Prove that $xy^2+y^3z^4+z^5x^6=1$ has a solution in an open neighborhood about the point $x_0=(0,1,-1)$
2
$begingroup$
Prove with implicit function theorem that $xy^2+y^3z^4+z^5x^6=1$ has a $C^1$ solution with a form of $(x,g(x,z),z)$ in an open neighborhood about the point $x_0=(0,1,-1)$.
What I have gotten so far:
Let $f(x,y,z)=xy^2+y^3z^4+z^5x^6-1$.
Now $f(0,1,-1)=0cdot1^2+1^3cdot(-1)^4+(-1)^5cdot0^6-1=0$.
Also $D_yf(x,y,z)=2xy+3z^4y^2$ and $D_yf(0,1,-1)=2cdot0cdot1+3cdot(-1)^4cdot1^2=3ne0$.
Now I can conclude with implicit function theorem that the function has a $C^1$ solution in an open neighborhood about the point $x_0=(0,1,-1)$ but how do I know it takes the form $(x,g(x,z),z)$?
calculus multivariable-calculus vector-analysis implicit-function-theorem
asked Dec 29 '18 at 9:13
JoeJoe
1795
$endgroup$
|
show 2 more comments
2
$begingroup$
Prove with implicit function theorem that $xy^2+y^3z^4+z^5x^6=1$ has a $C^1$ solution with a form of $(x,g(x,z),z)$ in an open neighborhood about the point $x_0=(0,1,-1)$.
What I have gotten so far:
Let $f(x,y,z)=xy^2+y^3z^4+z^5x^6-1$.
Now $f(0,1,-1)=0cdot1^2+1^3cdot(-1)^4+(-1)^5cdot0^6-1=0$.
Also $D_yf(x,y,z)=2xy+3z^4y^2$ and $D_yf(0,1,-1)=2cdot0cdot1+3cdot(-1)^4cdot1^2=3ne0$.
Now I can conclude with implicit function theorem that the function has a $C^1$ solution in an open neighborhood about the point $x_0=(0,1,-1)$ but how do I know it takes the form $(x,g(x,z),z)$?
calculus multivariable-calculus vector-analysis implicit-function-theorem
asked Dec 29 '18 at 9:13
JoeJoe
1795
$endgroup$
$begingroup$
Hint: consider when the condition does not hold.
$endgroup$
– Trebor
Dec 29 '18 at 9:19
3
$begingroup$
IFT specifically correlates the non-vanishing of the partial derivatie with respect to $y$ with the ability to "implicitly" solve $y$ locally as a $C^1$-function of the other coordinates. If you want to "solve" $z$ in terms of $x$ and $y$ instead of $y$, then you need to check that $D_zf$ does not vanish.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 9:30
$begingroup$
@JyrkiLahtonen Hi! Firstly, thank you for answering. However, I'm afraid I don't quite understand. How do I know I have to solve z in terms of x and y? Would the same statement hold if I choose to solve x in terms of y and z?
$endgroup$
– Joe
Dec 29 '18 at 10:24
1
$begingroup$
If you want to solve $x$, then you need to check that $D_x f$ does not vanish at the point of interest.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 10:27
2
$begingroup$
In a sense IFT is about linearizing the solution set (locally and approximately). So, when teaching, I emphasize the analogy with linear equations (and systems of such). From $$ax+by+cz=d$$ you can solve $x$ when $aneq0$, $y$ when $bneq0$ and $z$ when $cneq0$. And, guess what, those $a,b,c$ are exactly the partial derivatives of the function here! The same with several constraints, Jacobians and the relevant minors of matrices of the coefficients of a linear system of equations.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 10:31
|
show 2 more comments
2
2
2
$begingroup$
Prove with implicit function theorem that $xy^2+y^3z^4+z^5x^6=1$ has a $C^1$ solution with a form of $(x,g(x,z),z)$ in an open neighborhood about the point $x_0=(0,1,-1)$.
What I have gotten so far:
Let $f(x,y,z)=xy^2+y^3z^4+z^5x^6-1$.
Now $f(0,1,-1)=0cdot1^2+1^3cdot(-1)^4+(-1)^5cdot0^6-1=0$.
Also $D_yf(x,y,z)=2xy+3z^4y^2$ and $D_yf(0,1,-1)=2cdot0cdot1+3cdot(-1)^4cdot1^2=3ne0$.
Now I can conclude with implicit function theorem that the function has a $C^1$ solution in an open neighborhood about the point $x_0=(0,1,-1)$ but how do I know it takes the form $(x,g(x,z),z)$?
calculus multivariable-calculus vector-analysis implicit-function-theorem
asked Dec 29 '18 at 9:13
JoeJoe
1795
$endgroup$
Prove with implicit function theorem that $xy^2+y^3z^4+z^5x^6=1$ has a $C^1$ solution with a form of $(x,g(x,z),z)$ in an open neighborhood about the point $x_0=(0,1,-1)$.
What I have gotten so far:
Let $f(x,y,z)=xy^2+y^3z^4+z^5x^6-1$.
Now $f(0,1,-1)=0cdot1^2+1^3cdot(-1)^4+(-1)^5cdot0^6-1=0$.
Also $D_yf(x,y,z)=2xy+3z^4y^2$ and $D_yf(0,1,-1)=2cdot0cdot1+3cdot(-1)^4cdot1^2=3ne0$.
Now I can conclude with implicit function theorem that the function has a $C^1$ solution in an open neighborhood about the point $x_0=(0,1,-1)$ but how do I know it takes the form $(x,g(x,z),z)$?
calculus multivariable-calculus vector-analysis implicit-function-theorem
calculus multivariable-calculus vector-analysis implicit-function-theorem
asked Dec 29 '18 at 9:13
JoeJoe
1795
asked Dec 29 '18 at 9:13
JoeJoe
1795
asked Dec 29 '18 at 9:13
JoeJoe
1795
asked Dec 29 '18 at 9:13
JoeJoe
1795
asked Dec 29 '18 at 9:13
JoeJoe
1795
1795
$begingroup$
Hint: consider when the condition does not hold.
$endgroup$
– Trebor
Dec 29 '18 at 9:19
3
$begingroup$
IFT specifically correlates the non-vanishing of the partial derivatie with respect to $y$ with the ability to "implicitly" solve $y$ locally as a $C^1$-function of the other coordinates. If you want to "solve" $z$ in terms of $x$ and $y$ instead of $y$, then you need to check that $D_zf$ does not vanish.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 9:30
$begingroup$
@JyrkiLahtonen Hi! Firstly, thank you for answering. However, I'm afraid I don't quite understand. How do I know I have to solve z in terms of x and y? Would the same statement hold if I choose to solve x in terms of y and z?
$endgroup$
– Joe
Dec 29 '18 at 10:24
1
$begingroup$
If you want to solve $x$, then you need to check that $D_x f$ does not vanish at the point of interest.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 10:27
2
$begingroup$
In a sense IFT is about linearizing the solution set (locally and approximately). So, when teaching, I emphasize the analogy with linear equations (and systems of such). From $$ax+by+cz=d$$ you can solve $x$ when $aneq0$, $y$ when $bneq0$ and $z$ when $cneq0$. And, guess what, those $a,b,c$ are exactly the partial derivatives of the function here! The same with several constraints, Jacobians and the relevant minors of matrices of the coefficients of a linear system of equations.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 10:31
|
show 2 more comments
$begingroup$
Hint: consider when the condition does not hold.
$endgroup$
– Trebor
Dec 29 '18 at 9:19
3
$begingroup$
IFT specifically correlates the non-vanishing of the partial derivatie with respect to $y$ with the ability to "implicitly" solve $y$ locally as a $C^1$-function of the other coordinates. If you want to "solve" $z$ in terms of $x$ and $y$ instead of $y$, then you need to check that $D_zf$ does not vanish.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 9:30
$begingroup$
@JyrkiLahtonen Hi! Firstly, thank you for answering. However, I'm afraid I don't quite understand. How do I know I have to solve z in terms of x and y? Would the same statement hold if I choose to solve x in terms of y and z?
$endgroup$
– Joe
Dec 29 '18 at 10:24
1
$begingroup$
If you want to solve $x$, then you need to check that $D_x f$ does not vanish at the point of interest.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 10:27
2
$begingroup$
In a sense IFT is about linearizing the solution set (locally and approximately). So, when teaching, I emphasize the analogy with linear equations (and systems of such). From $$ax+by+cz=d$$ you can solve $x$ when $aneq0$, $y$ when $bneq0$ and $z$ when $cneq0$. And, guess what, those $a,b,c$ are exactly the partial derivatives of the function here! The same with several constraints, Jacobians and the relevant minors of matrices of the coefficients of a linear system of equations.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 10:31
$begingroup$
Hint: consider when the condition does not hold.
$endgroup$
– Trebor
Dec 29 '18 at 9:19
$begingroup$
Hint: consider when the condition does not hold.
$endgroup$
– Trebor
Dec 29 '18 at 9:19
3
3
$begingroup$
IFT specifically correlates the non-vanishing of the partial derivatie with respect to $y$ with the ability to "implicitly" solve $y$ locally as a $C^1$-function of the other coordinates. If you want to "solve" $z$ in terms of $x$ and $y$ instead of $y$, then you need to check that $D_zf$ does not vanish.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 9:30
$begingroup$
IFT specifically correlates the non-vanishing of the partial derivatie with respect to $y$ with the ability to "implicitly" solve $y$ locally as a $C^1$-function of the other coordinates. If you want to "solve" $z$ in terms of $x$ and $y$ instead of $y$, then you need to check that $D_zf$ does not vanish.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 9:30
$begingroup$
@JyrkiLahtonen Hi! Firstly, thank you for answering. However, I'm afraid I don't quite understand. How do I know I have to solve z in terms of x and y? Would the same statement hold if I choose to solve x in terms of y and z?
$endgroup$
– Joe
Dec 29 '18 at 10:24
$begingroup$
@JyrkiLahtonen Hi! Firstly, thank you for answering. However, I'm afraid I don't quite understand. How do I know I have to solve z in terms of x and y? Would the same statement hold if I choose to solve x in terms of y and z?
$endgroup$
– Joe
Dec 29 '18 at 10:24
1
1
$begingroup$
If you want to solve $x$, then you need to check that $D_x f$ does not vanish at the point of interest.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 10:27
$begingroup$
If you want to solve $x$, then you need to check that $D_x f$ does not vanish at the point of interest.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 10:27
2
2
$begingroup$
In a sense IFT is about linearizing the solution set (locally and approximately). So, when teaching, I emphasize the analogy with linear equations (and systems of such). From $$ax+by+cz=d$$ you can solve $x$ when $aneq0$, $y$ when $bneq0$ and $z$ when $cneq0$. And, guess what, those $a,b,c$ are exactly the partial derivatives of the function here! The same with several constraints, Jacobians and the relevant minors of matrices of the coefficients of a linear system of equations.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 10:31
$begingroup$
In a sense IFT is about linearizing the solution set (locally and approximately). So, when teaching, I emphasize the analogy with linear equations (and systems of such). From $$ax+by+cz=d$$ you can solve $x$ when $aneq0$, $y$ when $bneq0$ and $z$ when $cneq0$. And, guess what, those $a,b,c$ are exactly the partial derivatives of the function here! The same with several constraints, Jacobians and the relevant minors of matrices of the coefficients of a linear system of equations.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 10:31
|
show 2 more comments
0
active
oldest
votes
Your Answer
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
});
}
});
draft saved
draft discarded
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%2f3055672%2fprove-that-xy2y3z4z5x6-1-has-a-solution-in-an-open-neighborhood-about-t%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3055672%2fprove-that-xy2y3z4z5x6-1-has-a-solution-in-an-open-neighborhood-about-t%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
$begingroup$
Hint: consider when the condition does not hold.
$endgroup$
– Trebor
Dec 29 '18 at 9:19
3
$begingroup$
IFT specifically correlates the non-vanishing of the partial derivatie with respect to $y$ with the ability to "implicitly" solve $y$ locally as a $C^1$-function of the other coordinates. If you want to "solve" $z$ in terms of $x$ and $y$ instead of $y$, then you need to check that $D_zf$ does not vanish.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 9:30
$begingroup$
@JyrkiLahtonen Hi! Firstly, thank you for answering. However, I'm afraid I don't quite understand. How do I know I have to solve z in terms of x and y? Would the same statement hold if I choose to solve x in terms of y and z?
$endgroup$
– Joe
Dec 29 '18 at 10:24
1
$begingroup$
If you want to solve $x$, then you need to check that $D_x f$ does not vanish at the point of interest.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 10:27
2
$begingroup$
In a sense IFT is about linearizing the solution set (locally and approximately). So, when teaching, I emphasize the analogy with linear equations (and systems of such). From $$ax+by+cz=d$$ you can solve $x$ when $aneq0$, $y$ when $bneq0$ and $z$ when $cneq0$. And, guess what, those $a,b,c$ are exactly the partial derivatives of the function here! The same with several constraints, Jacobians and the relevant minors of matrices of the coefficients of a linear system of equations.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 10:31