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Every derivative of order $n+1$ of $\sin (2x)$ has absolute value at most $2^{n+1}$. The equally spaced nodes are symmetric about $1/2$. Pair each node $a\le 1/2$ with $1-a$; for $x\in [0,1]$,If an unpaired middle node remains, its factor satisfies $|x-1/2|\le 1/2$. Consequently the complete nodal product is at most $2^{-(n+1)}$. The error formula actually gives the stronger estimateThe first bound tends to zero, proving uniform convergence on the whole interval. There is no appeal to a general claim that equispaced interpolation always converges; the derivative and symmetric-product bounds are particular to this analytic function and these nodes.

Yes. A Cauchy sequence in the supremum norm converges in the given complete space $B([0,1])$ to a bounded function $f$. Supremum-norm convergence is uniform convergence. The uniform limit of continuous functions is continuous: for a point $x_0$ and tolerance $ϵ$, choose a continuous $f_n$ within $ϵ/3$ of $f$ uniformly, then use its continuity at $x_0$ and the triangle inequality. Hence $f\in C([0,1])$. Thus $C([0,1])$ is a closed, complete subspace of $B([0,1])$.

The compactness conclusion justified by the hypotheses is $u\in W^{1,1}$ with the prescribed traces; the printed claim $u\in W^{1,\infty }$ needs additional regularity. DefineThen $u\in W^{1,1}$ and $u^{\prime }=v$. Weak $L^1$ convergence tested against $1_{(a,x)}$ gives $u_{j_k}(x)\to u(x)$ for every $x$, including the endpoint trace at $b$. Uniform integrability of the derivatives makes the functions equicontinuous: their increment over an interval is bounded by the integral of $|u_j^{\prime }|$ on that interval, uniformly small for short intervals. Their fixed initial trace and common derivative $L^1$ bound also make them uniformly bounded. The Arzelà-Ascoli theorem therefore upgrades the convergence to uniform convergence, and in particularUniform integrability does not bound the derivative in $L^{\infty }$. The natural compact class here is absolutely continuous functions, rather than Lipschitz functions. This distinction can affect attainment: smooth strictly convex superlinear integrands can have non-Lipschitz minimizing limits, so it is not repaired merely by adding convexity in the last variable. If one additionally has a uniform Lipschitz bound on the minimizing sequence, weak testing on measurable subsets passes that bound to $v$ and proves the printed $u\in \mathcal{A}$. Alternatively one can formulate the direct method from the outset in the fixed-trace $W^{1,1}$ class. The proof above gives all the asserted convergence and trace conclusions without assuming the missing bound.

The affine competitor $1-x$ has finite energy and the energy is nonnegative. A minimizing sequence therefore has bounded $\int u_j^{\prime 2}$. Its fixed value at zero gives $|u_j(x)|\le 1+\Vert u_j^{\prime }{\Vert }_2$, so it is also bounded in $L^2$ and hence in $W^{1,2}$. The estimategives uniform equicontinuity. Extract a weakly convergent subsequence in $W^{1,2}$ and, by the Arzelà-Ascoli theorem, a uniformly convergent subsequence to the same limit $u$. The endpoint values are retained. Uniform convergence on a common bounded range gives $\int u_j^{2008}\to \int u^{2008}$, while the derivative norm is weakly lower semicontinuous. Consequentlyso the global minimum is attained in the stated $W^{1,2}$ class. In fact the gradient term is strictly convex on this fixed-endpoint affine class and $t^{2008}$ is convex, so the minimizer is unique as well.

Every derivative of order $n+1$ of $\sin 2x$ has magnitude at most $2^{n+1}$. Pair the symmetric nodes $a=i/n$ and $1-a$. For $0\le x,a\le 1$,If there is an unpaired middle node, its factor has magnitude at most $1/2$. Thus the full nodal product is at most $2^{-(n+1)}$, giving the stronger bound $\Vert f-p_n{\Vert }_{\infty }\le 1/(n+1)!$. In particular this implies the requestedThe bound tends to zero by factorial growth, so $\boxed{p_n\to f\text{ uniformly on }[0,1]}$. This conclusion uses smoothness of this particular function and does not assert uniform convergence of equispaced interpolation for arbitrary continuous data.

On the fixed parameter interval the integrand tends to $\cos \theta$ as $\kappa \to 0$ and is bounded by $\cos \theta$. The dominated convergence theorem, or uniform convergence here, givesThis equals Sam's time between departure and reunion. Physically, $dx/dt=-c\kappa \tan \theta$ tends uniformly to zero, so Alice's worldline approaches Sam's, and the time dilation disappears.

Subtract the particular plane Poiseuille flow profile from the full rectangular duct flow. The remainder is a harmonic function:The data are even in $x$ and $y$. The boundary value problem is unique: if two solutions differed by $v$, Green's first identity and the zero Dirichlet boundary condition would giveThus $v$ is constant, and its zero boundary values force $v=0$. Reflection of a solution in either coordinate axis therefore gives the same solution. We may seek even separated eigenfunctions.

Use separation of variables withEach summand solves the Laplace equation, vanishes at $y=\pm b$, and has the required evenness. The hyperbolic cosine ratio equals one at either side wall. The functions $\cos (k_{n}y)$ have orthogonality relationThey are the even Dirichlet eigenfunctions on $[-b,b]$, so their Fourier series represents the even boundary data. Projecting that data onto the $n$th eigenfunction yields the Fourier coefficientFor completeness, two integrations by parts give the quadratic arch expansion in half-integer cosine modes: the displayed integral is $4(-1{)}^n/k_n^3$, agreeing with the identity supplied in the paper.

Adding the particular solution givesThe coefficients are absolutely summable and the hyperbolic cosine ratio is at most one throughout the closed rectangle, giving uniform convergence and the prescribed boundary values. On compact subsets of the interior, the hyperbolic cosine ratio supplies exponential decay in $n$, justifying differentiation of the series and the Laplace equation. The side-wall correction slows the plane Poiseuille flow enough to satisfy all four no-slip boundary conditions.

The volumetric flow rate is the integral of axial velocity over the cross-section:The plane Poiseuille flow contribution isFor each correction eigenfunction,Uniform convergence permits termwise integration. Multiplying these two integrals by $A_n$ givesConsequentlyEvery summand in the subtracted quantity is positive. Thus the side walls reduce the flux below the infinite-width plane Poiseuille flow prediction. Its dimensions are volume per unit time, as required for a volumetric flow rate.

For the Fourier sine series write $u_0(x)={\sum }_{n\ge 1}{b}_n\sin (n\pi x/L)$. Orthogonality of trigonometric functions givesIn the second half of the integral put $z=L-x$; then $u_0(L-z)=\rho z$ and $\sin [n\pi (L-z)/L]=(-1{)}^{n+1}\sin (n\pi z/L)$. HenceEvery even coefficient is zero. For odd $n$, part (a) or the supplied integral yieldsPutting $n=2m+1$ therefore givesThis is the Fourier sine series of a symmetric tent. The absolute values of its terms are uniformly bounded by a summable sequence, giving absolute convergence and uniform convergence; the piecewise smooth odd extension and the Dirichlet convergence theorem for Fourier series identify its value with the continuous $u_0$, including the zero endpoints and the peak.

The printed lower limit $m=1$ is an error. The original PDF visibly has this lower limit. It omits the nonzero fundamental coefficient $F_0=4L\rho /{\pi }^2$. In particular, the printed series equals $u_0(x)-F_0\sin (\pi x/L)$, rather than $u_0(x)$. At $x=L/2$ it gives $L\rho /2-4L\rho /{\pi }^2$ instead of $L\rho /2$. The corrected lower limit is displayed explicitly above.

After bath cooling followed by insulated equilibration, the remaining mean temperature is conserved: integrating the insulated heat equation givesFor constant density, specific heat capacity and cross-section, this is conservation of energy. Its Fourier cosine series has a constant mode equal to the initial integral average; every nonconstant mode decays. Therefore the eventual uniform temperature isFor each mode with an odd integer index in part (c), $L^{-1}{\int }_0^L\sin ((2j+1)\pi x/L)dx=2/[(2j+1)\pi ]$. Integrating the bath-cooled series yieldsFor $t_f>0$, uniform convergence justifies termwise integration; the $t_f=0$ result also follows by dominated convergence of the summable positive coefficients. As a check, ${\sum }_{j\ge 0}(2j+1{)}^{-2}={\pi }^2/8$ gives $T_f=T_0$ for zero bath time, and $T_f\to 0$ as $t_f\to \infty$. More bath time strictly lowers the retained average temperature when $T_0>0$ and $\kappa >0$.

Let $U\subset \mathbb{R}\times {\mathbb{R}}^d$ be an open set, $d<\infty$, and $f:U\to {\mathbb{R}}^d$ a continuous function. Through every $(t_0,y_0)\in U$ there is a local solution of a differential equation: a continuously differentiable function $y:I\to {\mathbb{R}}^d$ on an open interval containing $t_0$, satisfying the displayed initial value problem pointwise. No Lipschitz condition is assumed.^[ref]

Continuity of a proposed rate law therefore guarantees at least one trajectory for a short time. For a physical model, this is a first consistency check: can the stated rule and initial condition actually be followed? Prediction also needs uniqueness. The delayed solutions of a square-root differential equation below show how the same rule and initial condition can leave a departure time undetermined. The Picard-Lindelöf theorem supplies a sufficient additional hypothesis for uniqueness; failure of that hypothesis alone does not prove nonuniqueness.

For the initial value problem $y^{\prime }=2\sqrt{|y|}$, $y(0)=0$, a forward solution of a differential equation can stay zero until any $c\ge 0$, then follow $y=(t-c{)}^2$. The delayed solutions of a square-root differential equation verify this freedom exactly.

Local existence gives neither a global solution of a differential equation nor a bound on all future values. The continuous function $f(t,y)=y^2$ already permits finite-time blowup; the finite-dimensional continuation criterion explains the additional issue of extending a trajectory within its domain.

Proof outline: choose a compact set $K=[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$. By the extreme value theorem, $M={\max }_K\Vert f\Vert <\infty$. Take $0<h\le a$ with $Mh\le r$. Forward Euler method polygons remain in the closed ball and satisfy $\Vert y_n(t)-y_n(s)\Vert \le M|t-s|$. The Arzelà-Ascoli theorem yields a convergent subsequence with uniform convergence. Uniform continuity of $f$ on $K$ makes their integral equation defect tend uniformly to zero, so the limit satisfiesThe fundamental theorem of calculus gives a continuously differentiable function. Repeat backward and join at $t_0$, where both derivatives equal $f(t_0,y_0)$.^[ref]

The compact set here uses a finite-dimensional vector space; bounded closed balls in an arbitrary Banach space need not be compact. When $M=0$, the constant solution of a differential equation $y=y_0$ works on the entire time interval in $K$. In general the proof selects a subsequence, without asserting that all polygons converge to one unique trajectory.

For $2\pi$-periodic continuous functions, convergence of positive linear operators on continuous functions to the identity on $1,\sin x,\cos x$ implies uniform convergence on every function. Indeed $1-\cos (t-x)$ is nonnegative and has just one zero on the circle, so the compact Korovkin test space criterion applies.

For a Schwartz function on the real numbers, its periodization $P_f$ is a smooth periodic function of period of a function one. Rapid decay gives uniform convergence on $[0,1]$ of the defining series and each differentiated series. Its $k$th Fourier coefficient isThe change of variable $u=x+n$ uses $e^{2\pi ikn}=1$. Rapidly decreasing Fourier coefficients give the absolutely and uniformly convergent Fourier series $P_f(x)={\sum }_k\hat{f}(k)e^{2\pi ikx}$. Evaluation at $x=0$ proves the Poisson summation formula.^[ref]

The same identity explains aliasing in the Nyquist–Shannon sampling theorem: applying the Poisson summation formula to the Schwartz function $x\mapsto f(x)e^{-2\pi i\xi x}$ givesThe sampled spectrum is periodic in frequency, and shifted copies of the continuous Fourier transform add together. When those copies overlap, the samples cannot separate their contributions. The Nyquist–Shannon sampling theorem adds a bandlimited function hypothesis to make reconstruction possible.

Picard iteration replaces a tentative trajectory by the accumulated integral of its proposed rate. For an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, define the Picard map and its starting function byHere $y_0(t)$ is the constant starting trajectory; the same symbol without its argument denotes the initial state. A fixed point of $\mathcal{P}$ solves this integral equation, and the fundamental theorem of calculus gives the original ordinary differential equation.

For $y^{\prime }=y$, $y(0)=1$, successive integrals giveAt $t=1$, the first five values are $1,2,2.5,2.66666667,2.70833333$, while $e\approx 2.71828183$. The fourth polynomial misses by about $0.00994850$. On $0\le t\le 1$, every omitted term of the exponential series is nonnegative and increasing, so the largest error occurs at $t=1$. The tail tends to zero, giving uniform convergence to the exponential function on this interval.

The beauty of Picard iteration is that each proposed trajectory supplies a better rate history for the next one. For this example, one integration adds exactly the next term of the exponential series. In general, the integrals may be difficult to compute, but the construction still explains local existence and uniqueness.

The short contraction mapping argument needs two independent controls. Choose a time-state cylinder $[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$ on whichThe second bound is the uniform local Lipschitz condition. Set $J=[t_0-h,t_0+h]$ with $0<h\le a$, $Mh\le r$ and $Lh<1$. In the function space of continuous functions $J\to {\mathbb{R}}^d$, use the supremum norm $\Vert u{\Vert }_{\infty }={\sup }_{t\in J}\Vert u(t)\Vert$. On its closed ball $\Vert u-y_0{\Vert }_{\infty }\le r$, direct integral estimates giveThus the map preserves the closed ball and is a contraction mapping there. This closed ball is a complete metric space, so the contraction mapping theorem gives convergence to exactly one fixed point. Any other local solution of a differential equation stays in the same ball after shortening its interval, and hence agrees there.

For $y^{\prime }=y$, take $r=1$, so $|y|\le 2$ in the state ball and $M=2$, $L=1$. Choosing $h=1/2$ gives $Mh=1$ and $Lh=1/2$, with the actual solution satisfying $|e^{1/2}-1|\approx 0.64872127<1$. The plotted interval $[0,1]$ is longer than this simple guarantee: the explicitly computed exponential series converges there anyway. The condition $Lh<1$ is sufficient for this one-step contraction mapping argument, rather than a necessary lifespan bound.

For a sequence of functions $f_n:X\to \mathbb{R}$, pointwise convergence to a function $f:X\to \mathbb{R}$ meansIt is the ordinary limit of a sequence at each fixed input. The threshold $N=N(x,\epsilon )$ may depend on the input. Uniform convergence instead requires one threshold for the entire domain of a function.^[ref]

Consider the polynomial functions $f_n(x)=x^n$, for positive integers $n$, on the interval $[0,1]$. For $0\le x<1$ the geometric sequence $(x^n)$ tends to zero, while $f_n(1)=1$. ThusEach $f_n$ is a continuous function, but its pointwise limit is discontinuous at one. Holding an input still hides what happens near that endpoint. The analogous even-power example on $[-1,1]$ appears in the source.^[ref]

For $0<x<1$ and $0<\epsilon <1$, taking the logarithm of $x^n<\epsilon$ and dividing by the negative real number $\log x$ reverses the inequality:The floor function followed by adding one handles the strict inequality even when the ratio is an integer. For tolerance $\epsilon =0.01$, the first successful indices at $x=0.5$, $0.8$, and $0.99$ are respectively $7$, $21$, and $459$. At $x=0.5$, $2^{-6}=0.015625$ fails and $2^{-7}=0.0078125$ succeeds. At $x=0$ and $x=1$, the error from this pointwise limit is already zero, so $N=1$ works. Nevertheless, $N(x,\epsilon )$ grows without bound as $x\to 1$ from below. A finite list of accurate test inputs cannot establish uniform convergence.

An even sharper warning concerns integration. On $[0,1]$, define the continuous functionsTheir supports are $[0,1/n]$, with height $n$ at $1/(2n)$ and value zero at both ends. Every fixed $x>0$ eventually lies beyond the support, and $g_n(0)=0$ always, so pointwise convergence gives the zero function. Nevertheless, each integral is the area of a triangle: ${\int }_0^1{g}_n(x)dx=\frac{1}{2}(1/n)n=\frac{1}{2}$. Shrinking width compensates increasing height. Hence the limit of a sequence of these integrals differs from the integral of the pointwise limit; interchanging them requires additional conditions, such as uniform convergence preserves integrals on a compact interval.

Starting with $p_0=0$ on $[-1,1]$, this polynomial recurrence gives $0\le p_r(x)\le |x|$ and $p_{r+1}(x)\ge p_r(x)$. Indeed the error satisfies $|x|-p_{r+1}=(|x|-p_r)[1-(|x|+p_r)/2]$. Its pointwise limit solves $p^2=x^2$, so it is $|x|$. Dini's theorem gives uniform convergence. Each continuous piecewise linear function is an affine function plus a finite linear combination of $(x-t{)}_{+}=(x-t+|x-t|)/2$. Scaling and translating the recurrence therefore gives polynomial approximation of every such function, and their density gives the Weierstrass approximation theorem.

For $b>0$ and $-b\le y\le b$, the even quadratic polynomial $b^2-y^2$ has the displayed expansion in even Dirichlet eigenfunctions. Put $k_n=(2n+1)\pi /(2b)$. Their squared norm on $[-b,b]$ is $b$. Two integrations by parts, using $\cos (k_{n}b)=0$, giveDivision by $b$ gives the displayed Fourier coefficients. The cubic decay gives uniform convergence. The half-integer frequencies arise from requiring both evenness and zero values at the endpoints, rather than from a periodic boundary condition. This expansion is useful in rectangular duct flow.

Real analysis studies limits, continuity, differentiation, integration, and convergence of sequences and functions taking values in the real numbers. Its central question is how finite approximations determine an exact answer: what does an arbitrarily small error mean, and which operations respect that error? A derivative is a limit of quotients, while a Riemann integral is obtained from increasingly fine Riemann sums. Both need conditions that make the answer independent of the approximations chosen.

Two organizing ideas are completeness of the real numbers, which ensures that suitable approximations have a destination, and uniform convergence, which controls approximation error over an entire domain. Pointwise convergence supplies separate guarantees at separate inputs; uniform convergence supplies one guarantee for all inputs. This distinction explains why replacing an approximating function by its limit can preserve continuity or integrals under appropriate hypotheses, yet fail without them.

For historical context, Cauchy's 1821 Cours d'analyse explicitly discusses convergence of series. Its Note VII treats double series and compares summing by rows and by columns under stated convergence assumptions: controlling an infinite process was already a concrete problem. This is historical context for questions about limits, rather than an attribution of the modern definitions below. ^[ref].

The residue of a function $f$ at an isolated singularity $a$ is its Laurent coefficient of degree $-1$. More explicitly, if $f$ is holomorphic for $0<|z-a|<\rho$, write its Laurent series asThis one Laurent coefficient determines the contour integral around a sufficiently small counterclockwise circle about $a$. A complicated local Laurent series can therefore leave a remarkably simple answer after one complete trip around the point.

For a simple pole, multiplying by $z-a$ removes the pole, leaving a holomorphic extension whose value at $a$ is the residue:There is a useful shortcut for a quotient $f=g/h$. If $g$ and $h$ are holomorphic near $a$, $h$ has a simple zero there, and $g(a)\ne 0$, then $f$ has a simple pole andThe first-order Taylor series of $h$ explains the denominator: dividing $h(z)$ by $z-a$ leaves its derivative at $a$ in the limit. This computes a residue without finding the full Laurent series, which is especially useful when the numerator or denominator is complicated. ^[ref]

For example, the rational function $f(z)=1/(z^2+1)$ has simple poles at $i$ and $-i$. Factoring the denominator makes both residues immediately visible:The signs matter: these residues cancel when both poles are enclosed, but either one contributes when enclosed alone. The residue theorem below turns this distinction into different contour integrals of the same function.

A residue is not the same thing as a pole, nor does it measure the order of a pole. The function $1/z^2$ has a double pole at zero but residue zero. Conversely,has an essential singularity at zero and residue one. Infinitely many negative powers cause no difficulty in reading off that particular Laurent coefficient. At a removable singularity the residue is zero. ^[ref]

Why does only this Laurent coefficient survive? Parametrize a counterclockwise circle by $z=a+r{e}^{it}$, $0\le t\le 2\pi$, with $0<r<\rho$. For every integer $n$,When $n\ne -1$, the complex exponential completes a whole number of turns and its contributions cancel; when $n=-1$, it is constant. Uniform convergence of the Laurent series on this circle justifies termwise integration, giving $\oint f(z)dz=2\pi i\mathrm{Res}(f,a)$. Thus the residue isolates exactly the local information that a closed contour integral detects. ^[ref]

This selection rule explains why a very large value near an isolated singularity need not produce a large contour integral. The double pole $1/z^2$ grows faster near zero than $1/z$, yet its contributions around a circle cancel completely. What matters is the arrangement of complex values along the path, including the differential $dz$, rather than their moduli alone. A residue records that surviving contribution, not the size of the function near the point.

A sequence whose terms are functions on a common domain. Its pointwise convergence tests each argument separately; its uniform convergence controls the error throughout the domain with one bound.