Notation: gradient as vector field
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Consider the tangent space $T_pmathbb{R}^n$, and suppose ${big(frac{partial }{partial x^i}big)_p}$ is a basis. So my textbook says that the gradient of a function $f$, $fin C^infty(U)$, $Usubseteqmathbb{R}^n$ with $U$ open, is defined to be: $$text{grad}(f):=sum_{i=1}^n frac{partial f}{partial x_i}$$ but I am failing to see why it would not be $$:=sum_{i=1}^n frac{partial f}{partial x^i}frac{partial }{partial x^i}$$
so that evaluated at $pin U$, we get the gradient vector at $p$.
In other words, how is $frac{partial f}{partial x^i}$ a vector field? Thanks
differential-geometry smooth-manifolds differential-forms tangent-bundle
asked Jan 7 at 19:23
Ac711Ac711
274
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add a comment |
$begingroup$
Consider the tangent space $T_pmathbb{R}^n$, and suppose ${big(frac{partial }{partial x^i}big)_p}$ is a basis. So my textbook says that the gradient of a function $f$, $fin C^infty(U)$, $Usubseteqmathbb{R}^n$ with $U$ open, is defined to be: $$text{grad}(f):=sum_{i=1}^n frac{partial f}{partial x_i}$$ but I am failing to see why it would not be $$:=sum_{i=1}^n frac{partial f}{partial x^i}frac{partial }{partial x^i}$$
so that evaluated at $pin U$, we get the gradient vector at $p$.
In other words, how is $frac{partial f}{partial x^i}$ a vector field? Thanks
differential-geometry smooth-manifolds differential-forms tangent-bundle
asked Jan 7 at 19:23
Ac711Ac711
274
$endgroup$
add a comment |
$begingroup$
Consider the tangent space $T_pmathbb{R}^n$, and suppose ${big(frac{partial }{partial x^i}big)_p}$ is a basis. So my textbook says that the gradient of a function $f$, $fin C^infty(U)$, $Usubseteqmathbb{R}^n$ with $U$ open, is defined to be: $$text{grad}(f):=sum_{i=1}^n frac{partial f}{partial x_i}$$ but I am failing to see why it would not be $$:=sum_{i=1}^n frac{partial f}{partial x^i}frac{partial }{partial x^i}$$
so that evaluated at $pin U$, we get the gradient vector at $p$.
In other words, how is $frac{partial f}{partial x^i}$ a vector field? Thanks
differential-geometry smooth-manifolds differential-forms tangent-bundle
asked Jan 7 at 19:23
Ac711Ac711
274
$endgroup$
Consider the tangent space $T_pmathbb{R}^n$, and suppose ${big(frac{partial }{partial x^i}big)_p}$ is a basis. So my textbook says that the gradient of a function $f$, $fin C^infty(U)$, $Usubseteqmathbb{R}^n$ with $U$ open, is defined to be: $$text{grad}(f):=sum_{i=1}^n frac{partial f}{partial x_i}$$ but I am failing to see why it would not be $$:=sum_{i=1}^n frac{partial f}{partial x^i}frac{partial }{partial x^i}$$
so that evaluated at $pin U$, we get the gradient vector at $p$.
In other words, how is $frac{partial f}{partial x^i}$ a vector field? Thanks
differential-geometry smooth-manifolds differential-forms tangent-bundle
differential-geometry smooth-manifolds differential-forms tangent-bundle
asked Jan 7 at 19:23
Ac711Ac711
274
asked Jan 7 at 19:23
Ac711Ac711
274
edited Jan 7 at 19:43
Ac711
asked Jan 7 at 19:23
Ac711Ac711
274
asked Jan 7 at 19:23
Ac711Ac711
274
asked Jan 7 at 19:23
Ac711Ac711
274
274
add a comment |
add a comment |
1 Answer
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I agree with your interpretation. Given the conventions I am familiar with,
$$ nabla f=sum_{i=1}^n frac{partial f}{partial x^i}cdotfrac{partial}{partial x^i}.$$
In this way, $nabla f$ lives in the tangent bundle. The expression
$$ sum_{i=1}^n frac{partial f}{partial x^i}bigg|_pin mathbb{R}$$
which is probably not what we want our gradient to be.
answered Jan 7 at 19:43
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
$endgroup$
$begingroup$
$nablacdotnabla f$ is a sum of second partial derivatives, not a sum of first partial derivatives.
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– Neal
Jan 7 at 20:16
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Sorry you're totally right, not sure why I wrote that :-) Wasn't relevant anyways.
$endgroup$
– Antonios-Alexandros Robotis
Jan 7 at 20:16
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
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votes
$begingroup$
I agree with your interpretation. Given the conventions I am familiar with,
$$ nabla f=sum_{i=1}^n frac{partial f}{partial x^i}cdotfrac{partial}{partial x^i}.$$
In this way, $nabla f$ lives in the tangent bundle. The expression
$$ sum_{i=1}^n frac{partial f}{partial x^i}bigg|_pin mathbb{R}$$
which is probably not what we want our gradient to be.
answered Jan 7 at 19:43
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
$endgroup$
$begingroup$
$nablacdotnabla f$ is a sum of second partial derivatives, not a sum of first partial derivatives.
$endgroup$
– Neal
Jan 7 at 20:16
$begingroup$
Sorry you're totally right, not sure why I wrote that :-) Wasn't relevant anyways.
$endgroup$
– Antonios-Alexandros Robotis
Jan 7 at 20:16
add a comment |
$begingroup$
I agree with your interpretation. Given the conventions I am familiar with,
$$ nabla f=sum_{i=1}^n frac{partial f}{partial x^i}cdotfrac{partial}{partial x^i}.$$
In this way, $nabla f$ lives in the tangent bundle. The expression
$$ sum_{i=1}^n frac{partial f}{partial x^i}bigg|_pin mathbb{R}$$
which is probably not what we want our gradient to be.
answered Jan 7 at 19:43
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
$endgroup$
$begingroup$
$nablacdotnabla f$ is a sum of second partial derivatives, not a sum of first partial derivatives.
$endgroup$
– Neal
Jan 7 at 20:16
$begingroup$
Sorry you're totally right, not sure why I wrote that :-) Wasn't relevant anyways.
$endgroup$
– Antonios-Alexandros Robotis
Jan 7 at 20:16
add a comment |
$begingroup$
I agree with your interpretation. Given the conventions I am familiar with,
$$ nabla f=sum_{i=1}^n frac{partial f}{partial x^i}cdotfrac{partial}{partial x^i}.$$
In this way, $nabla f$ lives in the tangent bundle. The expression
$$ sum_{i=1}^n frac{partial f}{partial x^i}bigg|_pin mathbb{R}$$
which is probably not what we want our gradient to be.
answered Jan 7 at 19:43
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
$endgroup$
I agree with your interpretation. Given the conventions I am familiar with,
$$ nabla f=sum_{i=1}^n frac{partial f}{partial x^i}cdotfrac{partial}{partial x^i}.$$
In this way, $nabla f$ lives in the tangent bundle. The expression
$$ sum_{i=1}^n frac{partial f}{partial x^i}bigg|_pin mathbb{R}$$
which is probably not what we want our gradient to be.
answered Jan 7 at 19:43
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
edited Jan 7 at 20:17
answered Jan 7 at 19:43
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
answered Jan 7 at 19:43
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
answered Jan 7 at 19:43
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,71741640
9,71741640
$begingroup$
$nablacdotnabla f$ is a sum of second partial derivatives, not a sum of first partial derivatives.
$endgroup$
– Neal
Jan 7 at 20:16
$begingroup$
Sorry you're totally right, not sure why I wrote that :-) Wasn't relevant anyways.
$endgroup$
– Antonios-Alexandros Robotis
Jan 7 at 20:16
add a comment |
$begingroup$
$nablacdotnabla f$ is a sum of second partial derivatives, not a sum of first partial derivatives.
$endgroup$
– Neal
Jan 7 at 20:16
$begingroup$
Sorry you're totally right, not sure why I wrote that :-) Wasn't relevant anyways.
$endgroup$
– Antonios-Alexandros Robotis
Jan 7 at 20:16
$begingroup$
$nablacdotnabla f$ is a sum of second partial derivatives, not a sum of first partial derivatives.
$endgroup$
– Neal
Jan 7 at 20:16
$begingroup$
$nablacdotnabla f$ is a sum of second partial derivatives, not a sum of first partial derivatives.
$endgroup$
– Neal
Jan 7 at 20:16
$begingroup$
Sorry you're totally right, not sure why I wrote that :-) Wasn't relevant anyways.
$endgroup$
– Antonios-Alexandros Robotis
Jan 7 at 20:16
$begingroup$
Sorry you're totally right, not sure why I wrote that :-) Wasn't relevant anyways.
$endgroup$
– Antonios-Alexandros Robotis
Jan 7 at 20:16
add a comment |
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