Trouble with Power Series and Closed Form Notation

**Carbon** · May 29th 2011, 07:47 PM

Problem
Find the
$(t^2-1)x''+(t+1)x=0$
$x(0)=1$
$x'(0)=0$

Using expanded notation for series I can get an answer that matches up with maple:
$1+(1/2)x^2+(1/6)x^3+(1/8)x^4+(1/12)x^5$

What I can't seem to do is find this with a closed form notation for series.

My attempt
$x=\sum_{n=0}^\infty a_n t^n$
$x'=\sum_{n=0}^\infty na_n t^{n-1}$
$x''=\sum_{n=0}^\infty n(n-1) a_n t^{n-2}$

First I substitute...
$\sum_{n=0}^\infty n(n-1) a_n t^{n} - \sum_{n=0}^\infty n(n-1) a_n t^{n-2}+\sum_{n=0}^\infty a_n t^{n+1}+\sum_{n=0}^\infty a_n t^n$

Then I try to get everything to be to power of n...
$\sum_{n=0}^\infty n(n-1) a_n t^{n} - \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} t^{n}+\sum_{n=1}^\infty a_{n-1} t^{n}+\sum_{n=0}^\infty a_n t^n$

Then I try to get everything to the same index by taking out n=0 elements where necessary...
$\sum_{n=1}^\infty n(n-1) a_n t^{n} - (2a_2) - \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} t^{n}+\sum_{n=1}^\infty a_{n-1} t^{n}+(a_0)+ \sum_{n=1}^\infty a_n t^n$

Using initial conditions:
$a_0=1 a_1=0$

The recurrence relationship I get is:

$a_{n+2}=\frac{n(n-1)a_n-a_{n-1}+a_n}{(n+2)(n+1)}$

But this is wrong because if I plug in n=2, I get

$a_{4}=\frac{3}{32}$

Where did I go wrong?

**chisigma** · May 30th 2011, 02:03 AM

Originally Posted by Carbon

Problem
Find the
$(t^2-1)x''+(t+1)x=0 x(0)=1 x'(0)=0$

...

Only a minor detail: why don't write...

$(t-1) x'' + x=0\ ,\ x(0)=1 \ ,\ x'(0)=0$

Kind regards

$\chi$ $\sigma$

**Carbon** · May 30th 2011, 10:06 AM

Sorry about that I changed it.. Do you see where I made a mistake with the closed notation?

**chisigma** · May 31st 2011, 06:33 AM

Because semplicity is always the best alternative, the initial values problem is

$(t-1)x''+x=0\ ,\ x(0)=1\ ,\ x'(0)=0$ (1)

If we suppose that x(t) is analytic in x=0 it will be...

$x(t)= \sum_{n=0}^{\infty} \frac{x^{(n)}(0)}{n!}\ x^{n} = \sum_{n=0}^{\infty} a_{n}\ x^{n}$ (2)

The initial conditions give us...

$a_{0}=1$

$a_{1}=0$

The (1) permits us to derive...

$- x''(0)+x(0)=0 \implies x''(0)=1 \implies a_{2}= \frac{1}{2}$ (2)

Deriving (1) we obtain...

$(t-1)\ x^{(3)} + x'' + x'=0 \implies x^{(3)}(0)=1 \implies a_{3}= \frac{1}{6}$ (3)

Deriving (3) we obtain...

$(t-1)\ x^{(4)} +2\ x^{(3)} + x''=0 \implies x^{(4)}(0)= 3 \implies a_{4}= \frac{1}{8}$ (4)

Deriving (4) we obtain...

$(t-1)\ x^{(5)} + 3\ x^{(4)} + x^{(3)}=0 \implies x^{(5)}(0)= 10 \implies a_{5}= \frac{1}{12}$ (5)

In general is...

$x^{(n)}= (n-2)\ x^{(n-1)} + x^{(n-2)} \implies a_{n}= \frac{n-2}{n}\ a_{n-1} + \frac{a_{n-2}}{n\ (n-1)}$ (6)

Kind regards

$\chi$ $\sigma$

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