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Thread: Difference equation

  1. #1
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    Difference equation

    How do I calculate:
    $\displaystyle y_{n+1}-y_{n}=\sin(n) $
    $\displaystyle n>=0, y_0=0$
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by a4swe View Post
    How do I calculate:
    $\displaystyle y_{n+1}-y_{n}=\sin(n) $
    $\displaystyle n>=0, y_0=0$
    Looks to me like:
    $\displaystyle y_0 = 0$
    $\displaystyle y_1 = y_0 + sin(1) = sin(1)$
    $\displaystyle y_2 = y_1 + sin(2) = sin(1) + sin(2)$

    etc.

    So
    $\displaystyle y_{n} = \sum_{i = 0}^n sin(i)$

    -Dan
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  3. #3
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    Yes, I would agree to that.
    But what then?
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  4. #4
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    You could start by building your iterations.

    $\displaystyle y_{n+1}-y_{n}=sin(n)\Rightarrow{y_{1}+y_{0}=sin(0)}\Righta rrow{y_{1}=0}$

    $\displaystyle y_{2}-y_{1}=sin(1)\Rightarrow{y_{2}=sin(1)}$

    $\displaystyle y_{3}-y_{2}=sin(2)\Rightarrow{y_{3}-sin(1)=sin(2)}$

    $\displaystyle y_{4}-y_{3}=sin(3)\Rightarrow{y_{4}-(sin(2)+sin(1))=sin(3)}$


    And so on and so on.
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  5. #5
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    I want to have a formula for ANY n, that method won't help me.
    I know there is one, and getting to that is my problem.

    $\displaystyle \frac{\sin(\frac{n}{2})*\sin(\frac{n-1}{2})}{\sin(\frac{1}{2})}$
    Is what I want it to be.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by a4swe View Post
    I want to have a formula for ANY n, that method won't help me.
    I know there is one, and getting to that is my problem.

    $\displaystyle \frac{\sin(\frac{n}{2})*\sin(\frac{n-1}{2})}{\sin(\frac{1}{2})}$
    Is what I want it to be.
    (Hmmm...It seems I was off by a term in my earlier post. Galactus has it right.)

    Have you tried an induction proof?

    For n = 1: $\displaystyle \frac{\sin(\frac{1}{2})*\sin(\frac{1-1}{2})}{\sin(\frac{1}{2})} = 0$
    Since $\displaystyle y_1 = 0$ this is good. (The formula also works for $\displaystyle y_0 = 0$.)

    So it works for some n. Let's see if it works for n + 1.
    Now, $\displaystyle y_{n+1} = \sum_{i = 0}^{n+1} sin(i) =(?) \frac{\sin(\frac{n+1}{2})*\sin(\frac{n}{2})}{\sin( \frac{1}{2})}$

    By hypothesis $\displaystyle y_{n+1} = y_n + sin(n)$ so
    $\displaystyle \frac{\sin(\frac{n+1}{2})*\sin(\frac{n}{2})}{\sin( \frac{1}{2})} = \frac{\sin(\frac{n}{2})*\sin(\frac{n-1}{2})}{\sin(\frac{1}{2})} + sin(n)$
    should be true.

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

    On the LHS:
    $\displaystyle \sin(\frac{n+1}{2})*\sin(\frac{n}{2}) = sin(\frac{n}{2} + \frac{1}{2}) * sin(\frac{n}{2})$

    = $\displaystyle \left [ sin(\frac{n}{2}) cos(\frac{1}{2}) + sin(\frac{1}{2}) cos(\frac{n}{2}) \right ] sin(\frac{n}{2})$

    = $\displaystyle sin^2(\frac{n}{2}) cos(\frac{1}{2}) + sin(\frac{1}{2}) sin(\frac{n}{2}) cos(\frac{n}{2})$

    = $\displaystyle sin^2(\frac{n}{2}) cos(\frac{1}{2}) + \frac{1}{2} sin(\frac{1}{2}) sin(n)$

    Similarly we may show that the RHS becomes
    $\displaystyle \sin(\frac{n}{2})*\sin(\frac{n-1}{2}) + sin(\frac{1}{2})sin(n) = $ $\displaystyle sin^2(\frac{n}{2}) cos(\frac{1}{2}) - \frac{1}{2} sin(\frac{1}{2}) sin(n) + sin(\frac{1}{2})sin(n)$

    Which again gives:
    $\displaystyle sin^2(\frac{n}{2}) cos(\frac{1}{2}) + \frac{1}{2} sin(\frac{1}{2}) sin(n)$

    So the two sides are equal.

    So the theorem is true for n and n + 1 etc.

    -Dan
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  7. #7
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    Ahhh, yes!
    Thank you tops.
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