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September 23rd, 2015, 10:46 AM   #1
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Fundamental Underlying Theorem of Implicit Functions

If f(x,y,z)=0, is $\displaystyle f_x=f_y=f_z$=0?

The reason I ask:

Taylor, Advanced Calculus, Implicit Function Theorem:

In an open set S, given f(x,y,z), $\displaystyle f_z$ continuous, $\displaystyle f(p_o)$=0 and $\displaystyle f_z|p_0\neq0$.
Then f(x,y,z)=0 in a neighborhood of $\displaystyle p_0$
But then $\displaystyle f_z|p_0=0$????
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September 23rd, 2015, 03:48 PM   #2
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Quote:
Originally Posted by zylo View Post
If f(x,y,z)=0, is $\displaystyle f_x=f_y=f_z$=0?
I presume you mean f(x,y,z)=0 at a particular point. If f(x,y,z)=0 for all points, then obviously all derivatives are 0.

Let f(x,y,z)=x+y+z, then f(x,y,z)=0 at (0,0,0), but all partials = 1.
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September 24th, 2015, 04:38 AM   #3
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It's a matter of precise interpretation in the context of the theorem.

Given f(x,y,z).
IF $\displaystyle f(p_0)$ = 0 and $\displaystyle f_z|p_0\neq0$
THEN f(x,y,z) = 0 in a neigborhood of $\displaystyle p_0$

f(x,y,z) = xy + 2z
f(0,0,0) = 0, $\displaystyle f_z|p_0$ = 2 ->
xy+2z = 0 in a neighborhood of (0,0,0)

It wasn't a trick question. Trying something else and got stuck on this point. Woke up and saw it.

Last edited by skipjack; September 24th, 2015 at 06:28 AM.
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September 24th, 2015, 06:29 AM   #4
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Where exactly in the book was this?
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September 28th, 2015, 06:17 AM   #5
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Quote:
Originally Posted by skipjack View Post
Where exactly in the book was this?
Taylor, Advanced Calculus, 1955, par 8.1, pg 241
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September 28th, 2015, 01:09 PM   #6
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Implicit Function Theorem

Quote:
Originally Posted by zylo
If f(x,y,z)=0, is $\displaystyle f_x=f_y=f_z$=0?
It depends.
$\displaystyle f_xdx+f_ydy+f_zdz=0$
If x,y,z are independent, $\displaystyle f_x=f_y=f_z=0$.
If f(x,y,z)=0 defines z=z(x,y),
$\displaystyle f_x+f_z\frac{\partial z}{\partial x}=0$
$\displaystyle f_y+f_z\frac{\partial z}{\partial y}=0$
This turns out to be crucial to a proof of Implicit Function Theorem based on Fundamental Theorem.

Fundamental Theorem (Taylor, Advanced Calculus)
Let $\displaystyle f(x_1,..,x_n)$ be defined in an open set S, have continuous first partial derivatives, $\displaystyle f(p_0)=0$ and $\displaystyle f_{x_n}(p_0)\neq0$. Then
f(p)=0 and $\displaystyle x_n=x_n(x_1,..,x_{n-1})$ in a neighborhood of $\displaystyle p_0$.

Implicit Function Theorem
Given f(x,y), condensed notation, with continuous partial derivatives in S, $\displaystyle f(p_0)=0$, and $\displaystyle |f_y|_{p_0}\neq0$. Then f(x,y)=0 and f(x,y(x))=0 in a neighborhood of $\displaystyle p_0$.

Suppose f(x,y) satisfying these conditions is
$\displaystyle f_1(x,y_1,y_2)\\
f_2(x,y_1,y_2)$
$\displaystyle |f_y|_{p_0}= \left .\frac{\partial( f_1,f_2)}{\partial (y_1,y_2) }\right |_{p_0}=
\begin{vmatrix}\frac{\partial f_1}{\partial y_1} &\frac{\partial f_1}{\partial y_2} \\ \frac{\partial f_2 }{\partial y_1} & \frac{\partial f_2}{\partial y_2}
\end{vmatrix}_{p_0}\neq 0
$
Then
$\displaystyle f_1(x,y_1,y_2)=0\\
f_2(x,y_1,y_2)=0$

If $\displaystyle \frac{\partial f_1}{\partial y_2} \neq 0$ then $\displaystyle y_2=y_2(x,y_1)$, (Fundamental Theorem) then
f$\displaystyle _2(x,y_1,y_2(x,y_1))=0$
If $\displaystyle \frac{\partial f_2}{\partial y_1} \neq 0, y_1=y_1(x),$
$\displaystyle \left .\frac{\partial f_2}{\partial y_1} \right |_{x}=
\left .\frac{\partial f_2}{\partial y_1}\right |_{x,x_2} +
\left .\frac{\partial f_2}{\partial y_2}\right |_{x,y_1} \left .\frac{\partial y_2}{\partial y_1} \right |_{x}$
and you need
$\displaystyle \left .\frac{\partial y_2}{\partial y_1} \right |_{x}$, which you get from top of this post.

given $\displaystyle f_1(x,y_1,y_2)=0$ and $\displaystyle y_2=y_2(x,y_1)$

$\displaystyle df_1=\frac{\partial f_1}{\partial x}dx+\frac{\partial f_1}{\partial y_1}dy_1+\frac{\partial f_1}{\partial y_2}dy_2
$
divide by $\displaystyle dy_1$ holding x constant:

$\displaystyle \frac{\partial f_1 }{\partial y_1}+\frac{\partial f_1}{\partial y_2}\frac{\partial y_2}{\partial y_1}=0\\
$
$\displaystyle \left .\frac{\partial f_2}{\partial y_1} \right |_{x}=\left .\frac{\partial f_2}{\partial y_1} \right |_{x.y_2}-
\left .\frac{\partial f_2}{\partial y_2} \right |_{x,y_1}\frac{\partial f_1}{\partial y_1}/\frac{\partial f_1 }{\partial f_2}
$
$\displaystyle \left .\frac{\partial f_2}{\partial y_1}\right |_x=\frac{\frac{\partial f_2}{\partial y_1}\frac{\partial f_1 }{\partial y_2}-\frac{\partial f_2}{\partial y_2}
\frac{\partial f_1}{\partial y_1}}{\frac{\partial f_1}{\partial y_2}}=\frac{J(f_1,f_2)}{J(y_1,y_2)}\neq 0
$

Therefore $\displaystyle f_2(x,y_1,y_2(x,y_1))=0$ has solution $\displaystyle y_1=y_1(x)$

There is some clean-up, but the type-setting is murderous. This could be pursued by induction, which may be Aris.

Last edited by zylo; September 28th, 2015 at 01:17 PM.
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October 1st, 2015, 11:08 AM   #7
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The previous post is just a special case of induction proof by Aris. It is actually Exersize B.1. Fortunately Aris is online:

http://claudiomelo.paginas.ufsc.br/f...-Mechanics.pdf

Appendix B, Implicit Function Theorem and Jacobians
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