Show that for every $x$ there exists a $n in mathbb{N}$ so that $f(x-(frac{1}{2})^{n} nabla f(x)) leq...
$begingroup$
Let $f: mathbb{R}^{n}rightarrowmathbb{R}$ be continuous and differentiable and positive.
As stated in the title show that for every $x$ there exists a $n in mathbb{N}$ so that
$f(x-(frac{1}{2})^{n} nabla f(x)) leq f(x)-(frac{1}{2})^{n+1} ||nabla f(x)||^2$
Given Hints: are Cauchy-Schwarz and the fundamental theorem of integral and differential calculus.
There is probably an equation of the function $f(x+y)$ so that you can rewrite it into an Integral from where I can work on, however nothing comes into my mind so far so that I could start. I hope someone will help me with this one!
inequality
asked Jan 20 at 18:18
babemcnuggetsbabemcnuggets
1129
$endgroup$
add a comment |
$begingroup$
Let $f: mathbb{R}^{n}rightarrowmathbb{R}$ be continuous and differentiable and positive.
As stated in the title show that for every $x$ there exists a $n in mathbb{N}$ so that
$f(x-(frac{1}{2})^{n} nabla f(x)) leq f(x)-(frac{1}{2})^{n+1} ||nabla f(x)||^2$
Given Hints: are Cauchy-Schwarz and the fundamental theorem of integral and differential calculus.
There is probably an equation of the function $f(x+y)$ so that you can rewrite it into an Integral from where I can work on, however nothing comes into my mind so far so that I could start. I hope someone will help me with this one!
inequality
asked Jan 20 at 18:18
babemcnuggetsbabemcnuggets
1129
$endgroup$
add a comment |
$begingroup$
Let $f: mathbb{R}^{n}rightarrowmathbb{R}$ be continuous and differentiable and positive.
As stated in the title show that for every $x$ there exists a $n in mathbb{N}$ so that
$f(x-(frac{1}{2})^{n} nabla f(x)) leq f(x)-(frac{1}{2})^{n+1} ||nabla f(x)||^2$
Given Hints: are Cauchy-Schwarz and the fundamental theorem of integral and differential calculus.
There is probably an equation of the function $f(x+y)$ so that you can rewrite it into an Integral from where I can work on, however nothing comes into my mind so far so that I could start. I hope someone will help me with this one!
inequality
asked Jan 20 at 18:18
babemcnuggetsbabemcnuggets
1129
$endgroup$
Let $f: mathbb{R}^{n}rightarrowmathbb{R}$ be continuous and differentiable and positive.
As stated in the title show that for every $x$ there exists a $n in mathbb{N}$ so that
$f(x-(frac{1}{2})^{n} nabla f(x)) leq f(x)-(frac{1}{2})^{n+1} ||nabla f(x)||^2$
Given Hints: are Cauchy-Schwarz and the fundamental theorem of integral and differential calculus.
There is probably an equation of the function $f(x+y)$ so that you can rewrite it into an Integral from where I can work on, however nothing comes into my mind so far so that I could start. I hope someone will help me with this one!
inequality
inequality
asked Jan 20 at 18:18
babemcnuggetsbabemcnuggets
1129
asked Jan 20 at 18:18
babemcnuggetsbabemcnuggets
1129
asked Jan 20 at 18:18
babemcnuggetsbabemcnuggets
1129
asked Jan 20 at 18:18
babemcnuggetsbabemcnuggets
1129
asked Jan 20 at 18:18
babemcnuggetsbabemcnuggets
1129
1129
add a comment |
add a comment |
1 Answer
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There is a really short answer to that. Recalling that $D_x f(v)=lim_{hrightarrow 0}frac{f(x)-f(x+hv)}{h}=nabla f(x)^Tv=$<$nabla f,v$> you can rewrite the inequality.
$f(x-(frac{1}{2})^{n} nabla f(x)) leq f(x)-(frac{1}{2})^{n+1} ||nabla f(x)||^2$
= $ (frac{1}{2}) ||nabla f(x)||^2 leq frac{f(x)-(x-(frac{1}{2})^{n} nabla f(x))}{frac{1}{2}^n}$ $(1)$
Let $h=(frac{1}{2})^{n}$ and $v=nabla f$. We see that for $lim_{hrightarrow 0}$ we get $frac{1}{2}||nabla f(x)||leq||nabla f(x)||^2$ in $(1)$. If we let $n$ be "big enough" the inequality therefore follows.
answered Feb 1 at 22:45
babemcnuggetsbabemcnuggets
1129
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
There is a really short answer to that. Recalling that $D_x f(v)=lim_{hrightarrow 0}frac{f(x)-f(x+hv)}{h}=nabla f(x)^Tv=$<$nabla f,v$> you can rewrite the inequality.
$f(x-(frac{1}{2})^{n} nabla f(x)) leq f(x)-(frac{1}{2})^{n+1} ||nabla f(x)||^2$
= $ (frac{1}{2}) ||nabla f(x)||^2 leq frac{f(x)-(x-(frac{1}{2})^{n} nabla f(x))}{frac{1}{2}^n}$ $(1)$
Let $h=(frac{1}{2})^{n}$ and $v=nabla f$. We see that for $lim_{hrightarrow 0}$ we get $frac{1}{2}||nabla f(x)||leq||nabla f(x)||^2$ in $(1)$. If we let $n$ be "big enough" the inequality therefore follows.
answered Feb 1 at 22:45
babemcnuggetsbabemcnuggets
1129
$endgroup$
add a comment |
$begingroup$
There is a really short answer to that. Recalling that $D_x f(v)=lim_{hrightarrow 0}frac{f(x)-f(x+hv)}{h}=nabla f(x)^Tv=$<$nabla f,v$> you can rewrite the inequality.
$f(x-(frac{1}{2})^{n} nabla f(x)) leq f(x)-(frac{1}{2})^{n+1} ||nabla f(x)||^2$
= $ (frac{1}{2}) ||nabla f(x)||^2 leq frac{f(x)-(x-(frac{1}{2})^{n} nabla f(x))}{frac{1}{2}^n}$ $(1)$
Let $h=(frac{1}{2})^{n}$ and $v=nabla f$. We see that for $lim_{hrightarrow 0}$ we get $frac{1}{2}||nabla f(x)||leq||nabla f(x)||^2$ in $(1)$. If we let $n$ be "big enough" the inequality therefore follows.
answered Feb 1 at 22:45
babemcnuggetsbabemcnuggets
1129
$endgroup$
add a comment |
$begingroup$
There is a really short answer to that. Recalling that $D_x f(v)=lim_{hrightarrow 0}frac{f(x)-f(x+hv)}{h}=nabla f(x)^Tv=$<$nabla f,v$> you can rewrite the inequality.
$f(x-(frac{1}{2})^{n} nabla f(x)) leq f(x)-(frac{1}{2})^{n+1} ||nabla f(x)||^2$
= $ (frac{1}{2}) ||nabla f(x)||^2 leq frac{f(x)-(x-(frac{1}{2})^{n} nabla f(x))}{frac{1}{2}^n}$ $(1)$
Let $h=(frac{1}{2})^{n}$ and $v=nabla f$. We see that for $lim_{hrightarrow 0}$ we get $frac{1}{2}||nabla f(x)||leq||nabla f(x)||^2$ in $(1)$. If we let $n$ be "big enough" the inequality therefore follows.
answered Feb 1 at 22:45
babemcnuggetsbabemcnuggets
1129
$endgroup$
There is a really short answer to that. Recalling that $D_x f(v)=lim_{hrightarrow 0}frac{f(x)-f(x+hv)}{h}=nabla f(x)^Tv=$<$nabla f,v$> you can rewrite the inequality.
$f(x-(frac{1}{2})^{n} nabla f(x)) leq f(x)-(frac{1}{2})^{n+1} ||nabla f(x)||^2$
= $ (frac{1}{2}) ||nabla f(x)||^2 leq frac{f(x)-(x-(frac{1}{2})^{n} nabla f(x))}{frac{1}{2}^n}$ $(1)$
Let $h=(frac{1}{2})^{n}$ and $v=nabla f$. We see that for $lim_{hrightarrow 0}$ we get $frac{1}{2}||nabla f(x)||leq||nabla f(x)||^2$ in $(1)$. If we let $n$ be "big enough" the inequality therefore follows.
answered Feb 1 at 22:45
babemcnuggetsbabemcnuggets
1129
answered Feb 1 at 22:45
babemcnuggetsbabemcnuggets
1129
answered Feb 1 at 22:45
babemcnuggetsbabemcnuggets
1129
answered Feb 1 at 22:45
babemcnuggetsbabemcnuggets
1129
1129
add a comment |
add a comment |
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