Mixing
location of vector Taylor expansion remainder
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I know in 1-dimension case, $$f(x)=f(x_0)+f'(tildex_0)(x-x_0)$$ where $tildex_0$ lies between $x$ and $x_0$.
What about in k-dimension case? $$mathbff(mathbfx)=mathbff(mathbfx_0)+mathbff'(tildemathbfx_0)(mathbfx-mathbfx_0)$$
Does $mathbftildex_0$ still lies between $mathbfx$ and $mathbfx_0$?
If $mathbfxrightarrowmathbfx_0$, does $mathbftildex_0rightarrowmathbfx_0$?
taylor-expansion vector-analysis
edited Jul 25 at 22:02
Bernard
110k635103
asked Jul 25 at 22:01
Xingzhe He
63
add a comment |Â
up vote
0
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I know in 1-dimension case, $$f(x)=f(x_0)+f'(tildex_0)(x-x_0)$$ where $tildex_0$ lies between $x$ and $x_0$.
What about in k-dimension case? $$mathbff(mathbfx)=mathbff(mathbfx_0)+mathbff'(tildemathbfx_0)(mathbfx-mathbfx_0)$$
Does $mathbftildex_0$ still lies between $mathbfx$ and $mathbfx_0$?
If $mathbfxrightarrowmathbfx_0$, does $mathbftildex_0rightarrowmathbfx_0$?
taylor-expansion vector-analysis
edited Jul 25 at 22:02
Bernard
110k635103
asked Jul 25 at 22:01
Xingzhe He
63
1
The 1D case is a form of the mean value theorem. What do you expect for $k$ dimensions? What will happen to the derivative? What could "between" mean, as $x$ and $x_0$ are now points in $mathbbR^k$.
– mvw
Jul 25 at 22:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know in 1-dimension case, $$f(x)=f(x_0)+f'(tildex_0)(x-x_0)$$ where $tildex_0$ lies between $x$ and $x_0$.
What about in k-dimension case? $$mathbff(mathbfx)=mathbff(mathbfx_0)+mathbff'(tildemathbfx_0)(mathbfx-mathbfx_0)$$
Does $mathbftildex_0$ still lies between $mathbfx$ and $mathbfx_0$?
If $mathbfxrightarrowmathbfx_0$, does $mathbftildex_0rightarrowmathbfx_0$?
taylor-expansion vector-analysis
edited Jul 25 at 22:02
Bernard
110k635103
asked Jul 25 at 22:01
Xingzhe He
63
I know in 1-dimension case, $$f(x)=f(x_0)+f'(tildex_0)(x-x_0)$$ where $tildex_0$ lies between $x$ and $x_0$.
What about in k-dimension case? $$mathbff(mathbfx)=mathbff(mathbfx_0)+mathbff'(tildemathbfx_0)(mathbfx-mathbfx_0)$$
Does $mathbftildex_0$ still lies between $mathbfx$ and $mathbfx_0$?
If $mathbfxrightarrowmathbfx_0$, does $mathbftildex_0rightarrowmathbfx_0$?
taylor-expansion vector-analysis
edited Jul 25 at 22:02
Bernard
110k635103
asked Jul 25 at 22:01
Xingzhe He
63
edited Jul 25 at 22:02
Bernard
110k635103
edited Jul 25 at 22:02
Bernard
110k635103
edited Jul 25 at 22:02
Bernard
110k635103
110k635103
asked Jul 25 at 22:01
Xingzhe He
63
asked Jul 25 at 22:01
Xingzhe He
63
asked Jul 25 at 22:01
Xingzhe He
63
63
1
The 1D case is a form of the mean value theorem. What do you expect for $k$ dimensions? What will happen to the derivative? What could "between" mean, as $x$ and $x_0$ are now points in $mathbbR^k$.
– mvw
Jul 25 at 22:12
add a comment |Â
1
The 1D case is a form of the mean value theorem. What do you expect for $k$ dimensions? What will happen to the derivative? What could "between" mean, as $x$ and $x_0$ are now points in $mathbbR^k$.
– mvw
Jul 25 at 22:12
1
1
The 1D case is a form of the mean value theorem. What do you expect for $k$ dimensions? What will happen to the derivative? What could "between" mean, as $x$ and $x_0$ are now points in $mathbbR^k$.
– mvw
Jul 25 at 22:12
The 1D case is a form of the mean value theorem. What do you expect for $k$ dimensions? What will happen to the derivative? What could "between" mean, as $x$ and $x_0$ are now points in $mathbbR^k$.
– mvw
Jul 25 at 22:12
add a comment |Â
1 Answer
1
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Consider $mathbb C$ which can be treated as $mathbb R^2$. Consider teh function $f(z)=e^z$. $e^0-e^2pi i=0$ cannot be written as $2pi i e^z$ for any $z$.
answered Jul 25 at 23:31
Kavi Rama Murthy
20k2829
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Consider $mathbb C$ which can be treated as $mathbb R^2$. Consider teh function $f(z)=e^z$. $e^0-e^2pi i=0$ cannot be written as $2pi i e^z$ for any $z$.
answered Jul 25 at 23:31
Kavi Rama Murthy
20k2829
add a comment |Â
up vote
0
down vote
Consider $mathbb C$ which can be treated as $mathbb R^2$. Consider teh function $f(z)=e^z$. $e^0-e^2pi i=0$ cannot be written as $2pi i e^z$ for any $z$.
answered Jul 25 at 23:31
Kavi Rama Murthy
20k2829
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Consider $mathbb C$ which can be treated as $mathbb R^2$. Consider teh function $f(z)=e^z$. $e^0-e^2pi i=0$ cannot be written as $2pi i e^z$ for any $z$.
answered Jul 25 at 23:31
Kavi Rama Murthy
20k2829
Consider $mathbb C$ which can be treated as $mathbb R^2$. Consider teh function $f(z)=e^z$. $e^0-e^2pi i=0$ cannot be written as $2pi i e^z$ for any $z$.
answered Jul 25 at 23:31
Kavi Rama Murthy
20k2829
answered Jul 25 at 23:31
Kavi Rama Murthy
20k2829
answered Jul 25 at 23:31
Kavi Rama Murthy
20k2829
answered Jul 25 at 23:31
Kavi Rama Murthy
20k2829
20k2829
add a comment |Â
add a comment |Â
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1
The 1D case is a form of the mean value theorem. What do you expect for $k$ dimensions? What will happen to the derivative? What could "between" mean, as $x$ and $x_0$ are now points in $mathbbR^k$.
– mvw
Jul 25 at 22:12