Mixing
Is $ f(x):= frac{x}{ |x|^n }$ a gradient field?
$begingroup$
for $ 2 leq n $ let be
$ f: mathbb{R}^n backslash { 0 } rightarrow mathbb{R}^n $
$ f(x):= frac{x}{|x|^n} $ , $ xin mathbb{R}^n backslash { 0 } $
Is f a gradient field?
A vectorfield $ f $ ist a gradient field, if there is a function
$ f= nabla v $
$ left[ nabla f (x) = (frac{ partial f}{ partial x_1} (x),...,frac{ partial f}{ partial x_n} (x)) right] $
So, I need to find a function $v$ so that $ f= nabla v $ right?
If I look for the case $n=2 $
so $ f(x):= frac{x}{|x|^2} in mathbb{R}^n backslash { 0 } $
let be $ partial_j v(x)= frac{x_j}{|x|}$
then $v(x)= ln(|x|) $ and
$$ partial_j v(x)= frac{1}{|x|} frac{x_j}{|x|} = frac{x_j}{|x^2|} $$
Is this the right idea?
If yes, how can I proceed for larger $n$?
If no, what would you suggest?
Appreciate any help !
real-analysis multivariable-calculus vector-analysis
asked Jan 14 at 12:21
constant94constant94
6310
$endgroup$
add a comment |
$begingroup$
for $ 2 leq n $ let be
$ f: mathbb{R}^n backslash { 0 } rightarrow mathbb{R}^n $
$ f(x):= frac{x}{|x|^n} $ , $ xin mathbb{R}^n backslash { 0 } $
Is f a gradient field?
A vectorfield $ f $ ist a gradient field, if there is a function
$ f= nabla v $
$ left[ nabla f (x) = (frac{ partial f}{ partial x_1} (x),...,frac{ partial f}{ partial x_n} (x)) right] $
So, I need to find a function $v$ so that $ f= nabla v $ right?
If I look for the case $n=2 $
so $ f(x):= frac{x}{|x|^2} in mathbb{R}^n backslash { 0 } $
let be $ partial_j v(x)= frac{x_j}{|x|}$
then $v(x)= ln(|x|) $ and
$$ partial_j v(x)= frac{1}{|x|} frac{x_j}{|x|} = frac{x_j}{|x^2|} $$
Is this the right idea?
If yes, how can I proceed for larger $n$?
If no, what would you suggest?
Appreciate any help !
real-analysis multivariable-calculus vector-analysis
asked Jan 14 at 12:21
constant94constant94
6310
$endgroup$
add a comment |
$begingroup$
for $ 2 leq n $ let be
$ f: mathbb{R}^n backslash { 0 } rightarrow mathbb{R}^n $
$ f(x):= frac{x}{|x|^n} $ , $ xin mathbb{R}^n backslash { 0 } $
Is f a gradient field?
A vectorfield $ f $ ist a gradient field, if there is a function
$ f= nabla v $
$ left[ nabla f (x) = (frac{ partial f}{ partial x_1} (x),...,frac{ partial f}{ partial x_n} (x)) right] $
So, I need to find a function $v$ so that $ f= nabla v $ right?
If I look for the case $n=2 $
so $ f(x):= frac{x}{|x|^2} in mathbb{R}^n backslash { 0 } $
let be $ partial_j v(x)= frac{x_j}{|x|}$
then $v(x)= ln(|x|) $ and
$$ partial_j v(x)= frac{1}{|x|} frac{x_j}{|x|} = frac{x_j}{|x^2|} $$
Is this the right idea?
If yes, how can I proceed for larger $n$?
If no, what would you suggest?
Appreciate any help !
real-analysis multivariable-calculus vector-analysis
asked Jan 14 at 12:21
constant94constant94
6310
$endgroup$
for $ 2 leq n $ let be
$ f: mathbb{R}^n backslash { 0 } rightarrow mathbb{R}^n $
$ f(x):= frac{x}{|x|^n} $ , $ xin mathbb{R}^n backslash { 0 } $
Is f a gradient field?
A vectorfield $ f $ ist a gradient field, if there is a function
$ f= nabla v $
$ left[ nabla f (x) = (frac{ partial f}{ partial x_1} (x),...,frac{ partial f}{ partial x_n} (x)) right] $
So, I need to find a function $v$ so that $ f= nabla v $ right?
If I look for the case $n=2 $
so $ f(x):= frac{x}{|x|^2} in mathbb{R}^n backslash { 0 } $
let be $ partial_j v(x)= frac{x_j}{|x|}$
then $v(x)= ln(|x|) $ and
$$ partial_j v(x)= frac{1}{|x|} frac{x_j}{|x|} = frac{x_j}{|x^2|} $$
Is this the right idea?
If yes, how can I proceed for larger $n$?
If no, what would you suggest?
Appreciate any help !
real-analysis multivariable-calculus vector-analysis
real-analysis multivariable-calculus vector-analysis
asked Jan 14 at 12:21
constant94constant94
6310
asked Jan 14 at 12:21
constant94constant94
6310
asked Jan 14 at 12:21
constant94constant94
6310
asked Jan 14 at 12:21
constant94constant94
6310
asked Jan 14 at 12:21
constant94constant94
6310
6310
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's suppose that there exists some $V(x)$ such that $frac{d}{dx}V(x)=f(x)$.
Therefore:
$$int_a^yfrac{d}{dx}V(x)dx =int_a^yf(x) dx$$
Applying the fundamental theorem of calculus gives:
$$V(y)=int_a^y frac{x}{|x|^n} dx$$
Now let's break into cases. Suppose we are only considering $x,a,y > 0$. Then, we are left with a really simple integral, namely:
$$V(y)=int_a^y frac{x}{x^n} dx = int_a^y frac{1}{x^{n-1}} dx.$$
Now, let's assume we are only considering $x, a, y < 0$. This also simplifies very nicely:
$$V(y)=int_a^y frac{x}{(-x)^n} dx = (-1)^nint_a^y frac{1}{x^{n-1}} dx.$$
So it seems that we have a definition for $V$.
begin{cases}
V(x) = frac{1}{2-n}x^{2-n} & x < 0 \
V(x) = frac{1}{2-n}(-1)^n x^{2-n} & x>0
end{cases}
answered Jan 14 at 15:46
CuhrazateeCuhrazatee
361110
$endgroup$
add a comment |
$begingroup$
Your $f$ is a radial field and it depends only on the magnitude of your vector $r = |vec{r}|$ so it is of the form $$f(vec{r}) = frac{vec{r}}{r^n} = F(r)hat{r}$$ where $F(r) = frac1{r^{n-1}}$ and $hat{r}$ is the unit radial vector.
Notice that for a scalar function $phi : mathbb{R}^n to mathbb{R}$ depending only on magnitude $r$ we have $$nabla phi(r) = phi'(r)nabla r = phi'(r)hat{r}$$
since $nabla r = hat{r}$.
Hence if $$phi(r) = int F(r),dr = -frac1{(n-2)r^{n-2}}$$
then $f = nabla phi$.
answered Jan 14 at 15:56
mechanodroidmechanodroid
27.6k62447
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3073173%2fis-fx-fracx-xn-a-gradient-field%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's suppose that there exists some $V(x)$ such that $frac{d}{dx}V(x)=f(x)$.
Therefore:
$$int_a^yfrac{d}{dx}V(x)dx =int_a^yf(x) dx$$
Applying the fundamental theorem of calculus gives:
$$V(y)=int_a^y frac{x}{|x|^n} dx$$
Now let's break into cases. Suppose we are only considering $x,a,y > 0$. Then, we are left with a really simple integral, namely:
$$V(y)=int_a^y frac{x}{x^n} dx = int_a^y frac{1}{x^{n-1}} dx.$$
Now, let's assume we are only considering $x, a, y < 0$. This also simplifies very nicely:
$$V(y)=int_a^y frac{x}{(-x)^n} dx = (-1)^nint_a^y frac{1}{x^{n-1}} dx.$$
So it seems that we have a definition for $V$.
begin{cases}
V(x) = frac{1}{2-n}x^{2-n} & x < 0 \
V(x) = frac{1}{2-n}(-1)^n x^{2-n} & x>0
end{cases}
answered Jan 14 at 15:46
CuhrazateeCuhrazatee
361110
$endgroup$
add a comment |
$begingroup$
Let's suppose that there exists some $V(x)$ such that $frac{d}{dx}V(x)=f(x)$.
Therefore:
$$int_a^yfrac{d}{dx}V(x)dx =int_a^yf(x) dx$$
Applying the fundamental theorem of calculus gives:
$$V(y)=int_a^y frac{x}{|x|^n} dx$$
Now let's break into cases. Suppose we are only considering $x,a,y > 0$. Then, we are left with a really simple integral, namely:
$$V(y)=int_a^y frac{x}{x^n} dx = int_a^y frac{1}{x^{n-1}} dx.$$
Now, let's assume we are only considering $x, a, y < 0$. This also simplifies very nicely:
$$V(y)=int_a^y frac{x}{(-x)^n} dx = (-1)^nint_a^y frac{1}{x^{n-1}} dx.$$
So it seems that we have a definition for $V$.
begin{cases}
V(x) = frac{1}{2-n}x^{2-n} & x < 0 \
V(x) = frac{1}{2-n}(-1)^n x^{2-n} & x>0
end{cases}
answered Jan 14 at 15:46
CuhrazateeCuhrazatee
361110
$endgroup$
add a comment |
$begingroup$
Let's suppose that there exists some $V(x)$ such that $frac{d}{dx}V(x)=f(x)$.
Therefore:
$$int_a^yfrac{d}{dx}V(x)dx =int_a^yf(x) dx$$
Applying the fundamental theorem of calculus gives:
$$V(y)=int_a^y frac{x}{|x|^n} dx$$
Now let's break into cases. Suppose we are only considering $x,a,y > 0$. Then, we are left with a really simple integral, namely:
$$V(y)=int_a^y frac{x}{x^n} dx = int_a^y frac{1}{x^{n-1}} dx.$$
Now, let's assume we are only considering $x, a, y < 0$. This also simplifies very nicely:
$$V(y)=int_a^y frac{x}{(-x)^n} dx = (-1)^nint_a^y frac{1}{x^{n-1}} dx.$$
So it seems that we have a definition for $V$.
begin{cases}
V(x) = frac{1}{2-n}x^{2-n} & x < 0 \
V(x) = frac{1}{2-n}(-1)^n x^{2-n} & x>0
end{cases}
answered Jan 14 at 15:46
CuhrazateeCuhrazatee
361110
$endgroup$
Let's suppose that there exists some $V(x)$ such that $frac{d}{dx}V(x)=f(x)$.
Therefore:
$$int_a^yfrac{d}{dx}V(x)dx =int_a^yf(x) dx$$
Applying the fundamental theorem of calculus gives:
$$V(y)=int_a^y frac{x}{|x|^n} dx$$
Now let's break into cases. Suppose we are only considering $x,a,y > 0$. Then, we are left with a really simple integral, namely:
$$V(y)=int_a^y frac{x}{x^n} dx = int_a^y frac{1}{x^{n-1}} dx.$$
Now, let's assume we are only considering $x, a, y < 0$. This also simplifies very nicely:
$$V(y)=int_a^y frac{x}{(-x)^n} dx = (-1)^nint_a^y frac{1}{x^{n-1}} dx.$$
So it seems that we have a definition for $V$.
begin{cases}
V(x) = frac{1}{2-n}x^{2-n} & x < 0 \
V(x) = frac{1}{2-n}(-1)^n x^{2-n} & x>0
end{cases}
answered Jan 14 at 15:46
CuhrazateeCuhrazatee
361110
edited Jan 14 at 16:19
answered Jan 14 at 15:46
CuhrazateeCuhrazatee
361110
answered Jan 14 at 15:46
CuhrazateeCuhrazatee
361110
answered Jan 14 at 15:46
CuhrazateeCuhrazatee
361110
361110
add a comment |
add a comment |
$begingroup$
Your $f$ is a radial field and it depends only on the magnitude of your vector $r = |vec{r}|$ so it is of the form $$f(vec{r}) = frac{vec{r}}{r^n} = F(r)hat{r}$$ where $F(r) = frac1{r^{n-1}}$ and $hat{r}$ is the unit radial vector.
Notice that for a scalar function $phi : mathbb{R}^n to mathbb{R}$ depending only on magnitude $r$ we have $$nabla phi(r) = phi'(r)nabla r = phi'(r)hat{r}$$
since $nabla r = hat{r}$.
Hence if $$phi(r) = int F(r),dr = -frac1{(n-2)r^{n-2}}$$
then $f = nabla phi$.
answered Jan 14 at 15:56
mechanodroidmechanodroid
27.6k62447
$endgroup$
add a comment |
$begingroup$
Your $f$ is a radial field and it depends only on the magnitude of your vector $r = |vec{r}|$ so it is of the form $$f(vec{r}) = frac{vec{r}}{r^n} = F(r)hat{r}$$ where $F(r) = frac1{r^{n-1}}$ and $hat{r}$ is the unit radial vector.
Notice that for a scalar function $phi : mathbb{R}^n to mathbb{R}$ depending only on magnitude $r$ we have $$nabla phi(r) = phi'(r)nabla r = phi'(r)hat{r}$$
since $nabla r = hat{r}$.
Hence if $$phi(r) = int F(r),dr = -frac1{(n-2)r^{n-2}}$$
then $f = nabla phi$.
answered Jan 14 at 15:56
mechanodroidmechanodroid
27.6k62447
$endgroup$
add a comment |
$begingroup$
Your $f$ is a radial field and it depends only on the magnitude of your vector $r = |vec{r}|$ so it is of the form $$f(vec{r}) = frac{vec{r}}{r^n} = F(r)hat{r}$$ where $F(r) = frac1{r^{n-1}}$ and $hat{r}$ is the unit radial vector.
Notice that for a scalar function $phi : mathbb{R}^n to mathbb{R}$ depending only on magnitude $r$ we have $$nabla phi(r) = phi'(r)nabla r = phi'(r)hat{r}$$
since $nabla r = hat{r}$.
Hence if $$phi(r) = int F(r),dr = -frac1{(n-2)r^{n-2}}$$
then $f = nabla phi$.
answered Jan 14 at 15:56
mechanodroidmechanodroid
27.6k62447
$endgroup$
Your $f$ is a radial field and it depends only on the magnitude of your vector $r = |vec{r}|$ so it is of the form $$f(vec{r}) = frac{vec{r}}{r^n} = F(r)hat{r}$$ where $F(r) = frac1{r^{n-1}}$ and $hat{r}$ is the unit radial vector.
Notice that for a scalar function $phi : mathbb{R}^n to mathbb{R}$ depending only on magnitude $r$ we have $$nabla phi(r) = phi'(r)nabla r = phi'(r)hat{r}$$
since $nabla r = hat{r}$.
Hence if $$phi(r) = int F(r),dr = -frac1{(n-2)r^{n-2}}$$
then $f = nabla phi$.
answered Jan 14 at 15:56
mechanodroidmechanodroid
27.6k62447
answered Jan 14 at 15:56
mechanodroidmechanodroid
27.6k62447
answered Jan 14 at 15:56
mechanodroidmechanodroid
27.6k62447
answered Jan 14 at 15:56
mechanodroidmechanodroid
27.6k62447
27.6k62447
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3073173%2fis-fx-fracx-xn-a-gradient-field%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
