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For $f(x,y)=y$ and $\beta =1/2$, the explicit midpoint method multiplies the current value by $R=1+h+h^2/2$. Starting from one, induction givesLet $A=e^h$ and $B=R$. The Taylor theorem with Lagrange remainder for the exponential function givesThe difference of powers identity gives $A^n-B^n=(A-B){\sum }_{k=0}^{n-1}{A}^{n-1-k}{B}^k$. Every summand lies between one and $A^{n-1}$. Since $nh=x_n$, the global error therefore satisfiesThe same inequalities hold at $n=0$, where the error vanishes. At the fixed endpoint $x_N=1$, the lower bound $h^2/6$ prevents any higher order of a numerical method, while the upper bound is $e{h}^2/6$. Thus the convergence order is exactly two.

Put $x=e^{-t}$. Thenwhere the second substitution is $u=(k+1)t$ and the final integral is the given gamma function value. The formula also holds for $k=0$, with value $1$.

For $0<x<1$, the exponential function expansion givesEvery term is nonnegative. By the monotone convergence theorem, integration can be interchanged with summation, and henceThis identity is one form of Sophomore's dream.

The ratio can be evaluated directly:The assumption $|f-g|\to 0$ gives $f-g\to 0$, so continuity of the exponential function givesThe additional premise $f\sim g$ is not needed for this implication. The useful requirement is additive control of the exponents, as explained by asymptotic comparison of exponential functions.

Dropping the small highest-derivative term gives $y_o^{\prime }+x{y}_o=0$. The outer solution satisfying the right endpoint condition isIt fails the left endpoint condition, so a boundary layer is needed at $x=0$. If $X=x/{ϵ}^{\gamma }$, diffusion and first derivative balance when ${ϵ}^{1-2\gamma }={ϵ}^{-\gamma }$, giving $\gamma =1$.

With $y(x)=Y(X)$, the exact inner expansion after multiplication by $ϵ$ isIts leading solution is $Y=A+B{e}^{-X}$. The zero boundary value gives $B=-A$, and asymptotic matching condition as $X\to \infty$, $x\to 0$ fixes $A=e^{1/2}$. ThusThe common overlap region value is $e^{1/2}$. Add the outer and inner approximations and subtract that overlap region:The decaying inner exponential function can accommodate the left condition, whereas a right-end inner coordinate would give a mode that grows into the interval. There is no right boundary layer. The composite asymptotic expansion satisfies the left condition exactly and the right one up to the exponentially small term $e^{1/2-1/ϵ}$; the remaining approximation error is algebraic and vanishes as $ϵ\downarrow 0$.

For $\mathrm{Im}k>0$, put $s=-ik$, so $\mathrm{Re}s>0$. The Laplace transform for the Gamma function, analytically continued from positive real $s$, givesChoose $\arg k\in (0,\pi )$ in the upper half-plane. Since $\arg (-ik)=\arg k-\pi /2$,This integral is absolutely convergent throughout the upper half-plane. On the nonzero real axis it converges conditionally by Dirichlet test for improper integrals, and its value is the upper boundary limit of the same branch. In particular negative real $k$ has argument $\pi$, not $-\pi$ in this formula.

At $k=0$ the integral diverges at infinity; in the lower half-plane its exponential function factor grows and the ordinary improper integral diverges. The origin itself causes no local-integrability problem for $x^{-1/4}$. A useful continuation of the chosen power uses a branch cut descending from $k=0$ with argument in $(-\pi /2,3\pi /2)$.

Take $x$ horizontal, $y$ upward and $g>0$. Resolve Newton's second law into components for projectile motion with linear drag:For $\alpha >0$, let ${\tau }_d=m/\alpha$. An integrating factor $e^{t/{\tau }_d}$ givesThe relaxation time is ${\tau }_d$, and the terminal velocity is $(0,-g{\tau }_d)$. For fixed finite $t$, expanding the exponential function as $\alpha \to 0$ gives the continuous limit $v_x=v_0\cos \theta$ and $v_y=v_0\sin \theta -gt$; the two terms proportional to ${\tau }_d$ cancel before taking the limit.

In a classically forbidden region, the Schrödinger equation has evanescent solutions rather than forcing the wavefunction to vanish. Matching at both sides of a finite barrier permits nonzero transmission. For a broad barrier the WKB approximation giveswhere $C$ is a matching prefactor. Transmission is exponentially sensitive to barrier width and height. A single decaying exponential function carries no probability current; the matched scattering solution includes the combination needed for transmitted current. energy is conserved throughout.

In alpha decay, a quasibound alpha particle penetrates the nuclear Coulomb barrier. Multiplying the penetration probability by an assault frequency estimates the decay rate; the strong dependence on alpha energy explains the wide range of lifetimes. In a scanning tunnelling microscope, a voltage drives electron quantum tunnelling between a conducting tip and sample. Approximately $I\propto V{e}^{-2\kappa d}$, so constant-current feedback resolves changes in gap width, with electronic density of states also affecting the image. Piezoelectric motion, vibration isolation and stable voltage are necessary; a bright feature is not automatically a taller atom. Atomic-scale imaging was demonstrated with a vacuum-tunnelling microscope in the original surface study.^[ref] In a Josephson junction, Cooper pair quantum tunnelling through a weak link gives $I=I_c\sin \phi$, allowing supercurrent without an applied voltage and phase-sensitive interference. These are distinct applications of barrier penetration and coupling with quantum coherence, not particles borrowing energy to climb over a barrier.

Let the infinite outer walls be at $x=0,D$, the finite steps at $x=a,b$, and the three potential energy values be $V_L,V_B,V_R$. The original drawing specifies the orderingThe energy eigenstates can be chosen real. The Dirichlet boundary conditions are $\psi (0)=\psi (D)=0$; wavefunction matching at a finite potential step requires both $\psi$ and ${\psi }^{\prime }$ to be continuous at $a,b$. In particular, neither finite step acts as an additional hard wall.

The time-independent Schrödinger equation givesThus a wavefunction is sinusoidal in a classically allowed region, with wavenumber $k=\sqrt{2m(E-V)}/\hslash$, and a linear combination of exponential functions in a classically forbidden region, with inverse decay length $q=\sqrt{2m(V-E)}/\hslash$. Its curvature changes sign relative to $\psi$ at a step crossing $E$; its slope does not jump.

By the Sturm oscillation theorem, the three energy eigenstates have exactly zero, one, and two interior nodes, respectively. Choose their global phases so that their slopes at the left wall are positive:

For $E_0,E_1$, the right-hand piece is proportional to $\sinh [q_R(D-x)]$, rather than a pure decaying exponential: this enforces the finite-distance hard wall exactly. The barrier pieces likewise contain both exponential solutions. Their small tails are an instance of quantum tunnelling.

This is an original illustrative calculation, with vertical lines marking only the region boundaries. It uses $D=1$, $a=0.4$, $b=0.6$, $V_L=0$, $V_R=0.62$, $V_B=0.82$, and ${\hslash }^2/(2m{D}^2)=0.002$ in arbitrary energy units. Its lowest energy eigenvalues are approximately $0.098,0.384,0.696$, so it has exactly the required ordering. Here both nodes of the third wavefunction happen to lie towards the left; its dominant lobe is on the right. These precise positions and relative amplitudes are illustrative: the original drawing supplies no numerical lengths, barrier heights, or mass, so it does not determine a unique quantitative wavefunction. The required sketch features are the matching, curvature, hard-wall zeros, and ordered node counts.

Let $e_v$ be the parcel's water vapor partial pressure. Fixed mole fraction before condensation means $e_v/p$ is constant. Its dew point $T_D$ is defined by $E(T_D)=e_v$, where $E$ is the saturation pressure. The Clausius-Clapeyron relation givesThe last approximation uses hydrostatic balance, the ideal gas law, and an environmental temperature close to the parcel's. Rearrangement yieldsFor a modest initial dew-point depression, $T_D/T\simeq 1$ in absolute temperature, soWith $L_v\simeq 2.5\times 10^6\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$, $R_d\simeq 287\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and $R_v\simeq 461.5\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, $H\simeq 1.58\times 10^5\mathrm{m}$. This dew point scale is much longer than the atmospheric pressure scale height.

The exponential function is an approximation, not the exact consequence of a dry adiabatic process. Retaining $T=T_s-{\Gamma }_{d}z$ gives $p/p_s=(T/T_s{)}^{c_p/R_d}$ and the useful comparisonIt also shows explicitly that expansion lowers the dew point even before any water vapor condenses.

Using the probability mass function of the Poisson distribution and the power series of the exponential function,This moment-generating function of a Poisson distribution is finite for every real $t$. It also covers $\mu =0$, when $X=0$ with probability one and the moment-generating function is identically one.

For fixed real $t$, Taylor's theorem for the exponential function givesSubstituting into the previous moment-generating function,The continuity theorem for moment-generating functions states that, if the moment-generating functions of $Z_n$ and $Z$ are finite on a common open interval containing zero and $M_{Z_n}(t)\to M_Z(t)$ throughout that interval, then $Z_n\stackrel{\mathrm{d}}{\to }Z$. Here finiteness holds for all real $t$, and the limit is the moment-generating function of a standard normal variable. ThereforeThis proves the required instance of the Poisson central limit theorem directly from moment-generating functions; no appeal to a general central limit theorem is needed.

Choose the zero of energy so that the translation-invariant vacuum state has $P^{\mu }|\Omega ⟩=0$. A one-particle state of rest mass $m_{\psi }$ obeys the relativistic energy-momentum relationThe translation covariance of a quantum field in c remains valid for the full interacting four-momentum operator. Acting its exponential functions on the two momentum eigenstates givesHere $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. A different additive vacuum energy must be subtracted consistently from both energy eigenvalues before identifying the physical $p^0$.

Expanding the exponential function produces the Laurent seriesIn particular its residue is $a_{-1}=n^{n+1}/(n+1)!$.

Expanding the exponential function produces the Laurent seriesIn particular its residue is $a_{-1}=n^{n+1}/(n+1)!$.

The exponential function eventually grows faster than the fixed polynomial. A direct justification uses the ratio $r_n=4^n/n^4$:For all sufficiently large $n$, this ratio exceeds $2$, so $r_n$ grows without bound. The proposed Big O notation bound is therefore impossible:

For functions $f,g:\mathbb{N}\to \mathbb{R}$, big O notation meansUsing the natural logarithm and exponential function, $3^{\log n}=n^{\log 3}$. Since $\log 3>1$, for every $n\ge 1$ we have $n\le n^{\log 3}$. The witnesses $C=1,N=1$ therefore prove

The chain rule givesThe exponential function never vanishes, so all critical points areIn particular $f^{\prime }(0)=0$.

The characteristic equation of the constant-coefficient linear ordinary differential equation iswhose roots of a polynomial are $r=-1\pm i/2$. Taking real and imaginary parts of the corresponding exponential functions gives the real general solution

For natural logarithms and sufficiently large $n$, take the logarithm of the positive expression:Set $t=\log \log n$. Then $\log n=e^t$. The exponential function grows faster than $t^2$: for example, its power series gives $e^t\ge t^3/6$ for $t>0$, so $t^2/e^t\le 6/t\to 0$. In particular, eventually $(\log \log n{)}^2\le \log n$. Exponentiating yieldsFor logarithms to a fixed base $b>1$, the natural logarithm of the left side is $(\log \log n-\log \log b{)}^2/\log b$, still smaller than $\log n$ eventually. The big O notation conclusion is therefore independent of that base choice.

Both injective and surjective. The natural logarithm is strictly increasing on the positive real numbers, so it is injective. For every $y\in \mathbb{R}$, $x=e^y>0$ satisfies $\ln x=y$; hence the function is surjective, with the exponential function as its inverse function.