shift to combine 2 summations. Example:
$ \displaystyle{ \sum_{n=0}^\infty nx^{n-1} + \sum_{n=0}^\infty x^n \underbrace{=}_{discard} \sum_{n=1}^\infty nx^{n-1} + \sum_{n=0}^\infty x^n }$

$ \displaystyle{ \underbrace{=}_{shift} \sum_{k=0}^\infty (k+1)x^k + \sum_{n=0}^\infty x^n } $

$ \displaystyle{ \underbrace{=}_{renaming} \sum_{n=0}^\infty (n+1)x^n + \sum_{n=0}^\infty x^n \underbrace{=}_{combining} \sum_{n=0}^\infty (n+2)x^n } $
December 16th 2010, 03:16 PM
liquidFuzz

Mhmm, I see. I guess I have to tinker a bit to totally get it. Thanks!
December 17th 2010, 12:55 AM
liquidFuzz

Ok, looking at it with fresh eyes it looks pretty straight forward. However, how do I know that I don't discard a non zero term?
December 17th 2010, 05:59 AM
snowtea

Just plug in the indices you want to discard.

Example:
$ \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1} $

because $nx^{n-1}$ is 0 when you plug in $n=0$
December 17th 2010, 08:44 AM
liquidFuzz

Ahh... and $n(n-1)x^{n-2}$ is zero if $n=0,1$ . Yey brilliant liquid...

I guess it's because I was feeling unsure of how to solve the equation in the first place. Thank you for explaining, and for having such patience.