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Thread: set solutions

  1. #1
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    set solutions

    the question is
    ''Find the solution to the equations.
    (i) $\displaystyle 25T_{n+2}=T_{n}\\T_{0}=125\\T_{1}=25$

    (ii) $\displaystyle 25T_{n+2}=-T_{n}\\T_{0}=125\\T_{1}=25$

    for i) i have got
    If $\displaystyle T_{n}=x^n$, so $\displaystyle 25x^{n+2}-x^n=0\\x^n(25x^2-1)=0\\x=0.2$
    general solution: $\displaystyle T_{n}=A(1/5)^n\\T_{0}=A=125\\A=125$
    therefore actual solution is $\displaystyle T_{n}=125*(1/5)^n$

    for ii) i have got
    If
    $\displaystyle T_{n}=x^n$,
    so
    $\displaystyle 25x^{n+2}+x^n=0\\x^n(25x^2+1)=0\\x=\pm 0.2i$

    general solution:
    $\displaystyle T_{n}=A\cos(0.2n)+B\sin(0.2n)$

    where A,B are constants.

    after using
    $\displaystyle T_{0}=125$,
    I've got A=125

    I'm wondering is my general solution incorrect and how can I find the constant B?
    Last edited by hazeleyes; May 4th 2011 at 01:31 AM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    The general solutions for (i) and (ii) are

    $\displaystyle (i)\quad T_n=A\left(\dfrac{1}{5}\right)^n+B\left(-\dfrac{1}{5}\right)^n$

    $\displaystyle (ii)\quad T_n=\left(\dfrac{1}{5}\right)^n\left(A\cos \dfrac{n\pi}{2}+B\sin \dfrac{n\pi}{2}\right)$

    In both cases substitute

    $\displaystyle T_0=125,\;T_1=25$

    and you'll find A and B by means of a 2 x 2 system.
    Last edited by FernandoRevilla; May 4th 2011 at 03:49 PM.
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  3. #3
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    Quote Originally Posted by FernandoRevilla View Post
    The general solutions for (i) and (ii) are

    $\displaystyle (i)\quad T_n=A\left(\dfrac{1}{5}\right)^n+B\left(-\dfrac{1}{5}\right)^n$

    $\displaystyle (ii)\quad T_n=\left(\dfrac{1}{5}\right)^n\left(A\cos \dfrac{m\pi}{2}+B\sin \dfrac{m\pi}{2}\right)$

    In both cases substitute

    $\displaystyle T_0=125,\;T_1=25$

    and you'll find A and B by means of a 2 x 2 system.
    Hi thanks for your replied, but what is m stand for?
    also,
    $\displaystyle (i)\quad T_n=A\left(\dfrac{1}{5}\right)^n+B\left(-\dfrac{1}{5}\right)^n$
    why does the second term has a negative sign? thanks
    Last edited by hazeleyes; May 4th 2011 at 01:13 PM.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by hazeleyes View Post
    Hi thanks for your replied, but what is m stand for?

    A typo. I've just corrected it ( n instead of m ).


    also,
    $\displaystyle (i)\quad T_n=A\left(\dfrac{1}{5}\right)^n+B\left(-\dfrac{1}{5}\right)^n$
    why does the second term has a negative sign? thanks

    The roots of the characteristic equation are 1/5 and -1/5.
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  5. #5
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    sorry for (i) I'm still confused about the roots, $\displaystyle 25x^{n+2}-x^n=0\\x^n(25x^2-1)=0\\then\\x=\sqrt{1/25}\\x=\frac{1}{5} $
    therefore why does second term has negative sign. thanks
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by hazeleyes View Post
    sorry for (i) I'm still confused about the roots, $\displaystyle 25x^{n+2}-x^n=0\\x^n(25x^2-1)=0\\then\\x=\sqrt{1/25}\\x=\frac{1}{5} $
    therefore why does second term has negative sign. thanks

    The characteristic equation of

    $\displaystyle T_{n+2}=T_n\quad [1]$

    is

    $\displaystyle 25x^2-1=0$

    whose solutions are

    $\displaystyle x=\sqrt {1/25}=\pm 1/5$

    According to a well known theorem, the general solution of [1] is

    $\displaystyle T_n=C_1\;(1/5)^n+C_2\;(-1/5)^n$
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