Taylor expansion on Images which are warped
$begingroup$
I'm currently reading following paper: https://www.ri.cmu.edu/pub_files/pub3/baker_simon_2001_2/baker_simon_2001_2.pdf
Here I struggle to follow the Taylor expansion from equation 2 to equation 3. I mean I know in general how Taylor expansion works, but in this case it is not clear to me.
Starting equation:
$$
sum_{x} [ mathit{I}(mathbf{W}(mathbf{x};mathbf{p} + Deltamathbf{p})) - mathit{T}(mathbf{x})]^2
$$
Taylor expansion
$$
sum_{x} [ mathit{I}(mathbf{W}(mathbf{x};mathbf{p} )) + nablamathit{I}frac{partialmathbf{W}}{partialmathbf{p}}Deltamathbf{p} - mathit{T}(mathbf{x})]^2
$$
Maybe somebody of you has more experience and can elaborate little more on this?
taylor-expansion
asked Dec 1 '18 at 17:42
antibusantibus
1083
$endgroup$
add a comment |
$begingroup$
I'm currently reading following paper: https://www.ri.cmu.edu/pub_files/pub3/baker_simon_2001_2/baker_simon_2001_2.pdf
Here I struggle to follow the Taylor expansion from equation 2 to equation 3. I mean I know in general how Taylor expansion works, but in this case it is not clear to me.
Starting equation:
$$
sum_{x} [ mathit{I}(mathbf{W}(mathbf{x};mathbf{p} + Deltamathbf{p})) - mathit{T}(mathbf{x})]^2
$$
Taylor expansion
$$
sum_{x} [ mathit{I}(mathbf{W}(mathbf{x};mathbf{p} )) + nablamathit{I}frac{partialmathbf{W}}{partialmathbf{p}}Deltamathbf{p} - mathit{T}(mathbf{x})]^2
$$
Maybe somebody of you has more experience and can elaborate little more on this?
taylor-expansion
asked Dec 1 '18 at 17:42
antibusantibus
1083
$endgroup$
add a comment |
$begingroup$
I'm currently reading following paper: https://www.ri.cmu.edu/pub_files/pub3/baker_simon_2001_2/baker_simon_2001_2.pdf
Here I struggle to follow the Taylor expansion from equation 2 to equation 3. I mean I know in general how Taylor expansion works, but in this case it is not clear to me.
Starting equation:
$$
sum_{x} [ mathit{I}(mathbf{W}(mathbf{x};mathbf{p} + Deltamathbf{p})) - mathit{T}(mathbf{x})]^2
$$
Taylor expansion
$$
sum_{x} [ mathit{I}(mathbf{W}(mathbf{x};mathbf{p} )) + nablamathit{I}frac{partialmathbf{W}}{partialmathbf{p}}Deltamathbf{p} - mathit{T}(mathbf{x})]^2
$$
Maybe somebody of you has more experience and can elaborate little more on this?
taylor-expansion
asked Dec 1 '18 at 17:42
antibusantibus
1083
$endgroup$
I'm currently reading following paper: https://www.ri.cmu.edu/pub_files/pub3/baker_simon_2001_2/baker_simon_2001_2.pdf
Here I struggle to follow the Taylor expansion from equation 2 to equation 3. I mean I know in general how Taylor expansion works, but in this case it is not clear to me.
Starting equation:
$$
sum_{x} [ mathit{I}(mathbf{W}(mathbf{x};mathbf{p} + Deltamathbf{p})) - mathit{T}(mathbf{x})]^2
$$
Taylor expansion
$$
sum_{x} [ mathit{I}(mathbf{W}(mathbf{x};mathbf{p} )) + nablamathit{I}frac{partialmathbf{W}}{partialmathbf{p}}Deltamathbf{p} - mathit{T}(mathbf{x})]^2
$$
Maybe somebody of you has more experience and can elaborate little more on this?
taylor-expansion
taylor-expansion
asked Dec 1 '18 at 17:42
antibusantibus
1083
asked Dec 1 '18 at 17:42
antibusantibus
1083
asked Dec 1 '18 at 17:42
antibusantibus
1083
asked Dec 1 '18 at 17:42
antibusantibus
1083
asked Dec 1 '18 at 17:42
antibusantibus
1083
1083
add a comment |
add a comment |
1 Answer
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$begingroup$
You just need to apply the expansion twice. First for ${bf W}$ up to first order in $Delta {bf p}$, we get
$$
{bf W}({bf x}; {bf p} + Delta {bf p}) approx {bf W}({bf x}; {bf p}) + frac{partial {bf W}}{partial {bf p}}Delta {bf p} equiv {bf W} + Delta{bf W}tag{1}
$$
Now for $I$ up to first order in $Delta {bf W}$,
$$
I({bf W} + Delta{bf W}) approx I({bf W}) + nabla I Delta{bf W} = I({bf W}) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p} tag{2}
$$
Putting everything together
$$
I({bf W}({bf x}; {bf p} + Delta {bf p})) approx I({bf W}({bf x}; {bf p})) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p}
$$
answered Dec 1 '18 at 21:17
caveraccaverac
14.2k21130
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
You just need to apply the expansion twice. First for ${bf W}$ up to first order in $Delta {bf p}$, we get
$$
{bf W}({bf x}; {bf p} + Delta {bf p}) approx {bf W}({bf x}; {bf p}) + frac{partial {bf W}}{partial {bf p}}Delta {bf p} equiv {bf W} + Delta{bf W}tag{1}
$$
Now for $I$ up to first order in $Delta {bf W}$,
$$
I({bf W} + Delta{bf W}) approx I({bf W}) + nabla I Delta{bf W} = I({bf W}) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p} tag{2}
$$
Putting everything together
$$
I({bf W}({bf x}; {bf p} + Delta {bf p})) approx I({bf W}({bf x}; {bf p})) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p}
$$
answered Dec 1 '18 at 21:17
caveraccaverac
14.2k21130
$endgroup$
add a comment |
$begingroup$
You just need to apply the expansion twice. First for ${bf W}$ up to first order in $Delta {bf p}$, we get
$$
{bf W}({bf x}; {bf p} + Delta {bf p}) approx {bf W}({bf x}; {bf p}) + frac{partial {bf W}}{partial {bf p}}Delta {bf p} equiv {bf W} + Delta{bf W}tag{1}
$$
Now for $I$ up to first order in $Delta {bf W}$,
$$
I({bf W} + Delta{bf W}) approx I({bf W}) + nabla I Delta{bf W} = I({bf W}) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p} tag{2}
$$
Putting everything together
$$
I({bf W}({bf x}; {bf p} + Delta {bf p})) approx I({bf W}({bf x}; {bf p})) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p}
$$
answered Dec 1 '18 at 21:17
caveraccaverac
14.2k21130
$endgroup$
add a comment |
$begingroup$
You just need to apply the expansion twice. First for ${bf W}$ up to first order in $Delta {bf p}$, we get
$$
{bf W}({bf x}; {bf p} + Delta {bf p}) approx {bf W}({bf x}; {bf p}) + frac{partial {bf W}}{partial {bf p}}Delta {bf p} equiv {bf W} + Delta{bf W}tag{1}
$$
Now for $I$ up to first order in $Delta {bf W}$,
$$
I({bf W} + Delta{bf W}) approx I({bf W}) + nabla I Delta{bf W} = I({bf W}) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p} tag{2}
$$
Putting everything together
$$
I({bf W}({bf x}; {bf p} + Delta {bf p})) approx I({bf W}({bf x}; {bf p})) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p}
$$
answered Dec 1 '18 at 21:17
caveraccaverac
14.2k21130
$endgroup$
You just need to apply the expansion twice. First for ${bf W}$ up to first order in $Delta {bf p}$, we get
$$
{bf W}({bf x}; {bf p} + Delta {bf p}) approx {bf W}({bf x}; {bf p}) + frac{partial {bf W}}{partial {bf p}}Delta {bf p} equiv {bf W} + Delta{bf W}tag{1}
$$
Now for $I$ up to first order in $Delta {bf W}$,
$$
I({bf W} + Delta{bf W}) approx I({bf W}) + nabla I Delta{bf W} = I({bf W}) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p} tag{2}
$$
Putting everything together
$$
I({bf W}({bf x}; {bf p} + Delta {bf p})) approx I({bf W}({bf x}; {bf p})) + nabla I frac{partial {bf W}}{partial {bf p}}Delta {bf p}
$$
answered Dec 1 '18 at 21:17
caveraccaverac
14.2k21130
answered Dec 1 '18 at 21:17
caveraccaverac
14.2k21130
answered Dec 1 '18 at 21:17
caveraccaverac
14.2k21130
answered Dec 1 '18 at 21:17
caveraccaverac
14.2k21130
14.2k21130
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