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Math Help - Exponential Function

  1. #1
    Super Member Bernhard's Avatar
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    Exponential Function

    On Page 78 of Lang - Undergraduate Analysis (2nd Ed) we find the following:
    "We assume that there is a function f defined for all numbers such that f ' = f
    and f(0) = 1.
    We note that f(x)is not equal to 0 for all x. Indeed differentiating the function
    f(x)f(-x) we find 0"

    Can anyone help me show rigorously that the derivative of f(x)f(-x) for the exponential function is zero - relying only on f ' = f and f(0) = 1.

    Help will be appreciated.
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    Quote Originally Posted by Bernhard View Post
    On Page 78 of Lang - Undergraduate Analysis (2nd Ed) we find the following:
    "We assume that there is a function f defined for all numbers such that f ' = f
    and f(0) = 1.
    We note that f(x)is not equal to 0 for all x. Indeed differentiating the function
    f(x)f(-x) we find 0"

    Can anyone help me show rigorously that the derivative of f(x)f(-x) for the exponential function is zero - relying only on f ' = f and f(0) = 1.

    Help will be appreciated.
    Use the product rule: the derivative of f(x)f(-x) is f'(x)f(-x)+ f(x)(f(-x))' which, by the chain rule is f'(x)f(-x)- f(x)f'(-x). Since we have that f'(x)= f(x), we also have that f'(-x)= f(-x) and so (f(x)f(-x))'= f(x)f(-x)- f(x)f(-x)= 0.

    (You don't need f(0)= 1. This is true for any exponential Ae^x. (Ae^x)(Ae^{-x})= A^2 which is a constant and has derivative 0.)
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  3. #3
    Super Member Bernhard's Avatar
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    Thanks HallsofIvy. appreciate your help! Can you help further?

    Thanks.

    I think my problem was how to differentiate f(-x) but I should have used chain rule for composite functions, so

    d/dx [f(-x)] = d/dx f(-x) . d/dx (-x)

    = d/dx f(x) . -1

    = - f ' (x)

    Then when you put this into the product rule the result follows

    BUT

    I am a bit uncomfortable with the formal step d/dx f(-x) = d/dx f(x)

    Can you explain how this follows exactly
    Last edited by Bernhard; January 16th 2010 at 02:33 PM.
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    Quote Originally Posted by Bernhard View Post
    Thanks.

    Can you explain how you show rigorously that f ' (-x) = f(-x). Why, exactly, does this follow?
     f'(x) = f(x) \ \forall x \in \mathbb{R}.
    Let  t = -x \Rightarrow f'(t) = f(t) by the above assumption, and thus  f'(-x) = f(-x)
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