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Prove this statement

  1. Sep 17, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    For [itex]y=sin(m \arcsin x) [/itex], prove that
    $$(1-x^2)y_{n+2} - (2n+1)xy_{n+1}+(m^2-n^2)y_n=0 $$

    2. Relevant equations

    3. The attempt at a solution
    My first approach would be to find a general expression for y_n. Starting with
    $$y_1=m\dfrac{\cos(m \arcsin x)}{\sqrt{1-x^2}} \\
    y_2 = m\dfrac{\cos(m \arcsin x)}{\sqrt{1-x^2}(1-x^2)} - m^2 \dfrac{\sin(m \arcsin x)}{1-x^2} $$

    The expression seems to get uglier as n increases and it's too difficult to calculate y_3. Also it's not easy to generalize an expression for y_n just by looking at the results obtained.
     
  2. jcsd
  3. Sep 17, 2014 #2

    PeroK

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    Have you thought of using induction?
     
  4. Sep 17, 2014 #3

    Ray Vickson

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    Please do not use ##y_n## for the ##n##th derivative of ##y##; the standard notation would be ##y^{(n)}##, and your use of other notation (without explanation) is too confusing: I could not figure out what your question was asking.
     
    Last edited: Sep 17, 2014
  5. Sep 17, 2014 #4

    utkarshakash

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    Let P(n) be the given expression.
    $$P(0)=(1-x^2)y_2-3xy_1+(m^2-1)y=0$$
    Let P(n) be true. Therefore,
    [itex](1-x^2)y_{n+2} - (2n+1)xy_{n+1}+(m^2-n^2)y_n=0\\
    P(n+1)=(1-x^2)y_{n+3} - (2n+3)xy_{n+2}+(m^2-(n+1)^2)y_{n+1}[/itex]

    How do I proceed further?
     
  6. Sep 18, 2014 #5

    PeroK

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    What is the most obvious thing you can do with P(n)?
     
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