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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.

Both segments start at $(1,1,1)$, so their direction vectors are $\mathbf{a}=(-1,-1,-1)$ and $\mathbf{b}=(1,1,0)$. Their dot product is $-2$, and their norms are $\sqrt{3}$ and $\sqrt{2}$. ThusThe angle between the rays lies in $[0,\pi ]$, so its sine is nonnegative:The negative cosine correctly identifies an obtuse angle. One also obtains the sine from $|\mathbf{a}\times \mathbf{b}|/(|\mathbf{a}||\mathbf{b}|)=\sqrt{2}/\sqrt{6}$, using the cross product.

The tangent half-angle substitution gives $\sin \theta =2t/(1+t^2)$ and $\cos \theta =(1-t^2)/(1+t^2)$. Dividing and then taking the reciprocal square of the cosine givesThese formulas apply where $t$ is defined and $t\ne \pm 1$; at $t=\pm 1$ the original tangent and secant function are undefined because $\cos \theta =0$.

Write $I=\int e^x\sin xdx$ and $J=\int e^x\cos xdx$. Two applications of integration by parts, first taking the sine and then the cosine as the differentiated factor, giveThus $I+J=e^x\sin x$ and $J-I=e^x\cos x$, in each case up to an arbitrary constant of integration. Keeping that freedom avoids identifying independently chosen antiderivatives too rigidly.

The Fourier coefficients follow by multiplying the expansion by $1$, by the appropriate cosine, or by the appropriate sine, then using orthogonality. In the convention with constant term $a_0/2$,The factor $1/L$ comes from the squared norms of the nonconstant sine and cosine functions: the integral of each square over $[-L,L]$ is $L$. The constant function has squared norm $2L$.

The addition formula for the sine givesEquating the sine and cosine coefficients gives $A\cos \varphi =\sqrt{3}$ and $A\sin \varphi =1$. Squaring and adding yields $A^2=4$, so $A=2$ because $A>0$. Both coefficients are positive, so $\varphi$ is in the first quadrant. Consequently

For $\theta \notin 2\pi \mathbb{Z}$, use the finite geometric series with ratio $e^{i\theta }$:The second equality follows by factoring $1-e^{it}=-2i{e}^{it/2}\sin (t/2)$ in numerator and denominator. Taking real parts proves the finite trigonometric sum identityWhen $\theta =2p\pi$, every cosine is $1$, so the sum is $\boxed{N}$. This also gives the continuous limiting value at the excluded arguments.

Use the real inner productOrthogonality means this integral is zero for any two distinct listed functions. The individual squared norms are positive: the constant function has squared norm $2L$, while every listed sine or cosine has squared norm $L$. Thus they are orthogonal but not normalized to unit length in this inner product.

Let $I_n$ denote the integral with the cosine factor and $J_n$ the one with the sine factor. In integration by parts, take $u=\cos (nx)$ and $dv=e^{x}dx$. ThenThe integer restriction gives $\cos (n\pi )=(-1{)}^n$.

A Fourier series represents a periodic function as a constant plus sine and cosine waves whose frequencies are integer multiples of its fundamental frequency. For period $2L$, use the conventionwhere the Fourier coefficients areThe coefficient formulas follow by multiplying by a basis sine or cosine and using their orthogonality over a full period. A Fourier series need not equal an arbitrary function at every point. The pointwise convergence of a piecewise smooth Fourier series theorem says that for a piecewise continuously differentiable function, the Fourier series converges at each point to the average of the two one-sided limits, and hence to $f(x)$ wherever $f$ is continuous.

At $x=\pi /2$, all odd-indexed cosine terms vanish, while $\cos (2k\pi /2)=(-1{)}^k$. Thus the Fourier series from part (c) givesDivide by $2(e^{\pi }-1)/\pi$ and rearrange. The hyperbolic sine identity $e^{\pi }-1=2e^{\pi /2}\sinh (\pi /2)$ yieldsThe evaluation is legitimate by the convergence established in part (c).

The double-angle identity for the cosine gives ${\cos }^{2}x-1=(\cos 2x-1)/2$. Termwise integration therefore givesHere $C$ is an arbitrary constant in the antiderivative.

The characteristic equation of the reduced homogeneous linear differential equation is $m^2-\beta =0$. Its polynomial roots change type with the sign of $\beta$, givingFor $\beta =0$, the zero polynomial root is repeated, so the second independent solution is $t$. For $\beta <0$, the polynomial roots are the complex conjugate pair $\pm i\sqrt{-\beta }$, and the displayed sine and cosine give a real basis of solutions. The constants are arbitrary real numbers. In every case $x(t)=y^{\prime }(t)-\cos t$ reconstructs the corresponding general solution of the coupled system.

Let $J=A/a$, so $J^2=-I$. The even and odd powers in the matrix exponential giveHere the two scalar sums are the Taylor series for cosine and sine. ConsequentlyWith the usual convention that positive angles turn anticlockwise, this rotation matrix rotates by $-a$; the parameter in the form printed in the paper is $\theta =a$.

For $n=0$, a constant solution and its normalization give $y_0=1$. For $n=1$, substitute $y=ax+b$: the equation gives $b=0$, and the normalization gives $a=1$. For $n=2$, put $y=a{x}^2+bx+c$. Comparing coefficients gives $b=0$ and $c=-a/2$; then $y(1)=1$ fixes $a=2$. ThusThese are the first three Chebyshev polynomials. For their squared weighted norms, use $x=\cos \theta$, which converts $dx/\sqrt{1-x^2}$ into $-d\theta$:More generally, $y_n(\cos \theta )=\cos (n\theta )$ makes their orthogonality the usual cosine orthogonality.

On the boundary, integrate the expansion from part (a) over one angular period. Every nonconstant sine and cosine has zero average, soAll positive-power radial terms vanish at $r=0$, leaving $\Psi (0)=A_0$. HenceThis is the mean value property for harmonic functions. It is an average over arclength as well as over angle, since $ds=Rd\varphi$. The usual continuity or integrable-boundary-data hypotheses justify taking this angular average.

Write the boundary Fourier series as $f=A_0+{\sum }_{n\ge 1}(a_n\cos n\varphi +b_n\sin n\varphi )$. Since $R=2$, its coefficients obey $a_n=2^n{A}_n$ and $b_n=2^n{B}_n$. The mean isFor $n\ge 1$, the integral of $\cos n\varphi$ on the first semicircle is zero. On the second semicircle, use ${\cos }^2\varphi =(1+\cos 2\varphi )/2$ and cosine orthogonality. This gives $a_2=1/4$ and every other $a_n=0$.

For the sine coefficients, every even $n$ gives zero on both semicircles. For odd $n$, product-to-sum formulas giveThus the requested radial-expansion coefficients are

The Cauchy integral formula for derivatives, with $z_0=0$, givesThe derivative cycles of sine and cosine give ${\sin }^{(n)}(0)=\sin (n\pi /2)$ and ${\cos }^{(n)}(0)=\cos (n\pi /2)$. ThereforeIn particular the sine integral vanishes for even $n$, and the cosine integral vanishes for odd $n$. For the remaining cases their signs alternate. The radius is irrelevant provided $R>0$, since the numerator is entire and the origin is enclosed once counterclockwise.

The initial displacement lies in an invariant subspace of axial motions. Let the three opposite vertex pairs have positionsFor a vertex on the positive $x$ axis, the two springs towards the $y$ vertices contribute $-k(q_x+q_y)$ in the $x$ direction; the two towards the $z$ vertices contribute $-k(q_x+q_z)$. Transverse components cancel in pairs. Cyclically,The matrix is $I+11^{\mathsf{T}}$. It has eigenvalue $4$ on the breathing vector $(1,1,1)$, and eigenvalue $1$ on the orthogonal plane $q_x+q_y+q_z=0$. This identifies the axial normal modes of an octahedral spring framework without a large characteristic determinant.

Put ${\omega }_0=\sqrt{k/m}$ and decompose the initial vector asSince all initial velocities are zero, only cosine time factors occur. The complete motion to first order in the displacements iswith the vertex positions given by the first displayed formula. No translation or rotation is excited: the initial displacements are orthogonal to those zero modes and their initial velocities vanish. The two checks ${ϵ}_1={ϵ}_2$ and ${ϵ}_1+2{ϵ}_2=0$ respectively isolate pure breathing and pure traceless axial oscillation. Nonlinear corrections beyond the small-displacement approximation have been omitted.

Write $A=a{e}^{i\varphi }$. The modulation equation gives $a_T=0$ and ${\varphi }_T=-5a^2/3$. Matching the leading initial conditions gives $a=1$, $\varphi (0)=0$. ThereforeThe leading amplitude remains two and has no slow growth or decay. The frequency changes from one to $1-5{ϵ}^2/3+O({ϵ}^3)$; expanding this shifted cosine recovers the second-order secular term from part (i).

There are bounded first-order waveform changes as well. If they are retained while matching $u(0)=2$, one may write $2\cos \tau +ϵ[2-(4/3)\cos \tau -(2/3)\cos 2\tau ]+\cdots$, where $\tau =(1-5{ϵ}^2/3)t$. Thus the first harmonic's coefficient is $2-4ϵ/3$, with a mean shift and a second harmonic. These bounded shape corrections are distinct from the constant leading amplitude in the zeroth-order multiple-scales solution.