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How can I determine the double derivative of a function $f:R^n to R$?
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A function $f:R^n to R$ belong to $C^2$. So $Df$ will be a matrix of order "1 by n". So now the range set of function $Df$ will be all row matrix of order "1 by n" while domain set will remain same(set of all column matrix of order "n by 1"). Then how can I handle this function? How would I determine $D^2f$? Can anyone please help me out?
calculus multivariable-calculus partial-derivative
asked Jul 24 at 15:15
cmi
6249
 |Â
show 10 more comments
up vote
0
down vote
favorite
A function $f:R^n to R$ belong to $C^2$. So $Df$ will be a matrix of order "1 by n". So now the range set of function $Df$ will be all row matrix of order "1 by n" while domain set will remain same(set of all column matrix of order "n by 1"). Then how can I handle this function? How would I determine $D^2f$? Can anyone please help me out?
calculus multivariable-calculus partial-derivative
asked Jul 24 at 15:15
cmi
6249
1
Have you ever done it for some $f:mathbb R^2 to mathbb R$? In any case, see the hessian matrix
– Andres Mejia
Jul 24 at 15:28
1
for functions $f:mathbb R^m to mathbb R^n$ you will probably need the concept of tensors
– Andres Mejia
Jul 24 at 15:30
1
$Df$ is a "linear approximation" of a function $mathbb R^n to mathbb R$. Try to compute if for $f(x,y)=x+y$. This is the "trivial case" where the function is already linear. Try then to compute $f(1,1)+Df(1,1)(mathbfx-(1,1))$. Next you can try a function $f(x)=x^2+y^2$ and do the same computation. Evaluate the difference between $f$ and the expression I give for $mathbfx=(1,1.1)$ and $mathbf x = (1.1,1)$. I'm not really a teacher, so I don't know the best way to internalize this kind of thing. Try here or Hubbard's bk.
– Andres Mejia
Jul 24 at 15:40
1
There is the following property of the standard derivative: $f(a+Delta x) approx f(a)+f^prime(a) cdot (Delta x)$. In this sense, $f^prime(a)$ is the best linear approximation near $a$ (since a one dimensional matrix is multiplication. The situation is similar in higher dimensions, but linear transformations now take the form of matrices. In your case $n times 1$ matrices applied to $n$-vectors
– Andres Mejia
Jul 24 at 15:43
1
The trouble with the second derivative is that in each co-ordinate direction, you now have to see how an $n times 1$ "row" is behaving, so for each $x_i$ you get a column, and hence at the end an $n times n$ matrix.
– Andres Mejia
Jul 24 at 15:45
 |Â
show 10 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A function $f:R^n to R$ belong to $C^2$. So $Df$ will be a matrix of order "1 by n". So now the range set of function $Df$ will be all row matrix of order "1 by n" while domain set will remain same(set of all column matrix of order "n by 1"). Then how can I handle this function? How would I determine $D^2f$? Can anyone please help me out?
calculus multivariable-calculus partial-derivative
asked Jul 24 at 15:15
cmi
6249
A function $f:R^n to R$ belong to $C^2$. So $Df$ will be a matrix of order "1 by n". So now the range set of function $Df$ will be all row matrix of order "1 by n" while domain set will remain same(set of all column matrix of order "n by 1"). Then how can I handle this function? How would I determine $D^2f$? Can anyone please help me out?
calculus multivariable-calculus partial-derivative
asked Jul 24 at 15:15
cmi
6249
asked Jul 24 at 15:15
cmi
6249
asked Jul 24 at 15:15
cmi
6249
asked Jul 24 at 15:15
cmi
6249
6249
1
Have you ever done it for some $f:mathbb R^2 to mathbb R$? In any case, see the hessian matrix
– Andres Mejia
Jul 24 at 15:28
1
for functions $f:mathbb R^m to mathbb R^n$ you will probably need the concept of tensors
– Andres Mejia
Jul 24 at 15:30
1
$Df$ is a "linear approximation" of a function $mathbb R^n to mathbb R$. Try to compute if for $f(x,y)=x+y$. This is the "trivial case" where the function is already linear. Try then to compute $f(1,1)+Df(1,1)(mathbfx-(1,1))$. Next you can try a function $f(x)=x^2+y^2$ and do the same computation. Evaluate the difference between $f$ and the expression I give for $mathbfx=(1,1.1)$ and $mathbf x = (1.1,1)$. I'm not really a teacher, so I don't know the best way to internalize this kind of thing. Try here or Hubbard's bk.
– Andres Mejia
Jul 24 at 15:40
1
There is the following property of the standard derivative: $f(a+Delta x) approx f(a)+f^prime(a) cdot (Delta x)$. In this sense, $f^prime(a)$ is the best linear approximation near $a$ (since a one dimensional matrix is multiplication. The situation is similar in higher dimensions, but linear transformations now take the form of matrices. In your case $n times 1$ matrices applied to $n$-vectors
– Andres Mejia
Jul 24 at 15:43
1
The trouble with the second derivative is that in each co-ordinate direction, you now have to see how an $n times 1$ "row" is behaving, so for each $x_i$ you get a column, and hence at the end an $n times n$ matrix.
– Andres Mejia
Jul 24 at 15:45
 |Â
show 10 more comments
1
Have you ever done it for some $f:mathbb R^2 to mathbb R$? In any case, see the hessian matrix
– Andres Mejia
Jul 24 at 15:28
1
for functions $f:mathbb R^m to mathbb R^n$ you will probably need the concept of tensors
– Andres Mejia
Jul 24 at 15:30
1
$Df$ is a "linear approximation" of a function $mathbb R^n to mathbb R$. Try to compute if for $f(x,y)=x+y$. This is the "trivial case" where the function is already linear. Try then to compute $f(1,1)+Df(1,1)(mathbfx-(1,1))$. Next you can try a function $f(x)=x^2+y^2$ and do the same computation. Evaluate the difference between $f$ and the expression I give for $mathbfx=(1,1.1)$ and $mathbf x = (1.1,1)$. I'm not really a teacher, so I don't know the best way to internalize this kind of thing. Try here or Hubbard's bk.
– Andres Mejia
Jul 24 at 15:40
1
There is the following property of the standard derivative: $f(a+Delta x) approx f(a)+f^prime(a) cdot (Delta x)$. In this sense, $f^prime(a)$ is the best linear approximation near $a$ (since a one dimensional matrix is multiplication. The situation is similar in higher dimensions, but linear transformations now take the form of matrices. In your case $n times 1$ matrices applied to $n$-vectors
– Andres Mejia
Jul 24 at 15:43
1
The trouble with the second derivative is that in each co-ordinate direction, you now have to see how an $n times 1$ "row" is behaving, so for each $x_i$ you get a column, and hence at the end an $n times n$ matrix.
– Andres Mejia
Jul 24 at 15:45
1
1
Have you ever done it for some $f:mathbb R^2 to mathbb R$? In any case, see the hessian matrix
– Andres Mejia
Jul 24 at 15:28
Have you ever done it for some $f:mathbb R^2 to mathbb R$? In any case, see the hessian matrix
– Andres Mejia
Jul 24 at 15:28
1
1
for functions $f:mathbb R^m to mathbb R^n$ you will probably need the concept of tensors
– Andres Mejia
Jul 24 at 15:30
for functions $f:mathbb R^m to mathbb R^n$ you will probably need the concept of tensors
– Andres Mejia
Jul 24 at 15:30
1
1
$Df$ is a "linear approximation" of a function $mathbb R^n to mathbb R$. Try to compute if for $f(x,y)=x+y$. This is the "trivial case" where the function is already linear. Try then to compute $f(1,1)+Df(1,1)(mathbfx-(1,1))$. Next you can try a function $f(x)=x^2+y^2$ and do the same computation. Evaluate the difference between $f$ and the expression I give for $mathbfx=(1,1.1)$ and $mathbf x = (1.1,1)$. I'm not really a teacher, so I don't know the best way to internalize this kind of thing. Try here or Hubbard's bk.
– Andres Mejia
Jul 24 at 15:40
$Df$ is a "linear approximation" of a function $mathbb R^n to mathbb R$. Try to compute if for $f(x,y)=x+y$. This is the "trivial case" where the function is already linear. Try then to compute $f(1,1)+Df(1,1)(mathbfx-(1,1))$. Next you can try a function $f(x)=x^2+y^2$ and do the same computation. Evaluate the difference between $f$ and the expression I give for $mathbfx=(1,1.1)$ and $mathbf x = (1.1,1)$. I'm not really a teacher, so I don't know the best way to internalize this kind of thing. Try here or Hubbard's bk.
– Andres Mejia
Jul 24 at 15:40
1
1
There is the following property of the standard derivative: $f(a+Delta x) approx f(a)+f^prime(a) cdot (Delta x)$. In this sense, $f^prime(a)$ is the best linear approximation near $a$ (since a one dimensional matrix is multiplication. The situation is similar in higher dimensions, but linear transformations now take the form of matrices. In your case $n times 1$ matrices applied to $n$-vectors
– Andres Mejia
Jul 24 at 15:43
There is the following property of the standard derivative: $f(a+Delta x) approx f(a)+f^prime(a) cdot (Delta x)$. In this sense, $f^prime(a)$ is the best linear approximation near $a$ (since a one dimensional matrix is multiplication. The situation is similar in higher dimensions, but linear transformations now take the form of matrices. In your case $n times 1$ matrices applied to $n$-vectors
– Andres Mejia
Jul 24 at 15:43
1
1
The trouble with the second derivative is that in each co-ordinate direction, you now have to see how an $n times 1$ "row" is behaving, so for each $x_i$ you get a column, and hence at the end an $n times n$ matrix.
– Andres Mejia
Jul 24 at 15:45
The trouble with the second derivative is that in each co-ordinate direction, you now have to see how an $n times 1$ "row" is behaving, so for each $x_i$ you get a column, and hence at the end an $n times n$ matrix.
– Andres Mejia
Jul 24 at 15:45
 |Â
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1
Have you ever done it for some $f:mathbb R^2 to mathbb R$? In any case, see the hessian matrix
– Andres Mejia
Jul 24 at 15:28
1
for functions $f:mathbb R^m to mathbb R^n$ you will probably need the concept of tensors
– Andres Mejia
Jul 24 at 15:30
1
$Df$ is a "linear approximation" of a function $mathbb R^n to mathbb R$. Try to compute if for $f(x,y)=x+y$. This is the "trivial case" where the function is already linear. Try then to compute $f(1,1)+Df(1,1)(mathbfx-(1,1))$. Next you can try a function $f(x)=x^2+y^2$ and do the same computation. Evaluate the difference between $f$ and the expression I give for $mathbfx=(1,1.1)$ and $mathbf x = (1.1,1)$. I'm not really a teacher, so I don't know the best way to internalize this kind of thing. Try here or Hubbard's bk.
– Andres Mejia
Jul 24 at 15:40
1
There is the following property of the standard derivative: $f(a+Delta x) approx f(a)+f^prime(a) cdot (Delta x)$. In this sense, $f^prime(a)$ is the best linear approximation near $a$ (since a one dimensional matrix is multiplication. The situation is similar in higher dimensions, but linear transformations now take the form of matrices. In your case $n times 1$ matrices applied to $n$-vectors
– Andres Mejia
Jul 24 at 15:43
1
The trouble with the second derivative is that in each co-ordinate direction, you now have to see how an $n times 1$ "row" is behaving, so for each $x_i$ you get a column, and hence at the end an $n times n$ matrix.
– Andres Mejia
Jul 24 at 15:45