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Away from zero, the product and composition of the displayed smooth functions are continuous and differentiable. At zero,so $f$ remains continuous. Hence the continuity set is $\boxed{\mathbb{R}}$.

The difference quotient at zero is $f(h)/h=\sin (1/h)$. For $h_n=(\pi /2+2\pi n{)}^{-1}$ it equals one, and for $k_n=(3\pi /2+2\pi n{)}^{-1}$ it equals minus one. Both sequences tend to zero, so no derivative exists there. For $x\ne 0$, the product rule and chain rule giveThus the differentiability set is $\boxed{\mathbb{R}\setminus \{0\}}$.

At zero, $|f(x)-f(0)|=|x|\to 0$, so the function is continuous. At a nonzero $x_0$, approach along rational and irrational sequences; density of the rational numbers and density of the irrational numbers give limiting values $x_0$ and $-x_0$, which are different. Thus the continuity set is $\boxed{\{0\}}$.

At zero, the difference quotient is one for rational nonzero increments and minus one for irrational increments, so the derivative does not exist. Elsewhere differentiability implies continuity, and continuity already fails. Thus the differentiability set is $\boxed{\varnothing }$.

For the first additional construction, modify the rational/irrational example toAt zero, $|g(h)/h|=|h|\to 0$, so $g^{\prime }(0)=0$. At any nonzero point, density of the rational numbers and density of the irrational numbers give incompatible limiting values $x^2$ and $-x^2$. Thus $g$ is discontinuous, and hence not differentiable, there. Its differentiability set is exactly $\{0\}$.

For the second construction, use a summable absolute-value cusp series:On $[-R,R]$, each summand is bounded by $2^{-n}(R+1/2)$. The Weierstrass M-test therefore gives uniform convergence on every bounded interval, and the uniform limit theorem makes $G$ continuous everywhere.

To justify differentiability, each summand's difference quotient has absolute value at most $2^{-n}$, by the reverse triangle inequality. The sum of these bounds is finite, uniformly in the increment. Thus the tail of the difference quotient is uniformly small, and limits may be passed through the sum by first retaining finitely many terms and then making the tail small. If $x\ne 1/n$ for every $n\ge 2$, each individual summand is differentiable, givingIn particular $G^{\prime }(0)=-{\sum }_{n=2}^{\infty }{2}^{-n}=-1/2$. The accumulation of cusp locations at zero does not destroy the derivative, because their weights are summable.

At $x=1/m$, the same tail argument applies to both one-sided derivatives. Every term except the $m$th has the same derivative from both sides; the $m$th has derivatives $-2^{-m}$ and $2^{-m}$. ConsequentlyTherefore $G$ fails to be differentiable exactly at $1/2,1/3,1/4,\dots$, and is differentiable everywhere else.

The Inada conditions give $U^{\prime }(0+)=\infty$ and $U^{\prime }(\infty )=0$. Differentiability and strict concavity make $U^{\prime }$ continuous and strictly decreasing, with range $(0,\infty )$. Hence the inverse marginal utility $I=(U^{\prime }{)}^{-1}$ exists. The unique maximizer in the utility conjugate is $x=I(y)$, andThe derivative formula does not require differentiability of $I$: compare the optimizing values at $y$ and $y+h$ to squeeze the difference quotient between $-I(y)$ and $-I(y+h)$, and use continuity of $I$. Thus the dual is continuously differentiable, strictly decreasing, and strictly convex, since $-I$ is strictly increasing.

For the requested second-derivative assertion, a curvature hypothesis is missing. Under the intended nondegeneracy $U^{\prime \prime }(x)<0$ for every $x>0$, inverse differentiation givesThis proves the intended dual differentiability with nonvanishing utility curvature. If $U$ is only twice differentiable, it gives pointwise twice differentiability of the dual; continuous second derivatives additionally follow when $U\in C^2$.

Literal strict concavity does not imply nonvanishing curvature. An explicit Inada utility with vanishing curvature isIts derivative is positive, strictly decreasing, tends to infinity at zero and to zero at infinity. It is smooth and strictly concave, but $U^{\prime \prime }(1)=0$. At $y_0=U^{\prime }(1)=1/3$, a finite derivative of $I$ would contradict differentiation of $U^{\prime }(I(y))=y$, giving $0\cdot I^{\prime }(y_0)=1$. Hence ${\hat{U}}^{\prime }$ is not differentiable there. As printed, the twice-differentiable-dual claim is false; it is valid with $U^{\prime \prime }<0$. The remaining differentiability, monotonicity and strict-convexity conclusions above hold under the printed hypotheses.

Fix nested relatively compact subsets of $\Omega$ and let $M$ bound the gradient on the larger one. The fundamental theorem of calculus along a line segment gives $|w_h|\le M$ on the smaller one. Part (i) gives coefficient bounds independent of small $h$, so the De Giorgi-Nash-Moser theorem gives uniform interior Hölder continuity estimates for these weak solutions. Since $Du$ is continuous, the difference quotients $w_h$ converge locally uniformly to $D_{\ell }u$. Passing to the limit in the Hölder seminorm estimate, and repeating for every coordinate, yieldsThe first derivatives are locally Hölder continuous. Here the initially obtained exponent depends on $n$ and the local gradient bound. This is the interior interpretation of the printed regularity conclusion; a uniform global estimate does not follow from $u\in C^1(\Omega )$ alone.

Write the minimal surface equation for a graph as $\mathrm{div}F(Du)=0$ in its weak formulation, where $F(p)=p/\sqrt{1+|p{|}^2}$. For negative increments use the same difference quotient convention, soFor $\zeta \in C_c^1({\Omega }^{\prime \prime })$, the function ${\delta }_{\ell ,h}\zeta$ is an admissible test function in $\Omega$. Commuting the first partial derivative with translation, then applying discrete integration by parts, givesThe fundamental theorem of calculus along a line segment gives the averaged linearization of a nonlinear divergence-form equationThus the backward difference quotient solves a linear equation in divergence form:To verify the ellipticity of the minimal surface flux, computeand henceThe eigenvalue parallel to $p$ is $(1+|p{|}^2{)}^{-3/2}$; every orthogonal eigenvalue is $(1+|p{|}^2{)}^{-1/2}$. The same lower and upper bounds pass to the averaged symmetric matrix $A_h$ whenever both endpoint gradients have norm at most $M$:The printed $C^1(\Omega )$ hypothesis supplies such an $M$ on each relatively compact subset, uniformly for sufficiently small $h$. It therefore establishes a locally uniformly elliptic operator. A single lower bound on all of ${\Omega }^{\prime \prime }$ additionally requires bounded gradients there and at the shifted points; this is automatic for bounded $\Omega$ and fixed $h$, but is not supplied on an arbitrary open set.

A wavelet with $p$ vanishing moments satisfies $x^k\psi \in L^1$ and $\int x^k\psi (x)dx=0$ for $0\le k<p$. The moment differentiation of the Fourier transform theorem states that these weighted integrability hypotheses make $\hat{\psi }\in C^{p-1}$, with ${\hat{\psi }}^{(k)}(\xi )=\int (-ix{)}^k\psi (x)e^{-ix\xi }dx$. Hence ${\hat{\psi }}^{(k)}(0)=0$. The MRA projection Fourier identity, together with density of the approximation spaces and continuity of $\hat{\phi }$ at zero, gives $|\hat{\phi }(0)|=1$. These are the two analytic theorems needed in addition to the quadrature mirror filter construction.

For the ordinary derivative conclusion, assume also that the MRA low-pass filter is $C^{p-1}$ near $\pi$ and $\hat{\phi }$ is $C^{p-1}$ near zero. These hypotheses hold, for example, for a compactly supported scaling function with a finite refinement filter. The wavelet identity givesThe denominator is nonzero near zero. Its reciprocal and the exponential are $C^{p-1}$, so the product rule and the zero Taylor polynomial of $\hat{\psi }$ give the smooth-mask vanishing-moment criterion

The printed integrability hypothesis alone does not guarantee ordinary higher derivatives. It gives only a continuous $\hat{\phi }$. Without added smoothness, Taylor theorem for $\hat{\psi }$ still gives $m(\pi +t)=o(|t{|}^{p-1})$ for the representative defined by the quotient: a Peano zero. This establishes the value at $\pi$, and for $p\ge 2$ the first derivative there, but it does not automatically establish iterated derivatives.

Here is a lacunary scaling-phase regularity counterexample for $p\ge 3$. Start with a compactly supported Daubechies wavelet having $p$ vanishing moments, with scaling function ${\phi }_0$ and finite MRA low-pass filter $m_0$. Define the periodic phaseThe Fourier coefficients of $\theta$ are absolutely summable. The Wiener algebra is closed under multiplication and the exponential series, so $a$ and $a^{-1}$ have absolutely summable Fourier coefficients. Thus $\phi$ is an absolutely summable combination of integer function translations of ${\phi }_0$, belongs to $L^1\cap L^2$, and generates the same $V_0$. Unimodularity preserves orthogonality and unit norm of its integer function translations; the inverse phase preserves their complete span. The same approximation spaces therefore give a multiresolution analysis.

The identity $\theta (2t)=\theta (t)+\theta (t+\pi )$ gives $a(2t)=a(t)a(t+\pi )$, so the periodic phase change of a scaling function produces $m(t)=a(t+\pi )m_0(t)$. In the high-pass formula the phases cancel:The associated wavelet is therefore exactly the original one, with unchanged vanishing moments.

At any dyadic point $t_0=2\pi k/2^j$, the difference quotient of the terms with $j\le n\le {\log }_2(1/|h|)$ is $1+O((2^{n}h{)}^2)$ per term. The sum of these errors and the remaining tail divided by $h$ are bounded; the finitely many earlier terms have finite limits. Hence $(\theta (t_0+h)-\theta (t_0))/h={\log }_2(1/|h|)+O(1)$, which has no finite limit. The exponential has the same failure of differentiability. Dyadic points are dense, and $m_0(\pi +t)$ is nonzero for sufficiently small $t\ne 0$. Thus $m$ is not differentiable on any neighborhood of $\pi$, so an ordinary second derivative at $\pi$, understood as the derivative of a locally defined first derivative, need not exist. The Peano zero remains valid. This separates the intended smooth-mask theorem from what the literal assumptions establish.

The space of test functions isA sequence ${\phi }_m$ converges to $\phi$ in $\mathcal{D}(\mathbb{R})$ when all supports lie in one compact set $K$ andfor every nonnegative integer $j$. The distribution space ${\mathcal{D}}^{\prime }(\mathbb{R})$ is the continuous dual space of $\mathcal{D}(\mathbb{R})$, and $u_m\to u$ in ${\mathcal{D}}^{\prime }$ meansfor every test function $\phi$.

If a linear form $u$ is continuous, it clearly maps every null sequence to a scalar sequence tending to zero. Conversely, suppose it has this sequential property. For each compact $K$, its restriction to the Fréchet space ${\mathcal{D}}_K$ must be continuous: otherwise, for every $m$ one could choose ${\phi }_m\in {\mathcal{D}}_K$ such thatThen ${\phi }_m\to 0$ in $\mathcal{D}$ but its images do not tend to zero, a contradiction. Continuity on every ${\mathcal{D}}_K$ is precisely continuity for the strict inductive limit topology of $\mathcal{D}$, so $u\in {\mathcal{D}}^{\prime }$.

Use the convention ${\tau }_h\phi (x)=\phi (x-h)$. Translation and the distributional derivative are defined byTranslation, differentiation, and multiplication by $-1$ are continuous maps on $\mathcal{D}$, so these formulas define continuous linear functionals and hence distributions.

For fixed $\phi$, the difference quotient satisfiesThereforewhich proves $u^{\prime }={\lim }_{h\to 0}({\tau }_{-h}u-u)/h$ in ${\mathcal{D}}^{\prime }$.

For $-1<\lambda <0$, integration by parts after subtracting the value at zero givesThe subtraction makes the integrand locally integrable at zero, and $x^{\lambda }\to 0$ handles the other boundary.

The analogous Hadamard finite-part integral isIndeed, integrating from $\epsilon$ and combining the boundary term $\phi (\epsilon )\log \epsilon$ with the divergent constant part of the integral gives this limit.

Because $x_{+}^{\lambda }$ is a locally integrable function, its first distributional derivative has order at most one. It is not of order zero. Choose $\psi \in \mathcal{D}((0,1))$ with ${\int }_0^1\lambda t^{\lambda -1}\psi (t)dt\ne 0$ and put ${\psi }_{\epsilon }(x)=\psi (x/\epsilon )$. The sup norms stay bounded whileis unbounded as $\epsilon \downarrow 0$. This contradicts the local sup-norm estimate required of an order-zero distribution. Hence $(x_{+}^{\lambda }{)}^{\prime }$ has order exactly $\boxed{1}$.

At $-a$, the prefactor $(x-a)/x$ tends to $2$, so the nonvanishing amplitude leaves the oscillatory discontinuity intact. At zero, use $\sin (\tan x)=x+O(x^3)$ to obtainThe limit is $-a$, whereas the assigned value is zero. This is a removable discontinuity, not a continuous crossing.

At $a$, write $x=a+h$. The estimateproves continuity at the assigned zero. But its difference quotient isThe sequences from part (i) make this approach $1/a$ and $-1/a$, so the derivative does not exist. This illustrates the distinct continuity and differentiability thresholds for a damped oscillation.

Therefore the discontinuities are $x=-\pi /2$ and $x=0$; the nondifferentiability points are $x=-\pi /2,0,\pi /2$. The function is smooth elsewhere.

For $x\ne 0$, induction giveswhere each $P_n$ is a polynomial. This recurrence follows from the product rule and $d(1/x)/dx=-1/x^2$. Since Gaussian decay dominates every power,These limits also describe the derivatives of the extension at zero. If $f^{(n)}(0)=0$, its next difference quotient is $P_n(1/h)e^{-1/h^2}/h$, which again tends to zero. Induction proves existence and continuity of every derivative, all zero there. This is a flat function, covered more generally by polynomial multiples of a Gaussian flat function.

Continuity. The function is continuous at $x_0$ when ${\lim }_{x\to x_0}f(x)=f(x_0)$. Equivalently, for every $\epsilon >0$ there exists $\delta >0$ such that $|x-x_0|<\delta$ within the domain of a function implies $|f(x)-f(x_0)|<\epsilon$; this is the epsilon-delta definition of continuity.

Differentiability. At an interior point $x_0$, the difference quotient must have a finite two-sided limit of a function:This defines differentiability and the derivative. At an endpoint one may instead state a one-sided definition relative to the interval. Differentiability implies continuity, because the increment is $h$ times a quotient with finite limit, but continuity alone does not guarantee differentiability.

The original PDF specifies $x\to a$. The quotient is the difference quotient for the function $f(x)=x^x=e^{x\ln x}$ at the positive point $a$. The chain rule gives $f^{\prime }(x)=x^x(\ln x+1)$, so

The difference quotient at zero isIf $p>1$, its absolute value is at most $|h{|}^{p-1}\to 0$, so $f$ is differentiable at zero with derivative $0$.

To exclude $p\le 1$, use $h_n=1/(2\pi n)$ and $k_n=1/(2\pi n+\pi /2)$. The difference quotient equals $(2\pi n{)}^{1-p}$ along the first sequence, and $0$ along the second. For $p=1$ the first value is $1$, while for $p<1$ it is unbounded. Hence a common finite limit is impossible. Therefore

The function $f$ is differentiable at $x_0$ if there exists a finite real number $L$ such thatThen $f^{\prime }(x_0)=L$. The puncture $0<|x-x_0|$ makes the difference quotient defined; requiring a finite real $L$ distinguishes an ordinary derivative from an infinite slope.

The difference quotient at zero isFor $a>1$, its absolute value is at most $|h{|}^{a-1}$. Given $\epsilon >0$, set $\delta ={\epsilon }^{1/(a-1)}$. Then $0<|h|<\delta$ makes the difference quotient have absolute value below $\epsilon$. Thus $f_a^{\prime }(0)=0$.

Conversely, along the second sequence the difference quotient is exactly zero. Any finite derivative would consequently have to be zero. Along the first positive sequence, however, the difference quotient equals $x_n^{a-1}$. It is one for $a=1$ and tends to $+\infty$ for $a<1$. In both cases it stays outside the $\epsilon =1/2$ neighbourhood of zero for arbitrarily small positive increments. Thus no finite derivative exists. We obtainIn particular, $0<a\le 1$ gives continuity without differentiability.

Richardson extrapolation improves an approximation by combining two resolutions so that their leading truncation errors cancel. The useful surprise is that the size of that leading error need not be known: its scaling under refinement is enough. This lets an existing finite difference method or quadrature rule produce a more accurate answer from two runs, without first deriving a new method from scratch.

Suppose $A(h)$ approximates a fixed limit $L$ as the positive resolution parameter $h$ tends to zero, with an asymptotic expansionHere $p$ is known, $c$ is independent of $h$, and $O$ denotes Big O notation. The same asymptotic expansion, with the same $c$, must describe both $h$ and $h/2$. Nested grids make it natural to reuse samples, although nesting itself does not establish this error model. Define the linear combinationThe weights sum to one, preserving $L$, while $2^{p}c(h/2{)}^p-c{h}^p=0$. Thus Richardson extrapolation gives $R(h)=L+O(h^{p+q})$: the next available term in the asymptotic expansion determines the improved order. Justin Solomon's numerical analysis text develops this cancellation for finite differences and applies it to numerical integration.^[ref]

Geometrically, regard the approximation as a function of $t=h^p$. Its first two terms are a straight line $L+ct$. The two computed points lie at positive $t$; extending the line through them to $t=0$ gives $R(h)$. This is why the operation is extrapolation: it estimates a value outside the sampled parameter interval. The remaining curvature corresponds to the uncancelled terms of the asymptotic expansion.

Worked example: a centered derivative of sine. For the smooth function $f(x)=\sin x$, the exact derivative at $x=0$ is $L=\cos 0=1$. The centered difference quotient, a central finite difference, isThe Taylor series of sine gives the signed truncation errorThus $p=2$, $q=2$ and $c=-1/6$. Halving $h$ divides the leading truncation error by four. Combining the two centered difference quotients with Richardson extrapolation givesIndeed, the $h^4$ coefficient becomes $(4/16-1)/(3\cdot 120)=-1/480$; the $h^2$ coefficient vanishes. This is the even-power case of Richardson extrapolation discussed in Persson's authored lecture notes.^[ref]

For $h=0.4$ and $h/2=0.2$, the actual computations areThe absolute error drops from about $2.6454\times 10^{-2}$ for $A(0.4)$ to $5.3080\times 10^{-5}$ for $R(0.4)$, an improvement by roughly a factor of 498. Refining again gives $R(0.2)\approx 0.999996670632607$, whose absolute error is about $3.3294\times 10^{-6}$. The refined truncation errors decrease by nearly four for $A$ and nearly sixteen for $R$, reflecting their second and fourth orders.

The logarithmic scale on each axis turns power laws into straight lines. The dashed reference curves are $h^2/6$ and $h^4/480$, the magnitudes of the leading truncation errors. The plotted errors are computed with 70-digit decimal arithmetic, rather than clipped to a display floor; this isolates the asymptotic expansion from roundoff error.

Worked example: integrating an exponential. Approximate the definite integral of the exponential function on $[0,1]$:For numerical integration, the composite trapezoidal rule adds the areas of trapezoids between successive samples; this is the quadrature use of the endpoint average that also occurs in the ODE trapezoidal rule. On $n$ equal intervals, with $h=1/n$, defineSumming the geometric series gives $T_n=(e-1)\frac{h}{2}\frac{e^h+1}{e^h-1}$. Expanding the exponential function near zero yieldsThe leading truncation error again has $p=2$, $q=2$. On the uniform nested grids with $n$ and $2n$ intervals, Richardson extrapolation therefore givesThis linear combination is exactly composite Simpson's rule on the finer grid, whose spacing is $h/2$. On each coarse panel, an old endpoint receives weight $h/6$ and the new midpoint weight $2h/3$; shared old endpoints receive weight $h/3$. These are the usual Simpson's rule weights $1,4,2,\dots ,4,1$ multiplied by $h/6$. The equivalence is explained directly in LeMesurier's university notes.^[ref]

For two coarse panels and four fine panels,The absolute errors are about $0.0356493$, $0.00894008$ and $0.0000370135$, respectively. Reusing the fine-grid samples makes Richardson extrapolation especially attractive here: the resulting Simpson's rule needs no samples beyond those already used by $T_{2n}$.

When the improvement is justified. Richardson extrapolation requires the stated asymptotic expansion and the correct positive $p$. If the actual leading exponent is $r\ne p$, its coefficient after combination is $c(2^{p-r}-1)/(2^p-1)$, so the leading truncation error generally remains. Merely knowing $A(h)-L=O(h^p)$ does not supply a common coefficient $c$ or a higher-order remainder. A discontinuity of a function, a singular endpoint, or unrelated changes between grids can invalidate the assumed asymptotic expansion; refinement alone does not guarantee improvement for every numerical problem.

Roundoff error and errors in the sampled values survive the cancellation of truncation error. If the coarse and fine values have absolute perturbations bounded by $\epsilon$, the perturbation in $R(h)$ is bounded by $(2^p+1)\epsilon /(2^p-1)$; for $p=2$ this is $5\epsilon /3$. In floating-point arithmetic, a centered difference quotient can also amplify sample errors through division by $h$, so an indefinitely smaller $h$ need not be more accurate. Persson's notes describe this balance between truncation error and roundoff error.^[ref]

The same cancellation idea appears in step-doubling error estimation. For an order-$p$ one-step method for an ordinary differential equation, two half steps each contribute local truncation error of order $h^{p+1}$, so their accumulated leading local error has ratio $2^{-p}$ to one full step. This explains the denominator $2^p-1$ there; the appropriate exponent follows from that local error model, rather than automatically from a generic use of Richardson extrapolation.

Rolle's theorem is the equal-endpoint case of the mean value theorem. Let $a<b$ and let the real-valued function $f:[a,b]\to \mathbb{R}$ be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. If $f(a)=f(b)$, there exists $c\in (a,b)$ with $f^{\prime }(c)=0$. Geometrically, a tangent line somewhere in the interior is horizontal, like the secant line joining the endpoints. ^[ref]

For the polynomial $f(x)=x(x-2)$ on $[0,2]$, both endpoint values are zero and the derivative is $f^{\prime }(x)=2x-2$. Thus $c=1$, where $f(1)=-1$, and the horizontal tangent line is $y=-1$. The horizontal secant line is $y=0$: equal endpoint values do not force the interior tangent line to pass through those endpoints.

The absolute value function $f(x)=|x|$ on $[-1,1]$ exposes the importance of differentiability. It is continuous and $f(-1)=f(1)=1$, so the secant line has slope zero. However,and the derivative does not exist at zero because the one-sided derivatives disagree. Every existing derivative in the interior is nonzero. The corner fails a hypothesis of both Rolle's theorem and the mean value theorem, so neither conclusion is guaranteed.

For a short proof, the extreme value theorem supplies a greatest and a least value on $[a,b]$. If $f$ is a constant function, every interior point works. Otherwise at least one of these values differs from the common endpoint value, so it is attained at an interior extremum. At an interior local maximum or local minimum, differentiability forces the derivative to be zero: the difference quotients from the two sides have opposite weak signs and the same limit.

This turns repeated values into information about derivatives. Two distinct roots of a polynomial force a root of its derivative between them; iterating this idea gives the Rolle root count with multiplicities. More generally, a differentiable function with nonzero derivative throughout a real interval cannot repeat a value there, hence is injective. The beauty is that knowing only two equal values guarantees an exact local feature somewhere between them.

A secant line is a straight line passing through two distinct points of a curve. For a graph $y=f(x)$ at $x=a$ and $x=a+h$, its slope is the difference quotient $(f(a+h)-f(a))/h$. As the second point approaches the first, a limiting secant line can give the tangent line.

For $f\in H^1(\mathbb{R})$, its difference quotient equals $D_{h}f={\int }_0^1{f}^{\prime }(\cdot +\theta h)d\theta$ in L2 space. Thus $\Vert D_{h}f{\Vert }_2\le \Vert f^{\prime }{\Vert }_2$, and translation of a function gives $D_{h}f\to f^{\prime }$ in L2 space. Conversely, bounded difference quotients have a weakly convergent subsequence as $h\to 0$. Against a test function, $\int D_{h}fv\to -\int f{v}^{\prime }$, so that weak limit is the weak derivative of $f$. It lies in L2 space, which is precisely the first-order Sobolev space condition.

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.