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Math Help - Series of function problem

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Series of function problem

    Find y=\Sigma c_n x^n, with the next conditions:

    1. y=1  when x=1

    2. y'=1  when x=1

    3. y''+y=0


    I need an answer without using the fact that the function: y=sinx+cosx satisfies all conditions above.

    What I'm asking is, how can I solve it strictly from those conditions...(without guessing)

    Thank you!
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  2. #2
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    Note that y = sinx + cosx only satisfies the ODE, not the initial conditions (1, 2).

    To generally solve this, consider

     y = \sum_{n=0}^{\infty}c_nx^n so
     y' = \sum_{n=1}^{\infty}nc_nx^{n-1}
    and  y'' = \sum_{n=2}^{\infty}n(n-1)c_nx^{n-2}

    Now put those two in the ODE ( y'' + y =0) and put them together under the same summation (consider the change of variable k = n-2 for y'')
    You'll get a recursive formula for c_n, from which you will be able to derive a general solution. After that, plug in the initial conditions to get the specific solution.
    Last edited by Defunkt; June 29th 2010 at 08:53 AM. Reason: clarity
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Find y=\Sigma c_n x^n, with the next conditions:

    1. y=1  when x=1

    2. y'=1  when x=1

    3. y''+y=0


    I need an answer without using the fact that the function: y=sinx+cosx satisfies all conditions above.

    What I'm asking is, how can I solve it strictly from those conditions...(without guessing)

    Thank you!
    First take two derivatives of the series

    y''=\sum_{n=2}^{\infty}n(n-1)c_nx^{n-2}

    and plug this into the ODE to get

    y''+y=\sum_{n=2}^{\infty}n(n-1)c_nx^{n-2}\sum_{n=0}^{\infty}c_{n}x^{n}=0

    Now re-index the the first sereis to start at 0.

    \sum_{n=0}^{\infty}(n+2)(n+1)c_{n+2}x^{n}+\sum_{n=  0}^{\infty}c_{n}x^{n}=0

    \sum_{n=0}^{\infty}\left[(n+2)(n+1)c_{n+2}+c_{n}\right]x^{n}=0

    This gives us a relationship between the coeffients of the power series.

    c_{n+2}=-\frac{c_{n}}{(n+1)(n+2)}

    Using this we get

    c_2=-\frac{c_{0}}{1\cdot 2}
    c_3=-\frac{c_{1}}{2\cdot 3}
    c_4=\frac{c_{2}}{3\cdot 4}=\frac{c_{0}}{4!}
    c_5=\frac{c_{1}}{4\cdot 5}=\frac{c_{1}}{5!}
    c_6=-\frac{c_{2}}{5\cdot 6}=\frac{c_{0}}{6!}
    .
    .
    .

    Notice the even c_n's give

    y=c_{0}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}=c_{0}\cos(x)

    Notice the odd c_n's give

    y=c_{1}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=c_{1}\sin(x)
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  4. #4
    MHF Contributor chisigma's Avatar
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    A Taylor expansion around x=1 is easy to find considering that is y(1)=1 , y^{'} (1) = 1 and that  y^{''} (x) = - y(x) , so that is...

    y^{''}(1) = -y(1) = -1

     y^{(3)} (1) = - y'(1) = -1

    y^{(4)} (1) = - y^{''} (1) = 1

    y^{(5)} (1) = - y^{(3)} (1) = 1

    \dots

    ... that brings to the Taylor expansion...

    y(x) = 1 + (x-1) - \frac{(x-1)^{2}}{2!} - \frac{(x-1)^{3}}{3!} + \frac{(x-1)^{4}}{4!} + \frac{(x-1)^{5}}{5!} - \dots (1)

    Kind regards

    \chi \sigma
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