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Thread: Challenge problem #6 [GLaw]

  1. #1
    Member GLaw's Avatar
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    Challenge problem #6 [GLaw]

    Let $a_1=b_1=1$ and define sequences $a_n$ and $b_n$ recursively by

    a_{n+1}\ =\ a_n-b_n

    b_{n+1}\ =\ a_n+b_n

    Thus

    a_2=a_1-b_1=0,\ b_2=a_1+b_1=2

    a_3=a_2-b_2=-2,\ b_3=a_2+b_2=2

    and so on.

    Find an explicit formula for $a_n$ and for $b_n$ in terms of $n$.
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  2. #2
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    Re: Challenge problem #6 [GLaw]

    Spoiler:
    $a_n = Re\{(1+i)^n\}$

    $b_n = Im \{(1+i)^n\}$
    Last edited by romsek; Apr 30th 2016 at 10:19 AM.
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  3. #3
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    Re: Challenge problem #6 [GLaw]

    Look at the sequence a_n = {1, 0, -2, -4, -4, 0, 8, 16, 16, 0, -32, -64,.....}
    we notice that it comes in blocks of four numbers each being (-4) times the corresponding number in the previous block
    this means

    a_{n+4}=-4a_n
    Proof:

    a_{n+2}=a_{n+1}-b_{n+1}=\left(a_n-b_n\right)-\left(a_n+b_n\right)=-2b_n

    a_{n+4}=-2b_{n+2}=-2\left(a_{n+1}+b_{n+1}\right)=-4a_n

    so that's one recurrence relation with one variable with four initial conditions

    solution: a_n=2^{n/2} \cos \left(\frac{n \pi }{4}\right)

    similarly b_n=2^{n/2} \sin \left(\frac{n \pi }{4}\right)
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  4. #4
    Member GLaw's Avatar
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    Re: Challenge problem #6 [GLaw]

    Great job. My solution:

    Spoiler:
    \text{Let}\ z_n=a_n+ib_n.

    \text{Then}\ z_{n+1}=\left(a_n-b_n\right)\,+\,i\left(a_n+b_n) = \left(a_n+ib_n\right)(1+i)=z_nz_1.

    \therefore\ z_n = \left(\frac{z_n}{z_{n-1}}\right)\left(\frac{z_{n-1}}{z_{n-2}}\right) \cdots \left(\frac{z_2}{z_1}\right)z_1 = z_1^n\ \text{for all}\ n=1,2,\ldots.

    \text{Now}\ z_1 = 1+i = \sqrt2e^{\frac{\pi}4i}.

    \therefore\ z_n = \left(\sqrt2\right)^ne^{\frac{n\pi}4i}

    \implies\ \boxed{a_n = \left(\sqrt2\right)^n\cos\frac{n\pi}4,\quad b_n = \left(\sqrt2\right)^n\sin\frac{n\pi}4}.
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