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 September 23rd, 2015, 11:46 AM #1 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 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, 04:48 PM   #2
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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, 05:38 AM #3 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 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 07:28 AM. September 24th, 2015, 07:29 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,103 Thanks: 2321 Where exactly in the book was this? September 28th, 2015, 07:17 AM   #5
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Quote:
 Originally Posted by skipjack Where exactly in the book was this?
Taylor, Advanced Calculus, 1955, par 8.1, pg 241 September 28th, 2015, 02:09 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.

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 02:17 PM. October 1st, 2015, 12:08 PM #7 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 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 Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Mr Davis 97 Calculus 1 April 16th, 2015 05:06 PM Omnipotent Calculus 6 November 30th, 2014 07:16 AM Omnipotent Calculus 1 November 27th, 2014 02:11 PM ray Algebra 6 April 22nd, 2012 04:50 AM Aurica Calculus 1 June 10th, 2009 06:39 PM

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