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September 23rd, 2015, 10:46 AM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100  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: 6,560 Thanks: 604  
September 24th, 2015, 04:38 AM  #3 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 
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: 19,297 Thanks: 1687 
Where exactly in the book was this?

September 28th, 2015, 06:17 AM  #5 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100  
September 28th, 2015, 01:09 PM  #6  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100  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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 
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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functions, fundamental, implicit, theorem, underlying 
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