Derivative of a function defined in terms of another function
$begingroup$
Let $F: mathbf{R^2} rightarrow mathbf{R}$ have continuous second order partial derivatives. Assume its gradient is $nabla f(0,0) = (1,2)$ and its Hessian matrix is
$$begin{pmatrix}
1 & 2 \
2 &4
end{pmatrix} = nabla ^2 f $$
Define $phi (t) = f(t, 2t)$. Find $phi ^ prime (0)$ and $phi ^ {prime prime} (0)$.
Any hints? I know how to do this when $f$ is explicitly defined, but not when just given the gradient/Hessian at a certain point.
real-analysis
asked Dec 17 '18 at 16:01
hopelessundergradhopelessundergrad
292
$endgroup$
add a comment |
$begingroup$
Let $F: mathbf{R^2} rightarrow mathbf{R}$ have continuous second order partial derivatives. Assume its gradient is $nabla f(0,0) = (1,2)$ and its Hessian matrix is
$$begin{pmatrix}
1 & 2 \
2 &4
end{pmatrix} = nabla ^2 f $$
Define $phi (t) = f(t, 2t)$. Find $phi ^ prime (0)$ and $phi ^ {prime prime} (0)$.
Any hints? I know how to do this when $f$ is explicitly defined, but not when just given the gradient/Hessian at a certain point.
real-analysis
asked Dec 17 '18 at 16:01
hopelessundergradhopelessundergrad
292
$endgroup$
add a comment |
$begingroup$
Let $F: mathbf{R^2} rightarrow mathbf{R}$ have continuous second order partial derivatives. Assume its gradient is $nabla f(0,0) = (1,2)$ and its Hessian matrix is
$$begin{pmatrix}
1 & 2 \
2 &4
end{pmatrix} = nabla ^2 f $$
Define $phi (t) = f(t, 2t)$. Find $phi ^ prime (0)$ and $phi ^ {prime prime} (0)$.
Any hints? I know how to do this when $f$ is explicitly defined, but not when just given the gradient/Hessian at a certain point.
real-analysis
asked Dec 17 '18 at 16:01
hopelessundergradhopelessundergrad
292
$endgroup$
Let $F: mathbf{R^2} rightarrow mathbf{R}$ have continuous second order partial derivatives. Assume its gradient is $nabla f(0,0) = (1,2)$ and its Hessian matrix is
$$begin{pmatrix}
1 & 2 \
2 &4
end{pmatrix} = nabla ^2 f $$
Define $phi (t) = f(t, 2t)$. Find $phi ^ prime (0)$ and $phi ^ {prime prime} (0)$.
Any hints? I know how to do this when $f$ is explicitly defined, but not when just given the gradient/Hessian at a certain point.
real-analysis
real-analysis
asked Dec 17 '18 at 16:01
hopelessundergradhopelessundergrad
292
asked Dec 17 '18 at 16:01
hopelessundergradhopelessundergrad
292
asked Dec 17 '18 at 16:01
hopelessundergradhopelessundergrad
292
asked Dec 17 '18 at 16:01
hopelessundergradhopelessundergrad
292
asked Dec 17 '18 at 16:01
hopelessundergradhopelessundergrad
292
292
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Just aplpy chain rule, nota that $phi=fcirc h$, with $h:mathbb{R}to mathbb{R}^{2}$, $h(t)=(t,2t)$. Then
begin{equation} phi '(t)=nabla f (h(t)) h'(t) hspace{3cm}(1)
end{equation}
then put $t=0$ (h' is a column vector, (1,2) as a column)
$$ phi '(0)=nabla f (h(0))h'(0) $$
$$ phi '(0)=nabla f (0,0)h'(0) $$
$$ phi '(0)=(1,2)(1,2)^{T} $$
$$ phi '(0)=1+4=5$$
Then derive again in (1) (which can be writenn as $[(nabla fcirc h)(t)](1,2)^{T}$ to use the chain rule again ) to find $phi''$ taking into account that $h'(t)=(1,2)^{T}$ (constant):
$$phi ''(t)=[nabla^{2} f (h(t))h'(t) ]left[{begin{array}{c}
1 \
2 \
end{array} } right] $$
so puting $t=0$, we have:
$$phi ''(0)=[nabla^{2} f (h(0))h'(0)] left[ {begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi ''(0)=[nabla^{2} f (0,0) h'(0)] left[ {begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi ''(0)={ left[ {begin{array}{cc}
1 & 2 \
2 & 4 \
end{array} } right] left[ {begin{array}{c}
1 \
2 \
end{array} } right] } left[ {begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi ''(0)=left[ {begin{array}{c}
5 \
10 \
end{array} } right] left[{begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi''(0)=25.$$
answered Dec 17 '18 at 16:08
Gabriel PalauGabriel Palau
1126
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Just aplpy chain rule, nota that $phi=fcirc h$, with $h:mathbb{R}to mathbb{R}^{2}$, $h(t)=(t,2t)$. Then
begin{equation} phi '(t)=nabla f (h(t)) h'(t) hspace{3cm}(1)
end{equation}
then put $t=0$ (h' is a column vector, (1,2) as a column)
$$ phi '(0)=nabla f (h(0))h'(0) $$
$$ phi '(0)=nabla f (0,0)h'(0) $$
$$ phi '(0)=(1,2)(1,2)^{T} $$
$$ phi '(0)=1+4=5$$
Then derive again in (1) (which can be writenn as $[(nabla fcirc h)(t)](1,2)^{T}$ to use the chain rule again ) to find $phi''$ taking into account that $h'(t)=(1,2)^{T}$ (constant):
$$phi ''(t)=[nabla^{2} f (h(t))h'(t) ]left[{begin{array}{c}
1 \
2 \
end{array} } right] $$
so puting $t=0$, we have:
$$phi ''(0)=[nabla^{2} f (h(0))h'(0)] left[ {begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi ''(0)=[nabla^{2} f (0,0) h'(0)] left[ {begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi ''(0)={ left[ {begin{array}{cc}
1 & 2 \
2 & 4 \
end{array} } right] left[ {begin{array}{c}
1 \
2 \
end{array} } right] } left[ {begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi ''(0)=left[ {begin{array}{c}
5 \
10 \
end{array} } right] left[{begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi''(0)=25.$$
answered Dec 17 '18 at 16:08
Gabriel PalauGabriel Palau
1126
$endgroup$
add a comment |
$begingroup$
Just aplpy chain rule, nota that $phi=fcirc h$, with $h:mathbb{R}to mathbb{R}^{2}$, $h(t)=(t,2t)$. Then
begin{equation} phi '(t)=nabla f (h(t)) h'(t) hspace{3cm}(1)
end{equation}
then put $t=0$ (h' is a column vector, (1,2) as a column)
$$ phi '(0)=nabla f (h(0))h'(0) $$
$$ phi '(0)=nabla f (0,0)h'(0) $$
$$ phi '(0)=(1,2)(1,2)^{T} $$
$$ phi '(0)=1+4=5$$
Then derive again in (1) (which can be writenn as $[(nabla fcirc h)(t)](1,2)^{T}$ to use the chain rule again ) to find $phi''$ taking into account that $h'(t)=(1,2)^{T}$ (constant):
$$phi ''(t)=[nabla^{2} f (h(t))h'(t) ]left[{begin{array}{c}
1 \
2 \
end{array} } right] $$
so puting $t=0$, we have:
$$phi ''(0)=[nabla^{2} f (h(0))h'(0)] left[ {begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi ''(0)=[nabla^{2} f (0,0) h'(0)] left[ {begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi ''(0)={ left[ {begin{array}{cc}
1 & 2 \
2 & 4 \
end{array} } right] left[ {begin{array}{c}
1 \
2 \
end{array} } right] } left[ {begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi ''(0)=left[ {begin{array}{c}
5 \
10 \
end{array} } right] left[{begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi''(0)=25.$$
answered Dec 17 '18 at 16:08
Gabriel PalauGabriel Palau
1126
$endgroup$
add a comment |
$begingroup$
Just aplpy chain rule, nota that $phi=fcirc h$, with $h:mathbb{R}to mathbb{R}^{2}$, $h(t)=(t,2t)$. Then
begin{equation} phi '(t)=nabla f (h(t)) h'(t) hspace{3cm}(1)
end{equation}
then put $t=0$ (h' is a column vector, (1,2) as a column)
$$ phi '(0)=nabla f (h(0))h'(0) $$
$$ phi '(0)=nabla f (0,0)h'(0) $$
$$ phi '(0)=(1,2)(1,2)^{T} $$
$$ phi '(0)=1+4=5$$
Then derive again in (1) (which can be writenn as $[(nabla fcirc h)(t)](1,2)^{T}$ to use the chain rule again ) to find $phi''$ taking into account that $h'(t)=(1,2)^{T}$ (constant):
$$phi ''(t)=[nabla^{2} f (h(t))h'(t) ]left[{begin{array}{c}
1 \
2 \
end{array} } right] $$
so puting $t=0$, we have:
$$phi ''(0)=[nabla^{2} f (h(0))h'(0)] left[ {begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi ''(0)=[nabla^{2} f (0,0) h'(0)] left[ {begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi ''(0)={ left[ {begin{array}{cc}
1 & 2 \
2 & 4 \
end{array} } right] left[ {begin{array}{c}
1 \
2 \
end{array} } right] } left[ {begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi ''(0)=left[ {begin{array}{c}
5 \
10 \
end{array} } right] left[{begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi''(0)=25.$$
answered Dec 17 '18 at 16:08
Gabriel PalauGabriel Palau
1126
$endgroup$
Just aplpy chain rule, nota that $phi=fcirc h$, with $h:mathbb{R}to mathbb{R}^{2}$, $h(t)=(t,2t)$. Then
begin{equation} phi '(t)=nabla f (h(t)) h'(t) hspace{3cm}(1)
end{equation}
then put $t=0$ (h' is a column vector, (1,2) as a column)
$$ phi '(0)=nabla f (h(0))h'(0) $$
$$ phi '(0)=nabla f (0,0)h'(0) $$
$$ phi '(0)=(1,2)(1,2)^{T} $$
$$ phi '(0)=1+4=5$$
Then derive again in (1) (which can be writenn as $[(nabla fcirc h)(t)](1,2)^{T}$ to use the chain rule again ) to find $phi''$ taking into account that $h'(t)=(1,2)^{T}$ (constant):
$$phi ''(t)=[nabla^{2} f (h(t))h'(t) ]left[{begin{array}{c}
1 \
2 \
end{array} } right] $$
so puting $t=0$, we have:
$$phi ''(0)=[nabla^{2} f (h(0))h'(0)] left[ {begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi ''(0)=[nabla^{2} f (0,0) h'(0)] left[ {begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi ''(0)={ left[ {begin{array}{cc}
1 & 2 \
2 & 4 \
end{array} } right] left[ {begin{array}{c}
1 \
2 \
end{array} } right] } left[ {begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi ''(0)=left[ {begin{array}{c}
5 \
10 \
end{array} } right] left[{begin{array}{c}
1 \
2 \
end{array} } right] $$
$$phi''(0)=25.$$
answered Dec 17 '18 at 16:08
Gabriel PalauGabriel Palau
1126
edited Dec 17 '18 at 22:36
answered Dec 17 '18 at 16:08
Gabriel PalauGabriel Palau
1126
answered Dec 17 '18 at 16:08
Gabriel PalauGabriel Palau
1126
answered Dec 17 '18 at 16:08
Gabriel PalauGabriel Palau
1126
1126
add a comment |
add a comment |
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