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Linear Independence of trigonometric functions

  1. Nov 16, 2015 #1
    1. The problem statement, all variables and given/known data
    There's no reason to give you the problem from scratch. I just want to show that 5 trigonometric functions are linearly independent to prove what the problem wants. These 5 functions are sin2xcos2x. sin2x, cos2x, sin2x and cos2x.

    2. Relevant equations
    s1sin2xcos2x+s2sin2x+s3cos2x+s4sin2x+s5cos2x=0
    I need to prove that all s1,s2,s3,s4 and s5 must be equal to zero for the above equation to be true.

    3. The attempt at a solution
    I used the trigonometric formulas and came to this:
    (s1/2 + s2)sin2x + (s3-s4/2 + s5/2)cos2x + [(s4+s5)/2] = 0.
    We usually use the derive here but it doesn't seem to help.

    Edit: Oh, yeah. We haven't been taught the matrix yet.
     
    Last edited: Nov 16, 2015
  2. jcsd
  3. Nov 16, 2015 #2

    Krylov

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    There are probably easier ways, but you could show that the associated Wronskian doesn't vanish identically on ##\mathbb{R}##.

    EDIT: Ok, I see you edited your message. In that case the Wronskian is probably not the way to go.
     
  4. Nov 16, 2015 #3

    Krylov

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    Hold on... They are not linearly independent, because ##\cos{2x} = \cos^2{x} - \sin^2{x}##.
     
  5. Nov 16, 2015 #4
    Well... I made that assumption because I can't think of another way to solve the problem.
    The problem asks to find the dimension of the subspace of space C(-π,π) that is produced by these functions.
    I thought that if these vectors-functions are linearly independent then they are a base of the subspace, and that's how I prove its dimension.
     
  6. Nov 16, 2015 #5

    Krylov

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    The hint I can give you is to set up the condition
    but leave the ##\cos{2x}## term out, because we already know that this is a linear combination of some of the other functions. The dimension of the subspace of interest is therefore at most..?

    Then evaluate this condition in some well-chosen points in ##(-\pi,\pi)## such as: ##x = 0, x = \frac{\pi}{2}## and some others. This will give you sufficiently many independent linear equations from which you can solve for the ##s_i## to draw your conclusion.
     
  7. Nov 16, 2015 #6
    It's late and I've been working on the problem for a long time now. I'll review your hint tomorrow and let you know. Thanks a lot!
    Although it doesn't look like the way we work, I'm going to use it as long as I understand it.
     
  8. Nov 16, 2015 #7

    Krylov

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    Ok, then we both go to sleep. Let me know how it worked, good luck!
     
  9. Nov 17, 2015 #8
    If I prove that at least one of them is a linear combination of the others, does this mean that the same subspace can be produced without this one vector?
     
  10. Nov 17, 2015 #9

    Krylov

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    Yes. Maybe after you have used this fact to solve your exercise, you could try to prove that as well.

    For that, you would assume that you have vectors ##\{v_1,\ldots,v_m\}## in a vector space and, furthermore, there is some ##1 \le k \le m## such that ##v_k## is a linear combination of the other vectors. Then it would be up to you to show that any linear combination of ##v_1,\ldots,v_m## can be written as a linear combination of ##v_1,\ldots,v_m## with ##v_k## excluded.

    If you find the different indices confusing, first try it with, say, three vectors, the last of them being a linear combination of the first two.
     
  11. Nov 17, 2015 #10
    I don't really mind the proofs right now - they are all written in my notebook. The course itself doesn't even mind the proofs.
    So, if I keep doing this (excluding a vector that is a linear combination of another) until I end up with n vectors that are linearly independent, I have a base of the subspace. And the number n is its dimension. Am I right?
     
  12. Nov 17, 2015 #11

    Krylov

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    Yes. Still, if you have time, I would recommend that you try to prove these things for yourself, from scratch. If the proofs are already in your (note)book, even better, then you can check yourself. When you practise these kinds of small proofs, it also becomes easier to do the regular exercises.
     
  13. Nov 17, 2015 #12
    Thanks for your answers and advise. I'll find some time on the weekend to prove it.
     
  14. Nov 18, 2015 #13
    Hmm... is sin2x a linear combination of cos2x? I know that cos2x=sin(2x+π/2) but I'm not sure if that's a linear combination. Same goes for √(1-sin²2x).
     
  15. Nov 18, 2015 #14

    Krylov

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    Carefully review the definition of "linear combination", this is important.

    I can already tell you that ##\sin{2x}## is not a linear combination (i.e. a multiple, because we only have one vector) of ##\cos{2x}## on ##(-\pi,\pi)##. Why not? Loosely speaking, because the graph of ##\sin{2x}## is not a scalar multiple of the graph of ##\cos{2x}##.Indeed, ##\sin{\frac{\pi}{2}} = 1## and ##\cos{\frac{\pi}{2}} = 0## so there can never be a scalar ##c## such that
    $$
    \sin{2x} = c\cos{2x} \qquad \forall\,x \in (-\pi,\pi) \qquad \text{(not true!)}
    $$
    For the other one, note that ##\sqrt{1 - \sin^2{2x}} = |\cos{2x}|## and this is not a linear combination of ##\cos{2x}## on ##(-\pi,\pi)##, nor of ##\sin{2x}##, nor the other way around.
     
  16. Nov 18, 2015 #15

    Mark44

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    I agree.
    It's very easy (in fact, almost trivial) to determine whether a vector v is a linear combination of another vector u: v will be a scalar multiple of u. It's not so easy to determine whether one vector is a linear combination of two or more other vectors.
     
  17. Nov 19, 2015 #16
    But I can't find any way to prove it using the methods we've been taught. I used again the above condition and its derivative with all possible combinations and none worked.
     
  18. Nov 19, 2015 #17

    HallsofIvy

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    Then what definition of "linear combination" have you been taught? That, together with trig identities, should be enough.
     
  19. Nov 19, 2015 #18

    Krylov

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    I gave you this hint a few days ago and I still think it may be the easiest way. I edited the quote a bit to make it more explicit (and also to fix the weird way formulas are sometimes copied into quotes).

    It is however important that you do not just carry this out, but also understand precisely what this has to do with linear independence and why it solves your problem.
     
  20. Nov 20, 2015 #19
    Okay, first of all, I made a mistake when I first copied the problem. The first function is sinxcosx, not sin2xcos2x.

    Now, let me take it step by step, because I've gone further than the condition you suggest I should prove, and that's probably a mistake. I also left out sinxcosx(=sin2x/2), as well as cos²x(=1-sin²x), exactly the way I described earlier
    and ended up with sin2x and cos2x. Now my goal was to prove that these two functions are linearly independent, that's where my message was referring to.

    Is this wrong anywhere?
     
  21. Nov 20, 2015 #20
    ##\cos 2x = \cos ^2x - \sin ^2 x##. System is linearly dependent.
     
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