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The characteristic function of a real random variable $X$ isIt determines the probability distribution of $X$.

Here $t\in \mathbb{R}$ and $i^2=-1$. By Euler's formula, its real part and imaginary part are $\mathbb{E}[\cos (tX)]$ and $\mathbb{E}[\sin (tX)]$. The complex exponential has absolute value one, so this expected value exists for every real random variable, even without finite absolute moments. Moreover,The bound uses the triangle inequality for a vector-valued integral; the last identity is complex conjugate symmetry.^[ref] Consequently the real part is an even function and the imaginary part is an odd function.

For a discrete probability distribution with values $x_j$ and probabilities $p_j$, the characteristic function is ${\sum }_j{p}_j{e}^{it{x}_j}$: a weighted average of points on the unit circle in the complex plane. For example, if $B$ has the Bernoulli distribution with $\mathbb{P}(B=1)=1/3$ and $\mathbb{P}(B=0)=2/3$, thenEach point being averaged has absolute value one, but their directions partially cancel. At $t=\pi$, the same average is $1/3$; at $t=2\pi$, it returns to one. The characteristic function is thus neither a probability density function nor a probability mass function: its argument is a frequency parameter, and its values can be complex numbers. This Bernoulli distribution has no probability density function with respect to Lebesgue measure, yet its characteristic function is defined at every frequency.

For a probability density function $f$, the same definition becomes ${\phi }_X(t)={\int }_{\mathbb{R}}{e}^{itx}f(x)dx$, the Fourier transform with the displayed sign convention. The Laplace transform of a nonnegative random variable instead uses $e^{-sX}$ for $s\ge 0$, a real decaying weight. For the Bernoulli distribution above, it is $2/3+e^{-s}/3$; neither transform is the probability density function. The uniqueness theorem for characteristic functions explains why the oscillating averages still retain the entire probability law, while the characteristic function of a sum of independent variables turns combining laws into multiplication. These two properties make the representation useful even when directly computing a sum's probability distribution is cumbersome.

For the constant-coefficient model $x^{\prime }(t)=-kx(t-\tau )$ with $k>0$ and $\tau \ge 0$, the zero equilibrium point is asymptotically stable if and only if $k\tau <\pi /2$. For positive $\tau$, perturbations are measured by the supremum norm of the initial history of delayed feedback; in the stable regime, every continuous history produces a solution tending to zero. At equality, nondecaying sinusoids prevent asymptotic stability. For $k\tau >\pi /2$, growing oscillatory solutions exist. These statements concern this ideal linear time-invariant system, rather than an arbitrary controlled plant. The corresponding linear equation and its threshold are developed in Cain and Reynolds, section 6.4, pages 174–177.

Substitute the exponential function $x(t)=e^{st}$, allowing a complex number $s$. The characteristic equation isFor positive feedback delay this is not a characteristic polynomial: it has infinitely many roots. A root's real part determines growth or decay of its exponential mode, and its imaginary part determines the angular frequency. To find the first boundary, set $s=i\omega$ with $\omega >0$ and use Euler's formula:The smallest positive crossing is $\omega =k$ and $\omega \tau =\pi /2$. Thus at ${\tau }_c=\pi /(2k)$ the complex conjugate roots are $s=\pm ik$. In fact, with the matching initial history of delayed feedback,solves the delay differential equation exactly: a quarter-cycle feedback delay turns the past displacement into precisely the current derivative. Its cycle lasts $2\pi /k=4{\tau }_c$. The amplitude is arbitrary; this linear model does not select a finite-amplitude oscillation.

The roots cross toward growth as feedback delay increases. Differentiating the characteristic equation at fixed $k$ givesThis calculation identifies the direction of the first destabilization. The full threshold also requires ruling out earlier unstable roots: if $s=\alpha +i\beta$ with $\alpha \ge 0$ and $\beta >0$, the characteristic equation implies $\cos (\beta \tau )\le 0$ and $0<\beta \le k$, hence $k\tau \ge \beta \tau \ge \pi /2$. A real nonnegative root is impossible because $s+k{e}^{-s\tau }>0$ there. This supplies the geometric reason the threshold is a quarter-cycle rather than an arbitrary fitting constant.

The tradeoff is concrete. Without feedback delay, increasing $k$ shortens the exponential decay time constant $1/k$. With known $\tau >0$, the same proportional control must satisfy $k<\pi /(2\tau )$. For $\tau =0.2\mathrm{s}$, the threshold is $k\approx 7.854{\mathrm{s}}^{-1}$: $k=5{\mathrm{s}}^{-1}$ is stable and $k=8{\mathrm{s}}^{-1}$ is unstable. A design close to the boundary leaves little room for additional feedback delay; reducing $k$ or reducing measurement latency increases that room. The zero-delay time constant is only a comparison scale, not the decay time of the delayed response. Frank discusses the gain–delay compromise in Control Theory Tutorial, section 13.4 and figure 13.3.

An entire function is a holomorphic function whose domain of a function is the whole complex plane. Its derivative exists at every $z\in \mathbb{C}$. Every polynomial, the complex exponential function $e^z$, the product $z{e}^z$, and the complex sine function $\sin z$ are entire functions. The Taylor series of an entire function about any finite centre converges throughout the complex plane.

The complex exponential function is a particularly revealing entire function. Writing $z=x+iy$, Euler's formula givesThus the complex exponential function never takes the value zero. Conversely, for any $w\ne 0$, choose an argument $\theta$ of $w$. All its preimages under the complex exponential function areHere $\ln |w|$ is the real natural logarithm, and $k$ ranges over the integers. Substitution gives $e^{z_k}=|w|e^{i\theta }=w$; equality of moduli and arguments shows these are all the solutions. The complex exponential function therefore has image exactly $\mathbb{C}\setminus \{0\}$, with infinitely many preimages for each nonzero value.

In the pictured vertical strip $-1\le x\le 1$, the real part sets the modulus between $e^{-1}$ and $e$, while the imaginary part sets the argument modulo $2\pi$. The two drawn bands each make one full turn around the same annulus; their corresponding points differ by $2\pi i$ and have identical images. Allowing all real $x$ gives every positive modulus, so the whole complex plane covers every nonzero output.

An entire function has exceptionally strong global constraints. The Liouville theorem says that a bounded function which is also an entire function must be a constant function. Accordingly, the nonconstant entire functions above are unbounded somewhere in the complex plane, even when their restrictions to the real axis are bounded: the complex sine function satisfies $\sin (iy)=i\sinh y$, growing through the hyperbolic sine. The Little Picard theorem adds a different constraint: an unbounded nonconstant entire function still cannot leave two distinct complex values out of its image. These results make entire functions useful when global bounds or restrictions on the image can establish rigidity without solving for the function explicitly.

In the opening of his 1880 memoir, Émile Picard credits Karl Weierstrass with the terminology of entire functions and describes them by everywhere-convergent power series. ^[ref]

For a smooth, locally invertible material map $\mathbf{x}=\mathbf{X}(\mathbf{a},t)$ with ${\mathbf{X}}_t=\mathbf{u}(\mathbf{X},t)$ and positive Jacobian determinant $J=\det (\partial \mathbf{X}/\partial \mathbf{a})$, Euler's volume identity states that $J_t=J\nabla \cdot \mathbf{u}$. Indeed, $F=\partial \mathbf{X}/\partial \mathbf{a}$ obeys $F_t=(\nabla \mathbf{u})F$, so the determinant derivative is $J_t=J\mathrm{tr}(F^{-1}{F}_t)=J\mathrm{tr}(\nabla \mathbf{u})$. This form of the Jacobian evolution of a smooth flow applies also to explicitly time-dependent velocity fields. It gives the material-volume Reynolds transport theorem by change of variables to the fixed particle labels. It is distinct from the complex-exponential Euler's formula.

Fourier analysis describes a function by the oscillations that compose it. A graph records values at successive positions or times; a frequency spectrum records how much of each frequency is present, together with its wave phase. The same complicated shape can become a short list of simple components. For periodic functions, the components form a Fourier series; on the whole real line, the Fourier transform uses a continuous range of frequencies, and the Fourier inversion theorem puts them back together.

For a concrete example, consider the trigonometric polynomialWith $t$ measured in seconds, its three Fourier harmonics have frequencies $1$, $3$, and $5$ cycles per second and amplitudes $1$, $1/2$, and $1/4$. Each cosine has zero wave phase. Adding them creates the sharper peaks in the figure. The frequency spectrum makes the recipe immediately visible; changing an amplitude changes the shape without changing the component's frequency. Fourier orthogonality lets us recover each Fourier coefficient by averaging against the corresponding sine or cosine, because the other Fourier harmonics cancel over a full period of a function.

The deeper reason these components are so useful is that they simplify operations. By Euler's formula, the exponential function $e^{i\omega x}$ packages a cosine and a sine into one Fourier mode. A translation or a derivative changes only its multiplier:Thus Fourier modes are simultaneous eigenfunctions of these operations. In Fourier series, this gives a literal diagonalization in the Fourier basis; on the real line, the Fourier transform gives the continuous counterpart, with each frequency handled separately. The Translation property of the Fourier transform and the Fourier transform of a derivative turn shifting and differentiating a whole function into multiplication of its Fourier transform. A convolution likewise becomes multiplication by the convolution theorem. This is beautiful because one description simplifies several apparently different operations at once.

For example, the heat equation $u_t=\kappa u_{xx}$, with thermal diffusivity $\kappa >0$, evolves a spatial Fourier mode asFine ripples have larger angular frequencies and decay faster: the formula explains why heating smooths a jagged temperature profile. For a general initial profile with sufficient regularity, the Fourier transform gives $\hat{u}(\omega ,t)=e^{-\kappa {\omega }^{2}t}\hat{u}(\omega ,0)$, and Fourier inversion recovers the temperature. Similarly, multiplying selected parts of a frequency spectrum suppresses unwanted components. Claude E. Shannon's Communication in the Presence of Noise (1949), Section III, describes ideal filtering as a projection that retains some frequency coordinates and removes others.^[ref]

The physical motivation is already visible in Joseph Fourier's Théorie analytique de la chaleur (1822). Chapter III constructs heat profiles using sums of sines and cosines; Chapter IX, articles 355–359, passes from Fourier series to integrals as the spatial interval becomes unbounded. These heat problems explain the motivation for the discrete and continuous descriptions. Their modern formulations specify which functions admit reconstruction and in what sense the expansions converge.^[ref][ref]

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].

Mathematics studies patterns in quantity, structure, space, and change by giving objects precise mathematical definitions and examining what follows from them. Arithmetic starts with operations on numbers; algebra organizes relations and transformations; geometry and topology studies shapes and the structures preserved by maps; analysis studies limits and continuous change. These fields connect because the same structure can appear in different forms.

A small example shows the value of changing viewpoint. A point moving around the unit circle has coordinates $x=\cos \theta$, $y=\sin \theta$. Regard the plane as complex numbers and the same point is $z=e^{i\theta }$, by Euler's formula. Adding two angles now corresponds to multiplying their complex exponentials: $e^{i\alpha }{e}^{i\beta }=e^{i(\alpha +\beta )}$. A geometric rotation has become an algebraic operation.

The connection reaches calculus too. Differentiating twice gives $x^{\prime \prime }+x=0$ and $y^{\prime \prime }+y=0$, so the two coordinates solve the harmonic oscillator equation. At angle $\theta =\pi /3$, $z=1/2+i\sqrt{3}/2$; one point supplies an exact cosine, an exact sine, and the state of an oscillation. The Pythagorean trigonometric identity records that the modulus stays one. Seeing these as descriptions of one object makes each formula easier to remember and reuse.

This is why a useful mathematical explanation provides both a reliable mathematical proof and a way to see what the result is doing. In his 1994 essay, William Thurston discusses mathematical progress in terms of advancing understanding, and uses the many interpretations of a derivative to illustrate this aim. ^[ref] A diagram, a computed example, and a concise argument can make different parts of the same idea accessible.

Applications ask us to select a mathematical model of a physical or other system. Its assumptions determine which conclusions transfer to that system. Within mathematics, the same practice of making assumptions explicit lets a result about groups, vector spaces, or probability distributions apply to many concrete examples. The beauty is the combination of precision and reuse: an argument can reveal more than the particular case that first suggested it.

Rationalizing the denominator givesHence its real part, imaginary part, complex modulus, and principal argument are respectivelyThe quadrant matters: the real part is positive and the imaginary part negative. For the exponential expression, Euler's formula gives $e^{2+i\pi /4}=e^2(1+i)/\sqrt{2}$, so the same four quantities areGeneral complex arguments are obtained from each stated principal argument by adding $2\pi k$, $k\in \mathbb{Z}$.

Factor out the same half-angle phase from the numerator and denominator:Here Euler's formula converts the bracketed sums to cosine and sine. Both sides require $\theta \notin 2\pi \mathbb{Z}$; at the excluded angles the original denominator vanishes.

Squaring the modulus and doubling the complex argument givesThis uses Euler's formula $e^{i\theta }=\cos \theta +i\sin \theta$.

Put $w=e^{i\theta }$. By Euler's formula, $w-w^{-1}=2i\sin \theta$. The binomial theorem givesGroup opposite powers and use $w^j-w^{-j}=2i\sin (j\theta )$:Since $(2i{)}^7=-128i$, division by $-2i$ yieldsThis derivation is an application of the same exponential representation underlying De Moivre's theorem.

To invert the complex cosine, let $z=\cos w$ and put $u=e^{iw}$. Since the exponential function is nonzero, $u\ne 0$, and Euler's formula gives $2z=u+u^{-1}$. Multiplying by $u$ yields the quadratic equation $u^2-2zu+1=0$. ThusChoosing either square root of $1-z^2$ merely exchanges the two signs. The two values have product one, so neither vanishes; taking a complex logarithm is legitimate.

For each choice of sign, all solutions of $e^{iw}=u$ satisfy $iw=\mathrm{Log}u+2\pi in$. Therefore the full inverse relation isConversely, exponentiating any displayed value recovers one root of the quadratic equation, and hence $(e^{iw}+e^{-iw})/2=z$, proving completeness. With a multivalued complex logarithm, the $2\pi n$ term is already included and this is the source's compact formula for inverse cosine.

The principal inverse cosine requires compatible choices of logarithm branch, square root and sign. Both signs cannot represent that same principal function indiscriminately: at $z=0$, the two principal-log values are $\pi /2$ and $-\pi /2$, whereas the principal $\arccos 0$ is $\pi /2$. Thus the printed $\pm$ is correct for the inverse relation, not a pair of interchangeable principal-value formulas.

Using Euler's formula, the difference of the two complex exponentials is $2i\sin (\pi /4)=i\sqrt{2}$. HenceThese are the real part and imaginary part, respectively.

Work first on $x>0$, where the proposed logarithm is defined. Substitution of $y=x^m$ into the Cauchy-Euler differential equation gives the indicial equationIts distinct indicial roots are $m=\pm i$. Using $x^{\pm i}=e^{\pm i\log x}$ and Euler's formula, a real basis of solutions is $\cos (\log x),\sin (\log x)$. Their Wronskian is $1/x$, so they are linearly independent and span the two-dimensional solution space of a homogeneous linear differential equation. ThereforeFor the requested change of variables, set $Y(z)=y(e^z)$. The chain rule givesThe transformed second-order linear differential equation is $Y^{\prime \prime }+Y=0$. Its characteristic roots are again $\pm i$, giving $Y(z)=A\cos z+B\sin z$. Substituting $z=\log x$ recovers precisely the previous general solution.

If solutions on $x<0$ are wanted, the same calculation uses $z=\log |x|$ and gives independent constants on that interval. Zero is a singular point of a differential equation; the logarithmic oscillations do not define a nonzero continuous solution through it.

By Euler's formula, $z=e^{i\theta }$. Factoring out the half-angle givesSince $0\le \theta <\pi$, the cosine is positive. ThereforeOn the Argand diagram, the points $0,1,z,1+z$ form a rhombus. Its diagonal from $0$ to $1+z$ bisects the angle between the unit vectors $1$ and $z$, so its direction is $\theta /2$; resolving either side along the diagonal gives its length $2\cos (\theta /2)$.

Using Euler's formula, the complex exponential function satisfiesHence

A subgroup $N\le G$ is normal whenIts left and right cosets then agree, and multiplicationis well defined on the cosets. The identity is $N$, the inverse of $gN$ is $g^{-1}N$, and associativity descends from $G$, so these cosets form the quotient group $G/N$.

For a group homomorphism $\theta :G\to H$, its kernel and image areThe kernel is normal becausewhenever $k\in \ker \theta$. Conversely, if $K⊴G$, the quotient mapis a homomorphism with kernel $K$. The image of any homomorphism is closed under products and inverses, so it is a subgroup of $H$. Finally, the mapis well defined and bijective and preserves multiplication. This is the first isomorphism theorem.

DefineEuler's formula shows that $\Phi$ is a homomorphism, its image is the complex unit circle, and its kernel is $\mathbb{Z}$. The first isomorphism theorem therefore givesThe image of $\mathbb{Q}/\mathbb{Z}$ consists exactly of the roots of unity: if $t=p/q$, then $\Phi (t{)}^q=1$, while every element of finite order on the unit circle has an argument that is a rational multiple of $2\pi$.

Euler's formula for a connected plane graph is $|V|-|E|+|F|=2$. If the graph is triangle-free, every face has boundary length at least four, soEuler's formula then gives $|E|\le 2|V|-4$, so the average degree is less than four. There is a vertex of degree at most three. Delete it, color the remaining graph inductively with four colors, and restore it using a color absent from its at most three neighbors. Thus

Using Euler's formula and the Inverse Fourier transforms of phase factors,andThese are distributional identities, with $\delta$ the Dirac delta function.

Using Euler's formula,ThusOrdering the modes as $(-1,0,1)$, let $a=1/2+i/4$. ThenConsequentlyThis matrix is triangular, so its eigenvalues are its diagonal entries: