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Homework Help: Prove that v is an eigenvector of operator B

  1. Jun 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Let B be the linear operator

    [itex](1-x^{2}) \frac{d^2}{dx^2}-x\frac{d}{dx}[/itex]

    Show that

    [itex]T_{4}(x) = 8x^{4} - 8x^{2} + 1[/itex]

    is an eigenvector of B, and find the corresponding eigenvalue.


    Righto, I find these rather difficult so a step by step solution would be nice but I know that's a lot of typing and formatting so I'm grateful for any corrections.

    Step 1: [itex]Bv_{4} = \lambda _{4} v_{4}[/itex]

    Step 2: [itex]((1-x^{2}) \frac{d^2}{dx^2}-x\frac{d}{dx})(8x^{4} - 8x^{2} + 1) = \lambda _{4} (8x^{4} - 8x^{2} + 1)[/itex]

    Step 3: This is the first point where I'm unsure of what comes next. Should I be doing something like this

    [itex]\frac{d^2}{dx^2}((1-x^{2})(8x^{4} - 8x^{2} + 1)) - x\frac{d}{dx}(8x^{4} - 8x^{2} + 1) = 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4}

    \\ \frac{d^2}{dx^2}(-8 x^6 +16 x^4 -9 x^2 + 1) - \frac{d}{dx}(8x^{5} - 8x^{3} + x) = 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4}

    \\ -240x^4 + 192x^2 - 18 - 40x^4 + 18x^2 - 1 = 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4}

    \\ -280x^4 + 216x^2 - 19 = 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4}[/itex]

    Is this all correct so far? If so, awesome. Now I don't really know how to go about solving for [itex]\lambda[/itex]

    In general, is it enough to say that [itex]\lambda = \frac{-280x^4 + 216x^2 - 19}{8x^{4}- 8x^{2} + 1}[/itex]
    Last edited: Jun 13, 2013
  2. jcsd
  3. Jun 13, 2013 #2


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    Science Advisor
    Homework Helper

    You can do that, but you don't need the right hand side. It's probably easier to start by calculating
    $$B v = \frac{d^2}{dx^2}((1-x^{2})(8x^{4} - 8x^{2} + 1)) - x\frac{d}{dx}(8x^{4} - 8x^{2} + 1)$$
    and after simplifying you should get a multiple of ##8x^{4} - 8x^{2} + 1## from which you can read off the value of ##\lambda##.

    No! You cannot move the x after the derivative operator:
    $$x \frac{d}{dx} x^n \neq \frac{d}{dx} x^{n+1}$$
  4. Jun 13, 2013 #3
    Ok great, thanks for the response. By that logic, wouldn't it imply that you can't move the [itex](1-x^2)[/itex] to the right side either? So you just take the double derivative of the eigenvector then multiply by [itex](1-x^2)[/itex] minus the first derivative of the eigenvector multiplied by x then solve for lambda. If I follow that process, I get a lambda of -16 =] so I guess that's the way to do it
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