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Set $w=e^z\ne 0$. The hyperbolic cosine equation becomes $w+w^{-1}=-4$, hence $w^2+4w+1=0$. Its two polynomial roots are $w=-2\pm \sqrt{3}$, both negative. Taking every value of the complex logarithm yieldsSince $(2-\sqrt{3})(2+\sqrt{3})=1$, their real logarithms have opposite signs. All solutions are thereforeThe positive real constant is also $\mathrm{arcosh}2$, the inverse hyperbolic cosine of $2$. Substitution gives $\cosh z=-\cosh (\ln (2+\sqrt{3}))=-2$ for both signs and every integer $k$.

Differentiate the defining function with the other coordinates independent. Put $C=\cosh (x+z)$ for brevity. ThenThe implicit differentiation ratios therefore giveEach expression applies on the implicit surface wherever its own denominator is nonzero. The hyperbolic cosine appears because the derivative of $\sinh (x+z)$ with respect to either $x$ or $z$ is $\cosh (x+z)$.

Separate one factor of the hyperbolic sine and use the hyperbolic Pythagorean identity ${\sinh }^{2}x={\cosh }^{2}x-1$. With $u=\cosh x$ and $du=\sinh xdx$, integration by substitution givesFor real $x$, the hyperbolic cosine never vanishes. Differentiating the result gives $\sinh x(1-{\mathrm{sech}}^{2}x)={\sinh }^{3}x/{\cosh }^{2}x$, checking the sign of the reciprocal term.

Multiply the Maclaurin series of hyperbolic cosine by the binomial series for the reciprocal square root:The quadratic terms cancel, while the quartic coefficient is $1/24-1/4+3/8=1/6$. HenceThe absence of odd powers follows from the function's even symmetry.

The exponential definition of hyperbolic sine gives $e^z-e^{-z}=0$, or $e^{2z}=1$. Taking every branch of the complex logarithm givesFor hyperbolic cosine, multiply $\cosh z+1=0$ by $2e^z$ to obtain $(e^z+1{)}^2=0$. Thus $e^z=-1$, giving

Use the Maclaurin series of the logarithm and hyperbolic cosine:To retain terms through degree six, only the constant and quadratic terms of the reciprocal factor can multiply the leading cubic term. ThusThe numerator in the original PDF is $\ln (1+x^3)$; the cubic power is missing from the local TeX transcription.

The exponential definitions of hyperbolic sine and hyperbolic cosine, together with Euler's formula, give $\sinh (iu)=i\sin u$ and $\cosh (iu)=\cos u$. The argument of the complex logarithm therefore reduces to $i\sin (\pi /2)+\cos (9\pi /2)=i$.

Using the principal logarithm, $\log i=i\pi /2$, soIf no branch of the complex logarithm is selected, its complete set of values is $i(\pi /2+2\pi k)$, $k\in \mathbb{Z}$. The real part remains zero, while the imaginary part is $\pi /2+2\pi k$. The original paper does not explicitly state a branch convention.

The Maclaurin series of the hyperbolic cosine and cosine areBoth are even, so write their quotient as $1+a{x}^2+b{x}^4+O(x^6)$. Multiplication by the denominator gives coefficients $a-1/2$ and $b-a/2+1/24$. Matching the numerator yields $a=1$ and $b=1/2$. ThusThe denominator is nonzero near the expansion point, so division of the power series is valid there.

For $x>0$, the definition of hyperbolic cosine simplifies the composite function exactly:Let $h=x-2$. The geometric series givesAdding $2+h$ and dividing by two gives the requested Taylor expansion:The underlying infinite geometric expansion has $|x-2|<2$.

For real $x$, $\cosh x>0$, so multiplication by the hyperbolic cosine is reversible. The identities $\cosh x-\sinh x=e^{-x}$ and $\mathrm{sech}x=1/\cosh x$ giveThus $e^{-2x}=3$. The exponential function is injective, so the unique real solution is

Set $w=e^z\ne 0$. The hyperbolic cosine identity gives $w+w^{-1}=2i$, hence the quadratic equationLet $L=\ln (1+\sqrt{2})>0$. The first root has modulus $1+\sqrt{2}$ and complex argument $\pi /2$; the second has modulus $\sqrt{2}-1=e^{-L}$ and complex argument $-\pi /2$. Taking all complex logarithm values givesThe two families are distinct because their real parts are respectively $L$ and $-L$. This exhausts the solutions because both roots for $w$ and all possible complex arguments have been included.

The Maclaurin series of the hyperbolic cosine is $\cosh x=1+x^2/2+x^4/24+O(x^6)$. The binomial series with exponent $-1/2$ gives $(1+u{)}^{-1/2}=1-u/2+3u^2/8+O(u^3)$. Substituting $u=x^2/2+x^4/24+O(x^6)$ yieldsThese are the first three nonzero terms; the remainder is written in Big O notation.

Use the exponential series to obtain the Maclaurin series of the hyperbolic sine and hyperbolic cosine:Dividing these power series gives the hyperbolic tangent expansionThe displayed remainder uses Big O notation as $x\to 0$.

Compute the complex derivative along real increments to get $f^{\prime }=u_x+i{v}_x$, and along imaginary increments to get $f^{\prime }=v_y-i{u}_y$. Equality gives the Cauchy-Riemann equations, $\boxed{u_x=v_y,u_y=-v_x}$.

For the specified real part, observe that $iz{e}^{-z}=e^{-x}[(x\sin y-y\cos y)+i(x\cos y+y\sin y)]$. Thus the most general holomorphic function with this real part on the connected plane isThe difference of two such holomorphic functions has zero real part, so the Cauchy-Riemann equations make its imaginary part constant.

Next, $\cosh z=\cosh x\cos y+i\sinh x\sin y$. Its real and imaginary parts satisfy $u_x=\sinh x\cos y=v_y$ and $u_y=-\cosh x\sin y=-v_x$. Their partial derivatives are continuous everywhere, proving that the complex hyperbolic cosine is entire.

For the Laurent series, set $h=z-1$. Since $z^2-1=h(2+h)$,The regular part at a simple pole therefore gives

For the displayed form take $c>0$ and $1+\beta x\ge c$. Substitute $s=(1+\beta x)/c$ in the first integral:Inverting the hyperbolic cosine gives the catenary light ray in a linear refractive indexThe minimum occurs at $y=y_0$ and $x=x_0=(c-1)/\beta$, so the integration constant has the specified meaning. The integral above describes the $y\ge y_0$ graph branch; the other sign of the square root continues it smoothly to the other side of the turning point in $x(y)$.

The signed-constant qualification matters: for $c<0$ in part (a), integration reverses horizontal direction, and the positive prefactor in the full catenary is $|c|/\beta$, with minimum height $(|c|-1)/\beta$. The printed expression presupposes $c>0$. The separate case $c=0$ gives $y$ constant, a vertical ray. At the turning point $dy/dx$ diverges, so $y(x)$ is not a single regular graph across it, although $x(y)$ is smooth.

Start with the exponential function definition of the hyperbolic sine and use Euler's formula:Identifying the hyperbolic sine and hyperbolic cosine in the final line gives

The equation requires $\sinh z=\pm i\cosh a$. Write $z=x+iy$, with real $x,y$. Comparing real parts and imaginary parts using part (b)(i) givesIf $x=0$, the second equality would require $|\sin y|=\cosh a>1$, impossible because $a>0$. Thus $\sinh x\ne 0$, forcing $\cos y=0$ andNow $|\sin y|=1$, so $\cosh x=\cosh a$. The hyperbolic cosine is even and is a strictly increasing function on $[0,\infty )$, hence $x=\pm a$. Conversely, substituting these values gives $\sinh (x+iy)=i(-1{)}^k\cosh a$, so every value found does satisfy the original equation. Therefore the complete set is

In the common $ct,x_1$ plane, multiplication of the two Lorentz boost matrices givesThis follows from the addition identities for hyperbolic sine and hyperbolic cosine. The transverse coordinates are unchanged by both transformations. Consequently the rapidity of $\overline{O}$ relative to $O$ isThe signed rapidities add because both boosts use the same spatial axis and the stated matrix convention.

For the sawtooth Ising chain, there is a source inconsistency: the printed Ising spin values are $\{-1,r+1\}$, with no definition of $r$. The requested transfer matrix follows for $r=0$, namely the usual Ising model with values $\{-1,+1\}$. For arbitrary $r$, summing an apex spin actually givesFor example, $r=1$, $s=s^{\prime }=-1$ and $\beta \bar{J}=1$ give the bracket $e^{-2}+e^4$, whereas the claimed hyperbolic cosine gives $e^{-2}+e^2$. Thus the printed identity is false literally. The rest of Question 1 uses the intended symmetric Ising spin values, $r=0$.

For a fixed configuration of base spins, each apex spin appears in only one local Boltzmann factor. The apex spins are therefore conditionally independent, and their finite sums factorize:Only the base spins carry information from one triangle to the next. The apex spins have already been summed out, so no apex index remains in the transfer matrix; the final trace implements the periodic boundary condition.

Subtract the particular plane Poiseuille flow profile from the full rectangular duct flow. The remainder is a harmonic function:The data are even in $x$ and $y$. The boundary value problem is unique: if two solutions differed by $v$, Green's first identity and the zero Dirichlet boundary condition would giveThus $v$ is constant, and its zero boundary values force $v=0$. Reflection of a solution in either coordinate axis therefore gives the same solution. We may seek even separated eigenfunctions.

Use separation of variables withEach summand solves the Laplace equation, vanishes at $y=\pm b$, and has the required evenness. The hyperbolic cosine ratio equals one at either side wall. The functions $\cos (k_{n}y)$ have orthogonality relationThey are the even Dirichlet eigenfunctions on $[-b,b]$, so their Fourier series represents the even boundary data. Projecting that data onto the $n$th eigenfunction yields the Fourier coefficientFor completeness, two integrations by parts give the quadratic arch expansion in half-integer cosine modes: the displayed integral is $4(-1{)}^n/k_n^3$, agreeing with the identity supplied in the paper.

Adding the particular solution givesThe coefficients are absolutely summable and the hyperbolic cosine ratio is at most one throughout the closed rectangle, giving uniform convergence and the prescribed boundary values. On compact subsets of the interior, the hyperbolic cosine ratio supplies exponential decay in $n$, justifying differentiation of the series and the Laplace equation. The side-wall correction slows the plane Poiseuille flow enough to satisfy all four no-slip boundary conditions.