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Take $\boxed{p(x)=x(L-x)}$. It vanishes at both endpoints. Its odd periodic extension is continuously differentiable across the joins, whereas its even extension has corners because $p^{\prime }(0)=L$ and $p^{\prime }(L)=-L$ are nonzero.

Two integrations by parts, with $q=n\pi /L$, giveThe Fourier coefficients are thereforeThe Fourier sine series has a uniform absolute tail $O(N^{-2})$, and each of its partial sums satisfies both zero boundary values exactly. The Fourier cosine series has an absolute tail bound $O(N^{-1})$; at $x=0$ its omitted even-index Fourier coefficients have the same sign and yield an error of order $N^{-1}$. This verifies a strictly faster uniform error bound for the sine approximation in this example.

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.

The Fourier cosine series and Fourier sine series use the respective formsTheir Fourier coefficients, obtained from Fourier orthogonality, areIn particular the constant term is $a_0/2={\pi }^{-1}{\int }_0^{\pi }f(t)dt$. The Fourier cosine series is the Fourier series of the even extension $f(|x|)$, made $2\pi$-periodic. The Fourier sine series is the Fourier series of the odd extension, equal to $f(x)$ for $0<x<\pi$ and $-f(-x)$ for $-\pi <x<0$, again made $2\pi$-periodic.

The truncated Fourier cosine series gives the better approximation on the closed interval, in the senses of uniform convergence and the asymptotic L2 norm error. To make the comparison precise, let $C_N$ and $S_N$ retain the frequencies $1\le n\le N$, with the constant term also included in $C_N$.

The cosine Fourier coefficients are $O(n^{-2})$. In fact, absolute convergence gives the uniform tail boundFor the Fourier sine series, every Fourier partial sum satisfies $S_N(0)=S_N(\pi )=0$. Thusso there is no uniform convergence to $f$ on the closed interval. Its odd extension has jump discontinuities, producing the Gibbs phenomenon near the endpoints; its Fourier coefficients decrease only like $1/n$.

The advantage persists in the L2 norm, which ignores individual endpoint values. By Parseval's identity and Fourier orthogonality,The nonzero odd cosine coefficients and both parity classes of sine coefficients also give matching lower bounds of these orders by comparison with the corresponding p-series. Hence their L2 norm errors decrease respectively like $N^{-3/2}$ and $N^{-1/2}$. This is a comparison of approximation quality in specified norms, not a claim that one finite sum has smaller error at every individual interior point.

For the Fourier cosine series, integration by parts givesand, for $n\ge 1$,Only odd Fourier harmonics remain, so

For the Fourier sine series, another integration by parts givesConsequentlyAt $x=0$ and $x=\pi$, the Fourier sine series instead has value zero.

To describe the requested graphs precisely, take $t\in [-\pi ,\pi )$ congruent to $x$ modulo $2\pi$. The Fourier cosine series converges to the continuous triangular periodic extensionIts minima occur at the even multiples of $\pi$, with value $1$, and its maxima at the odd multiples, with value $1+\pi$. The Fourier sine series converges toThe jump discontinuities at even multiples of $\pi$ go from $-1$ to $1$; those at odd multiples go from $1+\pi$ to $-(1+\pi )$. Each jump is assigned its midpoint, zero, by the Fourier series.

Take $\boxed{f(x)=\sin x}$. This function is not a polynomial: it is nonzero and satisfies $f^{\prime \prime }=-f$, whereas a nonzero polynomial cannot satisfy that identity, since taking two derivatives reduces its degree or gives zero.

By Fourier orthogonality, the Fourier sine series consists of the single term $\sin x$. Thus every truncation with $N\ge 1$ is exact, including at the endpoints. In contrast, using the product-to-sum trigonometric identity in its Fourier cosine coefficients givesFor example, ${\int }_0^{\pi }\sin x\cos (2mx)dx=-2/(4m^2-1)$ follows by integrating the two resulting sine terms. Therefore the Fourier cosine series isEvery finite cosine truncation omits infinitely many nonzero Fourier coefficients, so Parseval's identity gives it strictly positive squared L2 norm error. The one-term sine representation has zero error in both the L2 norm and the supremum norm, proving the requested advantage.

Set $u=X(x)Y(y)$ in Laplace's equation. Separation of variables givesThe Sturm-Liouville problem in $x$ has the constant mode $X_0=1$ with ${\lambda }_0=0$, and the eigenfunctions $X_n=\cos (nx)$ with ${\lambda }_n=n^2$, $n\ge 1$. Indeed, integration by parts makes $\lambda$ nonnegative; for $\lambda >0$ the first Neumann boundary condition removes the sine term and the second requires $\sqrt{\lambda }$ to be an integer. For $\lambda =0$, both derivative conditions make $X$ constant.

Expand the top data using a Fourier cosine series:The constant coefficient isFor $n\ge 1$, the constant part has zero integral against $\cos (nx)$, and two integrations by parts give

Since the bottom value is zero, a convenient factor depending on $y$ for the nonconstant modes is $\sinh (n(y+1))$, which vanishes at $y=-1$. Its value at $y=1$ is $\sinh (2n)$. The zero mode is linear in $y$, so matching both constant boundary values givesTo obtain exactly the paper's form, use the hyperbolic-function identityThus

Each product of hyperbolic sine or hyperbolic cosine with $\cos (nx)$ has $u_{xx}+u_{yy}=0$. The zero mode is also a harmonic function. On $-1\le y\le 1$, the factor $\sinh (n(y+1))/\sinh (2n)$ lies between zero and one. Since $\sum n^{-2}$ converges, the Weierstrass M-test gives uniform convergence on the closed rectangle. The series therefore gives zero at the bottom and the continuous Fourier cosine series of ${\pi }^2-x^2$ at the top, including its endpoints. On compact subsets with $y<1$, the factors decay exponentially with $n$, justifying termwise partial derivatives. In particular $u_x=0$ on both open vertical edges because every differentiated cosine factor vanishes there. The solution has finite Dirichlet energy: Fourier orthogonality gives the $n$th nonconstant mode's energy as $(\pi /2)n{g}_n^2\coth (2n)=O(n^{-3})$, whose sum is finite. Here $\coth$ is the hyperbolic cotangent. Thus a proves uniqueness in that class.

There is a corner compatibility condition for mixed boundary data in the printed wording. The top data have $g^{\prime }(\pi )=-2\pi$, but the paper writes $u_x(\pi ,y)=0$ even at $y=1$. The series solves the interior equation and all the open-edge boundary conditions; the literal derivative condition at $(\pi ,1)$ is incompatible with the top data. A function differentiable in $x$ along the top edge would necessarily have $u_x(\pi ,1)=g^{\prime }(\pi )=-2\pi$. Hence it cannot simultaneously satisfy the printed zero derivative at that corner, and there is no solution with all those literal corner requirements. The standard interpretation of this mixed boundary condition imposes the Neumann boundary condition on the open sides and the Dirichlet boundary condition continuously on the horizontal edges. Differentiating the series term by term on the top edge would be invalid at this corner; the uniform convergence of the undifferentiated series does not license that operation.

Using the usual piecewise smooth interpretation of the derivative hypothesis, Dirichlet's theorem on Fourier series gives the average of the two one-sided limits of the appropriate periodic extension. The even extension is continuous at $0$ and at the identified endpoints $\pm \pi$, so the Fourier cosine series converges toFor the odd extension, the one-sided limits at zero are $-f(0)$ and $f(0)$, while at $\pi$ the periodic one-sided limits are $f(\pi )$ and $-f(\pi )$. Hence the Fourier sine series converges toAt an endpoint this equals $f$ precisely when $f$ is zero there. One can also see the zero endpoint values directly: every partial sum consists of sines that vanish at both endpoints.

For the Fourier cosine series, first take the even extension $F_e(x)=f(|x|)$ on $[-\pi ,\pi ]$, and then its $2\pi$-periodic extension. Evenness eliminates the sine terms. With the constant term written as $a_0/2$, the Fourier coefficients and series areFor the Fourier sine series, take the odd extension $F_o(x)=f(x)$ for $0<x<\pi$ and $F_o(x)=-f(-x)$ for $-\pi <x<0$, and extend with period $2\pi$. Values assigned at isolated endpoints do not change the Fourier coefficients. Oddness eliminates the constant and cosine terms, leavingBoth formulas follow by orthogonality on $[-\pi ,\pi ]$ and the parity of the extensions.

The truncated Fourier cosine series gives the better approximation over the whole closed interval. Its Fourier coefficients obey $|a_n|\le 4/(\pi n^2)$, so absolute convergence and the tail boundgive uniform convergence. The even periodic extension is continuous; its slope changes only cause corners.

By contrast $b_n=(-1{)}^{n+1}/n+O(n^{-2})$. The odd periodic extension has a jump at $\pi$, giving slower coefficient decay and the Gibbs phenomenon near that endpoint. Every truncated Fourier sine series satisfies $S_N(\pi )=0$, soIt therefore cannot converge uniformly to $f$ on $[0,\pi ]$, although it converges at every point of $[0,\pi )$. This comparison concerns approximation across the interval; it does not assert that the cosine truncation has smaller error at every individual interior point for every $N$.

The Fourier cosine series converges to the continuous even $2\pi$-periodic extension. If $r\in [-\pi ,\pi ]$ represents $x$ modulo $2\pi$, its value isIt has value zero at $x=2j\pi$, rises linearly with distance from these points until height $\pi /2$, and then stays flat until the next descending ramp. Thus the sketch on $[-3\pi ,3\pi ]$ has minima at $-2\pi ,0,2\pi$, and plateaux centred at the odd multiples of $\pi$. Its large partial sums round the corners and closely follow the whole graph.

The Fourier sine series converges, away from the odd multiples of $\pi$, to the odd $2\pi$-periodic extensionAt every odd multiple of $\pi$ its value is zero, the midpoint of the jump from $\pi /2$ to $-\pi /2$. At even multiples of $\pi$ it passes continuously through zero. Thus on every interval $((2j-1)\pi ,(2j+1)\pi )$ it has a negative plateau, a rising linear segment through $2j\pi$, and a positive plateau. The jump size is $\pi$.

The blue curves are the partial sums through $n=60$. The open endpoint circles and filled zero dots in the lower panel distinguish the one-sided limits from the value of the Fourier sine series at each jump. Near these jumps the partial sums exhibit the Gibbs phenomenon: the oscillations become concentrated in narrower regions as the number of terms grows, while their limiting overshoot remains about $0.08949$ times the jump, here about $0.281$.

Put $a=\pi /2$. The constant Fourier coefficient isFor $n\ge 1$, integration by parts givesThe $n^{-1}$ terms cancel. ThusThis Fourier cosine series has $a_n=0$ when $n$ is divisible by four, $a_n=-4/(\pi n^2)$ when $n\equiv 2(\mathrm{mod}4)$, and $a_n=-2/(\pi n^2)$ when $n$ is odd.

Similarly,ThereforeThe Fourier sine series also equals $f(0)=0$ at zero, but has value zero at $\pi$, where $f(\pi )=\pi /2$.

Write $T=T_s+T_u$. The transient satisfies the heat equation and homogeneous mixed boundary conditionsThe left condition is a Neumann boundary condition and the right one a Dirichlet boundary condition. Separation of variables, $T_u=X(x)G(t)$, givesThe condition $X^{\prime }(0)=0$ removes the sine term. The condition $X(2\ell )=0$ then selectsThere is no nonzero mode with $\mu =0$, because a linear $X$ satisfying both conditions is zero. There is no growing mode either: integration by parts gives ${\int }_0^{2\ell }(X^{\prime }{)}^{2}dx={\mu }^2{\int }_0^{2\ell }{X}^{2}dx$, so every nonzero mode has ${\mu }^2>0$.

These Sturm-Liouville eigenfunctions have orthogonality and squared normExpand the initial transient in this half-integer Fourier cosine series. Its coefficients areThe integration formula follows by integration by parts; the cosine at the endpoint is zero. Thus the half-integer cosine expansion of a linear profile givesAt $t=0$ the Fourier series reproduces $-T_s$ everywhere, including the endpoints; its $n^{-2}$ coefficient decay gives uniform convergence. For every $t>0$ the exponential factors allow termwise differentiation and verify the heat equation and both boundary conditions. The sudden switch-on of nonzero heat flux is incompatible with the spatial derivative of the constant initial temperature at the corner $(0,0)$. Accordingly, the positive-time boundary derivative must not be inferred by differentiating the initial Fourier series term by term at that corner.

Evenness removes every sine coefficient. In the period-two Fourier cosine series convention $f=a_0/2+{\sum }_{n\ge 1}{a}_n\cos (n\pi x)$, the constant coefficient is $a_0=2{\int }_0^1(1-x)dx=1$. Integration by parts givesThus the Fourier cosine series of a triangular wave isThe periodic function is continuous and piecewise smooth, so the Dirichlet-Jordan convergence theorem gives its value at every point. The summable absolute coefficient bound also gives uniform convergence by the Weierstrass M-test.

Differentiation of the finite sum involves no convergence issue. The derivative has the Fourier cosine series

Assume $f\in L^2[-L,L]$, so the squared-error integral is finite. WriteAt a minimum, differentiating with respect to each coefficient givesBy Fourier orthogonality, the constant has squared norm $2L$, each nonconstant cosine has squared norm $L$, and all distinct modes are orthogonal. ThusThese are exactly the Fourier coefficients of the Fourier cosine series of the even function. To check that the stationary choice is a unique global minimum, perturb the coefficients by $\delta a_n$. The residual is orthogonal to every fitted mode, so the cross terms vanish andwith equality only when every perturbation is zero. The truncated series is the orthogonal projection onto a finite-dimensional subspace spanned by the constant and the first $N$ cosine modes.

The function is even, so each product with a sine is an odd function and all the Fourier coefficients $b_n$ vanish. Its constant coefficient isFor $n\ge 1$, part (a) gives the remaining Fourier coefficients:Consequently the Fourier cosine series isThe periodic extension is continuous, including at $\pm \pi$, and piecewise continuously differentiable. Thus the Fourier series equals $f$ everywhere. Its coefficients are also bounded in magnitude by a constant times $1/n^2$, which gives absolute convergence.

With constant positive thermal diffusivity $\nu$, no internal heat source and one spatial coordinate, the heat equation isSeparation of variables makes each spatial mode decay by $e^{-\nu k^{2}t}$. The endpoint conditions determine whether the initial triangular profile is expanded in a Fourier sine series or a Fourier cosine series.

For $y=\pm \sqrt{2/\pi }\sin x$, both the constraint and endpoint values hold, the extremal equation has $\lambda =1$, and direct integration gives $J[y]=1$.

For any admissible function, its derivative has Fourier cosine series coefficients $n{a}_n$ and zero constant coefficient. This follows by integration by parts: ${\int }_0^{\pi }{y}^{\prime }\cos (nx)dx=n{\int }_0^{\pi }y\sin (nx)dx$, and ${\int }_0^{\pi }{y}^{\prime }=0$. Thus the Parseval identity givesFor finite sums the same identity follows directly from the cosine product formula in the hint; the $L^2$ identity justifies the general case. Equality holds only if $a_n=0$ for all $n\ge 2$, and normalization then fixes $a_1=\pm \sqrt{2/\pi }$. ThereforeThis is the minimum Dirichlet energy at fixed L2 norm on an interval, equivalently the sharp Wirtinger inequality for this interval.

Put $\lambda ={\omega }^2/g$ andThis polynomial is a harmonic function: $Q_{xx}+Q_{zz}=0$. Also $Q_x(0,z)=i\omega c$, $Q_x(L,z)=0$ and $Q_z(x,-h)=0$. Thus $G$ satisfies Laplace's equation with homogeneous Neumann boundary conditions on both side walls and the bottom. Its Fourier cosine series consequently has the formThe surface condition becomes $G_z-\lambda G=H(x)$, whereFor a nonzero forcing frequency away from the natural frequencies, the coefficients of the oscillating-wall forcing of a rectangular tank areHere $k_0=0$ in the definition of $H_0$. In fact $H_n=2i\omega c\lambda /(L{k}_n^2)$ for $n\ge 1$, since ${\int }_0^L(L-x{)}^2\cos (k_{n}x)dx=2L/k_n^2$. The zero-mode coefficient ensures that the mean surface elevation is $hc/L$ in complex amplitude, exactly the first-order volume conservation adjustment to the moving wall.