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The even extension of a function $f:[0,L]\to \mathbb{R}$ is $F_e(x)=f(|x|)$ for $-L\le x\le L$. It is an even function. Its $2L$-periodic extension is the function expanded by the half-range Fourier cosine series.

The rate at which Fourier coefficients approach zero as their index increases. The differentiability of the periodic extension, including matching values and derivatives at its joins, controls this rate through integration by parts. An absolutely bounded tail $O(n^{-p})$ with $p>1$ gives a uniform truncation error $O(N^{1-p})$.

For $L>0$ and $0\le x\le L$, set ${\mu }_n=(2n+1)\pi /(2L)$. The cosine modes satisfy a homogeneous Neumann boundary condition at zero and a homogeneous Dirichlet boundary condition at $L$. Their squared norm is $L/2$, and integration by parts givesThus their Fourier coefficients are $2/(L{\mu }_n^2)$, proving the displayed formula by Sturm-Liouville theory. One can also obtain completeness by extending evenly across zero and oddly across $L$: the resulting continuous $4L$-periodic extension has only odd-index cosine modes. The $n^{-2}$ coefficient bound proves uniform convergence, including both endpoints. This expansion supplies the initial transient for a heat equation with mixed endpoint conditions.

To repeat integration by parts across whole periods, choose periodic primitives with zero mean. Set $h_0=f$, and recursively defineEvery $h_r$ has zero mean. Part (i) gives $h_{r+1}^{\prime }=-h_r$ almost everywhere, and $h_{r+1}(0)=h_{r+1}(1)$ because $R{h}_r$ vanishes at both endpoints. Its periodic extension is consequently locally an absolutely continuous function, with no jump at the integers. Part (ii) gives the uniform estimateThe weak derivative identity on the whole line is $D{\overline{h}}_{r+1}=-{\overline{h}}_r$. After scaling by the integer $j$ and repeating integration by parts $N$ times,The boundary terms vanish because $\varphi$ has compact support in $(0,1)$. The interval contains exactly $j$ periods, soTaking $N=m+1$ therefore yieldsThis rapid distributional decay of periodic oscillations uses only integrability of $f$ and smoothness of the test function.

Joseph Fourier's Théorie analytique de la chaleur by Joseph Fourier (1822) makes trigonometric series instruments for solving problems of heat conduction. A prescribed temperature distribution need not be given by one power series: it may describe different heating on different parts of a body. The extract occurs in Chapter III, §233, where the earlier assumption of an odd-power expansion is relaxed. The innovation concerns both representation and what can count as a function in an analytical calculation.^[ref]

Use a separate integration variable $t$ to unpack the printed notation. The proposed Fourier sine series has coefficientsThe integral signs in the extract denote definite integrals over $[0,\pi ]$, not antiderivatives; the author explicitly supplies these bounds in the surrounding sentence. Their absent printed bounds are a notational convention, not permission to choose arbitrary constants of integration.

The key calculation is the orthogonality relationThe product-to-sum formulas prove this by integrating cosines of integer multiples of $t$. If an expansion can be integrated term by term, multiplication by $\sin (mt)$ followed by integration selects exactly one coefficient. This derives the only possible coefficients, but does not by itself prove the expansion converges to the proposed function. That distinction is central to assessing the extract's claim.

For a revealing example take $f(x)=1$ on $0<x<\pi$. Direct integration gives $b_n=2(1-(-1{)}^n)/(\pi n)$, and thereforeEvery term is zero at $0$ and $\pi$, so the series cannot take the value $1$ at those endpoints. Its natural interpretation uses the odd extension followed by periodic extension; the endpoints become jump discontinuities. More generally, for a piecewise continuously differentiable function, the Fourier series converges at an interior jump to the average of the two one-sided limits, irrespective of an individually assigned value there. Changing one value also leaves the integrals unchanged. Consequently the unqualified assertion of representation for every arbitrary function cannot be a modern pointwise theorem. Dirichlet's theorem on Fourier series supplies sufficient conditions and the correct distinction between continuity points and jumps; its 1829 source explicitly addresses the proof problem.^[ref]

The physical payoff is explicit. For a rod scaled to $0<x<\pi$ with zero end temperature, separation of variables in the heat equation producesEach mode solves $u_{\tau }=\kappa u_{xx}$ and the end boundary conditions; the Fourier sine coefficients encode the initial temperature. This rod is an explanatory example, distinct from the stationary rectangular-solid problem opening Joseph Fourier's Chapter III. The factor $e^{-\kappa n^2\tau }$ damps high modes rapidly for positive time. The representation of arbitrary data makes the physical solution possible; the proof of that representation remains a separate task.

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 $L$.

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.

On $(-\pi ,\pi )$ the function is odd, so its mean and all its cosine Fourier coefficients vanish:The assigned endpoint value does not affect these integrals. The product of the two odd functions $x|x|$ and $\sin nx$ is even, hence the sine Fourier coefficients areTwo applications of integration by parts giveThereforeThe periodic extension has a jump discontinuity at every odd multiple of $\pi$: its left and right limits are ${\pi }^2$ and $-{\pi }^2$. Its assigned value at each such point is $-{\pi }^2$. Thus it is odd away from these points, but it is not a globally odd function pointwise. The Fourier series will take the midpoint value zero at these jumps.

By Dirichlet's theorem on Fourier series, a piecewise continuously differentiable periodic function has a Fourier series converging at each point to the average of its one-sided limits. In particular it converges to the function at every point of continuity. Our periodic extension satisfies these hypotheses, and $x=\pi /2$ is a point of continuity with $f(\pi /2)={\pi }^2/4$.

At this point $\sin (n\pi /2)=0$ for even $n$. For $n=2j-1$,Substituting in the Fourier series from part (c) provesThe sum is taken in its displayed order. Discarding its zero even-indexed terms is valid and does not rearrange the remaining terms.

The function $x\sin (px)$ is an even function, so every Fourier sine coefficient is zero. Introduce the auxiliary integralFor a nonzero integer $k$, integration by parts givesConsequentlyFor $n\ge 1$, a product-to-sum formula yieldsIf $n\ne p$, this becomesThe exceptional Fourier coefficient at $n=p$ must be evaluated before dividing by $p-n$:Thus the full Fourier series isThis equality holds on $[-\pi ,\pi ]$ and by periodicity everywhere. Indeed the periodic extension is continuous at the joined endpoints, since both endpoint values are zero, and is piecewise smooth. The theorem on pointwise convergence of a piecewise smooth Fourier series therefore applies, including at the endpoints. The Fourier coefficients also decay as $n^{-2}$, making the resulting series absolutely and uniformly convergent.

The theorem on pointwise convergence of a piecewise smooth Fourier series gives the average of the two one-sided values of the corresponding periodic extension. Under the usual piecewise smooth interpretation of the paper's derivative assumption, the even extension is continuous at $0$ and at the joined endpoints $\pm \pi$. ThusFor the odd extension, the values on either side of zero are $-f(0)$ and $f(0)$; those on either side of $\pi$ are $f(\pi )$ and $-f(\pi )$. ThereforeThe endpoint values are also immediate because each sine term vanishes there. If $f(0)$ or $f(\pi )$ is nonzero, the Fourier sine series does not reproduce that endpoint value. Merely assuming that the derivative is locally piecewise continuous in the open interval would be weaker than the usual piecewise smooth hypothesis; the convergence assertion uses the standard endpoint regularity implicit in that hypothesis.

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.

Using the usual piecewise smooth interpretation of the derivative hypothesis, Dirichlet's theorem on Fourier series gives the average of the two one-sided limits of the appropriate periodic extension. The even extension is continuous at $0$ and at the identified endpoints $\pm \pi$, so the Fourier cosine series converges toFor the odd extension, the one-sided limits at zero are $-f(0)$ and $f(0)$, while at $\pi$ the periodic one-sided limits are $f(\pi )$ and $-f(\pi )$. Hence the Fourier sine series converges toAt an endpoint this equals $f$ precisely when $f$ is zero there. One can also see the zero endpoint values directly: every partial sum consists of sines that vanish at both endpoints.

For the Fourier cosine series, first take the even extension $F_e(x)=f(|x|)$ on $[-\pi ,\pi ]$, and then its $2\pi$-periodic extension. Evenness eliminates the sine terms. With the constant term written as $a_0/2$, the Fourier coefficients and series areFor the Fourier sine series, take the odd extension $F_o(x)=f(x)$ for $0<x<\pi$ and $F_o(x)=-f(-x)$ for $-\pi <x<0$, and extend with period $2\pi$. Values assigned at isolated endpoints do not change the Fourier coefficients. Oddness eliminates the constant and cosine terms, leavingBoth formulas follow by orthogonality on $[-\pi ,\pi ]$ and the parity of the extensions.

The truncated Fourier cosine series gives the better approximation over the whole closed interval. Its Fourier coefficients obey $|a_n|\le 4/(\pi n^2)$, so absolute convergence and the tail boundgive uniform convergence. The even periodic extension is continuous; its slope changes only cause corners.

By contrast $b_n=(-1{)}^{n+1}/n+O(n^{-2})$. The odd periodic extension has a jump at $\pi$, giving slower coefficient decay and the Gibbs phenomenon near that endpoint. Every truncated Fourier sine series satisfies $S_N(\pi )=0$, soIt therefore cannot converge uniformly to $f$ on $[0,\pi ]$, although it converges at every point of $[0,\pi )$. This comparison concerns approximation across the interval; it does not assert that the cosine truncation has smaller error at every individual interior point for every $N$.

The Fourier cosine series converges to the continuous even $2\pi$-periodic extension. If $r\in [-\pi ,\pi ]$ represents $x$ modulo $2\pi$, its value isIt has value zero at $x=2j\pi$, rises linearly with distance from these points until height $\pi /2$, and then stays flat until the next descending ramp. Thus the sketch on $[-3\pi ,3\pi ]$ has minima at $-2\pi ,0,2\pi$, and plateaux centred at the odd multiples of $\pi$. Its large partial sums round the corners and closely follow the whole graph.

The Fourier sine series converges, away from the odd multiples of $\pi$, to the odd $2\pi$-periodic extensionAt every odd multiple of $\pi$ its value is zero, the midpoint of the jump from $\pi /2$ to $-\pi /2$. At even multiples of $\pi$ it passes continuously through zero. Thus on every interval $((2j-1)\pi ,(2j+1)\pi )$ it has a negative plateau, a rising linear segment through $2j\pi$, and a positive plateau. The jump size is $\pi$.

The blue curves are the partial sums through $n=60$. The open endpoint circles and filled zero dots in the lower panel distinguish the one-sided limits from the value of the Fourier sine series at each jump. Near these jumps the partial sums exhibit the Gibbs phenomenon: the oscillations become concentrated in narrower regions as the number of terms grows, while their limiting overshoot remains about $0.08949$ times the jump, here about $0.281$.

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

For an odd integrable function, both $f(x)$ and $f(x)\cos nx$ are odd. Their integrals over $[-\pi ,\pi ]$ therefore vanish, so the constant and cosine Fourier coefficients are zero. For $f(x)=\sin ax$, the remaining Fourier sine coefficients areThe product-to-sum identity in the first step reduces the integral to two cosines. Here $a$ is real and noninteger, so no denominator vanishes. Consequently the Fourier series of a noninteger-frequency sine isAt $x=\pm \pi$ every term is zero. The Fourier series converges there to the average of the two limits of the periodic extension, namely zero, rather than to the nonzero endpoint values $\pm \sin \pi a$.

For $h\in L^1(0,1)$ with integral zero, $Ph$ is the unique absolutely continuous function with $(Ph{)}^{\prime }=-h$, equal endpoint values, and integral zero. The first formula gives the derivative; zero integral of $h$ gives equal endpoint values, while the added constant makes the mean zero. Uniqueness follows because the difference of two such primitives has zero weak derivative and therefore is constant, then its mean is zero. Its periodic extension has distributional derivative $-\overline{h}$, with no endpoint Dirac delta distribution.

Writing $Rh(x)=x{\int }_0^{1}h-{\int }_0^{x}h$, the same operator on zero-mean inputs is $Ph=Rh-{\int }_0^{1}Rh$. For arbitrary $h\in L^1(0,1)$,The first inequality follows from the integral kernel $x-1_{\{t<x\}}$, whose absolute value has integral $(t^2+(1-t{)}^2)/2\le 1/2$ in $x$. The second follows by the triangle inequality. This bound permits arbitrarily many iterations without increasing the L1 space norm.

Target: Rapid distributional decay of periodic oscillations. Parent: Weak convergence of distributions. Intended file: distribution-theory.bigb.