How to prove that it is impossible to express one variable as function of others in a equation (implicit...
Consider implicit function theorem. Since the conditions the theorem gives are only sufficient and not necessary I would like to know how can one prove that it is impossible to express (locally) one variable as a function of other variables in a equation.
Here is what I found in 2D:
In 2D consider $f(x,y)=c$ (with $fin C^1$) and a point $(x_0,y_0)$ such that $f(x_0,y_0)=c$.
If $frac{partial f}{partial y}|_{{x_0,y_0}}=0$ this is not enough to prove that it is impossible to express $y=Psi(x)$ locally.
But if I find out that $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}neq0$ then I can conclude that is impossible to express $y=Psi(x)$ locally.
(EDIT: Here I'm assuming that in the calculations of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ I only need to know that, if it was possible to express $z$, then it would be $Psi (x_0)=y_0$.
In the expression of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ both $Psi (x_0)$ and $Psi' (x_0)$ would appear (and I don't know $Psi' (x_0)$) but I'm assuming $Psi' (x_0)$ vanishes and only $Psi (x_0)$ remains.)
But how can I extend this in 3D?
Consider $f(x,y,z)=c$ (with $fin C^1$) and a point $(x_0,y_0,z_0)$ such that $f(x_0,y_0,z_0)=c$.
If $frac{partial f}{partial z}|_{{x_0,y_0,z_0}}=0$ this is not enough to prove that it is impossible to express $z=Phi(x,y)$ locally.
But if I find out that $nabla f(x,y,Phi(x,y))|_{{x_0,y_0}}neq(0,0)$ then can I conclude that is impossible to express $z=Phi(x,y)$ locally?
(EDIT: Again I'm assuming that in the calculations of $nabla f(x,y,Phi(x,y))|_{{x_0,y_0}}$ I only need to know that, if it was possible to express $y$, then it would be $Phi (x_0,y_0)=z_0$.
In the expression of $nabla f(x,y,Phi(x,y))|_{{x_0,y_0}}$ there would be $Phi (x_0,y_0)=z_0$ but also the partial derivativers of $Phi$, (and I don't know them) but I'm assuming the partial derivatives vanish and only $Phi (x_0,y_0)$ remains.)
calculus multivariable-calculus derivatives implicit-function-theorem
asked Dec 9 at 11:52
Gianolepo
765918
add a comment |
Consider implicit function theorem. Since the conditions the theorem gives are only sufficient and not necessary I would like to know how can one prove that it is impossible to express (locally) one variable as a function of other variables in a equation.
Here is what I found in 2D:
In 2D consider $f(x,y)=c$ (with $fin C^1$) and a point $(x_0,y_0)$ such that $f(x_0,y_0)=c$.
If $frac{partial f}{partial y}|_{{x_0,y_0}}=0$ this is not enough to prove that it is impossible to express $y=Psi(x)$ locally.
But if I find out that $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}neq0$ then I can conclude that is impossible to express $y=Psi(x)$ locally.
(EDIT: Here I'm assuming that in the calculations of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ I only need to know that, if it was possible to express $z$, then it would be $Psi (x_0)=y_0$.
In the expression of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ both $Psi (x_0)$ and $Psi' (x_0)$ would appear (and I don't know $Psi' (x_0)$) but I'm assuming $Psi' (x_0)$ vanishes and only $Psi (x_0)$ remains.)
But how can I extend this in 3D?
Consider $f(x,y,z)=c$ (with $fin C^1$) and a point $(x_0,y_0,z_0)$ such that $f(x_0,y_0,z_0)=c$.
If $frac{partial f}{partial z}|_{{x_0,y_0,z_0}}=0$ this is not enough to prove that it is impossible to express $z=Phi(x,y)$ locally.
But if I find out that $nabla f(x,y,Phi(x,y))|_{{x_0,y_0}}neq(0,0)$ then can I conclude that is impossible to express $z=Phi(x,y)$ locally?
(EDIT: Again I'm assuming that in the calculations of $nabla f(x,y,Phi(x,y))|_{{x_0,y_0}}$ I only need to know that, if it was possible to express $y$, then it would be $Phi (x_0,y_0)=z_0$.
In the expression of $nabla f(x,y,Phi(x,y))|_{{x_0,y_0}}$ there would be $Phi (x_0,y_0)=z_0$ but also the partial derivativers of $Phi$, (and I don't know them) but I'm assuming the partial derivatives vanish and only $Phi (x_0,y_0)$ remains.)
calculus multivariable-calculus derivatives implicit-function-theorem
asked Dec 9 at 11:52
Gianolepo
765918
1
How can you calculate $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ if it is "impossible to express $y=Psi(x)$ locally"?
– smcc
Dec 9 at 12:04
Thanks for the comment I meant $neq$, I apologize, I edited the question
– Gianolepo
Dec 9 at 12:06
There was a typo in my comment. I meant how can you calculate that derivative at all? Perhaps you could give an example to show what you mean.
– smcc
Dec 9 at 12:07
Yes I'm sorry, you are right, but I'm assuming that in the calculations of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ I only need to know that, if it was possible to express $y$, then it would be $Psi (x_0)=y_0$. In the expression of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ both $Psi (x_0)$ and $Psi' (x_0)$ would appear (and I don't know $Psi' (x_0)$) but I'm assuming $Psi' (x_0)$ vanishes and only $Psi (x_0)$ remains.
– Gianolepo
Dec 9 at 12:17
add a comment |
Consider implicit function theorem. Since the conditions the theorem gives are only sufficient and not necessary I would like to know how can one prove that it is impossible to express (locally) one variable as a function of other variables in a equation.
Here is what I found in 2D:
In 2D consider $f(x,y)=c$ (with $fin C^1$) and a point $(x_0,y_0)$ such that $f(x_0,y_0)=c$.
If $frac{partial f}{partial y}|_{{x_0,y_0}}=0$ this is not enough to prove that it is impossible to express $y=Psi(x)$ locally.
But if I find out that $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}neq0$ then I can conclude that is impossible to express $y=Psi(x)$ locally.
(EDIT: Here I'm assuming that in the calculations of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ I only need to know that, if it was possible to express $z$, then it would be $Psi (x_0)=y_0$.
In the expression of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ both $Psi (x_0)$ and $Psi' (x_0)$ would appear (and I don't know $Psi' (x_0)$) but I'm assuming $Psi' (x_0)$ vanishes and only $Psi (x_0)$ remains.)
But how can I extend this in 3D?
Consider $f(x,y,z)=c$ (with $fin C^1$) and a point $(x_0,y_0,z_0)$ such that $f(x_0,y_0,z_0)=c$.
If $frac{partial f}{partial z}|_{{x_0,y_0,z_0}}=0$ this is not enough to prove that it is impossible to express $z=Phi(x,y)$ locally.
But if I find out that $nabla f(x,y,Phi(x,y))|_{{x_0,y_0}}neq(0,0)$ then can I conclude that is impossible to express $z=Phi(x,y)$ locally?
(EDIT: Again I'm assuming that in the calculations of $nabla f(x,y,Phi(x,y))|_{{x_0,y_0}}$ I only need to know that, if it was possible to express $y$, then it would be $Phi (x_0,y_0)=z_0$.
In the expression of $nabla f(x,y,Phi(x,y))|_{{x_0,y_0}}$ there would be $Phi (x_0,y_0)=z_0$ but also the partial derivativers of $Phi$, (and I don't know them) but I'm assuming the partial derivatives vanish and only $Phi (x_0,y_0)$ remains.)
calculus multivariable-calculus derivatives implicit-function-theorem
asked Dec 9 at 11:52
Gianolepo
765918
Consider implicit function theorem. Since the conditions the theorem gives are only sufficient and not necessary I would like to know how can one prove that it is impossible to express (locally) one variable as a function of other variables in a equation.
Here is what I found in 2D:
In 2D consider $f(x,y)=c$ (with $fin C^1$) and a point $(x_0,y_0)$ such that $f(x_0,y_0)=c$.
If $frac{partial f}{partial y}|_{{x_0,y_0}}=0$ this is not enough to prove that it is impossible to express $y=Psi(x)$ locally.
But if I find out that $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}neq0$ then I can conclude that is impossible to express $y=Psi(x)$ locally.
(EDIT: Here I'm assuming that in the calculations of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ I only need to know that, if it was possible to express $z$, then it would be $Psi (x_0)=y_0$.
In the expression of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ both $Psi (x_0)$ and $Psi' (x_0)$ would appear (and I don't know $Psi' (x_0)$) but I'm assuming $Psi' (x_0)$ vanishes and only $Psi (x_0)$ remains.)
But how can I extend this in 3D?
Consider $f(x,y,z)=c$ (with $fin C^1$) and a point $(x_0,y_0,z_0)$ such that $f(x_0,y_0,z_0)=c$.
If $frac{partial f}{partial z}|_{{x_0,y_0,z_0}}=0$ this is not enough to prove that it is impossible to express $z=Phi(x,y)$ locally.
But if I find out that $nabla f(x,y,Phi(x,y))|_{{x_0,y_0}}neq(0,0)$ then can I conclude that is impossible to express $z=Phi(x,y)$ locally?
(EDIT: Again I'm assuming that in the calculations of $nabla f(x,y,Phi(x,y))|_{{x_0,y_0}}$ I only need to know that, if it was possible to express $y$, then it would be $Phi (x_0,y_0)=z_0$.
In the expression of $nabla f(x,y,Phi(x,y))|_{{x_0,y_0}}$ there would be $Phi (x_0,y_0)=z_0$ but also the partial derivativers of $Phi$, (and I don't know them) but I'm assuming the partial derivatives vanish and only $Phi (x_0,y_0)$ remains.)
calculus multivariable-calculus derivatives implicit-function-theorem
calculus multivariable-calculus derivatives implicit-function-theorem
asked Dec 9 at 11:52
Gianolepo
765918
asked Dec 9 at 11:52
Gianolepo
765918
edited Dec 9 at 12:20
asked Dec 9 at 11:52
Gianolepo
765918
asked Dec 9 at 11:52
Gianolepo
765918
asked Dec 9 at 11:52
Gianolepo
765918
765918
1
How can you calculate $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ if it is "impossible to express $y=Psi(x)$ locally"?
– smcc
Dec 9 at 12:04
Thanks for the comment I meant $neq$, I apologize, I edited the question
– Gianolepo
Dec 9 at 12:06
There was a typo in my comment. I meant how can you calculate that derivative at all? Perhaps you could give an example to show what you mean.
– smcc
Dec 9 at 12:07
Yes I'm sorry, you are right, but I'm assuming that in the calculations of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ I only need to know that, if it was possible to express $y$, then it would be $Psi (x_0)=y_0$. In the expression of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ both $Psi (x_0)$ and $Psi' (x_0)$ would appear (and I don't know $Psi' (x_0)$) but I'm assuming $Psi' (x_0)$ vanishes and only $Psi (x_0)$ remains.
– Gianolepo
Dec 9 at 12:17
add a comment |
1
How can you calculate $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ if it is "impossible to express $y=Psi(x)$ locally"?
– smcc
Dec 9 at 12:04
Thanks for the comment I meant $neq$, I apologize, I edited the question
– Gianolepo
Dec 9 at 12:06
There was a typo in my comment. I meant how can you calculate that derivative at all? Perhaps you could give an example to show what you mean.
– smcc
Dec 9 at 12:07
Yes I'm sorry, you are right, but I'm assuming that in the calculations of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ I only need to know that, if it was possible to express $y$, then it would be $Psi (x_0)=y_0$. In the expression of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ both $Psi (x_0)$ and $Psi' (x_0)$ would appear (and I don't know $Psi' (x_0)$) but I'm assuming $Psi' (x_0)$ vanishes and only $Psi (x_0)$ remains.
– Gianolepo
Dec 9 at 12:17
1
1
How can you calculate $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ if it is "impossible to express $y=Psi(x)$ locally"?
– smcc
Dec 9 at 12:04
How can you calculate $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ if it is "impossible to express $y=Psi(x)$ locally"?
– smcc
Dec 9 at 12:04
Thanks for the comment I meant $neq$, I apologize, I edited the question
– Gianolepo
Dec 9 at 12:06
Thanks for the comment I meant $neq$, I apologize, I edited the question
– Gianolepo
Dec 9 at 12:06
There was a typo in my comment. I meant how can you calculate that derivative at all? Perhaps you could give an example to show what you mean.
– smcc
Dec 9 at 12:07
There was a typo in my comment. I meant how can you calculate that derivative at all? Perhaps you could give an example to show what you mean.
– smcc
Dec 9 at 12:07
Yes I'm sorry, you are right, but I'm assuming that in the calculations of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ I only need to know that, if it was possible to express $y$, then it would be $Psi (x_0)=y_0$. In the expression of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ both $Psi (x_0)$ and $Psi' (x_0)$ would appear (and I don't know $Psi' (x_0)$) but I'm assuming $Psi' (x_0)$ vanishes and only $Psi (x_0)$ remains.
– Gianolepo
Dec 9 at 12:17
Yes I'm sorry, you are right, but I'm assuming that in the calculations of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ I only need to know that, if it was possible to express $y$, then it would be $Psi (x_0)=y_0$. In the expression of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ both $Psi (x_0)$ and $Psi' (x_0)$ would appear (and I don't know $Psi' (x_0)$) but I'm assuming $Psi' (x_0)$ vanishes and only $Psi (x_0)$ remains.
– Gianolepo
Dec 9 at 12:17
add a comment |
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1
How can you calculate $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ if it is "impossible to express $y=Psi(x)$ locally"?
– smcc
Dec 9 at 12:04
Thanks for the comment I meant $neq$, I apologize, I edited the question
– Gianolepo
Dec 9 at 12:06
There was a typo in my comment. I meant how can you calculate that derivative at all? Perhaps you could give an example to show what you mean.
– smcc
Dec 9 at 12:07
Yes I'm sorry, you are right, but I'm assuming that in the calculations of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ I only need to know that, if it was possible to express $y$, then it would be $Psi (x_0)=y_0$. In the expression of $frac{mathrm{d}}{mathrm{d}x} f(x,Psi(x))|_{x_0}$ both $Psi (x_0)$ and $Psi' (x_0)$ would appear (and I don't know $Psi' (x_0)$) but I'm assuming $Psi' (x_0)$ vanishes and only $Psi (x_0)$ remains.
– Gianolepo
Dec 9 at 12:17