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Need to find a base to a certain space

  1. Aug 28, 2011 #1
    1. The problem statement, all variables and given/known data
    I'm probably incredibly rusty, but I can't seem to solve this.

    I've been given the space V = C[-[itex]\pi,\pi[/itex]] (continous functions on the closed segment) with the next linear transformation:

    T(f(x)) = g(x) = [itex]\int_{-\pi}^{\pi}[1+cos(x-t)]f(t)dt[/itex]

    I ought to prove that T(V), the range of V, has a finite dimension, and find an appropriate basis.

    2. Relevant equations

    Can't really think of any.

    3. The attempt at a solution

    I'm really simply blocked. How do I start getting the formation of functions in this space? Intuitively it seems like the addition of a certain number with a cos function, but I can't see any way to directly prove it, nor do I really have an obvious basis.

    I'm sorry if it's dumb :-)

    Thanks a lot,
    Tomer.
     
  2. jcsd
  3. Aug 28, 2011 #2

    tiny-tim

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    Hi Tomer! :smile:

    Have you tried starting with a basis for V ?
     
  4. Aug 28, 2011 #3
    It looks like V's dimension is infinite... I recall that functions on this segment can be described as infinite sums of trigonometric functions...
    :-)
     
  5. Aug 28, 2011 #4

    tiny-tim

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    ok, start there :smile:
     
  6. Aug 28, 2011 #5
    All I'm supposed to know up to now, if I'm not mistaken, is that the projection of a function f(x) on a subspace of V spanned by {1,sinx,sin2x,...sinnx,cosx,cos2x,...cosnx}, with the dimension of 2n+1, is the best approximation to f(x) in this subspace... (the "distance" between the two is the minimal)
    When I said "I recall", I meant that I remember it from previous studies (Fourier series). But this is not what I'm rehearsing right now - I'm rehearsing some linear algebra, and I therefore I don't think I'm meant to show f(x) as an infinite series or something like that.

    Any chance I'll get a thicker hint? :-)

    And thanks for replying!
     
  7. Aug 28, 2011 #6

    tiny-tim

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    That's correct :smile:

    {sin(nx)} and {cosnx} and {1} are a basis for V (they span V and they're linearly independent) …

    so find T of each of them …

    my guess is that you'll get only one or two independent results. :wink:
     
  8. Aug 28, 2011 #7
    Ok, I know what you mean, but the thing is, the actual fact that I can write: f(x) = (sum of trigonometric functions) is based on material which I'm pretty sure I'm not supposed to use here.
    It is mentioned nowhere in the book that any continuous function in a closed segment can be represented as this sum. They only talk about approximations to finite subspaces, and mention that in the finite subspace spanned by {1,sinx,sin2x....sinnx,cosx,cos2x...cosnx} (finite!) the projection of f(x) on it would be the best approximation to f(x)...

    I don't see how I can take it formally onwards. If you tell me I have to use the fact that any such function can be represented as an infinite sum, then I can solve it of course in no time.

    Thanks and sorry for bugging :-)
     
  9. Aug 29, 2011 #8

    tiny-tim

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    Hi Tomer! :smile:

    (sorry for not replying earlier :redface:)

    Try it either for cos(nt) and sin(nt), or just for tn, to see what results you get …

    you should find that the dimension (ie the number of independent solutions) is very low, and once you realise what the solutions are, you should be able to see a quick and simple way of proving it. :wink:

    What solutions do you get? :smile:
     
  10. Aug 29, 2011 #9
    The solutions I get are {1,sinx,cosx}. But I can't seem to be able to prove it without expanding f(x) as this infinite series which, as I've mentioned, I don't think I can. But maybe I'm wrong and I can? :)
     
  11. Aug 29, 2011 #10

    tiny-tim

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    That's what I got too. :smile:

    ok, ignore the {1} for the moment …

    what familiar equation has the general solution C1cosx + C2sinx ? :wink:
     
  12. Aug 29, 2011 #11
    f''(x) + f(x) = 0 ? :-)
     
  13. Aug 30, 2011 #12

    tiny-tim

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    (just got up :zzz: …)

    Yup! :smile:
    ok, so what is g''(x) ? :wink:

    (for a given f)
     
  14. Aug 30, 2011 #13
    I'd never manage to see it myself :-)
    so for a given f, g(x) must be a solution of the equation: g''(x) - g(x) = C. (C is the integral of f from -pi to pi, and is a constant).
    Therefore g has the form: g(x) = C1sinx + C2cosx + C3.
    Which of course holds as a proof... correct? :-)

    Thanks a lot!
     
  15. Aug 30, 2011 #14

    tiny-tim

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    Completely! :biggrin:

    The moral of all this is:

    in future similar questions, try differentiating wrt to x to get a simple equation

    and more generally always try a few simple cases to see if you can see a pattern! :wink:
     
  16. Aug 30, 2011 #15
    Thanks a lot :-) I'll remember that!
     
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