Mixing
Jacobian for function with couple of couple
Caclulate the jacobian :
$ f:mathbb{R}^2timesmathbb{R}^2tomathbb{R}^2 \
((x,y),(u,v))to(ux-3xv,yu)
$
and
$g:mathbb{R}^2timesmathbb{R}^2tomathbb{R}\
(U,V)to det(u,v)$
for $f$ I don't how to differantiate for a couple of couple. Can I consider it as map form $R^4$,
differential-equations multivariable-calculus partial-derivative
asked Nov 20 '18 at 20:10
Mary Maths
165
add a comment |
Caclulate the jacobian :
$ f:mathbb{R}^2timesmathbb{R}^2tomathbb{R}^2 \
((x,y),(u,v))to(ux-3xv,yu)
$
and
$g:mathbb{R}^2timesmathbb{R}^2tomathbb{R}\
(U,V)to det(u,v)$
for $f$ I don't how to differantiate for a couple of couple. Can I consider it as map form $R^4$,
differential-equations multivariable-calculus partial-derivative
asked Nov 20 '18 at 20:10
Mary Maths
165
add a comment |
Caclulate the jacobian :
$ f:mathbb{R}^2timesmathbb{R}^2tomathbb{R}^2 \
((x,y),(u,v))to(ux-3xv,yu)
$
and
$g:mathbb{R}^2timesmathbb{R}^2tomathbb{R}\
(U,V)to det(u,v)$
for $f$ I don't how to differantiate for a couple of couple. Can I consider it as map form $R^4$,
differential-equations multivariable-calculus partial-derivative
asked Nov 20 '18 at 20:10
Mary Maths
165
Caclulate the jacobian :
$ f:mathbb{R}^2timesmathbb{R}^2tomathbb{R}^2 \
((x,y),(u,v))to(ux-3xv,yu)
$
and
$g:mathbb{R}^2timesmathbb{R}^2tomathbb{R}\
(U,V)to det(u,v)$
for $f$ I don't how to differantiate for a couple of couple. Can I consider it as map form $R^4$,
differential-equations multivariable-calculus partial-derivative
differential-equations multivariable-calculus partial-derivative
asked Nov 20 '18 at 20:10
Mary Maths
165
asked Nov 20 '18 at 20:10
Mary Maths
165
asked Nov 20 '18 at 20:10
Mary Maths
165
asked Nov 20 '18 at 20:10
Mary Maths
165
asked Nov 20 '18 at 20:10
Mary Maths
165
165
add a comment |
add a comment |
1 Answer
1
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In the first one you have $f=(f_1,f_2)$ where $f_1:mathbb{R}^4tomathbb{R}$ is given by
$$f_1(x,y,u,v)=ux-3xv$$
and $f_2:mathbb{R}^4to mathbb{R}$ is given by
$$f_2(x,y,u,v)=yu.$$
Then $$begin{align*}
Df_1(x,y,u,v)&=begin{pmatrix}D_xf_1& D_yf_1 & D_uf_1 & D_vf_1end{pmatrix}=begin{pmatrix}u-3v & 0 & x & -3xend{pmatrix}\
Df_2(x,y,u,v)&=begin{pmatrix}D_xf_2& D_yf_2 & D_uf_2 & D_vf_2end{pmatrix}=begin{pmatrix}0 & u & y & 0end{pmatrix}
end{align*}$$
So $$Df(x,y,u,v)=begin{pmatrix}u-3v & 0 & x & -3x\0 & u & y & 0 end{pmatrix}$$
For the second one, denote $u=(u_1,u_2)$ and $v=(v_1,v_2)$. Then
$$g(u,v)=detbegin{pmatrix}u_1 & v_1 \ u_2 & v_2end{pmatrix}=u_1v_2-u_2v_1$$
so that $$Dg(u,v)=Dg(u_1,u_2,v_1,v_2)=begin{pmatrix}D_{u_1}g & D_{u_2}g & D_{v_1}g& D_{v_2}gend{pmatrix}=begin{pmatrix}v_2 & -v_1 & -u_2 & u_1end{pmatrix}$$
answered Nov 20 '18 at 20:54
smcc
4,297517
For $ f $ why do you consider it as $f(x,y,u,v)$ anf not$f((x,y),(u,v))$
– Mary Maths
Nov 20 '18 at 21:07
See here: math.stackexchange.com/questions/338319/…
– smcc
Nov 20 '18 at 22:29
add a comment |
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1 Answer
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1 Answer
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In the first one you have $f=(f_1,f_2)$ where $f_1:mathbb{R}^4tomathbb{R}$ is given by
$$f_1(x,y,u,v)=ux-3xv$$
and $f_2:mathbb{R}^4to mathbb{R}$ is given by
$$f_2(x,y,u,v)=yu.$$
Then $$begin{align*}
Df_1(x,y,u,v)&=begin{pmatrix}D_xf_1& D_yf_1 & D_uf_1 & D_vf_1end{pmatrix}=begin{pmatrix}u-3v & 0 & x & -3xend{pmatrix}\
Df_2(x,y,u,v)&=begin{pmatrix}D_xf_2& D_yf_2 & D_uf_2 & D_vf_2end{pmatrix}=begin{pmatrix}0 & u & y & 0end{pmatrix}
end{align*}$$
So $$Df(x,y,u,v)=begin{pmatrix}u-3v & 0 & x & -3x\0 & u & y & 0 end{pmatrix}$$
For the second one, denote $u=(u_1,u_2)$ and $v=(v_1,v_2)$. Then
$$g(u,v)=detbegin{pmatrix}u_1 & v_1 \ u_2 & v_2end{pmatrix}=u_1v_2-u_2v_1$$
so that $$Dg(u,v)=Dg(u_1,u_2,v_1,v_2)=begin{pmatrix}D_{u_1}g & D_{u_2}g & D_{v_1}g& D_{v_2}gend{pmatrix}=begin{pmatrix}v_2 & -v_1 & -u_2 & u_1end{pmatrix}$$
answered Nov 20 '18 at 20:54
smcc
4,297517
For $ f $ why do you consider it as $f(x,y,u,v)$ anf not$f((x,y),(u,v))$
– Mary Maths
Nov 20 '18 at 21:07
See here: math.stackexchange.com/questions/338319/…
– smcc
Nov 20 '18 at 22:29
add a comment |
In the first one you have $f=(f_1,f_2)$ where $f_1:mathbb{R}^4tomathbb{R}$ is given by
$$f_1(x,y,u,v)=ux-3xv$$
and $f_2:mathbb{R}^4to mathbb{R}$ is given by
$$f_2(x,y,u,v)=yu.$$
Then $$begin{align*}
Df_1(x,y,u,v)&=begin{pmatrix}D_xf_1& D_yf_1 & D_uf_1 & D_vf_1end{pmatrix}=begin{pmatrix}u-3v & 0 & x & -3xend{pmatrix}\
Df_2(x,y,u,v)&=begin{pmatrix}D_xf_2& D_yf_2 & D_uf_2 & D_vf_2end{pmatrix}=begin{pmatrix}0 & u & y & 0end{pmatrix}
end{align*}$$
So $$Df(x,y,u,v)=begin{pmatrix}u-3v & 0 & x & -3x\0 & u & y & 0 end{pmatrix}$$
For the second one, denote $u=(u_1,u_2)$ and $v=(v_1,v_2)$. Then
$$g(u,v)=detbegin{pmatrix}u_1 & v_1 \ u_2 & v_2end{pmatrix}=u_1v_2-u_2v_1$$
so that $$Dg(u,v)=Dg(u_1,u_2,v_1,v_2)=begin{pmatrix}D_{u_1}g & D_{u_2}g & D_{v_1}g& D_{v_2}gend{pmatrix}=begin{pmatrix}v_2 & -v_1 & -u_2 & u_1end{pmatrix}$$
answered Nov 20 '18 at 20:54
smcc
4,297517
For $ f $ why do you consider it as $f(x,y,u,v)$ anf not$f((x,y),(u,v))$
– Mary Maths
Nov 20 '18 at 21:07
See here: math.stackexchange.com/questions/338319/…
– smcc
Nov 20 '18 at 22:29
add a comment |
In the first one you have $f=(f_1,f_2)$ where $f_1:mathbb{R}^4tomathbb{R}$ is given by
$$f_1(x,y,u,v)=ux-3xv$$
and $f_2:mathbb{R}^4to mathbb{R}$ is given by
$$f_2(x,y,u,v)=yu.$$
Then $$begin{align*}
Df_1(x,y,u,v)&=begin{pmatrix}D_xf_1& D_yf_1 & D_uf_1 & D_vf_1end{pmatrix}=begin{pmatrix}u-3v & 0 & x & -3xend{pmatrix}\
Df_2(x,y,u,v)&=begin{pmatrix}D_xf_2& D_yf_2 & D_uf_2 & D_vf_2end{pmatrix}=begin{pmatrix}0 & u & y & 0end{pmatrix}
end{align*}$$
So $$Df(x,y,u,v)=begin{pmatrix}u-3v & 0 & x & -3x\0 & u & y & 0 end{pmatrix}$$
For the second one, denote $u=(u_1,u_2)$ and $v=(v_1,v_2)$. Then
$$g(u,v)=detbegin{pmatrix}u_1 & v_1 \ u_2 & v_2end{pmatrix}=u_1v_2-u_2v_1$$
so that $$Dg(u,v)=Dg(u_1,u_2,v_1,v_2)=begin{pmatrix}D_{u_1}g & D_{u_2}g & D_{v_1}g& D_{v_2}gend{pmatrix}=begin{pmatrix}v_2 & -v_1 & -u_2 & u_1end{pmatrix}$$
answered Nov 20 '18 at 20:54
smcc
4,297517
In the first one you have $f=(f_1,f_2)$ where $f_1:mathbb{R}^4tomathbb{R}$ is given by
$$f_1(x,y,u,v)=ux-3xv$$
and $f_2:mathbb{R}^4to mathbb{R}$ is given by
$$f_2(x,y,u,v)=yu.$$
Then $$begin{align*}
Df_1(x,y,u,v)&=begin{pmatrix}D_xf_1& D_yf_1 & D_uf_1 & D_vf_1end{pmatrix}=begin{pmatrix}u-3v & 0 & x & -3xend{pmatrix}\
Df_2(x,y,u,v)&=begin{pmatrix}D_xf_2& D_yf_2 & D_uf_2 & D_vf_2end{pmatrix}=begin{pmatrix}0 & u & y & 0end{pmatrix}
end{align*}$$
So $$Df(x,y,u,v)=begin{pmatrix}u-3v & 0 & x & -3x\0 & u & y & 0 end{pmatrix}$$
For the second one, denote $u=(u_1,u_2)$ and $v=(v_1,v_2)$. Then
$$g(u,v)=detbegin{pmatrix}u_1 & v_1 \ u_2 & v_2end{pmatrix}=u_1v_2-u_2v_1$$
so that $$Dg(u,v)=Dg(u_1,u_2,v_1,v_2)=begin{pmatrix}D_{u_1}g & D_{u_2}g & D_{v_1}g& D_{v_2}gend{pmatrix}=begin{pmatrix}v_2 & -v_1 & -u_2 & u_1end{pmatrix}$$
answered Nov 20 '18 at 20:54
smcc
4,297517
answered Nov 20 '18 at 20:54
smcc
4,297517
answered Nov 20 '18 at 20:54
smcc
4,297517
answered Nov 20 '18 at 20:54
smcc
4,297517
4,297517
For $ f $ why do you consider it as $f(x,y,u,v)$ anf not$f((x,y),(u,v))$
– Mary Maths
Nov 20 '18 at 21:07
See here: math.stackexchange.com/questions/338319/…
– smcc
Nov 20 '18 at 22:29
add a comment |
For $ f $ why do you consider it as $f(x,y,u,v)$ anf not$f((x,y),(u,v))$
– Mary Maths
Nov 20 '18 at 21:07
See here: math.stackexchange.com/questions/338319/…
– smcc
Nov 20 '18 at 22:29
For $ f $ why do you consider it as $f(x,y,u,v)$ anf not$f((x,y),(u,v))$
– Mary Maths
Nov 20 '18 at 21:07
For $ f $ why do you consider it as $f(x,y,u,v)$ anf not$f((x,y),(u,v))$
– Mary Maths
Nov 20 '18 at 21:07
See here: math.stackexchange.com/questions/338319/…
– smcc
Nov 20 '18 at 22:29
See here: math.stackexchange.com/questions/338319/…
– smcc
Nov 20 '18 at 22:29
add a comment |
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