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Theorem: Let $F:X\subseteq \mathbb{R^n}\rightarrow \mathbb{R}$ be of class of $C^1$ and let $a$ be a point of the level set $S=\{x\in\mathbb{R^n}|F(x)=c\}$. If $F_{x_n}(a)\neq 0$ then there is a neighborhood $U$ of $(a_1,a_1\dots a_{n-1})\in\mathbb{R^{n-1}}$, a neighborhood $V$ for of $a_n\in\mathbb{R}$ and a function $f:U\subseteq\mathbb{R^{n-1}}\rightarrow V$ of class $C^1$ such that if $(x_1,x_2,...x_{n-1})\in U$ and $x_n \in V$ satisfy $F(x_1,x_2...x_n)=c$ ,then $x_n=f(x_1,x_2,...x_n)$.

I can't quite understand statement " If $F_{x_n}(a)\neq 0$ then there is a neighborhood $U$ of $(a_1,a_1\dots a_{n-1})\in\mathbb{R^{n-1}}$, a neighborhood $V$ for of $a_n\in\mathbb{R}$ and a function $f:U\subseteq\mathbb{R^{n-1}}\rightarrow V$ of class $C^1$ such that if $(x_1,x_2,...x_{n-1})\in U$ and $x_n \in V$ satisfy $F(x_1,x_2...x_n)=c$, then $x_n=f(x_1,x_2,...x_n)$."Why the statement is true and what idea it talks about?

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So, what exactly gives you trouble here? –  fedja Jan 9 '12 at 12:16
    
As the topic,i dont quite understand the theorem why the statement is ture? –  johnny Jan 9 '12 at 12:19

1 Answer 1

up vote 3 down vote accepted

Proofs of the implicit function theorem are available everywhere on the internet, so I won't repeat a proof here, but I'll give you an idea of what it's talking about.

The essence of the statement of the theorem is that any nicely behaved subset is locally the graph of a function. (The idea of 'nicely behaved' is made precise in the theorem's statement).

So for example if you plot the unit circle $S^1 \subseteq \mathbb{R}^2$, at any point on the circle which doesn't lie on the $x$-axis, you can find a neighbourhood on which the circle is the graph of a function. So for example we can view the portion of $S^1$ lying in the upper-half-plane as the graph of $\sqrt{1-x^2}$, and the portion in the lower-half-plane as the graph of $-\sqrt{1-x^2}$. Here's a nice illustration from Wikipedia:

In this illustration, the point $A$ has a neighbourhood which projects onto the $x$-axis; and we can recover the neighbourhood of $A$ on the graph by taking the set of points $(x,\sqrt{1-x^2})$ for $x$ in this projection onto the $x$-axis. On the other hand, there is no neighbourhood of the point $B$ which is locally the graph of a function, because it lies on the $x$-axis; this is the significance of the stipulation $F_{x_n}(a) \ne 0$ in your statement of the theorem.

The implicit function theorem takes this idea and places it in the arbitrary finite-dimensional case of functions $\mathbb{R}^n \to \mathbb{R}^m$ and subsets of $\mathbb{R}^{n+m}$, rather than just the case of functions $\mathbb{R} \to \mathbb{R}$ and subsets of $\mathbb{R}^2$. (In fact, your case is slightly less general: it considers functions $\mathbb{R}^{n-1} \to \mathbb{R}$ and subsets of $\mathbb{R}^n$.)

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