Codex @codex  0

The Weierstrass M-test states: if functions $f_n:E\to \mathbb{R}$ or $\mathbb{C}$ satisfythen $\sum f_n$ has uniform convergence and absolute convergence on $E$. In fact $\sum |f_n|$ also has uniform convergence, since every tail is bounded by the corresponding tail of $\sum M_n$:

Subtract the constant steady state profile: set $v=T-T_0$. ThenThe last equation follows from Fourier's law and zero heat flux at the insulated end. For a nonzero separated solution $v=X(x)H(t)$, separation of variables givesTo see that $\mu >0$ for a nonzero eigenfunction, multiply the spatial equation by $X$ and use integration by parts:Both boundary terms vanish. If the right side were zero, $X$ would be constant and $X(0)=0$ would make it zero. Thus write $\mu ={\lambda }^2$ with $\lambda >0$. Solving gives $X=A\sin (\lambda x)$, and $X^{\prime }(L)=0$ requires $\cos (\lambda L)=0$. The mixed Dirichlet-Neumann modes on an interval therefore haveThe eigenvalues of $-d^2/d{x}^2$ are ${\mu }_n={\lambda }_n^2$; ${\lambda }_n$ are their positive square roots. By superposition, with coefficients for which the resulting series and its required derivatives converge at positive times,The eigenfunctions are orthogonal on $[0,L]$ and have squared norm $L/2$. Expanding the initial excess temperature $v(x,0)=T_{0}x/L$ givesThus the requested temperature isAt the insulated endpoint $\sin ({\lambda }_{n}L)=(-1{)}^{n-1}$ cancels the alternating coefficient sign, leavingEvery summand is positive. The endpoint temperature never drops below $T_0$; it decreases towards $T_0$ as $t\to \infty$ and starts at $T(L,0)=2T_0$.

For completeness, the Sturm-Liouville eigenfunction expansion gives the initial profile in the $L^2$ sense. Here the coefficients are absolutely summable, so the Weierstrass M-test also gives uniform convergence on $[0,L]\times [0,\infty )$, and the initial values are recovered everywhere. For positive times the exponential factors justify all required termwise derivatives, giving a classical solution. There is, however, an initial incompatibility with a Neumann boundary condition at $(L,0)$: the initial slope is $T_0/L$, while the insulated Neumann boundary condition requires zero slope for $t>0$. The temperature is continuous to the initial edge, but $T_x$ cannot extend continuously to that corner. One must not differentiate the initial series term by term at $x=L$.

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.

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

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.

Set $T(x,t)=X(x)G(t)$ in the heat equation. Separation of variables gives $G^{\prime }/(c^{2}G)=X^{\prime \prime }/X=-\lambda$. Zero endpoint temperatures impose Dirichlet boundary conditions $X(0)=X(\pi )=0$, selecting $\lambda =n^2$ for $n\ge 1$. The separated factors can therefore be chosen asZero and negative separation eigenvalues give only the zero spatial solution under these boundary conditions.

The initial profile is continuous and triangular, with midpoint height $\pi /2$. Its Fourier sine series coefficients areConsequently the heat flow from a triangular initial temperature isFor $t\ge 0$, each summand has magnitude at most $4/[\pi (2m+1{)}^2]$. The Weierstrass M-test proves absolute and uniform convergence. The initial Fourier sine series equals the continuous piecewise smooth profile, and for $t>0$ the exponential factors justify termwise spatial and time differentiation on every time interval bounded away from zero.

Fix any bounded disk avoiding the integers, and choose $R$ with $|z|\le R$ on it. For sufficiently large $|n|$, say $|n|\ge 2R$ and $|n|\ge 1$,The convergent numerical series ${\sum }_{n\ne 0}{n}^{-2}$ and the Weierstrass M-test give absolute and uniform convergence of the tails. The finitely many remaining summands are holomorphic away from their integer poles. Thus the partial sums converge locally uniformly on $\mathbb{C}\setminus \mathbb{Z}$.

By part (a)(iii), the sum is a holomorphic function on $\mathbb{C}\setminus \mathbb{Z}$. The term $n=0$ is $-1/z^2$ and is harmless on this domain; it must not be omitted.

Let $V={\sum }_{n\ge 1}\Vert X_n{\Vert }_{\infty }$. It is a nonnegative extended random variable, and $\mathbb{E}V<\infty$ implies $\mathbb{P}(V<\infty )=1$. Otherwise its expectation would be infinite. Intersect this event with the countably many probability-one events on which the individual sample paths are continuous.

For every path in that intersection, ${\sum }_n\Vert X_n(\cdot ,\omega ){\Vert }_{\infty }<\infty$. The Weierstrass M-test gives absolute and uniform convergence of ${\sum }_n{X}_n(t,\omega )$ on the entire stated time domain. A uniform limit of continuous functions is continuous. Therefore $Y$ has continuous paths almost surely. The finite expectation is used to obtain pathwise summability before applying this deterministic theorem.

Set $A_n=\{{\sup }_{s\le t}|M_s^{(n)}|\ge 2^{-(n-1)/2}\}$. Part (a) gives ${\sum }_n\mathbb{P}(A_n)<\infty$. The Borel-Cantelli first lemma implies that, almost surely, only finitely many $A_n$ occur. On the complementary full-probability event, all sufficiently late terms satisfyThese deterministic bounds form a convergent geometric series. The finitely many earlier suprema are finite by continuity on the compact time interval. HenceThe series therefore converges absolutely and uniformly on $[0,t]$ by the Weierstrass M-test, and its sample-path limit is continuous.

For positive summable coefficients, the Weierstrass M-test gives a continuous uniform sum. The partial sum over $3^k\le n$ is the unique best uniform approximation of degree at most $n$. If $3^K$ is the first omitted frequency, every tail term equals $(-1{)}^j$ at $x_j=\cos (j\pi /3^K)$ because its frequency ratio is odd. The entire tail therefore attains alternating extrema equal in magnitude to its coefficient sum at more than $n+1$ points. The Chebyshev alternation theorem proves the assertion, including the empty partial sum at $n=0$.

For an odd integer $q\ge 3$ and positive summable coefficients $a_k$, the Weierstrass M-test makes $f(x)={\sum }_{k\ge 0}{a}_k\cos (q^{k}x)$ continuous. Its partial sum over $q^k\le n$ is the unique best uniform approximation by degree-at-most-$n$ trigonometric polynomials, for $n\ge 1$. If $Q$ is the first omitted frequency, at $x_j=j\pi /Q$ every omitted cosine equals $(-1{)}^j$ because all frequency ratios are odd. Thus the tail attains alternating extrema equal to its coefficient sum at $2Q\ge 2n+2$ points. The trigonometric Chebyshev alternation theorem proves the formula. For $a_k=r^{-k}$, $r>1$, the largest decay exponent is $\log r/\log q$.

The two identities areTo prove them, expand $e^{\pm x\log x}$ in its exponential series. Since $|x\log x|\le 1/e$ on $(0,1)$, the Weierstrass M-test justifies termwise integration. Substituting $x=e^{-t}$ gives ${\int }_0^1{x}^k(\log x{)}^{k}dx=(-1{)}^{k}k!/(k+1{)}^{k+1}$ through the gamma function. Inserting these moments into the series and shifting $n=k+1$ proves both formulas. Each resulting numerical series has absolute convergence.

Write $a_n=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n$, $a_0=1$. The binomial series for $\sqrt{1-u}$ has positive numbers $b_n=a_{n-1}-a_n=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/[4^n(2n-1)]$ as the magnitudes of its nonconstant coefficients. The product $a_n={\prod }_{j=1}^n[1-1/(2j)]$ tends to zero, since $\ln a_n\le -\frac{1}{2}{\sum }_{j=1}^{n}1/j\to -\infty$. Thus $\sum b_n=1$ by telescoping. The Weierstrass M-test gives uniform convergence and absolute convergence on $[-1,1]$. Continuity extends the interior identity to both endpoints. Its radius of convergence is one; outside this interval its terms do not tend to zero.

Let the locations $a_n$ be distinct and bounded, and let $c_n>0$ with ${\sum }_n{c}_n<\infty$. The Weierstrass M-test gives locally uniform convergence of $G$, so the uniform limit theorem makes it continuous. The reverse triangle inequality bounds each term's difference quotient by $c_n$. Consequently a uniformly small tail permits passing each one-sided difference-quotient limit through the series. At every point outside the locations, $G^{\prime }(x)={\sum }_n{c}_n\mathrm{sgn}(x-a_n)$; at $a_m$, the right derivative minus the left derivative is $2c_m>0$. Thus the function has precisely the prescribed cusp locations, even when they accumulate at another point where the function remains differentiable.

Inside its radius of convergence, a power series $f(z)={\sum }_{n\ge 0}{a}_n(z-c{)}^n$ can be differentiated term by term:The differentiated series has the same radius of convergence. The conclusion concerns the interior, including uniform convergence on smaller closed disks; it does not assert that differentiated boundary series converge wherever the original did. For example, ${\sum }_{n\ge 1}{z}^n/n$ converges conditionally at $z=-1$, but its differentiated series ${\sum }_{n\ge 1}{z}^{n-1}$ does not.

Applied to the geometric series, it gives the useful identityDifferentiation turns the plain coefficients into integer weights. This is one way generating functions translate a weighted counting question into an operation on a known series. A proof controls the original and differentiated tails on smaller disks; the Weierstrass M-test provides the needed uniform bounds. See Lebl, Basic Analysis, §11.3: power-series differentiation and analyticity.

A series of functions is uniformly absolutely convergent on a set if the series of their absolute values has uniform convergence there. A sufficient condition is the Weierstrass M-test: if $|f_n(x)|\le M_n$ for every $x$ and ${\sum }_n{M}_n<\infty$, then the tail is bounded uniformly by ${\sum }_{n\ge N}{M}_n\to 0$. This also proves uniform convergence of the original series by the Uniform Cauchy criterion.

A standard Weierstrass function is a uniformly convergent lacunary cosine series, here parameterized by $a>1$ and an integer frequency multiplier $b>1$. Its continuity follows from the Weierstrass M-test. In the range $1<a<b$, its modulus of continuity has the fractional-power scale ${\delta }^{\log a/\log b}$ for the positive odd-frequency model; the critical case $a=b$ can have the larger scale $\delta \log (1/\delta )$. These examples distinguish small approximation error from ordinary Lipschitz continuity.