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Away from zero and the zeros of $\cos (\pi /x)$, the sign function is locally constant and the absolute value function is differentiable at the value of the cosine. The product rule and chain rule giveThus the triangle inequality and the bounds on sine and cosine imply $|f^{\prime }(x)|\le 2|x|+\pi$. Part (c) rules out differentiability at every nonzero cosine zero, and part (b) gives $f^{\prime }(0)=0$, so this estimate covers every point where the derivative exists.

For a finite interval $I$, choose $L<\infty$ such that $|x|\le L$ throughout $I$. ThenThe apparent $x^{-2}$ factor in the derivative of the oscillation cancels against the multiplying $x^2$, which is what makes a uniform bound possible even near zero.

A positive-eigenvalue normal mode satisfies $K{W}_n=A{k}_n^4{W}_n$. Solving the constant-coefficient equation gives the sine, cosine, hyperbolic sine and hyperbolic cosine family. To avoid confusing modal coefficients with the filament bending modulus, write those coefficients as $c_1,c_2,c_3,c_4$.

The free conditions at zero require $c_3=c_1$ and $c_4=c_2$. With $q=kL$, the remaining conditions areIts determinant is $2(1-\cos q\cosh q)$. Therefore the free-free bending spectrum isA convenient corresponding eigenfunction iswhere $N_n$ gives unit $L^2$ norm. The stable characteristic equation for free-free bending modes is $\cos q=\mathrm{sech}q$. Intersections or numerical bracketing giveFor the entire positive sequence, let $q_n^0=(n+\frac{1}{2})\pi$. Expanding $\cos q_n$ near its zero and using $\mathrm{sech}{q}_n\sim 2e^{-q_n}$ givesThere are also two zero-energy filament modes, not captured by substituting $k=0$ into the finite-coefficient trigonometric expression. Solving $W^{(4)}=0$ with free endpoints gives the affine kernel. An orthonormal basis of that kernel isThe positive normal modes are orthogonal to both. Including these two rigid-motion modes is essential for completeness; the regular self-adjoint operator on a finite interval has a complete discrete eigenbasis.

Using the exponential definition of the hyperbolic cosine and the complex exponential function,ThusFor real $x$, the hyperbolic sine vanishes only at $x=0$. Also $|y|\le 1<\pi$, so the only zero of sine available to $y$ is $y=0$. The product therefore vanishes exactly when $b=0$ or $4-b^2=0$:There are three possible parameter values. At $b=0$ the roots are $\pm i$ and $\cosh (\pm i)=\cos 1$ is real; at $b=\pm 2$ the roots themselves are real.

Let $\mathbf{n},\mathbf{w},\mathbf{u}$ be orthonormal unit vectors north, west and upward. A direction of altitude $\gamma$ and westward azimuth $\eta$ isThe north celestial pole and the upper equatorial meridian direction areIn the equatorial coordinate system, the same unit vector is $\mathbf{x}=\sin \delta \mathbf{p}+\cos \delta (\cos h\mathbf{e}+\sin h\mathbf{w})$. Taking dot products with $\mathbf{p},\mathbf{w},\mathbf{e}$ respectively givesThis is an orthogonal transformation between two bases, so it preserves the length of the direction vector. Recover $h$ from both its sine and cosine to keep the correct quadrant. At the zenith, azimuth itself is undefined; the vector equations remain meaningful by continuity. Using the more usual eastward azimuth changes the sign of the second equation.

Write spatial frequency in cycles across the optical pupil as $u=D/\Lambda$. The previous result gives a maximum axial spatial frequency $N/2$. If one counts one representative per positive-quadrant integer lattice point inside an isotropic disk $u_x^2+u_y^2\le (N/2{)}^2$, the continuum area estimate isThis is the geometric counting convention producing the printed formula. Lattice boundaries, the zero-frequency piston and axis points give lower-order corrections, so the formula is not an exact integer count.

The mode definition in the PDF is $D=j\Lambda /2$, hence $u=j/2$ and $j_{\max }=N$. It does not itself specify a two-dimensional lattice, its orientation, its independent wave phase components or a circular rather than square spatial frequency cutoff. Full-cycle integer frequencies correspond only to the even values of $j$ in that definition. A general real wavefront error also needs both sine and cosine quadratures, and distinct oblique orientations cannot all be identified with a single positive quadrant: the Fourier transform of a real wavefront has conjugate coefficients at $\mathbf{u}$ and $-\mathbf{u}$, rather than identifying every independent sign change of a component.

For comparison, a circular spatial frequency disk has approximately $\pi N^2/4$ signed full-cycle lattice points, or $\pi N^2/8$ conjugate pairs; two real wave phase components per pair restore approximately $\pi N^2/4$ real degrees of freedom. A square spatial frequency support gives a different count. The circular physical optical pupil similarly has only about $\pi N^2/4$ illuminated actuators. The printed $M$ is a quadrant area estimate under an extra counting convention; it is not a uniquely determined count of all independently correctable wavefront modes from the stated one-dimensional mode condition.

Put $t=\log (1+\sqrt{2})=\mathrm{arsinh}(1)$, so $c_K=2t/\pi$. The power series coefficients of the hyperbolic sine preprocessing areThus $f_j=|g_j|$ for every $j\ge 0$, and the series converge on the full interval. Applying coefficient-dominated entrywise positivity gives $Y⪰0$.

Every diagonal entry belongs to one of the diagonal blocks and equals $f(X_{ii})=f(1)=\sinh t$. Because $e^t=1+\sqrt{2}$ and $e^{-t}=\sqrt{2}-1$, this is $(e^t-e^{-t})/2=1$. ThereforeThe matrix lies in the elliptope and admits a Gram matrix representation by unit vectors. This preprocessing is the Krivine rounding scheme; the equality of absolute coefficients is what preserves positive semidefiniteness even though the cross-block sine coefficients alternate in sign.

Let $A$ be the interior-grid matrix representing ${\Gamma }_9$, with zero boundary error. It is real symmetric and has the stated sine eigenbasis. Put $s={\sin }^2(k\pi h/2)$ and $t={\sin }^2(\ell \pi h/2)$; the negative of its eigenvalue isThis increases in each variable on $[0,1{]}^2$. With $s_1={\sin }^2(\pi h/2)\ge h^2$, the smallest absolute eigenvalue is at least $8s_1-(8/3)s_1^2\ge (16/3)h^2$. Hence its Euclidean norm operator norm satisfiesThe corrected scheme has an unscaled residual $r$ whose individual entries are uniformly $O(h^6)$. There are $m^2$ entries and $m<h^{-1}$, so $\Vert r{\Vert }_2=O(h^5)$. Subtracting the exact nodal equations from the numerical equations gives $Ae=-r$ (with sign depending on the residual convention), and thereforeChoose a constant slightly larger than the resulting bound to obtain the requested strict $\Vert e\Vert <c{h}^3$ for all sufficiently small $h$. Uniform smoothness is needed for an $h$-independent constant; unspecified smoothness only in the interior would not by itself guarantee it.

Since the odd sine part of the complex exponential integrates to zero,Part (e)(ii) therefore yieldsin the sense of distributions. This is the Fourier representation of the Dirac delta function.

Write the second form of the functional equation asFor $-2\le \sigma \le 2$, the elementary exponential formula for the sine gives, uniformly for $t\ge 4$,The stated Stirling formula givesuniformly on the same strip. The bounded factors $2^{\sigma }{\pi }^{\sigma -1}$ and the cancelling exponentials therefore show thatTaking absolute values in the functional equation proves the Vertical-strip factor in the Riemann zeta functional equation:

The Fourier cosine series on $[0,L]$ isIt is the Fourier series of the even extension to $[-L,L]$, repeated with period $2L$. The Fourier sine series isIt comes from the odd extension and its $2L$-periodic extension. These Fourier coefficient formulas follow by Fourier orthogonality, since every nonconstant sine or cosine has squared integral $L/2$ on $[0,L]$, while the constant function has squared integral $L$.

For separation of variables, put $U=X(x)H(t)$ and set $X^{\prime \prime }/X=H^{\prime }/(\kappa H)=-\lambda$. The boundary conditions are $X(0)=0$, $X^{\prime }(L)=0$. Multiplying $-X^{\prime \prime }=\lambda X$ by $X$ and using integration by parts givesA nonzero solution therefore has $\lambda >0$: the zero case forces $X$ constant, and then $X(0)=0$ forces $X=0$. Writing $\lambda ={\mu }^2$, the eigenfunctions and their temporal factors areThese sine functions are orthogonal on $[0,L]$, with squared norm $L/2$. Expanding the initial condition and multiplying by $\sin ({\mu }_{m}x)$ gives, using the permitted interchange of the sum and integral,Here $\cos ({\mu }_{n}L)=0$ and $\sin ({\mu }_{n}L)=(-1{)}^n$. Thus the Fourier series solution isThe exponential decay permits differentiation of the series for $t>0$, and each term satisfies both boundary conditions. At $t=0$ it is the mixed-end Fourier expansion of $-T_{0}x/L$.

The gained heat is the increase in internal energy. Since the paper specifies no cross-sectional area, first give the answer per unit cross-sectional area; multiply by $A$ for a rod of cross-sectional area $A$. The initial temperature has integral $T_{0}L/2$, whereas ${\int }_0^L\sin ({\mu }_{n}x)dx=1/{\mu }_n$. At $t_{\ast }=L^2/\kappa$ this givesNumerically $\Delta Q/A\simeq 0.45623855\rho c{T}_{0}L$. This is positive and less than the eventual gain $\rho c{T}_{0}L/2$, as expected from relaxation towards the uniform steady state.

For separation of variables, set $u=X(x)Y(y)$, with $X^{\prime \prime }/X=-Y^{\prime \prime }/Y=-\lambda$. The side Neumann boundary conditions give $X^{\prime }(0)=X^{\prime }(\pi )=0$. As in the standard Sturm-Liouville problem, integration by parts shows $\lambda \ge 0$. The zero eigenvalue gives constant $X$, and $Y^{\prime \prime }=0$ together with $Y^{\prime }(\pi )=0$ makes $Y$ constant too.

For positive eigenvalues, the side conditions give $\lambda =n^2$, $X_n=\cos (nx)$, $n\ge 1$. The top condition chooses the hyperbolic cosine factor $\cosh (n(\pi -y))$. Normalize it to one at $y=0$, obtainingThe orthogonality of integer Fourier modes and the permitted interchange of summation and integration giveFor the remaining Fourier coefficients, integrate by parts twice. The sine boundary term vanishes, and with $f=x(\pi -x)$,The last integral vanishes because $f^{\prime \prime }=-2$ and $n$ is a positive integer. Thus $a_n=-2(1+(-1{)}^n)/n^2$: only even Fourier modes survive, as expected from reflection symmetry about $x=\pi /2$. Writing $n=2m$ givesThe factor involving hyperbolic cosine lies between zero and one on the closed rectangle. Comparison with $\sum m^{-2}$ proves uniform convergence, so the solution is continuous on the entire boundary and takes the prescribed bottom values. On every strip $y\ge \epsilon >0$, the Fourier coefficients have additional exponential decay, allowing differentiation and verifying Laplace's equation and all three Neumann boundary conditions.

To justify the endpoint calculation in b, differentiate at $0<x<\pi$, $y=0$ in the interior-edge sense:Subtracting $1$ from the hyperbolic tangent leaves an absolutely convergent series. For the remaining part, the power series $-\log (1-z)={\sum }_{m\ge 1}{z}^m/m$ gives, by taking real parts at $z=r{e}^{2ix}$ and then $r\uparrow 1$,The same formula follows as the limit of the differentiated interior solution, since its leading multiplier is $e^{-2my}$. This establishes the boundary normal derivative used in b. It also explains why the failure of the corner compatibility condition for mixed boundary data affects first derivatives at the bottom corners without spoiling continuity of $u$ itself.

On $(-\pi ,\pi )$ the function is odd, so its mean and all its cosine Fourier coefficients vanish:The assigned endpoint value does not affect these integrals. The product of the two odd functions $x|x|$ and $\sin nx$ is even, hence the sine Fourier coefficients areTwo applications of integration by parts giveThereforeThe periodic extension has a jump discontinuity at every odd multiple of $\pi$: its left and right limits are ${\pi }^2$ and $-{\pi }^2$. Its assigned value at each such point is $-{\pi }^2$. Thus it is odd away from these points, but it is not a globally odd function pointwise. The Fourier series will take the midpoint value zero at these jumps.

For a nonzero integer $k$, direct integration giveswhereas ${\int }_{-\pi }^{\pi }1dx=2\pi$. The product-to-sum formulastherefore establish Fourier orthogonality: for positive integers $m,n$,Here ${\delta }_{mn}$ is the Kronecker delta. In particular the constant function is orthogonal to every nonconstant sine and cosine in this system. Integrating the given Fourier series, then multiplying it by $\cos (mx)$ or $\sin (mx)$ and integrating, isolates its Fourier coefficients. Using the permitted interchange of the sum and integral givesIf $f$ is an even function, then $f(x)\sin (nx)$ is an odd function, while $f(x)\cos (nx)$ is an even function. ThusThe Fourier series consequently reduces to a Fourier cosine series.

For a nonzero separable solution of the Laplace equation in polar coordinates, write $T(r,\theta )=R(r)\Theta (\theta )$ with $r>0$. Multiplying the partial differential equation by $r^2$ and dividing where $R\Theta \ne 0$ giveswhere $\lambda$ is a real separation constant. The resulting ordinary differential equations areThe equations continue across zeros of the factors by continuity. The radial equation is a Cauchy-Euler equation; putting $s=\log r$ changes it to $d^{2}R/d{s}^2-\lambda R=0$.

For positive separation constant, write $\lambda =k^2$ with $k>0$. The radial factors are powers, while the angular factors are sine and cosine:For zero separation constant, each transformed equation has zero second derivative, givingFor negative separation constant, write $\lambda =-k^2$ with $k>0$. The radial factors oscillate in $\log r$, and the angular factors are exponential functions:These exhaust the nonzero real separated solutions before imposing angular periodicity or regularity at the origin. The zero solution is also allowed. Since Laplace's equation is linear, sums of these separated solutions are solutions wherever the sums and required derivatives converge appropriately.

For the first sequence, $|a_n|=n^{-3}\to 0$. The squeeze theorem gives $a_n\to 0$, and every subsequence of a convergent sequence has the same limit of a sequence. Thus its set of subsequential limits is $\boxed{\{0\}}$.

For the second sequence, the four possible sine values giveThe even-indexed subsequence is constantly one, and $b_{4k+3}\to 0$, so both zero and one occur as subsequential limits. A convergent subsequence is bounded, so it contains only finitely many terms with $n\equiv 1(\mathrm{mod}4)$. Of the remaining three residue classes, at least one occurs infinitely often. The corresponding further subsequence has limit of a sequence zero or one; the subsequence rule forces the original convergent subsequence to have that same limit of a sequence. Therefore the exact answer is $\boxed{\{0,1\}}$. Divergence to $+\infty$ along $n\equiv 1(\mathrm{mod}4)$ is not a finite subsequential limit.

We verify that the augmented partition of an interval in part i has vanishing mesh of a partition. Define ${\varphi }_n(t)=t^2/(t^2+n)$ for $t\ge 0$. Its derivative isSince $2u\le 1+u^2$ for $u\ge 0$, we have $0\le {\varphi }_n^{\prime }(t)\le 1/\sqrt{n}$. The mean value theorem therefore givesThe appended interval has length $1/(n+1)$, so $\Vert {\mathcal{P}}_n\Vert \le \max \{n^{-1/2},(n+1{)}^{-1}\}\to 0$ for the augmented partition of an interval.

The sine is continuous, hence Riemann integrable on $[0,1]$. By the Riemann sum convergence theorem, $T_n\to {\int }_0^1\sin tdt=1-\cos 1$, using the fundamental theorem of calculus and the cosine antiderivative. As $T_n-S_n=\sin 1/(n+1)\to 0$, the answer is

SetThe points strictly increase from zero to $x_{n,n}=n/(n+1)$. With $f(t)=\sin t$, the given expression is exactly the midpoint Riemann sumfor the partition of an interval ${\mathcal{P}}_n=(x_{0,n},\dots ,x_{n,n})$ of $[0,n/(n+1)]$. The sine argument is precisely the midpoint tag, since it is half the sum of the consecutive endpoints.

For a fixed integration interval $[0,1]$, append $x_{n+1,n}=1$ and choose the final tag ${\xi }_{n,n}=1$. The resulting Riemann sum $T_n$ satisfiesThis makes explicit the small final interval that the original sum omits.

First find the fixed axis of rotation by solving $(M-I)\mathbf{n}=0$. After removing a nonzero common factor, the linear equations areThey give $n_2=n_3=-n_1$. Choose the following unit vectors:Their dot products show that they form an orthonormal basis; hence $P=({\mathbf{p}}_1{\mathbf{p}}_2{\mathbf{p}}_3)$ is an orthogonal matrix. Since $M$ is an orthogonal matrix fixing ${\mathbf{p}}_1$, it preserves the orthogonal complement of ${\mathbf{p}}_1$. Its restriction there has determinant $1$, so it is a planar rotation.

The trace gives $1+2\cos \theta =\mathrm{tr}M=9/7$, so $\cos \theta =1/7$. To determine the sign of the sine, rather than just its magnitude, calculateThus a suitable angle and change of basis areIn fact direct matrix multiplication gives

The theorem on pointwise convergence of a piecewise smooth Fourier series gives the average of the two one-sided values of the corresponding periodic extension. Under the usual piecewise smooth interpretation of the paper's derivative assumption, the even extension is continuous at $0$ and at the joined endpoints $\pm \pi$. ThusFor the odd extension, the values on either side of zero are $-f(0)$ and $f(0)$; those on either side of $\pi$ are $f(\pi )$ and $-f(\pi )$. ThereforeThe endpoint values are also immediate because each sine term vanishes there. If $f(0)$ or $f(\pi )$ is nonzero, the Fourier sine series does not reproduce that endpoint value. Merely assuming that the derivative is locally piecewise continuous in the open interval would be weaker than the usual piecewise smooth hypothesis; the convergence assertion uses the standard endpoint regularity implicit in that hypothesis.