
Real Analysis Real Analysis Math Forum 
 LinkBack  Thread Tools  Display Modes 
September 23rd, 2015, 10:46 AM  #1 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  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 $\displaystyle f_zp_0\neq0$. Then f(x,y,z)=0 in a neighborhood of $\displaystyle p_0$ But then $\displaystyle f_zp_0=0$???? 
September 23rd, 2015, 03:48 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,705 Thanks: 670  
September 24th, 2015, 04:38 AM  #3 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
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 $\displaystyle f_zp_0\neq0$ THEN f(x,y,z) = 0 in a neigborhood of $\displaystyle p_0$ f(x,y,z) = xy + 2z f(0,0,0) = 0, $\displaystyle f_zp_0$ = 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: 20,375 Thanks: 2010 
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: 124  
September 28th, 2015, 01:09 PM  #6  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Implicit Function Theorem Quote:
$\displaystyle f_xdx+f_ydy+f_zdz=0$ 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 $\displaystyle x_n=x_n(x_1,..,x_{n1})$ 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 cleanup, but the typesetting 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: 124 
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 

Tags 
functions, fundamental, implicit, theorem, underlying 
Thread Tools  
Display Modes  

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