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Notations in the proof that $forall x in mathbbR^d f(x+h) = f(x) + langlenabla f(x), hrangle + o(|textbfh|)$

Gradient vector function using sum and scalarWhat is the name for a vector field that is both divergence-free and curl-free?Inner product identityJacobian of Stabilized Eikonal Equation $| nabla u| = 1$Proving $textdiv(textcurl(textbfF))=0$What is the relationship between the level curve of a surface and vectors tangent to the function?How to see the following Taylor expansion?Strict monotonicity of the partial derivatives of a strictly convex functionGiven function $f$ find directional derivative of $lVert nabla f rVert$ in direction given by $nabla f$Using a component-wise approach (using partial derivatives), show that the gradient of $f$ is given by $nabla f(x) = frac12(A +A^T)x + b$

0

$begingroup$

Check that the gradient does fulfill the following fundamental property

$$
forall x in mathbbR^d f(x+h) = f(x) + langlenabla f(x), hrangle + o(|textbfh|)^*1 (textbfhin mathbbR^d)
$$
Let $xin mathbbR^d, hin mathbbR^d, hne O_d^*2$
beginalign*
f(x+h) &= frac12(x+h)^T A(x+h) + b^T(x+h) \
&=underbracefrac12x^T Ax + b^Tx_f(x) + underbracefracx^TAh2+frach^TAx2_frac(x^T Ah)^T+h^TAx2+underbraceb^Th_h^Tb + frac12h^TAh\
&textand we can write: fracx^TAh2+frach^TAx2 + h^Tb = underbraceh^Tbig[(A^T + A)x + bbig]_langle h, nabla f(x) rangle
endalign*

We need to prove that: $frac12h^T Ah = o(|h|) Leftrightarrow exists epsilon(h) xrightarrow[hrightarrow 0] 0^*3$, $|h|epsilon (h) = frach^TAh2$
$$
frac12h^TAh = frac^22big(frachhbig)^T Abig(frachhbig) =underbrace _epsilon(h)
$$

and $big(frachhbig)^T Abig(frachhbig)$ is bounded for all $h$ by $|||A|||^*4 = max_xin mathbbR^d, x^T A x$
$$
epsilon (h) = |h|cdotbig|frac12big(frachhbig)^T A big(frachhbig)big| leq |h| cdot |||A|||
$$
Therefore $frac12h^TAh = |h|cdot epsilon(h) xrightarrow[hrightarrow 0] 0= o(|h|)$

$forall x in mathbbR^d f(x+h) = f(x) + langlenabla f(x), hrangle + o(|textbfh|)$

I do the following notations mean (noted by $^*n$):

• $o(|h|)$


• $O_d$


• $epsilon(h) xrightarrow[hrightarrow 0] 0$


• $|||A|||$

partial-derivative vector-analysis

share|cite|improve this question

asked yesterday

ecjbecjb

2688

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  1)Little-o, 2)d-dimensional zero vector, 3) check little-o notation, 4) matrix norm.
  $endgroup$
  – lightxbulb
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  many thanks for your helpful comment @lightxbulb
  $endgroup$
  – ecjb
  yesterday
  

  
  
  
  

add a comment | 

0

$begingroup$

Check that the gradient does fulfill the following fundamental property

$$
forall x in mathbbR^d f(x+h) = f(x) + langlenabla f(x), hrangle + o(|textbfh|)^*1 (textbfhin mathbbR^d)
$$
Let $xin mathbbR^d, hin mathbbR^d, hne O_d^*2$
beginalign*
f(x+h) &= frac12(x+h)^T A(x+h) + b^T(x+h) \
&=underbracefrac12x^T Ax + b^Tx_f(x) + underbracefracx^TAh2+frach^TAx2_frac(x^T Ah)^T+h^TAx2+underbraceb^Th_h^Tb + frac12h^TAh\
&textand we can write: fracx^TAh2+frach^TAx2 + h^Tb = underbraceh^Tbig[(A^T + A)x + bbig]_langle h, nabla f(x) rangle
endalign*

We need to prove that: $frac12h^T Ah = o(|h|) Leftrightarrow exists epsilon(h) xrightarrow[hrightarrow 0] 0^*3$, $|h|epsilon (h) = frach^TAh2$
$$
frac12h^TAh = frac^22big(frachhbig)^T Abig(frachhbig) =underbrace _epsilon(h)
$$

and $big(frachhbig)^T Abig(frachhbig)$ is bounded for all $h$ by $|||A|||^*4 = max_xin mathbbR^d, x^T A x$
$$
epsilon (h) = |h|cdotbig|frac12big(frachhbig)^T A big(frachhbig)big| leq |h| cdot |||A|||
$$
Therefore $frac12h^TAh = |h|cdot epsilon(h) xrightarrow[hrightarrow 0] 0= o(|h|)$

$forall x in mathbbR^d f(x+h) = f(x) + langlenabla f(x), hrangle + o(|textbfh|)$

I do the following notations mean (noted by $^*n$):

• $o(|h|)$


• $O_d$


• $epsilon(h) xrightarrow[hrightarrow 0] 0$


• $|||A|||$

partial-derivative vector-analysis

share|cite|improve this question

asked yesterday

ecjbecjb

2688

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  1)Little-o, 2)d-dimensional zero vector, 3) check little-o notation, 4) matrix norm.
  $endgroup$
  – lightxbulb
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  many thanks for your helpful comment @lightxbulb
  $endgroup$
  – ecjb
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Check that the gradient does fulfill the following fundamental property

$$
forall x in mathbbR^d f(x+h) = f(x) + langlenabla f(x), hrangle + o(|textbfh|)^*1 (textbfhin mathbbR^d)
$$
Let $xin mathbbR^d, hin mathbbR^d, hne O_d^*2$
beginalign*
f(x+h) &= frac12(x+h)^T A(x+h) + b^T(x+h) \
&=underbracefrac12x^T Ax + b^Tx_f(x) + underbracefracx^TAh2+frach^TAx2_frac(x^T Ah)^T+h^TAx2+underbraceb^Th_h^Tb + frac12h^TAh\
&textand we can write: fracx^TAh2+frach^TAx2 + h^Tb = underbraceh^Tbig[(A^T + A)x + bbig]_langle h, nabla f(x) rangle
endalign*

We need to prove that: $frac12h^T Ah = o(|h|) Leftrightarrow exists epsilon(h) xrightarrow[hrightarrow 0] 0^*3$, $|h|epsilon (h) = frach^TAh2$
$$
frac12h^TAh = frac^22big(frachhbig)^T Abig(frachhbig) =underbrace _epsilon(h)
$$

and $big(frachhbig)^T Abig(frachhbig)$ is bounded for all $h$ by $|||A|||^*4 = max_xin mathbbR^d, x^T A x$
$$
epsilon (h) = |h|cdotbig|frac12big(frachhbig)^T A big(frachhbig)big| leq |h| cdot |||A|||
$$
Therefore $frac12h^TAh = |h|cdot epsilon(h) xrightarrow[hrightarrow 0] 0= o(|h|)$

$forall x in mathbbR^d f(x+h) = f(x) + langlenabla f(x), hrangle + o(|textbfh|)$

I do the following notations mean (noted by $^*n$):

• $o(|h|)$


• $O_d$


• $epsilon(h) xrightarrow[hrightarrow 0] 0$


• $|||A|||$

partial-derivative vector-analysis

share|cite|improve this question

asked yesterday

ecjbecjb

2688

$endgroup$

share|cite|improve this question

asked yesterday

ecjbecjb

2688

share|cite|improve this question

asked yesterday

ecjbecjb

2688

asked yesterday

ecjbecjb

2688

asked yesterday

ecjbecjb

2688