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Thread: Series of function problem

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

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

    1. $\displaystyle y=1 $ when $\displaystyle x=1$

    2. $\displaystyle y'=1 $ when $\displaystyle x=1$

    3. $\displaystyle y''+y=0$


    I need an answer without using the fact that the function: $\displaystyle y=sinx+cosx$ 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 $\displaystyle y = sinx + cosx$ only satisfies the ODE, not the initial conditions (1, 2).

    To generally solve this, consider

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

    Now put those two in the ODE ($\displaystyle y'' + y =0$) 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 $\displaystyle c_n$, 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; Jun 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 $\displaystyle y=\Sigma c_n x^n$, with the next conditions:

    1. $\displaystyle y=1 $ when $\displaystyle x=1$

    2. $\displaystyle y'=1 $ when $\displaystyle x=1$

    3. $\displaystyle y''+y=0$


    I need an answer without using the fact that the function: $\displaystyle y=sinx+cosx$ 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

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

    and plug this into the ODE to get

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

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

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

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

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

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

    Using this we get

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

    Notice the even $\displaystyle c_n$'s give

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

    Notice the odd $\displaystyle c_n$'s give

    $\displaystyle y=c_{1}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=c_{1}\sin(x)$
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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 $\displaystyle y(1)=1$ , $\displaystyle y^{'} (1) = 1$ and that $\displaystyle y^{''} (x) = - y(x) $ , so that is...

    $\displaystyle y^{''}(1) = -y(1) = -1$

    $\displaystyle y^{(3)} (1) = - y'(1) = -1$

    $\displaystyle y^{(4)} (1) = - y^{''} (1) = 1$

    $\displaystyle y^{(5)} (1) = - y^{(3)} (1) = 1$

    $\displaystyle \dots$

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

    $\displaystyle 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

    $\displaystyle \chi$ $\displaystyle \sigma$
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