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Math Help - Frechet Derivative

  1. #1
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    Frechet Derivative

    f:\Re^2\longrightarrow\Re, \;\; f[x,y]^T=x^2y\;\;
    Prove from
    \lim_{\vec{h}\rightarrow 0}<br />
\;=\;\frac{f(\vec{a} + \vec{h})-f(\vec{a})-L(\vec{h})}{|\vec{h}|}=0 that f is differentiable at a general point \vec{a}=[a,b]^T\;\;\text{and find the Frechet derivative of f at } \vec{a}.<br />
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  2. #2
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    Quote Originally Posted by bigdoggy View Post
    f:\Re^2\longrightarrow\Re, \;\; f[x,y]^T=x^2y\;\;
    Prove from
    \lim_{\vec{h}\rightarrow 0}<br />
\; \frac{f(\vec{a} + \vec{h})-f(\vec{a})-L(\vec{h})}{|\vec{h}|}=0 that f is differentiable at a general point \vec{a}=[a,b]^T\;\;\text{and find the Frechet derivative of f at } \vec{a}.<br />
    for given \vec{a}=[r,s]^T \in \mathbb{R}^2, define L: \mathbb{R}^2 \longrightarrow \mathbb{R} by L([x,y]^T)=[2rs \ \ r^2][x,y]^T=2rsx + r^2y. now if [h_1,h_2]^T=\vec{h} \rightarrow \vec{0}, then:

    \frac{|f(\vec{a} + \vec{h})-f(\vec{a})-L(\vec{h})|}{|\vec{h}|}=\frac{|h_1^2h_2 + h_1^2s + 2rh_1h_2|}{\sqrt{h_1^2+h_2^2}} \leq |h_1|(|h_2| + |s| + 2|r|), and obviously |h_1|(|h_2| + |s| + 2|r|) \rightarrow 0, as \vec{h} \rightarrow \vec{0}.

    this also shows that the derivative of f at \vec{a} is L=[2rs \ \ r^2].
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    for given \vec{a}=[r,s]^T \in \mathbb{R}^2, define L: \mathbb{R}^2 \longrightarrow \mathbb{R} by L([x,y]^T)=[2rs \ \ r^2][x,y]^T=2rsx + r^2y. now if [h_1,h_2]^T=\vec{h} \rightarrow \vec{0}, then:

    \frac{|f(\vec{a} + \vec{h})-f(\vec{a})-L(\vec{h})|}{|\vec{h}|}=\frac{|h_1^2h_2 + h_1^2s + 2rh_1h_2|}{\sqrt{h_1^2+h_2^2}} \leq |h_1|(|h_2| + |s| + 2|r|), and obviously |h_1|(|h_2| + |s| + 2|r|) \rightarrow 0, as \vec{h} \rightarrow \vec{0}.

    this also shows that the derivative of f at \vec{a} is L=[2rs \ \ r^2].
    I'm a little confused as to what is happening above, please could you explain, line by line, if possible? Thanks very much
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