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Math Help - derivative problem

  1. #1
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    derivative problem

    If f(x-y)=f(x)g(y)-g(x)f(y) and g(x-y)=g(x)g(y)+f(x)f(y) for all x,y \in \mathbb{R} and the right hand derivative of f(x) at zero exists. Find g'(x) at x=0
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    MHF Contributor Bruno J.'s Avatar
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    g(-x) = g(0-x) = g(x)

    Assuming g is differentiable at 0, the limit

    \lim_{d \rightarrow 0} \frac{g(d)-g(-d)}{2d} converges and is equal to g'(0); which yields g'(0)=0.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Bruno J. View Post
    g(-x) = g(0-x) = g(x)

    Assuming g is differentiable at 0, the limit

    \lim_{d \rightarrow 0} \frac{g(d)-g(-d)}{2d} converges and is equal to g'(0); which yields g'(0)=0.
    You need to provide the explanation of why g is symmetric

    CB
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    MHF Contributor Bruno J.'s Avatar
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    I thought I had!

    g(x-0) = g(x)g(0)+f(x)f(0)
    g(0-x) = g(0)g(x)+f(0)f(x)
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  5. #5
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    query

    Its not given that g is differentiable
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  6. #6
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    query

    Its not given that g is differentiable, so u cant assume g' exists
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  7. #7
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Chandru1 View Post
    Its not given that g is differentiable, so u cant assume g' exists
    Here we go then!

    You have g(x) = g(x)g(0)+f(x)f(0) = g(x)g(0). Hence either g(0) = 1 or g(x) is identically zero which instantly yields g'(0)=0. Hence suppose g(0)=1.

    f(0) = 0 is easy; set x=y.
    f(x)=-f(-x) is also easy.

    f(x+y)=f(x)g(-y)-g(x)f(-y) = f(x)g(y)+g(x)f(y) by the above.

    g(x+y)=g(x)g(-y)+f(x)f(-y) = g(x)g(y)-f(x)f(y)

    Now consider the complex numbers

    m(x) = g(x)+if(x).

    We have m(x)m(y) = g(x+y)+if(x+y) = m(x+y).

    In particular m(0)=1, and m(x)m(-x)=1.


    Hence the function m : \mathbb{R} \rightarrow \mathbb{C} is a morphism of groups m : \mathbb{R}^+ \rightarrow \mathbb{C}^\times. Hence we have m(x)=\exp(tx) for some t=a+bi \in \mathbb{C}. Clearly a=0 since m(0)=1. Hence m(x)=\exp(xbi)

    Hence g(x) = \Re \exp(xbi) = \cos(bx),
    so that g'(0) = 0.
    Last edited by Bruno J.; June 16th 2009 at 04:02 PM.
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  8. #8
    MHF Contributor Bruno J.'s Avatar
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    Just a little extra :
    In case you are wondering why expressing m as an exponential map is possible, consider

    k=m(1)

    Then for all integers n:
    m(n)=m(1+...+1)=k^n
    k=m(1)=m(1/n+...+1/n)=m(1/n)^n
    so that m(1/n) = k^{1/n}

    from this it follows easily that for all rational x
    m(x) = k^{x}.

    Hence if m is continuous on any interval, it coincides everywhere on that interval with k^{x} (irrationals included).
    But f is certainly continuous at 0 (it's differentiable there!). To see that g is continuous at 0, we have

    g(x/2 + x/2) = g(x/2)^2-f(x/2)^2

    i.e.
     g(x/2)^2-g(x) = f(x/2)^2

    and from this it is easy to show (with sequences for example) that g is continuous at 0 (using the fact that f(x/2)^2 is continuous at 0 and is equal to 0)
    Last edited by Bruno J.; June 16th 2009 at 06:19 PM.
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