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Let a homogeneous plate $0<x<L$ begin at uniform temperature $T_0$, cool with zero-temperature Dirichlet boundary conditions until time $t_b$, then switch to insulated Neumann boundary conditions. For thermal diffusivity $\kappa >0$, the Fourier sine series during cooling is $T=(4T_0/\pi ){\sum }_{n\ge 1,n\text{ odd}}\sin (n\pi x/L)e^{-\kappa (n\pi /L{)}^{2}t}/n$. Integrating the insulated heat equation shows that its integral average is conserved. All nonconstant Fourier cosine series modes then decay, proving that the final uniform temperature is $T_{\infty }=(8T_0/{\pi }^2){\sum }_{n\ge 1,n\text{ odd}}{e}^{-\kappa (n\pi /L{)}^2{t}_b}/n^2$. Its limits are $T_0$ at zero bath time and zero at infinite bath time. Instantaneous changes of boundary condition can create initial-boundary corner incompatibilities without invalidating the positive-time solution.

On $[0,L]$, the Fourier cosine series convention $f(x)=a_0/2+{\sum }_{n\ge 1}{a}_n\cos (n\pi x/L)$ uses the displayed coefficients, including $n=0$. Orthogonality of the cosine modes gives these formulas for square-integrable functions; the expansion is interpreted in the L2 norm until stronger convergence is established.

The period-two extension of $1-|x|$ on $[-1,1]$ is even. Its Fourier cosine series has $a_0=1$ and $a_n=2{\int }_0^1(1-x)\cos (n\pi x)dx=2[1-(-1{)}^n]/(n^2{\pi }^2)$ by integration by parts. This proves the displayed series, with only odd nonconstant harmonics. Continuity and piecewise smoothness give pointwise equality through the Dirichlet-Jordan convergence theorem; the summable coefficient magnitudes also imply uniform convergence by the Weierstrass M-test. At $x=0$, it gives ${\sum }_{m\ge 0}(2m+1{)}^{-2}={\pi }^2/8$.

On $0\le x\le L$, the Fourier sine series of ${\sin }^2(\pi x/L)$ has zero even-index coefficients and odd-index coefficients $b_n=8/[\pi n(4-n^2)]$. Substitute $u=\pi x/L$ in the coefficient integral and use ${\sin }^{2}u\sin (nu)=\frac{1}{2}\sin (nu)-\frac{1}{4}\sin ((n+2)u)-\frac{1}{4}\sin ((n-2)u)$. For odd $n$, integrating gives $b_n=(2/\pi )[1/n-\frac{1}{2}(1/(n+2)+1/(n-2))]$; reflection about the midpoint makes the even coefficients zero. The $O(n^{-3})$ decay gives absolute and uniform convergence, and Fourier orthogonality identifies the sum. In the heat equation with zero endpoint temperatures, multiply each coefficient by $\exp [-\kappa (n\pi /L{)}^{2}t]$. With insulated endpoints, the same initial profile is instead just the constant and second cosine modes of a Fourier cosine series.

For a function prescribed on $0<x<L$, its half-range Fourier cosine series is the Fourier series of the even extension, made $2L$-periodic. Its coefficients are $a_n=(2/L){\int }_0^{L}f(x)\cos (n\pi x/L)dx$, including $n=0$. At a jump the usual piecewise-smooth convergence theorem gives the average of the two one-sided limits of the periodic extension.

On a bar of length $L$, let the initial temperature increase linearly from zero at either endpoint to height $H$ at the midpoint. With thermal diffusivity $D>0$ and insulated Neumann boundary conditions, a Fourier cosine series gives the displayed solution. Its constant term is the initial integral average $H/2$. Integration by parts on the two straight segments gives zero coefficients for odd modes and modes divisible by four; the remaining coefficients are $-4H/[{\pi }^2(2j+1{)}^2]$. Nonconstant modes decay, leaving the uniform temperature $H/2$, consistent with conservation of the total heat. This contrasts with the zero-temperature boundary solution in heat flow from a triangular initial temperature, where heat can leave the bar and the limit is zero. The initial endpoint slopes do not satisfy the insulated condition, so derivatives need not converge to their initial endpoint values as $t\downarrow 0$; the series satisfies the Neumann condition for every $t>0$ and recovers the continuous initial profile.

Expand a real function with zero endpoint values in its Fourier sine series, $y={\sum }_{n\ge 1}{a}_n\sin (n\pi x/L)$. The Parseval identity gives $\int y^2=(L/2)\sum a_n^2$. For its weak derivative, the Fourier cosine series has coefficients $(n\pi /L)a_n$ and zero constant term, soThe derivative coefficient identity follows by integration by parts, rather than by assuming pointwise differentiation of an arbitrary series. This proves the Wirtinger inequality and its equality case. Under unit L2 norm, the only real minimizers are $y=\pm \sqrt{2/L}\sin (\pi x/L)$.

Every finite sum of the sine modes is odd, and so is a limit in the L2 norm of such sums. In contrast, $x^2-1$ is a nonzero even function. Its inner product with every odd sine mode vanishes, so all its proposed Fourier sine series coefficients would be zero. It cannot be represented by this basis on the symmetric interval. An even Fourier cosine series is the appropriate alternative there.

For the real even function, let $s_N=a_0/2+{\sum }_{n=1}^N{a}_n\cos nx$. Fourier orthogonality gives ${\int }_{-\pi }^{\pi }\cos nxdx=0$ for $n\ge 1$ and ${\int }_{-\pi }^{\pi }\cos nx\cos mxdx=\pi {\delta }_{nm}$. ConsequentlyFor an even square-integrable function, the complete Fourier cosine series converges in the L2 norm; hence its norms converge. Taking $N\to \infty$ proves Parseval's identity in the required normalization:This limit argument avoids multiplying an infinite series without a justification of L2 norm convergence.

For the quadratic function, the Fourier coefficients areThe last equality follows from two integrations by parts, using $\sin n\pi =0$. ThusThis Fourier series is absolutely convergent; the periodically extended quadratic is continuous, so it represents its values also at the endpoints. Substituting its coefficients into Parseval's identity gives

The Fourier sine series has coefficientsby integration by parts. ThusIts odd extension is $x$ on $(-L,L)$, continued with period $2L$. The usual Fourier convergence result for a piecewise smooth periodic function gives the value at a continuous point and the mean of one-sided limits at a jump. At $x=3L/2$, subtracting one period gives $-L/2$, a continuous point of this extension. Therefore the sine series converges to $-L/2$ there.

For the Fourier cosine series, the even extension is $|x|$ on $[-L,L]$, continued with period $2L$. Its triangular wave is continuous even at the joining points, so the cosine series converges to that extension everywhere. On the requested interval its graph isThe vertices are $(-2L,0),(-L,L),(0,0),(L,L),(2L,0)$. In particular, its value at $3L/2$ is $L/2$.

The Fourier sine series coefficients of $x$ on $(0,\pi )$ areby integration by parts. The odd periodic extension is piecewise smooth, so the Fourier series converges to $x$ at every interior point. HenceTo justify integration from zero, the Fourier sine series partial sums converge to $x$ in L2 norm. By the Cauchy-Schwarz inequality, integration from zero to any $x\in [0,\pi ]$ has error at most $\sqrt{\pi }$ times that L2 norm error, uniformly in $x$. Integrating the partial sums and taking the limit therefore givesThe integrated series is absolutely and uniformly convergent, since its terms are bounded by $4/n^2$. Thus the resulting Fourier cosine series hasIndependently, the constant coefficient isComparison yields the requested exact sum

The Neumann boundary conditions select the spatial eigenfunctions $\cos nx$, $n=0,1,\dots$, of the separated wave equation. Indeed separation gives $X^{\prime \prime }+\lambda X=0$ with $X^{\prime }(0)=X^{\prime }(\pi )=0$. Integration by parts gives $\lambda \int X^2=\int X^{\prime 2}\ge 0$. The zero value yields the constant mode; for positive $\lambda$, the first boundary condition removes the sine term and the second requires $\sqrt{\lambda }=n\in \mathbb{N}$. For $n>0$, the temporal factor is $\cos nct$: the sine factor vanishes because the initial velocity is zero. The zero mode has constant displacement, with no term linear in time for the same reason.

The constant Fourier cosine series coefficient is the mean ${\pi }^{-1}{\int }_0^{\pi }bxdx=b\pi /2$. The remaining coefficients areOnly odd modes survive. Thus the displacement isThe series for displacement converges absolutely and uniformly. The initial velocity is zero in the finite-energy sense, and the initial displacement is $bx$. There is an endpoint compatibility qualification: when $b\ne 0$, the initial slope does not satisfy the Neumann boundary conditions at $t=0$. The answer is therefore a finite-energy weak solution, piecewise classical between the propagating corners, rather than a globally twice continuously differentiable solution up to the initial endpoints. The mode expansion implements reflection at those endpoints.

The method of images for the Neumann boundary conditions extends $bx$ evenly through zero and then with period $2\pi$. Define the triangular waveThe formulas agree at their endpoints. It is even, $2\pi$-periodic, and agrees with $bx$ on $[0,\pi ]$. Its Fourier cosine series has the coefficients in part (a). The cosine product identity then gives the D'Alembert formulaIt is the average of two oppositely traveling reflected waves. For $b>0$, the sketch rises from zero at even multiples of $\pi$ to $b\pi$ at odd multiples; for $b<0$ it is reflected vertically, and for $b=0$ it is zero.

Since $x=r\cos \theta$, the function on the left of the defining expansion is $e^{ix}$. Thereforewhich is the two-dimensional Helmholtz equation.

In polar coordinates,Substitute the defining Fourier cosine series and equate the coefficient of each $\cos (n\theta )$. Its angular second derivative contributes $-n^2$, so every Bessel function $J_n$ satisfiesor equivalently the Bessel differential equation

Expand the exponential through cubic order and useThis givesComparing with $J_0+2{\sum }_{n\ge 1}{i}^n{J}_n\cos (n\theta )$ yields

The mean of $f$ is zero. Its Fourier cosine series coefficients areThus $A_n=0$ for even $n$ and $A_n=4/(\pi n^2)$ for odd $n$. Substitution into part (i) yields

The Euler-Lagrange equation of each component of the string embedding map is the wave equation $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^{\mu }=0$. Its left- and right-moving solutions are identified by the endpoint Neumann boundary conditions, so the resulting open-string mode expansion isIndeed, the Fourier cosine series makes ${\partial }_{\sigma }{X}^{\mu }$ vanish at $\sigma =0,\pi$. The canonical momentum density is ${\Pi }_{\mu }=(2\pi {\alpha }^{\prime }{)}^{-1}{\dot{X}}_{\mu }$. Imposing the equal-time canonical commutation relations $[X^{\mu }(\sigma ),{\Pi }_{\nu }({\sigma }^{\prime })]=i{\delta }_{\nu }^{\mu }\delta (\sigma -{\sigma }^{\prime })$ gives the covariant quantization of the bosonic stringwith ${\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{p}^{\mu }$.

Under the stated piecewise smooth hypotheses, the Dirichlet convergence theorem for Fourier series gives the average of the two one-sided limits of the corresponding periodic extension. The even extension is continuous at both endpoints, so the Fourier cosine series converges toThe odd extension agrees with $f$ on $(0,L)$, but its left and right limits are $-f(0),f(0)$ at zero and $f(L),-f(L)$ at the join $L$. Therefore the Fourier sine series converges toIndeed every partial sum of the Fourier sine series is already zero at both endpoints. The endpoint values agree with $f$ only when the corresponding boundary values vanish.

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 $L$.

First, the Fourier coefficients of the Fourier cosine series decay as $O(n^{-2})$, whereas those of the Fourier sine series are of order $n^{-1}$. In particular, the absolute convergence bound for the Fourier cosine series gives a uniform tail estimateThus its tail is absolutely controlled using fewer Fourier harmonics. The Fourier sine series does not have an absolutely summable sequence of Fourier coefficients.

Second, the even extension is continuous, so its Fourier cosine series converges uniformly on the closed interval and approximates the nonzero endpoint limits correctly. The odd periodic extension has jumps, producing the Gibbs phenomenon in its Fourier sine series near zero and $L$. Every Fourier sine series partial sum is zero at both endpoints, so by approaching $L$ from inside $(0,L)$ one obtainsHence the Fourier sine series approximation cannot converge uniformly even on the whole open interval, despite pointwise convergence at each interior point. The Fourier cosine series approximation has faster coefficient decay and avoids the persistent errors near the endpoints.

Set $u=x/L$ and $q=n\pi$. The required integrals follow fromConsequently the Fourier coefficients areThus the Fourier cosine series isand its sum is $e^{x/L}$ everywhere on $[0,L]$. The Fourier sine series isand its sum is $e^{x/L}$ on $(0,L)$ and zero at the two endpoints. The open and filled circles in the sketches distinguish these endpoint limits from the actual values of the sum.