Codex @codex  0

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.

Put $u(r,\theta )=R(r)\Theta (\theta )$ in Laplace's equation and multiply by $r^2/(R\Theta )$ on intervals where the factors are nonzero. Separation of variables givesand henceThe angular Dirichlet boundary conditions are $\Theta (0)=\Theta (\pi /2)=0$. Multiplication of the angular equation by $\Theta$, followed by integration by parts, givesFor a nonzero angular eigenfunction this forces $\lambda >0$: equality would make $\Theta$ constant and then zero. Write $\lambda ={\mu }^2$. The first angular condition removes the cosine term, and the second requires $\sin (\mu \pi /2)=0$. ThusThe radial Cauchy-Euler equation has the two solutions $r^{2n}$ and $r^{-2n}$. By superposition, the general separated series is thereforeThe coefficients are arbitrary subject to convergence with enough partial derivatives on compact subsets to give a harmonic function and to the prescribed zero values on the two open radial edges. Completeness of the sine eigenfunctions gives the usual series representation for solutions smooth up to those edges on closed radial subintervals. There is no constant or logarithmic zero mode because its angular factor would be linear in $\theta$ and would vanish at both ends only if it were zero.

The negative powers must be retained: the stated domain omits $r=0$, and no boundedness at that vertex is prescribed. For example, $r^{-2}\sin (2\theta )$ is a harmonic function satisfying both edge conditions. An additional boundedness condition as $r\to 0$ would remove all $B_n$; an additional value on the circular arc would then determine the remaining $A_n$. With only the conditions in the paper, neither restriction is available and the solution is not unique.

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.

For the exponential distribution with rate $\theta$, the change of variables $u=\theta x$ givesThe integrals are one and two by integration by parts. Hence $\boxed{\mathbb{E}[X]=1/\theta ,\mathrm{Var}(X)=1/{\theta }^2}$.

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.

Use separation of variables to seek a normal mode $y(x,t)=X(x)\cos (\omega t+\delta )$, with angular frequency $\omega >0$ and wavenumber $k=\omega /c$. The endpoint boundary conditions are built intoLet $z=k\ell$ and denote the mass's displacement amplitude by $X_0$. Continuity and the mass equation becomeThe two possibilities must be separated before dividing by $\sin z$.

Antisymmetric modes. If $\sin z=0$, then $z=n\pi$, $n=1,2,\dots$, and $X_0=0$. The force equation gives $A+B=0$, since $\cos z\ne 0$. Thus $X(-x)=-X(x)$ and the mass stays at rest. These allowed angular frequencies satisfyThe restoring forces on the mass cancel even though the string vibrates.

Symmetric modes. If $\sin z\ne 0$, continuity forces $A=B$. Cancelling the nonzero amplitude in the mass equation and using ${\omega }^2=(T/\rho )k^2$ givesThese modes have $X(-x)=X(x)$ and move the mass. In dimensionless form their equation is $\cot z=\alpha z$, where $\alpha =m/(2\rho \ell )>0$. This is the spectral splitting for a string with a central point mass.

For the antisymmetric family, the graph of $\sin z$ crosses zero at every positive multiple of $\pi$. For the symmetric family, on each interval $(n\pi ,n\pi +\pi /2)$, $n=0,1,\dots$, the graph of $\cot z$ decreases continuously from $+\infty$ to zero while $\alpha z$ increases. By the intermediate value theorem and strict monotonicity, there is exactly one intersection in each interval. On the remaining half-intervals $\cot z<0<\alpha z$, so there are no further positive roots. This proves graphically and analytically that both families contain infinitely many normal modes. The figure uses $\alpha =1$ for illustration.

The algebraic case $k=0$ is not a mode: a zero-frequency profile would be linear on each half, continuous and zero at the endpoints; the zero-acceleration mass equation also makes its slopes agree, forcing the entire profile to be zero. More generally, integration by parts on the two halves and the matching condition givefor a nonzero real mode. Thus no nonzero static or exponentially growing mode has been omitted.

For this rate parametrization of the Laplace distribution, $\mu$ is its mean and median. By symmetry, $\mathbb{E}(X-\mu )=0$, while $|X-\mu |$ has exponential distribution with rate $\lambda$, soThe integral follows by two applications of integration by parts. The central limit theorem therefore givesWith known $\lambda$ and the supplied cutoff 2, an approximate 95% confidence interval isThis known-rate Laplace location confidence interval uses the sample mean; the later maximum-likelihood estimator of location need not be used to construct this valid asymptotic confidence interval.

For the standard normal density, $P^{\prime }(x)=-xP(x)$. Using integration by parts,The boundary term vanishes because Gaussian decay dominates every power. Since the given mean is zero and variance is one, $\mathbb{E}[X^2]=1$, yielding the Gaussian fourth moment

On the basic interval the graph is the pair of straight segments joining $(-\pi ,0)$, $(0,1)$ and $(\pi ,0)$. Its periodic extension is an even triangular wave.

The Fourier coefficients satisfy $b_n=0$ by evenness, andFor the second formula, integration by parts gives ${\int }_0^{\pi }x\cos (nx)dx=[(-1{)}^n-1]/n^2$. Only odd indices survive, giving this rescaled Fourier cosine series of a triangular wave:The pointwise convergence of a piecewise smooth Fourier series applies throughout the interval, including the continuous joined endpoints. The coefficients also imply uniform convergence by the Weierstrass M-test. At $x=0$, the value is one, so

The periodically extended $x^2$ is even, so every sine coefficient is zero. With the convention $f=a_0/2+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)$, the Fourier coefficients areFor the second integral, two applications of integration by parts give ${\int }_0^{\pi }{x}^2\cos nxdx=2\pi (-1{)}^n/n^2$. ConsequentlyThe periodic extension is continuous even at the joined endpoints, where both one-sided values equal ${\pi }^2$. The pointwise convergence of a piecewise smooth Fourier series therefore gives the displayed values throughout the interval; the coefficient bound also gives uniform convergence by the Weierstrass M-test. Setting $x=\pi$ and then $x=0$ yields

Let $I={\int }_0^{\infty }{e}^{-x^2}dx$. Positivity allows the squared integral to be written as a first-quadrant double integral. In polar coordinates the Jacobian determinant is $r$, soUsing integration by parts, with the boundary term $x{e}^{-x^2}$ vanishing at both endpoints,The substitution $u=x^2$ also gives the two antiderivativesThese values also follow from the Gaussian moment recurrence and can be checked by differentiation.

Use integration by parts with $u=\ln (1+x^2)$ and $dv=(1+3x^2)dx$, giving $v=x+x^3=x(1+x^2)$. The remaining integrand simplifies because $du=2xdx/(1+x^2)$:The real formula is valid for every real number $x$.

Two applications of integration by parts give $I=\cosh x\sin x-\int \sinh x\sin xdx$ and $\int \sinh x\sin xdx=-\sinh x\cos x+I$. Thus an antiderivative is $[\cosh x\sin x+\sinh x\cos x]/2$. Evaluating its endpoints gives

The expectation is $\mathbb{E}[X]=k{\int }_0^{\infty }t{e}^{-kt}dt$. An integration by parts, with the boundary term $t{e}^{-kt}$ tending to zero, gives

For $1<x<e$, the natural logarithm satisfies $0<\ln x<1$. Squaring therefore decreases its value, while $\ln (x^2)=2\ln x$. Pointwise comparison gives $(\ln x{)}^2<\ln x<\ln (x^2)$, soAs a check, integration by parts gives $I_1=1$, $I_2=2$ and $I_3=e-2$, consistent with this order.

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.

The probability density function is normalized because ${\int }_0^{\infty }\alpha e^{-\alpha t}dt=1$. Integration by parts gives $\mathbb{E}[T]={\int }_0^{\infty }t\alpha e^{-\alpha t}dt=1/\alpha$ and $\mathbb{E}[T^2]=2/{\alpha }^2$. ThereforeThese are the mean and variance of the exponential distribution with rate $\alpha$.

In first-quadrant polar coordinates, $x=r\cos \theta$, $y=r\sin \theta$, and the Jacobian determinant contributes $r$. The integrand and area element together become $r^2{e}^{-r^2}\sin \theta {\cos }^2\theta drd\theta$. HenceThe radial integral follows by integration by parts from the Gaussian integral. The isolated undefined ratio at the origin does not affect this convergent double integral.

Integration by parts with $u=\ln x$ and $dv=xdx$ givesThe remaining integral is $\frac{1}{2}\int xdx$.