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Math Help - Probably quite straight forward...

  1. #1
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    Probably quite straight forward...

    I haven't done mv calculus in a while ... I'm stuck trying to understand a line of a solution I've been given. For this question T is a constant and x and t are variables.

    If f'(x)=g(x)

    \frac{\partial}{\partial x}f(\frac{x}{\sqrt{T-t}}) = \frac{1}{\sqrt{T-t}}g(\frac{x}{\sqrt{T-t}}) - I'm OK with this

    \frac{\partial ^2}{\partial x^2}f(\frac{x}{\sqrt{T-t}}) = \frac{1}{T-t}g'(\frac{x}{\sqrt{T-t}}) - And this

    = \frac{x}{(T-t)^{3/2}}g(\frac{-x}{\sqrt{T-t}}) using g'(x) = -xg(x)

    I can see how the last equality holds given the "using...", but I have no idea where that came from. Any help?

    Possibly there is some bad notation going on with the first and last statements - it should be f'(x*) = g(x*) and g'(x*) = -x*g(x*) or something, maybe...
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  2. #2
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    Derivatives

    Hello RanDom
    Quote Originally Posted by RanDom View Post
    I haven't done mv calculus in a while ... I'm stuck trying to understand a line of a solution I've been given. For this question T is a constant and x and t are variables.

    If f'(x)=g(x)

    \frac{\partial}{\partial x}f(\frac{x}{\sqrt{T-t}}) = \frac{1}{\sqrt{T-t}}g(\frac{x}{\sqrt{T-t}}) - I'm OK with this

    \frac{\partial ^2}{\partial x^2}f(\frac{x}{\sqrt{T-t}}) = \frac{1}{T-t}g'(\frac{x}{\sqrt{T-t}}) - And this

    = \frac{x}{(T-t)^{3/2}}g(\frac{-x}{\sqrt{T-t}}) using g'(x) = -xg(x)

    I can see how the last equality holds given the "using...", but I have no idea where that came from. Any help?

    Possibly there is some bad notation going on with the first and last statements - it should be f'(x*) = g(x*) and g'(x*) = -x*g(x*) or something, maybe...
    I have no idea where g'(x) = -xg(x) comes from either. This will only be the case if g(x) is of the form:

    g(x) = Ae^{-\tfrac{1}{2}x^2}

    Do you have anything to indicate that this might be the case?

    Grandad
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  3. #3
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    Ohhhhhhh.... Yes I do! f is the standard normal distribution hahahaha..... What an idiot.

    Thanks!
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