1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

    Attempt

    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

    CompuChip

    User Avatar
    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Prove that v is an eigenvector of operator B
Loading...