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{{Use dmy dates|date=December 2023}}
{{Calculus}}
In [[
In [[multivariable calculus]], this theorem can be generalized to any [[continuously differentiable]], [[vector-valued function]] whose [[Jacobian determinant]] is nonzero at a point in its ___domain, giving a formula for the [[Jacobian matrix]] of the inverse. There are also versions of the inverse function theorem for [[holomorphic function]]s, for differentiable maps between [[manifold]]s, for differentiable functions between [[Banach space]]s, and so forth.▼
The theorem applies verbatim to [[complex-valued function]]s of a [[complex number|complex variable]]. It generalizes to functions from
''n''-[[tuples]] (of real or complex numbers) to ''n''-tuples, and to functions between [[vector space]]s of the same finite dimension, by replacing "derivative" with "[[Jacobian matrix]]" and "nonzero derivative" with "nonzero [[Jacobian determinant]]".
▲
The theorem was first established by [[Émile Picard|Picard]] and [[Édouard Goursat|Goursat]] using an iterative scheme: the basic idea is to prove a [[fixed point theorem]] using the [[contraction mapping theorem]].
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There are two variants of the inverse function theorem.<ref name="Hörmander" /> Given a continuously differentiable map <math>f : U \to \mathbb{R}^m</math>, the first is
*The derivative <math>f'(a)</math> is surjective (i.e., the Jacobian matrix representing it has rank <math>m</math>) if and only if there exists a continuously differentiable function <math>g</math> on a neighborhood <math>V</math> of <math>b = f(a)</math> such that <math>f \circ g = I</math> near <math>b</math>,
and the second is
*The derivative <math>f'(a)</math> is injective if and only if there exists a continuously differentiable function <math>g</math> on a neighborhood <math>V</math> of <math>b = f(a)</math> such that <math>g \circ f = I</math> near <math>a</math>.
In the first case (when <math>f'(a)</math> is surjective), the point <math>b = f(a)</math> is called a [[regular value]]. Since <math>m = \dim \ker(f'(a)) + \dim \operatorname{im}(f'(a))</math>, the first case is equivalent to saying <math>b = f(a)</math> is not in the image of [[Critical point (mathematics)#Critical point of a differentiable map|critical points]] <math>a</math> (a critical point is a point <math>a</math> such that the kernel of <math>f'(a)</math> is nonzero). The statement in the first case is a special case of the [[submersion theorem]].
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\end{bmatrix}.
</math>
The Jacobian matrix of it at <math>(x, y)</math> is:
:<math>
\begin{bmatrix}
{e^x \cos y} & {-e^x \sin y}\\
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\end{bmatrix}
</math>
with
:<math>
\det
e^{2x} \cos^2 y + e^{2x} \sin^2 y=
e^{2x}.
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Yet another proof uses [[Newton's method]], which has the advantage of providing an [[effective method|effective version]] of the theorem: bounds on the derivative of the function imply an estimate of the size of the neighborhood on which the function is invertible.<ref name="hubbard_hubbard">{{cite book |first1=John H. |last1=Hubbard |author-link=John H. Hubbard |first2=Barbara Burke |last2=Hubbard|author2-link=Barbara Burke Hubbard |title=Vector Analysis, Linear Algebra, and Differential Forms: A Unified Approach |edition=Matrix |year=2001 }}</ref>
=== Proof for
We want to prove the following: ''Let <math>D \subseteq \R</math> be an open set with <math>x_0 \in D, f: D \to \R</math> a continuously differentiable function defined on <math>D</math>, and suppose that <math>f'(x_0) \ne 0</math>. Then there exists an open interval <math>I</math> with <math>x_0 \in I</math> such that <math>f</math> maps <math>I</math> bijectively onto the open interval <math>J = f(I)</math>, and such that the inverse function <math>f^{-1} : J \to I</math> is continuously differentiable, and for any <math>y \in J</math>, if <math>x \in I</math> is such that <math>f(x) = y</math>, then <math>(f^{-1})'(y) = \dfrac{1}{f'(x)}</math>.''
We may without loss of generality assume that <math>f'(x_0) > 0</math>. Given that <math>D</math> is an open set and <math>f'</math> is continuous at <math>x_0</math>, there exists <math>r > 0</math> such that <math>(x_0 - r, x_0 + r) \subseteq D</math> and<math display="block">|f'(x) - f'(x_0)| < \dfrac{f'(x_0)}{2} \qquad \text{for all } |x - x_0| < r.</math>
In particular,<math display="block">f'(x) > \dfrac{f'(x_0)}{2} >0 \qquad \text{for all } |x - x_0| < r.</math>
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To check that <math>g=f^{-1}</math> is C<sup>1</sup>, write <math>g(y+k) = x+h</math> so that
<math>f(x+h)=f(x)+k</math>. By the inequalities above, <math>\|h-k\| <\|h\|/2</math> so that <math>\|h\|/2<\|k\| < 2\|h\|</math>.
On the other hand, if <math>A=f^\prime(x)</math>, then <math>\|A-I\|<1/2</math>. Using the [[geometric series]] for <math>B=I-A</math>, it follows that <math>\|A^{-1}\| < 2</math>. But then
:<math> {\|g(y+k) -g(y) - f^\prime(g(y))^{-1}k \| \over \|k\|}
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=== Over a real closed field ===
The inverse function theorem also holds over a [[real closed field]] ''k'' (or an [[
The usual proof of the IFT uses Banach's fixed point theorem, which relies on the Cauchy completeness. That part of the argument is replaced by the use of the [[extreme value theorem]], which does not need completeness. Explicitly, in {{section link||A_proof_using_the_contraction_mapping_principle}}, the Cauchy completeness is used only to establish the inclusion <math>B(0, r/2) \subset f(B(0, r))</math>. Here, we shall directly show <math>B(0, r/4) \subset f(B(0, r))</math> instead (which is enough). Given a point <math>y</math> in <math>B(0, r/4)</math>, consider the function <math>P(x) = |f(x) - y|^2</math> defined on a neighborhood of <math>\overline{B}(0, r)</math>. If <math>P'(x) = 0</math>, then <math>0 = P'(x) = 2[f_1(x) - y_1 \cdots f_n(x) - y_n]f'(x)</math> and so <math>f(x) = y</math>, since <math>f'(x)</math> is invertible. Now, by the extreme value theorem, <math>P</math> admits a minimal at some point <math>x_0</math> on the closed ball <math>\overline{B}(0, r)</math>, which can be shown to lie in <math>B(0, r)</math> using <math>2^{-1}|x| \le |f(x)|</math>. Since <math>P'(x_0) = 0</math>, <math>f(x_0) = y</math>, which proves the claimed inclusion. <math>\square</math>
Alternatively, one can deduce the theorem from the one over real numbers by [[Tarski's principle]].{{citation needed|date=December 2024}}
==See also==
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