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Math Help - Trouble with Power Series and Closed Form Notation

  1. #1
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    Trouble with Power Series and Closed Form Notation

    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?
    Last edited by Carbon; May 30th 2011 at 10:46 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Carbon View Post
    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
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  3. #3
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    Sorry about that I changed it.. Do you see where I made a mistake with the closed notation?
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  4. #4
    MHF Contributor chisigma's Avatar
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    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
    Last edited by chisigma; May 31st 2011 at 08:43 AM.
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