My Math Forum Fundamental Underlying Theorem of Implicit Functions
 User Name Remember Me? Password

 Real Analysis Real Analysis Math Forum

 September 23rd, 2015, 10:46 AM #1 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 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 $\displaystyle f_z|p_0=0$????
September 23rd, 2015, 03:48 PM   #2
Global Moderator

Joined: May 2007

Posts: 6,787
Thanks: 708

Quote:
 Originally Posted by zylo 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.

 September 24th, 2015, 04:38 AM #3 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 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.
 September 24th, 2015, 06:29 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,831 Thanks: 2161 Where exactly in the book was this?
September 28th, 2015, 06:17 AM   #5
Banned Camp

Joined: Mar 2015
From: New Jersey

Posts: 1,720
Thanks: 125

Quote:
 Originally Posted by skipjack Where exactly in the book was this?
Taylor, Advanced Calculus, 1955, par 8.1, pg 241

September 28th, 2015, 01:09 PM   #6
Banned Camp

Joined: Mar 2015
From: New Jersey

Posts: 1,720
Thanks: 125

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.

 October 1st, 2015, 11:08 AM #7 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 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

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Mr Davis 97 Calculus 1 April 16th, 2015 04:06 PM Omnipotent Calculus 6 November 30th, 2014 06:16 AM Omnipotent Calculus 1 November 27th, 2014 01:11 PM ray Algebra 6 April 22nd, 2012 03:50 AM Aurica Calculus 1 June 10th, 2009 05:39 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top