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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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


















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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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




















  • $begingroup$
    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












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