Derivative in mean
I have one question about calculation rules for means.
Say I have an equation of the form
begin{align*}
langle H(a,b) rangle = langle H_1(a,b) rangle + langle H_2(a,b)rangle
end{align*}
where $langle ... rangle$ is a mean (physical, i.e. integral over time / space).
Is then
begin{align*}
langle frac{partial H(a,b)}{partial a} rangle = langle frac{partial H_1(a,b)}{partial a} rangle + langle frac{partial H_2(a,b)}{partial a}rangle
end{align*}
also true? (a is a variable, but not the one considered in the mean integral, i.e not time / space).
statistics derivatives means
asked Nov 21 at 21:04
Q.stion
31
add a comment |
I have one question about calculation rules for means.
Say I have an equation of the form
begin{align*}
langle H(a,b) rangle = langle H_1(a,b) rangle + langle H_2(a,b)rangle
end{align*}
where $langle ... rangle$ is a mean (physical, i.e. integral over time / space).
Is then
begin{align*}
langle frac{partial H(a,b)}{partial a} rangle = langle frac{partial H_1(a,b)}{partial a} rangle + langle frac{partial H_2(a,b)}{partial a}rangle
end{align*}
also true? (a is a variable, but not the one considered in the mean integral, i.e not time / space).
statistics derivatives means
asked Nov 21 at 21:04
Q.stion
31
add a comment |
I have one question about calculation rules for means.
Say I have an equation of the form
begin{align*}
langle H(a,b) rangle = langle H_1(a,b) rangle + langle H_2(a,b)rangle
end{align*}
where $langle ... rangle$ is a mean (physical, i.e. integral over time / space).
Is then
begin{align*}
langle frac{partial H(a,b)}{partial a} rangle = langle frac{partial H_1(a,b)}{partial a} rangle + langle frac{partial H_2(a,b)}{partial a}rangle
end{align*}
also true? (a is a variable, but not the one considered in the mean integral, i.e not time / space).
statistics derivatives means
asked Nov 21 at 21:04
Q.stion
31
I have one question about calculation rules for means.
Say I have an equation of the form
begin{align*}
langle H(a,b) rangle = langle H_1(a,b) rangle + langle H_2(a,b)rangle
end{align*}
where $langle ... rangle$ is a mean (physical, i.e. integral over time / space).
Is then
begin{align*}
langle frac{partial H(a,b)}{partial a} rangle = langle frac{partial H_1(a,b)}{partial a} rangle + langle frac{partial H_2(a,b)}{partial a}rangle
end{align*}
also true? (a is a variable, but not the one considered in the mean integral, i.e not time / space).
statistics derivatives means
statistics derivatives means
asked Nov 21 at 21:04
Q.stion
31
asked Nov 21 at 21:04
Q.stion
31
asked Nov 21 at 21:04
Q.stion
31
asked Nov 21 at 21:04
Q.stion
31
asked Nov 21 at 21:04
Q.stion
31
31
add a comment |
add a comment |
1 Answer
1
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oldest
votes
Sure, if
$$
langle f(a)rangle = int{rm d^3}{bf x}~{rm d}t ~f(a, {bf x}, t)
$$
then
$$
frac{{rm d}}{{rm d}a}langle f(a)rangle = frac{{rm d}}{{rm d}a} int{rm d^3}{bf x}~{rm d}t ~f(a, {bf x}, t) = int{rm d^3}{bf x}~{rm d}t ~frac{partial}{partial a}f(a, {bf x}, t)
$$
answered Nov 21 at 22:01
caverac
13k21028
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sure, if
$$
langle f(a)rangle = int{rm d^3}{bf x}~{rm d}t ~f(a, {bf x}, t)
$$
then
$$
frac{{rm d}}{{rm d}a}langle f(a)rangle = frac{{rm d}}{{rm d}a} int{rm d^3}{bf x}~{rm d}t ~f(a, {bf x}, t) = int{rm d^3}{bf x}~{rm d}t ~frac{partial}{partial a}f(a, {bf x}, t)
$$
answered Nov 21 at 22:01
caverac
13k21028
add a comment |
Sure, if
$$
langle f(a)rangle = int{rm d^3}{bf x}~{rm d}t ~f(a, {bf x}, t)
$$
then
$$
frac{{rm d}}{{rm d}a}langle f(a)rangle = frac{{rm d}}{{rm d}a} int{rm d^3}{bf x}~{rm d}t ~f(a, {bf x}, t) = int{rm d^3}{bf x}~{rm d}t ~frac{partial}{partial a}f(a, {bf x}, t)
$$
answered Nov 21 at 22:01
caverac
13k21028
add a comment |
Sure, if
$$
langle f(a)rangle = int{rm d^3}{bf x}~{rm d}t ~f(a, {bf x}, t)
$$
then
$$
frac{{rm d}}{{rm d}a}langle f(a)rangle = frac{{rm d}}{{rm d}a} int{rm d^3}{bf x}~{rm d}t ~f(a, {bf x}, t) = int{rm d^3}{bf x}~{rm d}t ~frac{partial}{partial a}f(a, {bf x}, t)
$$
answered Nov 21 at 22:01
caverac
13k21028
Sure, if
$$
langle f(a)rangle = int{rm d^3}{bf x}~{rm d}t ~f(a, {bf x}, t)
$$
then
$$
frac{{rm d}}{{rm d}a}langle f(a)rangle = frac{{rm d}}{{rm d}a} int{rm d^3}{bf x}~{rm d}t ~f(a, {bf x}, t) = int{rm d^3}{bf x}~{rm d}t ~frac{partial}{partial a}f(a, {bf x}, t)
$$
answered Nov 21 at 22:01
caverac
13k21028
answered Nov 21 at 22:01
caverac
13k21028
answered Nov 21 at 22:01
caverac
13k21028
answered Nov 21 at 22:01
caverac
13k21028
13k21028
add a comment |
add a comment |
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