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Determine if the set of functions is linearly independent

  1. Oct 28, 2014 #1

    QuantumCurt

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    1. The problem statement, all variables and given/known data

    Determine if the given set of functions is linearly independent or linearly dependent.


    2. Relevant equations

    $$S=x~sin~x, ~ x~cos~x$$


    3. The attempt at a solution

    My first instinct was to use the Wronskian.

    $$W[y_1(x), y_2(x)]=\begin{vmatrix}
    x~sin~x & x~cos~x\\
    x~cos~x+sin~x & cos~x-x~sin~x
    \end{vmatrix}$$

    Now I take the determinant and I get -

    $$W=x~cos~x(cos~x-x~sin~x)-x~cos~x(x~cos~x+sin~x)$$
    $$W=x~cos~x~sin~x-x^2~sin^2~x-x^2~cos^2~x-x~cos~x~sin~x$$
    $$W=-x^2~sin^2~x-x^2~cos^2~x$$
    $$W=-x^2(sin^2~x+cos^2~x)$$
    $$W=-x^2$$

    Now, since ##-x^2\neq0##, can I conclude that the functions are linearly independent?

    I've done some examples in my textbook that involved determining linear independence on the interval of all real numbers, and there were some instances when the Wronskian didn't equal 0, but the answer key said that they were linearly dependent. I understood that the Wronksian was to determine linear independence or dependence. Am I correct, or do I misunderstand the Wronksian?
     
    Last edited by a moderator: Oct 28, 2014
  2. jcsd
  3. Oct 28, 2014 #2

    Mark44

    Staff: Mentor

    Yes, since the Wronskian isn't identically zero, the two functions are linearly independent.
     
  4. Oct 28, 2014 #3

    Dick

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    Homework Helper

    If the Wronskian is nonzero at any point on an interval, then the functions are linearly independent on that interval. The reverse isn't quite true. There are cases where the Wronskian is zero but the functions are still linearly independent. See http://en.wikipedia.org/wiki/Wronskian
     
  5. Oct 28, 2014 #4

    HallsofIvy

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    It seems to me simpler to just use the definition of "independent". The two given functions are independent if and only if the only way we can have Ax sin(x)+ Bx cos(x)= 0, for all x is if A= B= 0. If that is to be true for all x, in particular it is true for [itex]x= \pi/2[/itex] so that [itex]A(\pi/2) sin(\pi/2)+ B (\pi/2) cos(\pi/2)= A(\pi/2)= 0[/itex] so A= 0. But then we must have Bx cos(x)= 0 for all x. Choose any non-zero x such that cos(x) is not 0 to see that B= 0 also.
     
  6. Oct 28, 2014 #5

    QuantumCurt

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    Is there a formal way of showing the work using that method, or is it more along the lines of analytical guesswork?

    I guess what I'm really asking is if there's an algebraic way of proving that there are no non-trivial solutions.
     
  7. Oct 28, 2014 #6

    Mark44

    Staff: Mentor

    Using the definition of linear independence is a formal way. There was not any guesswork in what HallsOfIvy did.
     
    Last edited: Oct 28, 2014
  8. Oct 28, 2014 #7

    QuantumCurt

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    Guesswork wasn't the best way of putting it. Clearly when pi/2 is substituted in, the B term will go to zero and the A term will just go to A*pi/2. This is a fairly simple example, but with more complex sets of functions it may not always be so clear. I posted a set of 5 functions the other day that had to be tested for linear independence. The solutions weren't so immediately clear.
     
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