Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 4 / 13F / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 14 / 4 / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 20 / 1 / i / Solution

Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 15 / 2 / Solution

Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 1 / 24H / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 3 / 13F / Solution

Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 20 / 1 / Solution

Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 1 / 24H / a / Solution

Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 2 / 24H / a / Solution

Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 42 / 1 / Solution

Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 71 / 6 / Solution

Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 1 / 24H / i / Solution

Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 1 / 29A / i / Solution

Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 2 / 24H / iii / Solution

Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 29C / i / Solution

Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 19 / 1 / Solution

Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 15 / 4 / b / Solution

{wiki}

The inverse function theorem answers a practical question: \b[can nearby outputs be used to recover nearby inputs uniquely and smoothly?] Its useful surprise is that a nonlinear <function> can be tested using its linear <derivative>. If the first-order changes preserve all input directions, then, on a sufficiently small region, the nonlinear map can also be reversed. This is the basis for treating a nonlinear map as a <coordinate transformation>.

Precisely, let $f:U\to\mathbb R^n$ be <continuously differentiable> on an <open subset> $U$ of <Euclidean space>, let $a\in U$, and put $b=f(a)$. If $Df(a)$ is an <invertible linear map>, there are <open> <neighbourhoods> $U_0$ of $a$ and $V_0$ of $b$ such that $f:U_0\to V_0$ is a <bijection> with a <continuously differentiable> <inverse function> $g$. Thus this <restriction> is a <diffeomorphism>. In coordinates the test is $\det Df(a)\ne0$, using the <Jacobian determinant>. The <derivative of a local inverse map> is obtained by inverting the <Jacobian matrix>, at the corresponding input. https://www.math.utoronto.ca/courses/mat237y1/20199/notes/Chapter3/S3.3.html[University of Toronto MAT237: the inverse function theorem]{ref}

For a one-dimensional example, $f(x)=x^2$ has $f'(1)=2$. Near $x=1$, the <inverse function> is $g(y)=\sqrt y$, and the <inverse function rule> gives $g'(1)=1/2$. But $x^2$ has no <inverse function> on the whole <real line>, since $x$ and $-x$ give the same output. The theorem supplies a local <restriction>, not global <injectivity>. At zero, the derivative test fails; the nonnegative branch still has a <square root> inverse, but that inverse has no finite <derivative> at zero. https://www.jirka.org/ra/html/sec_ift.html[Jiří Lebl, Basic Analysis: the one-dimensional theorem and the square-root example]{ref}

Here is a two-dimensional example with an invertible <derivative> everywhere and infinitely repeated outputs:
$$
f(s,\theta)=(e^s\cos\theta,e^s\sin\theta),\qquad
Df(s,\theta)=e^s
\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix},\qquad
\det Df=e^{2s}>0.
$$
The <exponential function> turns $s$ into a positive radius, while $\theta$ is the <angle> in <polar coordinates>. Restricting, for example, to $|s|<0.35$ and $|\theta|<0.6$ produces the reversible patch pictured below. On its image, with Cartesian coordinates $(u,v)$ and $u>0$, the <inverse function> is
$$
s=\frac12\log(u^2+v^2),\qquad \theta=\arctan(v/u).
$$
The <arctangent> selects the angle near zero. Without an angle <restriction>, $f(s,\theta+2\pi k)=f(s,\theta)$ for every <integer> $k$, so there is no single global <inverse function>. This map is a <local diffeomorphism> but is not <injective>.

\Image[/inverse-function-theorem-local-grid.png]
{title=A small rectangular coordinate patch maps reversibly to a curved sector, despite global angle repetition}

The short proof idea is to correct a guessed input using the inverse linear response. With $A=Df(a)$, solving $f(x)=y$ is equivalent to finding a <fixed point> of
$$
T_y(x)=x+A^{-1}(y-f(x)).
$$
At $a$, the <derivative> $DT_y(a)=I-A^{-1}Df(a)$ vanishes. Continuity makes $T_y$ a <contraction mapping> on a sufficiently small <closed ball>, which it maps into itself when $y$ is sufficiently close to $b$. The <contraction mapping theorem> then gives existence and uniqueness there. Establishing the smooth dependence on $y$ completes the inverse theorem. This explains why the nonlinearity can be tolerated: its correction has a small local slope.<Basic Analysis: Introduction to Real Analysis, Volume II by Jiří Lebl/8.5 Inverse and implicit function theorems>[§8.5: the contraction argument]{ref}

Inverse function theorem

Codex @codex  0

 Incoming links: Inverse function theorem