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Linear Dependence and Span Question

  1. Jan 21, 2014 #1
    1. The problem statement, all variables and given/known data
    Is the following set linearly dependent or independent? And does this set span the given space?
    {eX, e-x}[itex]\in[/itex]C(R)

    2. Relevant equations



    3. The attempt at a solution

    So, if it's linearly independent, then:
    k1ex +k2e-x = 0 where k1,k2=0 and only 0. But if you let k1= -1/(e^x) and k2 = 1, then you also get 0, so this set is linearly dependent. I believe I am correct with this logic.

    Now how do I go about checking whether the set is in the span of the vector space?
     
  2. jcsd
  3. Jan 21, 2014 #2

    hilbert2

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    The coefficients in the linear combination must be numbers, not functions of ##x##. You should approach the problem by making the assumption that for some numbers ##a,b##, the function ##f(x) = ae^{x}+be^{-x}## is zero for any value of ##x##. Then derive a contradiction, the function must be nonzero at some point.

    The set ##C^{\infty}(\mathbb{R})## is infinite dimensional. You can't span it with only two functions.
     
  4. Jan 21, 2014 #3
    Ahh, ok, that makes sense. Thank you for the help. I'll see what I can do!
     
  5. Jan 21, 2014 #4
    So, deriving f(x)=ae^x+be^-x I get f'(x) = ae^x -be^-x, and f''(x) = ae^x+b^-x, so I'm back to the original function with the second derivative, and so forth. When proving linear dependence with functions, I understand that I need to show f(x) = 0 for any x, and where a,b≠0. Is that correct?
     
  6. Jan 21, 2014 #5

    hilbert2

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    If ##a## and ##b## are nonzero numbers, one of the terms ##ae^{x}##, ##be^{-x}## grows without bound in absolute value and the other one approaches zero when ##x \rightarrow \infty##. From this it follows that the function ##f(x)## must be nonzero for some ##x##. You just have to express this in a more exact way to make it a real mathematical proof.
     
  7. Jan 21, 2014 #6
    Gotcha. I follow. Thanks for your help, hilbert2.
     
  8. Jan 21, 2014 #7
    One more question: is C(R) represents the vector space of functions with infinitely continuous derivatives, then why wouldn't the above functions in that set span that vector space? Don't those functions have infinitely continuous derivatives?
     
  9. Jan 21, 2014 #8

    hilbert2

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    Saying that a set of functions {##f_{1},f_{2},f_{3},\dots##} spans the space ##A## means that any function in ##A## can be represented as a linear combination of functions ##f_{i}##. This is clearly not the case here. You need a set of infinitely many functions to span ##C^{\infty}(\mathbb{R})##.
     
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