# Frechet Derivative

• April 18th 2009, 01:47 PM
bigdoggy
Frechet Derivative
$f:\Re^2\longrightarrow\Re, \;\; f[x,y]^T=x^2y\;\;$
Prove from
$\lim_{\vec{h}\rightarrow 0}
\;=\;\frac{f(\vec{a} + \vec{h})-f(\vec{a})-L(\vec{h})}{|\vec{h}|}=0$
that f is differentiable at a general point $\vec{a}=[a,b]^T\;\;\text{and find the Frechet derivative of f at } \vec{a}.
$
• April 18th 2009, 03:22 PM
NonCommAlg
Quote:

Originally Posted by bigdoggy
$f:\Re^2\longrightarrow\Re, \;\; f[x,y]^T=x^2y\;\;$
Prove from
$\lim_{\vec{h}\rightarrow 0}
\; \frac{f(\vec{a} + \vec{h})-f(\vec{a})-L(\vec{h})}{|\vec{h}|}=0$
that f is differentiable at a general point $\vec{a}=[a,b]^T\;\;\text{and find the Frechet derivative of f at } \vec{a}.
$

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 $L=[2rs \ \ r^2].$
• April 19th 2009, 06:38 AM
bigdoggy
Quote:

Originally Posted by NonCommAlg
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