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Eigenfunction Do I just plug in the f(x) equation into T(f)?

  1. Nov 5, 2012 #1
    The problem is attached.

    I am not quite sure how to do the 3 parts.

    for part a:

    Do I just plug in the f(x) equation into T(f)?
    If so that gives me k2ekx-2*kekx-3ekx
    I'm just not sure what I'm supposed to be doing.
     

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    Last edited by a moderator: Feb 6, 2013
  2. jcsd
  3. Nov 5, 2012 #2
    Re: Eigenfunction

    What does it mean to be an eigenfunction? If you apply a linear differential operator to one of its eigenfunctions, what should the result be?
     
  4. Nov 5, 2012 #3

    Mute

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    Re: Eigenfunction

    The question defines an operator, T, which acts on functions f(x), which have two derivatives. If you apply the operator T to some function g(x) and you find that ##Tg(x) = \lambda g(x)## for some number ##\lambda##, then g(x) would be called an eigenfunction of the operator T, because the operator acts on g(x) and gives back g(x) times a constant.

    The question tells you to apply the operator T to the function f(x) = exp(kx), and asks you to show that it is an eigenfunction. You applied T to the function f(x) and got the result that you listed in your post; now you just have to manipulate that expression into the form ##\mbox{(some constant that depends on}~k) \times f(x)##.
     
  5. Nov 5, 2012 #4
    Re: Eigenfunction

    All eigenvalue problems have the form T f(x)= t f(x), where t is a constant. If you have an operator and it acts on a function in this way, you have an eigenfunction and eigenvalue.

    So for a.) you have to show that the eigenvalue exists for all values of k in the eigenfunction.

    Part b.) kind of follows from part a.) at least I think, unless I'm missing something.

    Part c.) involves some guessing. Find a function that satisfies that equation.
     
  6. Nov 5, 2012 #5
    Re: Eigenfunction

    Thanks so for k2ekx-2*kekx-3ekx
    this would be ekx(k^2-2k-3)=ekx(k-3)(k+1)
    But hwo does that "show for each value of k, the function f(x) is an eigenfunction for T?"
     
  7. Nov 5, 2012 #6
    Re: Eigenfunction

    So your eigenvalue for that function is (k-3)(k+1). For any value of k this is a well-defined function for the eigenvalue. Thus e^kx is an eigenfunction for all k.
     
  8. Nov 5, 2012 #7
    Re: Eigenfunction

    I'm not quite sure if i understand why this means for all values of k, this is a well defined function for the eigenvalue.
     
  9. Nov 5, 2012 #8
    Re: Eigenfunction

    So I labeled my eigenvalue in the first post as t, a number. You found t(k)=(k-3)(k+1). This function is well-defined for all k.

    Lets say that you found t(k)=(k-3)/(k+1). At k=-1 you have t=-4/0 which is infinite or undefined.

    The eigenvalue you found has no problems in its range like this. So it is well-defined for all values of k.
     
  10. Nov 5, 2012 #9
    Re: Eigenfunction

    Oh I see I think I understand this, thanks. I'll take a look at the other parts now.
     
  11. Nov 7, 2012 #10
    Re: Eigenfunction

    So for part b)
    Since e^kx(k^2-2k-3)=e^kx(k-3)(k+1)

    Does that mean the eigen values are just k=3 and k=-1?
     
  12. Nov 7, 2012 #11

    vela

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    Re: Eigenfunction

    No. You need to learn what an eigenvalue and an eigenfunction are instead of just guessing. What does your book say?
     
  13. Nov 7, 2012 #12
    Re: Eigenfunction

    There's one section that says:
    Let V be a vector space
    and T:V −→ V be a linear operator. A number λ is an eigenvalue of T provided
    that there is a nonzero vector v in V such that T (v) = λv. Every nonzero vector that
    satisfies this equation is an eigenvector of T corresponding to the eigenvalue λ.

    Nothing is said about eigenfunctions, so I guess I have to apply this definition.


    T(v)=e^kx(k^2-2k-3)=e^kx(k-3)(k+1)=λe^kx
    Canceling out e^kx
    gives λ=(k-3)(K+1)

    Doesn't that mean the eigenvalues are λ=3, -1?
     
  14. Nov 7, 2012 #13

    vela

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    Re: Eigenfunction

    No, why would it? How are you going from ##\lambda = (k-3)(k+1)## to ##\lambda=3, -1##?
     
    Last edited: Nov 7, 2012
  15. Nov 7, 2012 #14
    Re: Eigenfunction


    hmmm, I wish setting lambda=0, which is incorrect.
    Hmmm λ=(k-3)(k+1)=k^2-2k-3

    Does this mean lamba is all real numbers?

    So when it says Find the corresponding eigenvalues for each
    eigenfunction f (x) = e^kx.

    Basically lamda is all real values?
     
  16. Nov 7, 2012 #15

    vela

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    Re: Eigenfunction

    Remember k represents a specific value. For each possible value of k, ekx is an eigenfunction of T. For example, if k=2, then you have f(x)=e2x, and the corresponding eigenvalue is ##\lambda = -3##. Your answer to (b) is what you wrote: λ=(k-3)(k+1).

    If you want to know what the possible eigenvalues of T are, think about what the allowed values of k are and what values of ##\lambda## these correspond to. Hint: If you plotted ##\lambda## vs. k, what kind of curve would you get?
     
    Last edited: Nov 7, 2012
  17. Nov 7, 2012 #16
    Re: Eigenfunction

    Did you mean lambda=-3?



    I would be a parabolic curve. k can be all real #s as a previous user stated. Lambda would be all real #s. But I am unsure how I would describe lambda for each eigenfunction? Lambda is a parabolic function of all real k values.
     
  18. Nov 7, 2012 #17

    vela

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    Re: Eigenfunction

    Oops, yes.

    I edited my previous post, and it probably answers your question now.
     
  19. Nov 8, 2012 #18
    Re: Eigenfunction

    Thanks.

    I guess this λ=(k-3)(k+1)=k^2-2k-3
    Pretty much answers part a and b.

    Since it's saying that for all k, the function f(x)=e^(kx) is an eigenfunction because k can be any real #.

    For part b, lambda=k^2-2k-3

    for part c, is that just a differential equation? (First time seeing a diff eq in linear algebra)

    If so then the characteristic equation for it is r^2-2r-3=0
    (r-3)(r+1)=0
    r=-1, 3

    so the 2 functions are
    y(t)=Ae^(-t)
    G(t)=Be^(3t)
    Where A and B are constants.
     
  20. Nov 8, 2012 #19

    vela

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    Re: Eigenfunction

    You can find the answer to (c) that way, but it's probably not what was intended. How might you answer (c) based on the other parts of the problem?
     
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