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Suppose $f$ is continuously differentiable near $a$ with $f^{\prime }(a)\ne 0$. It has a local inverse function $g$ near $b=f(a)$, andThe chain rule applied to $g(f(x))=x$ gives the formula. The nonzero derivative condition also ensures local invertibility; it does not assert a global inverse function on an arbitrary larger domain of a function. For $f(x)=x^2$ at $a=2$, the corresponding output is $b=4$, and the square root's derivative there is $1/4$. At $a=0$, this division is unavailable: the nonnegative restriction still has a square root inverse, but its finite derivative at zero does not exist. ^[ref].

The inverse function theorem answers a practical question: 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 $\det Df(a)\ne 0$, using the Jacobian determinant. The derivative of a local inverse map is obtained by inverting the Jacobian matrix, at the corresponding input. ^[ref]

For a one-dimensional example, $f(x)=x^2$ has $f^{\prime }(1)=2$. Near $x=1$, the inverse function is $g(y)=\sqrt{y}$, and the inverse function rule gives $g^{\prime }(1)=1/2$. 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. ^[ref]

Here is a two-dimensional example with an invertible derivative everywhere and infinitely repeated outputs: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 isThe 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.

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 ofAt $a$, the derivative $D{T}_y(a)=I-A^{-1}Df(a)$ 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.^[ref]

The inverse of the tangent restricted to $(-\pi /2,\pi /2)$, taking real values in that interval. Its derivative is $(1+x^2{)}^{-1}$. A formula involving an inverse tangent of a ratio may need a continuous branch of a multivalued function to describe a smooth field through a pole of that ratio.

For an arbitrary square matrix $A$, the derivative of the determinant in direction $H$ is the displayed expression. Differentiating the determinant's multilinear expansion gives the sum of entries of $H$ weighted by their cofactors, equal to the displayed trace. If $A$ is invertible, $\mathrm{adj}(A)=(\det A)A^{-1}$ gives $(\det A)\mathrm{tr}(A^{-1}H)$, and division by $\det A$ gives the Jacobi determinant derivative formula for the logarithmic derivative. The adjugate expression remains valid at singular matrices.

This asks whether the Jones polynomial equal to one characterizes the unknot among knots. A zero breadth of a Laurent polynomial makes $V_K$ a monomial; the identities $V_K(1)=1$ and $V_K^{\prime }(1)=0$ then force that monomial to be one. Thus using Jones breadth as a strictly positive complexity for nontrivial knots requires this detection property. To prove the derivative identity, differentiate the Jones polynomial skein relation at one for a crossing change in a knot. Both crossed diagrams have value one, while the smoothed two-component link has value minus two by Jones polynomial evaluation at one, so $V_{K_{+}}^{\prime }(1)=V_{K_{-}}^{\prime }(1)$. Crossing changes reduce to the unknot, whose derivative is zero. The corresponding Alexander polynomial detection property is false, as nontrivial untwisted Whitehead doubles demonstrate. The Jones question is discussed in research on closed four-strand braids.

This derivative component is carried by the jump set of a BV function. Swapping the two BV traces on a hypersurface and reversing the normal leaves it unchanged. In one dimension it consists of the point measure atoms, each weighted by the difference between the right and left limits.

The jump set consists of points where a BV space function has distinct one-sided approximate values across an approximate normal. In the plane its derivative contribution is $[u]{\nu }_u{\mathcal{H}}^1\lfloor J_u$. The Mumford–Shah functional penalizes its length rather than the jump amplitude.

A Killing vector field describes a direction in which geometry has a continuous symmetry: moving points together along its flow of a vector field preserves the metric tensor, including its pairing of tangent vectors. Such a symmetry gives a geodesic conserved quantity from a Killing vector, often a component of momentum or angular momentum. The direction may vary with position: rotation provides the basic example.

On a Riemannian manifold or Lorentzian manifold with metric tensor $g$, a smooth vector field $K$ is a Killing vector field when its Lie derivative of $g$ vanishes. If ${\Phi }_s$ is its local flow of a vector field, the equivalent statements areThe middle equality uses the pullback of a covariant tensor: the local flow maps are isometries on their domains. The last equality is the Killing equation, using the Levi-Civita connection. The infinitesimal condition holds throughout an open domain and integrates to exact local isometries; it is stronger than making the first derivatives of $g$ vanish at a single point. See Tong, section 4.3.1, equation (4.29).

For example, on the interval $0<x<1$ with Euclidean metric $d{x}^2$, $K={\partial }_x$ translates points by $x\mapsto x+s$. This is a local isometry, but a point reaches the end of the interval in finite flow parameter. A local flow of a vector field can be incomplete; see Killing fields on incomplete manifolds need not generate global isometries. The full-plane translation and rotation examples under Euclidean Killing vector field instead have flows for every real $s$.

If $f$ is a univalent function on the unit disc, then $B(f(0),|f^{\prime }(0)|/4)\subseteq f(\mathbb{D})$. The constant $1/4$ is sharp, as shown by the Koebe function. This gives useful comparisons between the derivative of a conformal map, its conformal radius, and its distance to the domain boundary.

For ${\eta }_k(z)=|z-k|$, the displayed flux is continuous at $z=k$ and has derivative $\mathrm{sgn}(z-k)f^{\prime }(z)$ almost everywhere. On a bounded state range with $|f^{\prime }|\le M$, its symmetric two-state version $Q(a,b)$ obeys $|Q(a,b)|\le M|a-b|$ and is Lipschitz in both arguments. The vanishing factor at $a=b$ removes any sign-function jump contribution.

LetThe Fourier coefficients of $\theta$ are absolutely summable, so $a$ belongs to the Wiener algebra, as does $a^{-1}$. The identity $\theta (2t)=\theta (t)+\theta (t+\pi )$ implies $a(2t)=a(t)a(t+\pi )$. Starting with a compactly supported Daubechies wavelet having $p\ge 3$ vanishing moments, the periodic phase change of a scaling function gives $m(t)=a(t+\pi )m_0(t)$, while the canonical high-pass construction leaves $\hat{\psi }(2t)=e^{-it}\overline{m_0(t+\pi )}{\hat{\phi }}_0(t)$ unchanged.

At a dyadic point $t_0=2\pi k/2^j$, the terms of the difference quotient with $n\ge j$ and $2^n|h|\le 1$ each contribute $1+O((2^{n}h{)}^2)$. Their number tends to infinity, their total error is bounded, and the remaining tail contributes a bounded amount. Hence $(\theta (t_0+h)-\theta (t_0))/h={\log }_2(1/|h|)+O(1)$, so neither $\theta$ nor $a$ has a finite derivative there. These points are dense. Away from the isolated zero of $m_0$ near $\pi$, multiplication by its nonzero smooth value cannot remove this nondifferentiability. Thus the new MRA low-pass filter is not differentiable throughout any neighborhood of $\pi$, despite integrability of the scaling function and unchanged vanishing moments. Ordinary higher derivatives at $\pi$ cannot be inferred, although the corresponding Peano zero survives.

An approximate analytic fixed point of the period-doubling renormalization operator can be certified by a frozen Newton correction for a fixed-point equation. A coefficient Banach space, rigorous residual and derivative bounds, and an a posteriori contraction ball give existence and local uniqueness in the full function space. Interval arithmetic controls finite computations; analytic estimates control the infinite coefficient tail.

// Target: dynamical-systems.bigb

A linear differential-equation solution may be sought as ${\int }_C{e}^{xt}f(t)dt$. Integration by parts transfers the factor $x$ to derivatives of $f$, giving an equation for the amplitude and boundary terms. Admissible contours must cancel those terms and make the integral converge.

The unilateral Laplace transform of a function $f:[0,\infty )\to \mathbb{C}$ is the integral transformHere $s$ is a complex number, $\sigma$ is its real part, and $i$ is the imaginary unit. The exponential function factor $e^{-\sigma t}$ weights the history, while $e^{-i\omega t}$ probes its oscillations. The integral starts at zero, so it uses only the history from that starting time onward. This section concerns ordinary functions; their value at the single endpoint does not change the integral. See MIT 18.03SC, definition and elementary examples.

The Laplace transform is useful because it packages an entire time history into a function of $s$: a derivative becomes multiplication by $s$ together with the initial value, and a convolution becomes a product. Thus a linear differential equation can become an algebraic equation that already remembers its initial value problem. The worked example below shows both descriptions of the same response.

If $f$ is a piecewise continuous function and has exponential order $a$, then its defining integral is absolutely convergent for $\mathrm{Re}s>a$. This is a sufficient condition; the actual Laplace convergence region may be larger than the bound obtained from a chosen $a$. ^[ref].

The bound has a short explanation. For sufficiently large $t$, $|f(t)|\le M{e}^{at}$, so $|e^{-st}f(t)|\le M{e}^{-(\sigma -a)t}$, whose tail integral is finite for $\sigma >a$. The finite initial interval is covered by piecewise continuity.

Three calculations make the convergence condition concrete. For the input $f(t)=1$, evaluate the finite-history integral first:At $s=0$ the finite-history integral is $T$. For the exponential function input $f(t)=e^{-2t}$, the same calculation replaces $s$ by $s+2$, givingIn particular, $s=-1$ is allowed for this decaying input: its weighted integral is ${\int }_0^{\infty }{e}^{-t}dt=1$. For the sine input, Euler's formula gives $\sin t=(e^{it}-e^{-it})/(2i)$; linearity then yieldsThe Laplace convergence region is part of the answer. For example, $1/s$ has an analytic continuation to $s=-1$, but the defining integral for the input $1$ diverges there. The sine calculation is also developed in MIT 18.03SC, further elementary transforms.

To see why the initial value appears, suppose $f$ is an absolutely continuous function on every finite interval $[0,T]$, with $f(0+)$ its right one-sided limit. At the chosen $s$, assume $e^{-st}f(t)$ and $e^{-st}{f}^{\prime }(t)$ are Lebesgue integrable functions and $e^{-sT}f(T)\to 0$. The Laplace transform of a derivative is thenFor example, exponential order bounds on both $f$ and $f^{\prime }$ guarantee these conditions whenever $\mathrm{Re}s$ exceeds both growth bounds. The initial value is the lower-end boundary contribution of integration by parts.

Consider the initial value problemWrite $Y=\mathcal{L}\{y\}$. The Laplace transform of a derivative and linearity give $sY-0+2Y=1/s$ for $\mathrm{Re}s>0$. A partial fraction decomposition therefore givesWe recognize the two elementary Laplace transforms already computed. Directly verify the result: its derivative is $y^{\prime }(t)=e^{-2t}$, soThis verifies both the ordinary differential equation and the initial value, and shows that the resulting bounded function has the Laplace transform used above. The response rises from zero toward $1/2$: at that value the input $1$ balances the term $2y$. The general method is described in MIT 18.03SC, solving initial value problems.

The same solution is a convolution. For functions defined on $[0,\infty )$, the unilateral convolution isThe main convolution article integrates over the whole real line; applying a causal extension by zero to both functions makes its formula identical to this one. Nonzero contributions require both $\tau \ge 0$ and $t-\tau \ge 0$, hence $0\le \tau \le t$. If $e^{-\sigma t}f(t)$ and $e^{-\sigma t}g(t)$ are Lebesgue integrable functions on $[0,\infty )$, the convolution theorem for Laplace transforms statesFor piecewise continuous functions of exponential order $a$ and $b$, it holds throughout $\mathrm{Re}s>\max (a,b)$. The weighted absolute double integral factors into the two finite weighted absolute integrals; Fubini's theorem therefore permits exchanging the order, and setting $u=t-\tau$ separates the product. ^[ref].

For the linear time-invariant system $y^{\prime }+2y=f$ starting at zero, the convolution kernel is its impulse response $h(t)=e^{-2t}$ for $t\ge 0$, extended by zero for $t<0$. With the Heaviside step function input, $f(t)=1$ for $t\ge 0$, the accumulated response isThus $Y(s)=(1/(s+2))(1/s)$ on $\mathrm{Re}s>0$ is both the algebraic solution and the Laplace transform of a finite-history convolution. Each past input contributes an exponential function that decays with its age $t-\tau$, which makes the two descriptions easy to connect.

Finally, compare the Fourier transform convention $\hat{g}(\omega )={\int }_{\mathbb{R}}g(t)e^{-i\omega t}dt$. If $f_0$ is the causal extension by zero of $f$, define $g_{\sigma }(t)=e^{-\sigma t}{f}_0(t)$. Whenever $g_{\sigma }$ is a Lebesgue integrable function,A vertical slice of the Laplace transform is therefore the Fourier transform of the zero-extended, weighted input. Positive $\sigma$ damps its positive-time history. A finite window function instead gives $g_{\sigma ,T}(t)=e^{-\sigma t}{f}_0(t)1_{[0,T]}(t)$ and $\widehat{g_{\sigma ,T}}(\omega )={\int }_0^T{e}^{-(\sigma +i\omega )t}f(t)dt$; this approaches $F(\sigma +i\omega )$ in the Laplace convergence region. Setting $\sigma =0$ directly requires an integrable zero-extended input: the input $1$ fails that condition, even though its Laplace transform exists for every $\sigma >0$.

The history of the Laplace transform spans several stages. Schiff's historical summary notes that Laplace's 1812 treatise acknowledged earlier work on integral transforms and describes later development of the modern Laplace transform method. ^[ref].