Exponential function ($e^x$)

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 fixed real $x$, the base $1+x/n$ is positive for sufficiently large $n$. Applying Taylor's theorem to the natural logarithm at one gives $n\log (1+x/n)=x+O(n^{-1})$. Continuity of the exponential function then proves the displayed limit. In particular $(1-z/n{)}^n\to e^{-z}$ for fixed $z\ge 0$, a basic source of exponential distribution limits in probability theory.

The positive tail after term $n$ in the reciprocal-factorial series for Euler's number satisfies $1/(n+1)!<e-{\sum }_{i=0}^{n}1/i!\le (n+2)/((n+1)(n+1)!)$. The upper bound replaces each subsequent denominator factor by $n+2$ and sums the resulting geometric series; the lower bound retains only the first omitted term. This supplies an explicit error certificate and, with Stirling's formula, a term count of order $d/\log d$ for $d$ decimal digits. The arithmetic evaluation also has its own rounding error, distinct from this truncation bound.

A positive quantity undergoes exponential growth at constant rate $r>0$ when $y^{\prime }=ry$. For $y_0>0$, the solution of a differential equation is $y(t)=y_0{e}^{rt}$. Every interval of duration $\Delta t$ multiplies the value by the same factor $e^{r\Delta t}$, independently of its starting value.

The interval required to double a quantity under constant-rate exponential growth. Solving $e^{r{t}_2}=2$ with the natural logarithm gives $t_2=\ln 2/r$. At $r=0.2$ per time unit, it is approximately $3.46574$ time units; every subsequent doubling takes the same interval.

Euler's number is the value $\exp (1)={\sum }_{n=0}^{\infty }1/n!$ of the exponential function, approximately $2.71828$. It is the base of the natural logarithm. Its factorial-series tail gives an elementary proof of the irrationality of e. The compounding limit for e shows how the same constant arises from repeated small multiplications.

Start at one and apply $n$ updates, each multiplying the current value by $1+1/n$. The final value is $(1+1/n{)}^n$: one update gives $2$, two give $2.25$, ten give approximately $2.59374$, and one hundred give approximately $2.70481$. Increasing the number of steps, while reducing each increment proportionally, approaches Euler's number rather than producing unbounded accumulation.

The natural logarithm makes the limit transparent. For $u>0$, its integral definition gives $u/(1+u)\le \ln (1+u)\le u$, by bounding $1/t$ on $[1,1+u]$. With $u=1/n$,The squeeze theorem gives a logarithmic limit of one; continuity of the exponential function gives $(1+1/n{)}^n\to e$. This links a discrete multiplicative rule with continuous exponential growth, without assuming that a finite number of steps equals the limiting value.

If Euler's number were $a/b$, choose $n\ge \max (b,2)$ and multiply its factorial-series tail by $n!$. This produces an integer, but the tail is positive and smaller than ${\sum }_{j\ge 1}(n+1{)}^{-j}=1/n<1$. The contradiction proves that the number is an irrational number. The argument rules out a hypothetical rational remainder by placing it in an impossible integer interval.

The exponential function has the everywhere-convergent power series

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.

The complex exponential is the entire functionIt maps $\mathbb{C}$ onto ${\mathbb{C}}^{\ast }$ and has kernel $2\pi i\mathbb{Z}$.

The complex complex exponential function has nonzero derivative and maps the strip onto the annulus, with fibers given by translations by $2\pi i$. It is locally conformal, but not injective. No biholomorphism from the full strip onto an annulus exists because the strip is simply connected space and the annulus has a loop of nonzero winding number. The complex exponential function is a covering map realizing that topological distinction.

For real $x,y$, the complex exponential function has $e^{x+iy}=e^x(\cos y+i\sin y)$. Since the trigonometric factor lies on the unit circle,Thus the real part of the exponent controls its modulus, while the imaginary part controls its direction on the complex plane.

The complex sine is the entire functionFor real $x,y$, $\sin (x+iy)=\sin x\cosh y+i\cos x\sinh y$.

The kernel $2\pi i\mathbb{Z}$ of the complex exponential function givesThe same period appears as the translation generated by the monodromy of the complex inverse sine.

For a positive real base $b\ne 1$, the logarithm is the inverse function of the exponential function $t\mapsto b^t$:It asks which exponent produces a given positive number. Thus ${\log }_{2}8=3$ and ${\log }_2(1/8)=-3$. The base restrictions matter: $b=1$ would produce only one value, while a negative base does not define $b^t$ as a real-valued function for every real exponent.

The useful transformation isFor example, $8\cdot 16=128$ corresponds to $3+4=7$ in base two. Products become sums and ratios become differences. The logarithm does not generally convert a sum into a sum of logarithms. Base conversion uses the natural logarithm: ${\log }_{b}x=\ln x/\ln b$. It also gives $d({\log }_{b}x)/dx=1/(x\ln b)$, positive when $b>1$ and negative when $0<b<1$.

Historically, John Napier's 1614 Mirifici logarithmorum canonis descriptio presented logarithmic tables and their use, documented in the Smithsonian Libraries' digitized original. Tables made multiplication-heavy calculations more manageable by replacing products with addition and a lookup. The 1619 Constructio title page below belongs to a later publication from the same work on logarithms. These historical publications preceded the modern inverse function and integral presentation used here.

For integers $b>1$ and $n>0$, the integer logarithm is the largest nonnegative integer $i$ such that $b^i\le n$. Equivalently,The equivalence follows from the monotonicity of exponentiation in its exponent and the definition of the floor function. For example, $3^4=81\le 100<243=3^5$, so the base-three integer logarithm of $100$ is $4$.

A multiplication loop maintains $p=b^i\le n$ and accepts another factor $b$ while $pb\le n$. Repeated squaring makes the search faster: precompute $b^{2^k}$ starting at $k=0$, and consider these factors in decreasing $k$. Include a factor when the accumulated product remains at most $n$. If $L=\lfloor {\log }_{b}n\rfloor$, the loop invariant before considering $k$ is $i\le L<i+2^{k+1}$. Accepting $2^k$ exactly when $i+2^k\le L$ halves that uncertainty, proving the greedy algorithm returns $L$. This takes $O(1+\log (1+{\log }_{b}n))$ arithmetic operations; it is not the same claim as this bound on bit complexity.

For positive operands in bounded integer arithmetic, test $p\le n\mathrm{DIV}q$ before multiplying by $q$. Integer division makes this equivalent to $pq\le n$ while preventing integer overflow in a rejected candidate.

For positive $x$, start at $i=0,y=1$ and repeatedly increment $i$ and double $y$ while $y<x$. In exact integer arithmetic, the loop invariant $y=2^i$ together with $i=0$ or $2^{i-1}<x$ yields $i=\lceil {\log }_{2}x\rceil$ at exit. For fixed-width arithmetic, this proof requires every useful doubling to be representable. A signed 32-bit doubling variable becomes negative at $2^{31}$ and then zero, so the loop need not terminate for inputs above $2^{30}$. Widening the variable to a signed 64-bit type repairs the algorithm for positive 32-bit inputs.

For the natural logarithm, define $F(u)=u-1-\log u$. Its derivative is $1-1/u$, negative below one and positive above one. Thus its global minimum is $F(1)=0$, proving the bound, with equality only at one. Equivalently $\log (1+x)\le x$ for $x>-1$; this also follows from concavity and the tangent line at one.

For $0<u<1$, the power series for the logarithm gives $-\log (1-u)-u={\sum }_{n=2}^{\infty }{u}^n/n<\frac{1}{2}{\sum }_{n=2}^{\infty }{u}^n=u^2/[2(1-u)]$. The inequality is strict because $1/n<1/2$ for every $n\ge 3$.

For integers $a,b>1$, rationality of ${\log }_{a}b$ is equivalent to equality of some positive integer powers $a^p=b^q$. Exponentiation proves one direction and taking logarithms proves the other. Thus multiplicatively independent integers give an irrational number as their logarithm ratio. For distinct prime bases and arguments, unique factorization supplies that independence.

The logarithm is strictly concave on positive real numbers, as its second derivative is negative. Thus the Jensen inequality gives $\int \log fd\mu \le \log \int fd\mu$ for a probability measure and an integrable positive $f$, whenever the left side is defined. It also controls the negative logarithmic part of convex combinations of strictly positive functions.

The Cauchy principal value integral $\mathrm{li}(x)=\mathrm{PV}{\int }_0^{x}dt/\log t$ is the logarithmic integral. The offset version $\mathrm{Li}(x)={\int }_2^{x}dt/\log t$ differs by a constant and is often used with the prime-counting function. It has $\mathrm{Li}(x)\sim x/\log x$.

The natural logarithm $\ln x$ is the logarithm with base Euler's number, and the inverse function of the real exponential function. A particularly geometric definition isFor $x>1$, it is the area under the hyperbola $y=1/t$ from $1$ to $x$. For $0<x<1$, the reversed integration limits give a negative value. The fundamental theorem of calculus gives $(\ln x{)}^{\prime }=1/x$, so the natural logarithm is increasing, while $(\ln x{)}^{\prime \prime }=-1/x^2$ gives concavity.

The change of variables $t=au$ shows ${\int }_a^{ab}dt/t={\int }_1^{b}du/u=\ln b$ for $a,b>0$. Thus the intervals from $1$ to $2$ and from $2$ to $4$ each contribute area $\ln 2$, explaining $\ln 4=2\ln 2$. Equal ratios give equal increments on a logarithmic scale, even though the ordinary differences grow. This makes the natural logarithm useful for comparing multiplicative changes and solving for time in exponential growth or exponential decay. The integral construction and inverse function properties are discussed in OpenStax, the natural logarithm as an integral.

Integrating the geometric series for $1/(1+x)$ from zero to $x$ gives the displayed power series for $|x|<1$. It has radius of convergence one. At $x=1$ the alternating series test gives conditional convergence to $\ln 2$; at $x=-1$ the negative harmonic series diverges. Replacing $x$ by $-w$ gives $\ln (1-w)=-{\sum }_{n\ge 1}{w}^n/n$. An expression formed by substituting a nonlinear function for $w$ need not have the same convergence range after it is re-expanded as a Taylor series in the new variable.

The natural logarithm has the parameter-integral representationFor $0<t<1$, the identity $(t^x-1)/\ln t={\int }_0^x{t}^{s}ds$ shows convergence: the absolute value is at most $|x|t^{\min (x,0)}$, which is integrable when $x>-1$. At $t=1$, the apparent singularity is removable, with limiting value $x$.

Write the left side as $I(x)$. On a compact parameter interval $c\le x\le d$ with $c>-1$, its differentiated integrand $t^x$ is bounded by the integrable function $t^c$. The dominated convergence theorem justifies differentiation under the integral sign, giving $I^{\prime }(x)={\int }_0^1{t}^{x}dt=1/(1+x)$. Since $I(0)=0$, the fundamental theorem of calculus gives $I(x)=\ln (1+x)$. At $x=-1$, the original improper integral diverges as $-{\int }_{0}dt/[t|\ln t|]$, and it also diverges for $x<-1$.

A logarithmic scale places a positive quantity $x$ according to a logarithm, so equal ratios occupy equal distances. For example, $1,2,4,8$ are equally spaced under the natural logarithm because every successive difference is $\ln 2$. An ordinary linear scale instead gives equal distances to equal differences: $1,2,3,4$.

For a base $b>1$ and a positive reference $x_0$, define the axis positionThe reference has the same units as $x$, making $x/x_0$ a dimensionless quantity. The distance between two positions isMultiplying both values by the same positive factor preserves that distance. Replacing $x_0$ by $c{x}_0$ translates every position by $-{\log }_{b}c$; changing the base rescales positions through ${\log }_{b}z=\ln z/\ln b$. Neither change alters which ratios receive equal gaps.

In base ten, the values $1,10,100,1000$ occupy positions $0,1,2,3$ when $x_0=1$. The differences between neighboring values grow from nine to ninety to nine hundred, but every step spans one order of magnitude. The values $2,20,200$ occupy positions $0.30103,1.30103,2.30103$: the same spacing shifted by ${\log }_{10}2$. This lets one plot small and large positive quantities together while keeping equal proportional changes equally visible.

A logarithmic scale is a rule for position, not an instruction to relabel the tick marks with their logarithms. A mark labelled $100$ still means the original value one hundred. Ordinary real logarithms exclude zero and negative inputs. Omitting such values or clipping them to a plotting boundary does not give them genuine logarithmic positions; the plotting choice needs to be clear.^[ref]

For positive quantities, a difference of one order of magnitude means a factor of ten; a difference of $n$ orders of magnitude means a factor of $10^n$. The logarithmic difference ${\log }_{10}(x_2/x_1)$ records this comparison, without subtracting the quantities themselves. For example, $1000$ is three orders of magnitude above one, regardless of a common change of units. Roughly assigning a single value to an order of magnitude also requires a rounding convention; powers of ten are the reference levels.

For positive endpoints $a,b$, halfway along a logarithmic scale meansHere $\beta >1$ is the base. Equivalently, using the natural logarithm, $2\ln (m/x_0)=\ln (a/x_0)+\ln (b/x_0)$. The product of numbers identity for logarithms gives $m^2=ab$, so the midpoint is the geometric mean $\sqrt{ab}$. The two endpoints lie at equal multiplicative distances because $m/a=b/m$.

For $a=10$ and $b=100$, this midpoint is $\sqrt{1000}\approx 31.6228$, giving a factor of $\sqrt{10}$ on each side. The arithmetic mean $55$ is halfway on a linear scale, but its two ratios are $55/10=5.5$ and $100/55\approx 1.81818$. Thus reading halfway between the marks $10$ and $100$ as $55$ on a logarithmic scale gives the wrong value.

A semi-log plot uses a logarithmic scale on one axis and a linear scale on the other. Which axis is logarithmic matters. Plotting $x$ against $\log y$ makes positive exponential functions appear as straight lines; plotting $\log x$ against $y$ instead makes an affine function of a logarithm appear straight. The plotting operations change positions while retaining the original tick values.^[ref]

Take the original dimensionless mathematical model $y(t)=3e^{0.4t}$ for $0\le t\le 8$. It has exponential growth, with $y(0)=3$, $y(4)\approx 14.8591$, and $y(8)\approx 73.5976$. Taking a base-ten logarithm givesThe slope in these transformed positions is $0.4/\ln 10\approx 0.173718$, although the original relative growth rate $y^{\prime }/y$ is $0.4$. Equal steps of two in $t$ multiply $y$ by $e^{0.8}\approx 2.22554$. The corresponding ordinary gains increase, so the left plot bends while the right plot is straight.

One practical use is checking a proposed exponential distribution for failure times. For rate $\lambda >0$ and $t\ge 0$, its cumulative distribution function is $F(t)=1-e^{-\lambda t}$, so the reciprocal survival function $y=1/[1-F(t)]$ satisfies ${\log }_{10}y=\lambda t/\ln 10$. Estimated probabilities can therefore be placed on a logarithmic vertical axis and compared with a straight line. Agreement supports compatibility with the chosen statistical model; a roughly straight finite sample does not establish a unique physical mechanism.^[ref]

A log-log plot uses logarithmic scales on both axes. For a positive power law $y=C{x}^p$ with $x,C>0$, positive references $x_0,y_0$ giveIts transformed slope is the exponent $p$ when both axes use the same logarithm base and the slope is measured in logarithmic coordinate units. For example, $y=2x^2$ takes the values $2,8,32$ at $x=1,2,4$. Every doubling of $x$ quadruples $y$, giving slope two on a log-log plot.

This differs from a semi-log plot: a power law has constant proportional response to a proportional change of input, whereas exponential growth has constant proportional response to an additive change of input. Both can appear straight with appropriate axes, so one must read the scale labels before interpreting a slope.

For $x>0$, concavity of the natural logarithm givesEquivalently, ${\log }_{b}x\le ({\log }_{b}e)(x-1)$ for $b>1$.

For $t>0$, the natural logarithm satisfiesTo prove it, put $h(t)=\ln (1+t)-2t/(2+t)$. Then $h(0)=0$ and the derivative is $h^{\prime }(t)=t^2/[(1+t)(2+t{)}^2]>0$. More precisely,This positive integral controls errors in symmetric approximations to the logarithm. After writing $x=1+t$, the bound is $\ln x>2(x-1)/(x+1)$ for $x>1$; the inequality reverses for $0<x<1$ by replacing $x$ with $1/x$.

For positive-definite Hermitian operators,The same statement on supports follows by regularizing with a positive multiple of the identity and taking a limit.

A Gaussian function has the form $A\exp [-a(x-b{)}^2]$ with $a>0$. Its integral and moments reduce by translation and scaling to the Gaussian integral.

Multiply the two exponential series in $e^{-(x-a{)}^2}=e^{-a^2}{e}^{2ax}{e}^{-x^2}$ to obtain the displayed Taylor expansion. Its first three nonzero terms depend on the real parameter $a$. For $a\ne 0,\pm 1/\sqrt{2}$, they end at degree two. At $a=0$, odd coefficients vanish and the first three terms are $1-x^2+x^4/2$. At $a=\pm 1/\sqrt{2}$, the quadratic coefficient vanishes, leaving $e^{-1/2}(1+2ax-4a{x}^3/3)$ as the first three nonzero terms. Parameter-dependent vanishing coefficients must be checked when a question counts nonzero terms rather than prescribing an expansion order.

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 fixed $c,h>0$, $L\ge 1$ and $\epsilon \ge 0$, complete the square in the exponent. A translated Gaussian function summed on a fixed-spaced lattice has mass $O(\sqrt{L})$ uniformly in the translate, by comparison with its integral and its maximum. This keeps a square-root logarithm, rather than the full number of dyadic intervals, when summing exponentially damped exponential sum bounds.

Hyperbolic functions combine the exponential function to describe a hyperbola much as trigonometric functions describe a circle. Their two basic members are the hyperbolic cosine and hyperbolic sine, defined for every real number $t$ byTheir derivatives exchange the two functions:These formulas follow by applying differentiation to the exponential function definitions. ^[ref]

Adding and subtracting gives $\cosh t+\sinh t=e^t$ and $\cosh t-\sinh t=e^{-t}$. Multiplying these equalities proves the hyperbolic Pythagorean identity:Consequently the parametric equation $(x,y)=(\cosh t,\sinh t)$ traverses the entire right branch of the unit hyperbola $x^2-y^2=1$, where $x\ge 1$. For example, $t=0$ gives $(1,0)$, while $t=\pm 1$ gives approximately $(1.543,\pm 1.175)$. Compare $(\cos \theta ,\sin \theta )$ on the unit circle, where the squared coordinates add to one. The minus sign changes the geometry from a closed circle to an unbounded hyperbola.

The other members are ratios and reciprocals: hyperbolic tangent is $\tanh t=\sinh t/\cosh t$, hyperbolic cotangent is $\coth t=\cosh t/\sinh t$, and hyperbolic secant is $\mathrm{sech}t=1/\cosh t$; denominators must be nonzero. The hyperbolic sine and hyperbolic cosine both solve the second-order linear differential equation $f^{\prime \prime }=f$. In fact $f(t)=A\cosh t+B\sinh t$ has $f(0)=A$ and $f^{\prime }(0)=B$, making this basis convenient for an initial value problem. Their geometry also makes hyperbolic substitution useful for integrals containing $\sqrt{1+u^2}$, and the hyperbolic cosine gives the physical catenary of a hanging chain.

A hyperbolic angle $t$ is twice the signed area $A$ enclosed by the two rays from the origin to $(1,0)$ and $(\cosh t,\sinh t)$ and the intervening arc of the unit hyperbola. Counterclockwise traversal is positive: $t>0$ gives an ordinary area $t/2$ above the horizontal axis, and $t<0$ gives signed area $t/2$ below it. This parallels the unit circle, whose sector has area $\theta /2$ for an angle $\theta$ measured in radians.

To check the area interpretation, write $x(u)=\cosh u$ and $y(u)=\sinh u$. Green's theorem expresses the signed area as a line integral; the straight radial edges contribute zero, leavingThus $t$ is an area parameter, rather than the ordinary angle of the ray. For $t>0$, that ordinary angle is $\arctan (\tanh t)$; it stays below $\pi /4$ even as the hyperbolic angle grows without bound.

The inverse hyperbolic functions invert the hyperbolic functions, with real branches on their standard ranges and complex branches obtained through analytic continuation. For example, $\mathrm{arsinh}z=\log (z+\sqrt{z^2+1})$ on a chosen branch; the multivalued inverse hyperbolic sine records all inverse values.

The real inverse of the hyperbolic sine is defined on all real numbers and has derivative $1/\sqrt{1+x^2}$. It appears in arc length and in integrals involving square roots of sums of squares, where a hyperbolic substitution can simplify the expression.

The hyperbolic cosine is the even functionIt has derivative $\sinh x$ and second derivative $\cosh x>0$, so it is a strictly convex function. Its unique global minimum is $\cosh 0=1$, and its image is $[1,\infty )$. For example, $\cosh (-1)=\cosh 1\approx 1.543081$. On the unit hyperbola, the hyperbolic cosine is the horizontal coordinate; as a graph against $x$, it is a symmetric bowl.

Its Taylor series beginsThe positive fourth-order term explains why the bowl resembles a parabola near its minimum yet differs from one away from it. Restricting to $x\ge 0$ gives the inverse hyperbolic cosine.

A catenary is the static shape of an ideal flexible, inextensible chain hanging under its own gravitational weight in a uniform gravitational field, with uniform linear mass density along its arc length. With the lowest point above $x=0$, one convenient vertical origin givesHere $a$ has units of length. The supports and the chain's arc length determine the scale and position; horizontal and vertical shifts give the general catenary. ^[ref]

For $a=1$, the three points at $x=-1,0,1$ have heights $\cosh 1\approx 1.543081$, $1$, and $\cosh 1\approx 1.543081$. If heights should instead be measured from the lowest point, subtract $a$:The same three heights are then approximately $0.543081$, $0$, and $0.543081$. The shift changes the vertical origin and preserves the catenary shape.

The scale also measures tension: if $H>0$ is its constant horizontal component, $\mu$ is the linear mass density, and $g$ is the magnitude of gravitational acceleration, then $a=H/(\mu g)$. A short force balance explains this. For the portion from the lowest point to $x\ge 0$, with arc length $s(x)$, vertical tension balances its gravitational weight and horizontal tension givesApplying differentiation yields $a{y}^{\prime \prime }=\sqrt{1+y^{\prime 2}}$. The proposed hyperbolic cosine solves this differential equation, because $y^{\prime }=\sinh (x/a)$ and $a{y}^{\prime \prime }=\cosh (x/a)=\sqrt{1+{\sinh }^2(x/a)}$. This is the mechanism behind the shape, under the stated ideal chain assumptions.

The same identity simplifies the arc length between symmetric supports at $x=\pm b$:For $a=b=1$, $L\approx 2.350402$, exceeding the horizontal span of two. A larger $a$ at fixed $b$ gives a shallower sag: the Taylor series gives $a(\cosh (x/a)-1)=x^2/(2a)+x^4/(24a^3)+\cdots$. Thus a parabola is a local approximation, while the exact ideal chain shape is a catenary. The useful connection is that the same hyperbolic Pythagorean identity controls its coordinates, force balance, and arc length.

For a fixed positive integer $q$, $\cosh (1/q)={\sum }_{k\ge 0}{q}^{-2k}/(2k)!$ is irrational. If it were rational with denominator $b$, choose $N$ large enough that $b\mid (2N)!$. After multiplication by $q^{2N}(2N)!$, the finite partial sum is an integer and the remaining positive tail is bounded by a geometric series whose ratio is at most $1/[q^2(2N+1)(2N+2)]$. For sufficiently large $N$, this tail is strictly between zero and one, contradicting its integer value. Only a finite integer partial sum is used; convergence of an infinite series of rationals does not itself prove irrationality.

For $x\ge 1$, the inverse hyperbolic cosine on its nonnegative real branch is $\mathrm{arcosh}x=\log (x+\sqrt{x^2-1})$. It is the real cosine member of the inverse hyperbolic functions.

For $a>0$ and $|x|>a$, the displayed real antiderivative equals $\mathrm{arcosh}(x/a)$ on the right component and $-\mathrm{arcosh}(-x/a)$ on the left, up to constants. Differentiating the absolute logarithm gives $1/\sqrt{x^2-a^2}$ on both components. The constants are independent because the domain is disconnected. The integrals from either finite singular boundary are improper integrals that converge, since their integrands diverge only like the inverse square root of distance to that boundary.

The hyperbolic sine is the odd functionIts derivative is $\cosh x\ge 1$, so it increases strictly from $-\infty$ to $\infty$ and is a bijection of $\mathbb{R}$ onto itself. For example, $\sinh 0=0$, $\sinh 1\approx 1.175201$, and $\sinh (-1)\approx -1.175201$. It is the vertical coordinate on the unit hyperbola parametrized by $(\cosh x,\sinh x)$.

The Taylor seriesshows that $\sinh x\approx x$ near zero. Its tangent line there is $y=x$; unlike the ordinary sine, the hyperbolic sine grows without bound. Its antiderivative is $\cosh x+C$. In the catenary $y=a\cosh (x/a)$, the hyperbolic sine gives both the slope, through the derivative $y^{\prime }=\sinh (x/a)$, and the arc length $s(x)=a\sinh (x/a)$ from the lowest point for $x\ge 0$.

For real $x,y$, the exponential function definitions give $\sinh (x+iy)=\sinh x\cos y+i\cosh x\sin y$. Taking the squared complex modulus and using ${\cosh }^{2}x-{\sinh }^{2}x=1$ gives the identity. In particular $|\sinh (x+iy)|\ge |\sinh x|$, uniformly in $y$. This controls the vertical sides of rectangular contours in residue theorem arguments with hyperbolic functions.

The hyperbolic cotangent is $\coth x=\cosh x/\sinh x$ where $\sinh x\ne 0$.

The hyperbolic-sine infinite product gives, by taking its logarithmic derivative,The identity holds away from the poles, with locally convergent sums. For real $z>0$, it makes $(z\coth z-1)/z^2$ positive and strictly decreasing. It is useful for Hartmann flow flux monotonicity and small-argument expansions of hyperbolic functions.

The hyperbolic secant is

For real $|a|<\pi$, integrate $g(z)=e^{az}/\cosh \pi z$ around a rectangle of height $i$ and width $2R$. Only the simple pole $i/2$ lies inside, with residue $e^{ia/2}/(\pi i)$. Since $g(x+i)=-e^{ia}g(x)$, the reversed top edge adds $e^{ia}$ times the bottom integral. The two vertical edges vanish as $R\to \infty$, using $|a|<\pi$. The residue theorem gives $(1+e^{ia})J=2e^{ia/2}$, proving the formula. The condition is also the exact convergence range: the two tails decay with exponents $\pi -a$ and $\pi +a$.