$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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Implicit Function Theorem

Let $U\subseteq\mathbb{R}^{n+m}$ be nonempty and open, and let $\phi\in C^1(U, \mathbb{R}^m)$. Suppose there exists $\vec{x}_0\in\mathbb{R}^n$ and $\vec{y}_0\in\mathbb{R}^m$ with $(\vec{x}_0, \vec{y}_0)\in U$ satisfying $\phi(\vec{x}_0, \vec{y}_0)=0$, and $\text{det}(D_{\vec{y}}\phi(\vec{x}_o, \vec{y}_0))\neq 0$. Then, there exists $a, b>0$ such that $B_{a}(\vec{x}_0)\times B_b(\vec{y}_0)\subseteq U$ and there exists a unique function $f\in C^1(B_a(\vec{x}_0), B_b(\vec{y}_0))$ such that $\phi(\vec{x}, f(\vec{x}))=\vec{0}$ for all $\vec{x}\in B_a(\vec{x}_0)$.

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Implicit Function Theorem

Let $U\subseteq\mathbb{R}^{n+m}$ be nonempty and open, and let $\phi\in C^1(U, \mathbb{R}^m)$. Suppose there exists $\vec{x}_0\in\mathbb{R}^n$ and $\vec{y}_0\in\mathbb{R}^m$ with $(\vec{x}_0, \vec{y}_0)\in U$ satisfying $\phi(\vec{x}_0, \vec{y}_0)=0$, and $\text{det}(D_{\vec{y}}\phi(\vec{x}_o, \vec{y}_0))\neq 0$. Then, there exists $a, b>0$ such that $B_{a}(\vec{x}_0)\times B_b(\vec{y}_0)\subseteq U$ and there exists a unique function $f\in C^1(B_a(\vec{x}_0), B_b(\vec{y}_0))$ such that $\phi(\vec{x}, f(\vec{x}))=\vec{0}$ for all $\vec{x}\in B_a(\vec{x}_0)$.

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concepts
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proof
FullPage
result
concepts
hypothesis
implications
proof

Implicit Function Theorem

Let $U\subseteq\mathbb{R}^{n+m}$ be nonempty and open, and let $\phi\in C^1(U, \mathbb{R}^m)$. Suppose there exists $\vec{x}_0\in\mathbb{R}^n$ and $\vec{y}_0\in\mathbb{R}^m$ with $(\vec{x}_0, \vec{y}_0)\in U$ satisfying $\phi(\vec{x}_0, \vec{y}_0)=0$, and $\text{det}(D_{\vec{y}}\phi(\vec{x}_o, \vec{y}_0))\neq 0$. Then, there exists $a, b>0$ such that $B_{a}(\vec{x}_0)\times B_b(\vec{y}_0)\subseteq U$ and there exists a unique function $f\in C^1(B_a(\vec{x}_0), B_b(\vec{y}_0))$ such that $\phi(\vec{x}, f(\vec{x}))=\vec{0}$ for all $\vec{x}\in B_a(\vec{x}_0)$.

Concepts

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Hypothesis

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Results

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Proof

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Implicit Function Theorem

Let $U\subseteq\mathbb{R}^{n+m}$ be nonempty and open, and let $\phi\in C^1(U, \mathbb{R}^m)$. Suppose there exists $\vec{x}_0\in\mathbb{R}^n$ and $\vec{y}_0\in\mathbb{R}^m$ with $(\vec{x}_0, \vec{y}_0)\in U$ satisfying $\phi(\vec{x}_0, \vec{y}_0)=0$, and $\text{det}(D_{\vec{y}}\phi(\vec{x}_o, \vec{y}_0))\neq 0$. Then, there exists $a, b>0$ such that $B_{a}(\vec{x}_0)\times B_b(\vec{y}_0)\subseteq U$ and there exists a unique function $f\in C^1(B_a(\vec{x}_0), B_b(\vec{y}_0))$ such that $\phi(\vec{x}, f(\vec{x}))=\vec{0}$ for all $\vec{x}\in B_a(\vec{x}_0)$.

Concepts

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Hypothesis

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Results

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Proof

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result
concepts
hypothesis
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proof