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The real exponential function $\exp (x)=e^x$ converts addition into multiplication:Its values are positive, and it defines a bijection from the real numbers to the positive real numbers. The natural logarithm is its inverse function. These properties connect arithmetic with calculus: adding an amount to the input multiplies the output by a fixed factor, and the local rate of change equals the current value. The exponential series gives one construction; an equivalent characterization is the unique solution of a differential equation $y^{\prime }=y$, $y(0)=1$.

For a concrete example, $y(t)=10e^{0.2t}$ starts at $10$ and satisfies $y^{\prime }=0.2y$. The absolute derivative increases with $y$, while the relative rate $y^{\prime }/y$ stays $0.2$. At $t=5$, $y=10e\approx 27.1828$ and $y^{\prime }\approx 5.43656$. The exponential function therefore models accumulation proportional to what is already present, rather than addition of a fixed amount at every step. A positive constant rate gives exponential growth; a negative one gives exponential decay.

Every positive real base $b\ne 1$ fits the same construction through $b^x=e^{x\ln b}$. Thus its derivative is $(\ln b)b^x$: base $e$ is special because $\ln e=1$. The exponential function is also unchanged in form by differentiation and integration, helping explain its repeated appearance in linear ordinary differential equations. Its real definition, inverse relationship, and derivative rules are developed in OpenStax, integrals, exponential functions, and logarithms.

For a positive integer $k$, multiply the exponential series by $t^{k-1}$ and integrate term by term from zero to $x$. Its uniform convergence on every compact interval justifies the operation and proves the title identity. The resulting power series converges absolutely for every real or complex $x$, by the ratio test. Repeated integration by parts evaluates the right side as an exponential times a polynomial, with its constant chosen to make the integral vanish at zero.

For Picard iteration $y_{n+1}(x)=b+{\int }_a^{x}f(s,y_n(s))ds$, $y_0=b$, assume $|f|\le M$ and a Lipschitz constant $L$ on a rectangle containing every iterate. The first difference is at most $M|x-a|$. Integrating the preceding bound multiplied by $L$ proves the displayed estimate by mathematical induction. Its uniform bounds sum to a convergent exponential series for every finite $Lh$, so the Weierstrass M-test gives uniform convergence without requiring $Lh<1$.

For two arbitrary initial functions $y,z$ in a function set preserved by the Picard iteration operator $T$, the same induction givesThus some iterate is a contraction mapping on any fixed finite interval for which every iterate stays in that set. The iterated contraction argument gives a unique fixed point of $T$ on a nonempty complete metric space, without requiring $Lh<1$. For a rectangle with field bound $M$ and vertical radius $k$, $Mh\le k$ supplies the required self-mapping property.

For every fixed $m\ge 0$, choose an integer $N>m/2$. The positive exponential series gives $e^{t^2}\ge t^{2N}/N!$, so $|t{|}^m{e}^{-t^2}\le N!|t{|}^{m-2N}\to 0$. Equivalently, $|x{|}^{-m}{e}^{-1/x^2}\to 0$ as $x\to 0$.

For $N\sim \mathrm{Pois}(\lambda )$ with $\lambda \ge 0$, insert its probability mass function into the expected value:The exponential series shows finiteness for every real $t$. Differentiating this moment-generating function gives mean and variance both $\lambda$. Taking the logarithm gives the cumulant-generating function $\lambda (e^t-1)$, so every positive-order cumulant is also $\lambda$.

Rewrite the linear equation as $(x^{2}y{)}^{\prime }=0$. The initial value gives the unique solutionThe Picard iteration is its integral equationThe proposed formula holds for $n=0$. If it holds at $n$, termwise integration of the finite sum and $d(\log t)/dt=1/t$ giveThis proves the expression by induction. The exponential series then gives $y_n(x)\to e^{-2\log x}=x^{-2}$. Convergence is also uniform on every compact subinterval of $(0,\infty )$, since $|\log x|$ is bounded there and the exponential majorant series converges. This is Picard iteration for a simple-pole linear differential equation.

Near $x=0$, separation of variables gives $dy/y=dx/(1-x)$, so $\log |y|=-\log (1-x)+C$. The initial condition selectsDirect substitution verifies both the ordinary differential equation and the initial condition. The endpoint $x=1$ is excluded because the coefficient and this solution are singular there; the displayed interval is the maximal connected real interval containing the initial point.

For $|x|\le h<1$ and any two admissible values $y,\tilde{y}$,This is a Lipschitz condition in the dependent variable, uniform in $x$, with Lipschitz constant $L=(1-h{)}^{-1}$. Also $|f(x,y)|\le (1+k)/(1-h)$ on the prescribed rectangle, so the usual local hypotheses of the Picard-Lindelöf theorem hold.

Integrating from the initial point gives the integral equationConversely, differentiating this equation using the fundamental theorem of calculus recovers the initial value problem. Start Picard iteration with the constant initial approximation $y_0(x)=1$ and setThe first three approximations areTo prove the general expression by mathematical induction, put $\ell (x)=\log (1-x)$. The initial approximation gives the formula for $n=0$. If it holds for $n$, then the substitution rulegivesThus the formula follows at every stage.

For every fixed real $x<1$, these are partial sums of the exponential series evaluated at the finite real number $-\log (1-x)$. ThereforeThis range is larger than $|x|<1$: the expansion is in powers of $\log (1-x)$, not a Taylor series in $x$. For example it converges at $x=-2$. It also converges uniformly on every compact interval contained in $(-\infty ,1)$, because $|\log (1-x)|$ is bounded there and the exponential series supplies a uniform summable bound. At $x=1$ the approximations are undefined, while for $x>1$ the real integral from zero crosses the singularity. This is Picard iteration for a simple-pole linear differential equation.

For the Weierstrass M-test proof of the Picard-Lindelöf theorem, the required bound isTo see this, use Picard iteration starting with $y_0(x)=b$ andIf $y_n$ takes its values in $[b-k,b+k]$, then $|y_{n+1}-b|\le M|x-a|\le Mh\le k$. Thus every Picard iterate remains in the rectangle where the hypotheses hold. Successive applications of the Lipschitz condition and integration give the factorial estimate for Picard iterationFor example, the $n=0$ estimate is $M|x-a|$, and integrating the $n$th estimate multiplied by $L$ proves the next one. The sum of the uniform bounds isfor every finite $Lh$, by the exponential series. Hence the telescoping series of Picard iterates converges uniformly by the Weierstrass M-test. The uniform limit stays in the closed rectangle, and the Lipschitz condition permits passage to the limit in the integral equation. The fundamental theorem of calculus then gives the desired solution of a differential equation on $[a-h,a+h]$.

The exponential series gives $x^x=e^{x\log x}={\sum }_{k=0}^{\infty }(x\log x{)}^k/k!$. Since $|x\log x|\le 1/e$ on $(0,1)$, the Weierstrass M-test proves uniform absolute convergence, using the summable bounds $e^{-k}/k!$. Therefore one can integrate the series term by term over this finite interval. Using part i,This is the alternating identity in sophomore's dream. The same estimate also proves absolute convergence of the integrated series; no cancellation is needed to justify the interchange.

On bounded intervals the exponential series converges uniformly, allowing termwise integration. Since $e^{-t^2}=1-t^2+t^4/2+O(t^6)$,The requested first two nonzero terms are thereforeThe function is the error function, with $I(0)=0$ and derivative $2e^{-x^2}/\sqrt{\pi }$.

A Poisson distribution with parameter $\lambda >0$ assigns probability mass functionThe exponential series makes these probabilities sum to one. Shifting the indices in the absolutely convergent sums givesHence $E{Z}^2={\lambda }^2+\lambda$ andFor independent random variables $Z_1,\dots ,Z_n$ of parameter one, let $S_n={\sum }_j{Z}_j$. Summing their product probability mass function over all nonnegative $k_1+\cdots +k_n=k$ givesThe final equality follows from the multinomial theorem applied to $(1+\cdots +1{)}^k$. Thus $S_n$ has the Poisson distribution of parameter $n$.

Each $Z_j$ has expectation and variance one, so the central limit theorem gives convergence in distributionZero is a continuity point of the standard normal cumulative distribution function. ThereforeThe discrete atom at $S_n=n$ causes no difficulty: the central limit theorem applies to the entire event with threshold zero.

For the Poisson process of claims and relative safety loading $\rho$, the premium rate in the classical risk model is $c=(1+\rho )\lambda \mu$. Dividing the adjustment coefficient equation $\lambda (M(r)-1)=cr$ by $\lambda$ explains why the arrival rate cancels. Interpret $r_{\infty }$ as the upper endpoint of the finite moment-generating function domain, so $M(r)<\infty$ for $0\le r<r_{\infty }$ and its stated divergence occurs at that endpoint.

Set $F(r)=M(r)-1-(1+\rho )\mu r$. ThenDifferentiation is valid inside the finite-transform domain, and its positive neighbourhood gives finite moments of every order. Thus $F$ is strictly convex and negative just to the right of zero. If $r_{\infty }<\infty$, divergence of $M(r)$ makes $F(r)\to \infty$ at the endpoint. If $r_{\infty }=\infty$, choose $a>0$ with $\mathbb{P}(X_1\ge a)>0$; then $M(r)\ge \mathbb{P}(X_1\ge a)e^{ar}$, which eventually exceeds every linear function. Again $F(r)\to \infty$. The intermediate value theorem gives a positive zero.

There is at most one positive zero: strict convexity and $F(0)=0$ make the secant slope $F(r)/r$ strictly increasing for $r>0$. It starts at $-\rho \mu$ and crosses zero exactly once. HenceThis proves the secant-slope existence criterion for an adjustment coefficient in the present setting.

Put $m_j=\mathbb{E}{X}_1^j$. For $u>0$, the positive remainder of the exponential series gives $e^u>1+u+u^2/2$. Apply it to $u=R{X}_1$ and take expectations:Dividing by $R>0$ yieldsKeeping the cubic term similarly givesThe polynomial $g(r)=m_3{r}^2+3m_{2}r-6\rho \mu$ is strictly increasing for $r\ge 0$, is negative at zero and tends to infinity. Let $r_2$ be its unique positive zero. The last inequality says $g(R)<0$, soFinally, $3m_2{r}_1=6\rho \mu$, whence $g(r_1)=m_3{r}_1^2>0$. Monotonicity gives $r_2<r_1$. Thus the requested upper bounds can in fact be sharpened to the strict ordering $0<R<r_2<r_1$. The bounds themselves need not lie in the finite-transform domain; the argument only evaluates the actual transform at $R$ and polynomial moments elsewhere. These are polynomial moment bounds for the adjustment coefficient.

For rate-$\alpha$ exponential distribution claims,The positive root satisfies $1/(\alpha -R)=(1+\rho )/\alpha$ after dividing out $R$, soThe cubic-bound equation reduces to $r_2^2+\alpha r_2-\rho {\alpha }^2=0$, selecting the displayed positive root. In particular $R<\alpha$, as required by the transform domain, for every $\rho >0$.

Taking $n=e_i$ gives ${\sigma }_i^2=I$. Taking $n=(e_i+e_j)/\sqrt{2}$ for $i\ne j$ and expanding the square gives ${\sigma }_i{\sigma }_j+{\sigma }_j{\sigma }_i=0$. Thus the anticommutator and commutator together implySince $n\cdot \sigma$ is Hermitian and $\theta$ is real, the exponent is anti-Hermitian. Its exponential has inverse equal to its adjoint and is therefore unitary. Splitting the exponential series into even and odd powers, using $(n\cdot \sigma {)}^2=I$, givesPut $N=n\cdot \sigma ,M=m\cdot \sigma$ with $m\perp n$. Then $NM=i(n\times m)\cdot \sigma$, $MN=-NM$, and $NMN=-M$. Expanding $(c-iNs)M(c+iNs)$ gives $(c^2-s^2)M-2icsNM$. ConsequentlyThe spin-half angular momentum is $S_i=\hslash {\sigma }_i/2$. The commutators then become $[S_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$, and $(n\cdot S{)}^2={\hslash }^{2}I/4$ expresses the two possible directional spin values $\pm \hslash /2$. Also $S^2=3{\hslash }^{2}I/4$, appropriate to spin one half. The exponential represents a spatial rotation on spin states.

Choose $n=(0,1,0)$ and initially ${\sigma }_3|\chi ⟩=|\chi ⟩$. The conjugation formula sends ${\sigma }_3$ to ${\sigma }_3\cos \theta +{\sigma }_1\sin \theta$. ThereforeThus $U|\chi ⟩$ is spin up along $(\sin \theta ,0,\cos \theta )$, as required.

Fix a Hankel contour with a positive-radius circle around zero and a branch $-\pi <\arg t<\pi$, traversing the lower bank inwards and the upper bank outwards. Write its numerator as $I(z)={\int }_H{t}^{z-1}{e}^{t}dt$. For fixed circle radius, $I$ is an entire function of $z$: the finite contour pieces avoid zero, and the tails converge uniformly on compact $z$ sets by exponential decay, also after differentiation with respect to $z$.

At $z=-n$, the two straight-bank contributions cancel because $t^{-n-1}{e}^t$ is single-valued. The remaining circle is positively oriented, so the residue theorem and the exponential series giveSince $\sin (\pi z)=(-1{)}^n\pi (z+n)+O((z+n{)}^3)$, the quotient has the simple poleFor cancellation of positive-integer gamma singularities on a Hankel contour, at a positive integer $n$, the integrand $t^{n-1}{e}^t$ is entire and its bank contributions again cancel; its circular integral is zero. Hence $I(n)=0$. The numerator is analytic near $n$, while the denominator has a simple zero with nonzero derivative. Factoring $I(z)=(z-n)J(z)$ therefore cancels that zero, proving there is no pole at a positive integer using only this representation. Cancellation of the numerator, not merely the zero of the sine, decides the issue. One can also shrink the contour for $\Re z>0$, obtaining $I(z)=2i\sin (\pi z){\int }_0^{\infty }{u}^{z-1}{e}^{-u}du$ and hence the finite value $(n-1)!$.

The exponential series at $1$ givesFor every positive integer $q$, multiply by the factorial $q!$ and separate the integer part:Every term in $R_q$ is positive. Each denominator is at least $(q+1{)}^j$, and the inequality is strict when $j\ge 2$. Comparison with the geometric series therefore givesIf $e=p/q$ with integers $p$ and $q\ge 1$, then $q!e=p(q-1)!$ would be an integer. Since $I_q$ is also an integer, $R_q$ would be an integer strictly between zero and one, which is impossible. HenceThis proves the irrationality of e by using its rapidly decreasing series tail.

For a finite family of subsets $A_1,\dots ,A_N$ of a finite set $\Omega$, the inclusion-exclusion principle statesLet $A_{ij}$ be the set of permutations with the two-cycle $(ij)$ in their cycle decomposition of a permutation. Two distinct specified pairs sharing a label cannot both occur, so their intersection is empty. An intersection containing $r$ pairwise disjoint specified pairs fixes those $2r$ labels and leaves $(n-2r)!$ choices for the other labels.

The number of ways to choose $r$ disjoint unordered pairs is $n!/[2^{r}r!(n-2r)!]$: first order the selected labels, then divide by the two orders within each pair, by the $r!$ orders of the pairs, and by the orders of the unused labels. Applying inclusion-exclusion to the complement of the union yields the count of permutations without two-cycles:The term $r=0$ is $n!$. Fixed points and cycles of length at least three are allowed; the restriction concerns two-cycles, not whether a permutation can be expressed as a product of transpositions.

Dividing by $n!$ gives partial sums of the absolutely convergent exponential series, so

The definitions in a unital C-star algebra areThus a Hermitian element of a C-star algebra is self-adjoint, a Unitary element of a C-star algebra has inverse $x^{\ast }$, and a Normal element of a C-star algebra commutes with its adjoint. We derive the needed C-star algebra facts directly from $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$, as required.

First $\Vert 1\Vert =\Vert 1{\Vert }^2$ gives $\Vert 1\Vert =1$, andgives $\Vert a\Vert \le \Vert a^{\ast }\Vert$; applying the same argument to $a^{\ast }$ gives equality. Thus the involution is an isometry. For a Unitary element of a C-star algebra $u$,The general Banach algebra spectrum bound gives $|\lambda |\le 1$ for $\lambda \in \sigma (u)$. Such a $\lambda$ is nonzero, and the inverse spectral mapping theorem gives ${\lambda }^{-1}\in \sigma (u^{-1})$, so also $|\lambda {|}^{-1}\le 1$. Hence

If $h=h^{\ast }$, the convergent exponential series and the isometric involution give $(e^{ith}{)}^{\ast }=e^{-ith}$. These commuting exponentials multiply to $1$, so $e^{ith}$ is unitary for real $t$. By the exponential spectral mapping theorem, if $\lambda \in \sigma (h)$ then $e^{it\lambda }\in \sigma (e^{ith})\subseteq \mathbb{T}$. Taking $t=1$ gives $e^{-\mathrm{Im}\lambda }=1$, soOnly general Banach algebra spectral mapping theorem and the defining C-star identity were used.

Now prove spectral permanence for C-star algebras. Let $B\subseteq A$ be the norm-closed C-star subalgebra with the same identity. The algebraic inclusion always gives ${\sigma }_A(b)\subseteq {\sigma }_B(b)$. For $h=h^{\ast }\in B$, both spectra lie in $\mathbb{R}$. If $\lambda \notin {\sigma }_A(h)$, choose nonreal ${\lambda }_n\to \lambda$. Since ${\sigma }_B(h)\subseteq \mathbb{R}$, each $({\lambda }_{n}1-h{)}^{-1}$ belongs to $B$. By continuity of inversion in a Banach algebra, these inverses converge in $A$ to $(\lambda 1-h{)}^{-1}$, which belongs to $B$ because $B$ is closed. Thus $\lambda \notin {\sigma }_B(h)$, proving equality for hermitian elements.

For the requested normal $x\in B$, suppose $b=x-\lambda 1$ is invertible in $A$. The element $b^{\ast }b$ is hermitian and invertible in $A$, so the equality just proved places $(b^{\ast }b{)}^{-1}$ in $B$. ConsequentlySince $b$ already has a two-sided inverse in $A$, multiplying $yb=1$ by that inverse gives $y=b^{-1}$. Hence its inverse belongs to $B$, and

We need one more elementary consequence of the C-star identity: the spectral radius norm equality for normal elements. If $h=h^{\ast }$, then $\Vert h^2\Vert =\Vert h{\Vert }^2$. If $a$ is normal, commutativity of $a,a^{\ast }$ givesEvery power of $a$ is a Normal element of a C-star algebra. Induction gives $\Vert a^{2^n}\Vert =\Vert a{\Vert }^{2^n}$. The general spectral radius formula therefore implies

The continuous functional calculus is the unique unital $\ast$-homomorphism $C(K)\to \mathcal{B}(H)$ taking the coordinate function to $T$. For bounded linear operators on a Hilbert space, the adjoint identity gives $\Vert S\Vert ={\sup }_{\Vert v\Vert =\Vert w\Vert =1}|⟨Sv,w⟩|=\Vert S^{\ast }\Vert$. ThereforeThis proves the C-star identity for $\mathcal{B}(H)$ directly; completeness and submultiplicativity come from the operator norm. LetBecause $T$ is a normal operator, this is a commutative unital C-star algebra. Every element of $C$ is a Normal element of a C-star algebra, so the preceding norm identity says its Gelfand transform is isometric:Here $\Delta (C)$ is the compact character space from the general Gelfand representation theorem for commutative Banach algebras.

Each algebra character preserves the involution. Indeed, write $a=h+ik$ with $h,k$ hermitian; $\chi (h)\in {\sigma }_C(h)\subseteq \mathbb{R}$ and similarly for $k$, so $\chi (a^{\ast })=\overline{\chi (a)}$. Therefore the continuous mapis injective: its value determines $\chi (T^{\ast })$ and hence its value on all the dense polynomials. It is surjective because the general character description of the spectrum gives $\{\chi (T):\chi \in \Delta (C)\}={\sigma }_C(T)$, and spectral permanence for C-star algebras identifies this with ${\sigma }_{\mathcal{B}(H)}(T)=K$. A continuous function that is a bijection from a compact space to a Hausdorff space is a homeomorphism, so we identify $\Delta (C)$ with $K$.

Under this identification, the Gelfand transform sends $T$ to $u(z)=z$ and $T^{\ast }$ to $\overline{u}$. Its image is an isometric, and therefore closed, unital self-adjoint subalgebra of $C(K)$. It separates points because it contains $u$. The complex Stone-Weierstrass theorem makes the image dense, hence equal to all of $C(K)$. Inverting the Gelfand transform and including $C$ into $\mathcal{B}(H)$ givesan isometric unital C-star homomorphism with $u(T)=T$.

To prove uniqueness without assuming automatic continuity of C-star homomorphisms, let $\Psi$ be any other such map. Since every $f\in C(K)$ is normal, $\Psi (f)$ is normal. A unital algebra homomorphism preserves inverses, soUsing the normal-element norm identity yields $\Vert \Psi (f)\Vert =r(\Psi (f))\le \Vert f{\Vert }_{\infty }$. Thus $\Psi$ is contractive. It agrees with $\Phi$ on polynomials in $u,\overline{u}$, since both send them to the same polynomials in $T,T^{\ast }$. These polynomials are dense by the Stone-Weierstrass theorem, so continuity proves $\Psi =\Phi$. This establishes the continuous functional calculus with no unproved theorem specific to C-star algebras.

A disconnected spectrum gives a nontrivial closed invariant subspace. Write $K=K_1\bigsqcup K_2$ with $K_1,K_2$ nonempty and both open and closed in $K$. The indicator function $f=1_{K_1}$ is continuous on $K$, even though no such continuity is needed across the gap outside $K$. Let $P=f(T)$. The continuous functional calculus givesIts isometry gives $\Vert P\Vert =1$ and $\Vert I-P\Vert =1$, so $P\ne 0,I$. The linear projection has closed range $Y=PH=\ker (I-P)$. Its range is nonzero and proper, and $T(Px)=P(Tx)\in Y$ proves invariance. Since it also commutes with $T^{\ast }$, the same subspace is reducing. This proves disconnected spectrum gives a reducing subspace.

The figure can be realized without any eigenvalues: take $H=L^2(K,dA)$, where $dA$ is planar area on the two closed disks, and let $T$ multiply by $z$. Its adjoint multiplies by $\overline{z}$, so it is a normal operator. An eigenvector for $\lambda$ would be supported on the area-zero singleton $\{\lambda \}$, hence would be zero in $L^2$. Outside $K$, multiplication by $1/(z-\lambda )$ is a bounded inverse to $T-\lambda I$. For $\lambda \in K$, normalized indicator functions of $K\cap \{|z-\lambda |<\epsilon \}$ have $\Vert (T-\lambda I)h\Vert \le \epsilon$, excluding a bounded inverse. Thus its spectrum is exactly $K$. The linear projection in the figure multiplies by $1_{K_1}$, and its range consists of functions supported on the left disk.

For fixed $k$ and sufficiently large $N$,The finite product and last factor tend to one, while $(1-\lambda /N{)}^N\to e^{-\lambda }$. Thus the Poisson limit theorem givesThese numbers form a probability mass function because they are nonnegative and their sum is $e^{-\lambda }{\sum }_{k\ge 0}{\lambda }^k/k!=1$, by the exponential series. They define the Poisson distribution of parameter $\lambda$. Only $N\ge \lambda$ is needed to make the binomial parameter valid, which is sufficient for the limit.

LetAssume for a contradiction that $S=a/b$ is rational. For every $N\ge b$, both $N!S$ andare integers, because the $m_n$ are integers. Their positive differencemust therefore be a positive integer.

On the other hand, concavity of $x^{\alpha }$ for $0\le \alpha <1$ gives $(N+k{)}^{\alpha }\le N^{\alpha }+k^{\alpha }\le N^{\alpha }+k$. Since $(N+1)\cdots (N+k)\ge (N+1{)}^k$,Eventually $0<R_N<1$, contradicting its integrality. Hence

The integrality assumption is essential. For example, take $\alpha =0$, $A=2$, andThen $1<m_n<2$, but the exponential series givesThus the result does not remain true for arbitrary real $m_n$.

The first series is telescoping, sinceIts $N$th partial sum is $1-1/(N+1)$, soa rational number.

The second series converges absolutely, for example by the ratio test, and is the odd part of the exponential series:This value is irrational. Indeed, if $a=\sinh 1$ were algebraic, then $e$ would satisfymaking $e$ algebraic over the algebraic numbers and hence algebraic, contrary to the Hermite theorem on the transcendence of e. Thus the second sum is in fact transcendental.