The Inverse Function Theorem
$\begin{cases}y_1 = f_1(x_1, x_2, ..., x_n) \\
y_2 = f_2(x_1, x_2, ..., x_n) \\
\cdots \\
y_n = f_n(x_1, x_2, ..., x_n)
\end{cases}$
$f_i \in C^1$ $\quad \forall i$ 라고 하자.
$(\mathbf{x}_0, \mathbf{y}_0)$에 대하여 $f(\mathbf{x}_0, \mathbf{y}_0) = \mathbf{0}$ 이라고 하자.
Jacobian $\hspace{0.3cm}$ $J = \frac{\partial (f_1, f_2, ..., f_n)}{\partial (x_1, x_2, ..., x_n)} = \begin{vmatrix}
\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \cdots & \frac{\partial f_1}{\partial x_n} \\
\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \cdots & \frac{\partial f_2}{\partial x_n} \\
\vdots & \vdots & \ddots & \\
\frac{\partial f_n}{\partial x_1} & \frac{\partial f_n}{\partial x_2} & \cdots & \frac{\partial f_n}{\partial x_n} \\
\end{vmatrix} \neq 0$ 이라고 하자.
Define
$\begin{cases}
F_1(y_1, y_2, ..., y_n, x_1, x_2, ..., x_n) = f_1(x_1, x_2, ..., x_n) - y_1\\
F_2(y_1, y_2, ..., y_n, x_1, x_2, ..., x_n) = f_2(x_1, x_2, ..., x_n) - y_2\\
\cdots\\
F_n(y_1, y_2, ..., y_n, x_1, x_2, ..., x_n) = f_n(x_1, x_2, ..., x_n) - y_n\\
\end{cases}$
By the implicit function theorem, we have unique functions $\quad x_i = g_i(y_1, y_2, ..., y_n)$ $\ $ near $\ $ $\mathbf{y}_0\ $ which are $\ C^1$.
$\begin{cases}
x_1 = g_1(y_1, y_2, ..., y_n) \\
x_2 = g_2(y_1, y_2, ..., y_n) \\
\cdots\\
x_n = g_n(y_1, y_2, ..., y_n)
\end{cases}$
다른 말로, $\boldsymbol{(\mathbf{x}_0, \mathbf{y}_0)}$ 근방에서
$\begin{align*}
\begin{cases}y_1 = f_1(x_1, x_2, ..., x_n) \\
y_2 = f_2(x_1, x_2, ..., x_n) \\
\cdots \\
y_n = f_n(x_1, x_2, ..., x_n)
\end{cases}
\Leftrightarrow
\begin{cases}
x_1 = g_1(y_1, y_2, ..., y_n) \\
x_2 = g_2(y_1, y_2, ..., y_n) \\
\cdots\\
x_n = g_n(y_1, y_2, ..., y_n)
\end{cases}
\end{align*}$
가 성립한다.
'Math > Calculus' 카테고리의 다른 글
Cauchy's mean value theorem (0) | 2022.04.17 |
---|---|
Finding Extrema (0) | 2022.04.16 |
직선의 direction vector 찾기 (0) | 2022.04.16 |
The Implicit Function Theorem (0) | 2022.04.15 |
Total derivative (0) | 2022.04.15 |