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The function is even, so its Fourier series has $b_n=0$. In the convention $f\sim a_0/2+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)$,Twice applying integration by parts gives, for $n\ge 1$,Therefore the periodic Fourier representation iswith periodic continuation. The periodic extension is continuous, including the identified endpoints, and piecewise smooth, so the Fourier series converges to it. The coefficients also give absolute convergence and uniform convergence by comparison with $\sum n^{-2}$.

At $x=\pi$, the cosine cancels the alternating coefficient sign; at $x=0$, it is identically one. These evaluations yieldThe first identity solves the Basel problem.

A flat torus is the quotient ${\mathbb{R}}^d/\Lambda$, where $\Lambda$ is a full-rank Euclidean lattice, with the Riemannian metric induced from the Euclidean metric. Translation by a lattice vector is a Riemannian isometry, so this metric is well defined and has zero curvature. Its volume is the covolume of $\Lambda$.

The dual lattice is ${\Lambda }^{\vee }=\{w:⟨w,v⟩\in \mathbb{Z}\text{ for all }v\in \Lambda \}$. For $w\in {\Lambda }^{\vee }$, the function $e_w(x)=\exp (2\pi i⟨w,x⟩)$ descends to the flat torus, and direct differentiation givesConversely, write $\Lambda =B{\mathbb{Z}}^d$ and $x=By$. Ordinary Fourier series on ${\mathbb{R}}^d/{\mathbb{Z}}^d$ have frequencies $n\in {\mathbb{Z}}^d$, corresponding to $w=B^{-\mathsf{T}}n\in {\Lambda }^{\vee }$. After normalization they form a complete orthonormal basis in $L^2$, and the Fourier coefficients of a smooth function decay rapidly. Applying the Laplace-Beltrami operator term by term shows that a smooth eigenfunction with eigenvalue $\lambda$ has nonzero coefficients only where $4{\pi }^2|w{|}^2=\lambda$. Thus the spectrum of a flat torus isThe formula counts multiplicity over either the real or complex numbers: the two frequencies $w,-w$ give the real sine and cosine functions.

For the rigidity assertion in dimension two, it suffices to reconstruct a rank-two Euclidean lattice $L={\Lambda }^{\vee }$ from its vector-length multiset. The multiset first determines its covolume $A$. Indeed, a bounded fundamental parallelogram, and comparison of the cells meeting a disk with slightly larger and smaller disks, give the lattice-point asymptotics by fundamental cellsThe spectrum determines $N_L(R)$, including multiplicities, so it determines $A$.

Let $a$ be the smallest nonzero vector length, and choose $v\in L$ with $|v|=a$. This vector is primitive: $v=ku$ with $|k|>1$ would contradict minimality. From the length multiset subtract the two vectors $kv,-kv$ at every positive length $ka$, for $k=1,2,\dots$. The shortest remaining length $b$ is exactly the length of a shortest vector $w\notin \mathbb{Z}v$. This subtraction is legitimate even if several directions have length $a$: it subtracts just two copies, and then $b=a$. It also proves that $b$ is independent of our choice of shortest direction. Replace $w$ by $w-kv$ to arrangeThe replacement stays outside $\mathbb{Z}v$ and cannot be shorter than $b$, so a reduced choice still has length $b$.

The shortest-independent-vector basis lemma says that $v,w$ are a basis of $L$. To prove it, choose coordinates with $v=(a,0)$. Because $v$ is primitive, $L/\mathbb{Z}v\cong \mathbb{Z}$; let $h>0$ be the smallest positive vertical coordinate in $L$. Write $w=(x,mh)$, with $m\ge 1$ after changing its sign. If $m\ge 2$, a vector $u$ of vertical coordinate $h$, reduced horizontally modulo $v$, hasThis contradicts the definition of $b$, since $u\notin \mathbb{Z}v$. Hence $m=1$ and $v,w$ generate $L$.

The Gram matrix of this basis is $\left(\begin{array}{cc}{a}^2& c\\ c& b^2\end{array}\right)$, with $c=v\cdot w$, and its determinant is $A^2$. ThereforeChanging the sign of one basis vector allows $c\ge 0$. The numbers $a,b,A$, all audible from the spectrum, consequently determine the Gram matrix, and hence $L$, up to an orthogonal transformation. Taking dual lattices gives the same conclusion for $\Lambda$. Isospectral flat two-dimensional tori are isometric. The argument uses the special basis property of rank-two Euclidean lattices, so it does not claim higher-dimensional rigidity.

Hecke operators give a commuting family of arithmetic symmetries of modular forms. They preserve the weight and cusp forms, and diagonalizing them turns the Fourier coefficients of a modular form into multiplicative arithmetic data. Odd weight spaces are zero by the action of $-I$; in weight zero the operators act on constants by $T_{n}1={\sigma }_{-1}(n)$. For the substantive theory, fix an even positive weight $k$ and $\Gamma =S{L}_2(\mathbb{Z})$, and use the following normalization consistently.

Extend the slash operator for modular forms to $G{L}_2^{+}(\mathbb{Q})$ byIt is a right action and positive scalar matrices act trivially. Let ${\mathcal{D}}_n$ be the integral matrices of determinant $n>0$. The determinant-n matrix representatives for Hecke operators for $\Gamma \backslash {\mathcal{D}}_n$ areInteger row operations reduce the first column to $(a,0)$, after which a row shear reduces $b$ modulo $d$; uniqueness follows from these same conditions. The Hecke operator isRight multiplication by $\Gamma$ permutes the left cosets, so $T_{n}f$ is again invariant under the weight-$k$ slash operator for modular forms. Each summand is a holomorphic function on the complex upper half-plane, and the coefficient formula below proves holomorphy at a cusp. Thus $T_n$ acts on $M_k(\Gamma )$ and preserves $S_k(\Gamma )$. Geometrically, these operators sum over index-$n$ sublattices, or equivalently the associated finite-degree correspondences between complex tori; the factors above compensate for the weight's response to lattice scaling.

Write $f={\sum }_{r\ge 0}{a}_r{q}^r$. The sum over $b$ is zero unless $d\mid r$, in which case it is $d$. Substituting into the displayed definition proves the Fourier coefficients of a composite-index Hecke operator formulawith $\gcd (n,0)=n$. In particular, $a_0(T_{n}f)={\sigma }_{k-1}(n)a_0(f)$ and $a_1(T_{n}f)=a_n(f)$. At a prime this becomeswhere $a_{r/p}=0$ if $p\nmid r$.

The Hecke multiplication relations areThey can be checked by applying the coefficient formula twice and regrouping the divisors. More explicitly, coprime indices factor independently, so $T_m{T}_n=T_{mn}$ if $(m,n)=1$. At a prime, separating the terms in the divisor sum according to whether one more factor $p$ is available givesIteration gives $T_{p^a}{T}_{p^b}={\sum }_{j=0}^{\min (a,b)}{p}^{j(k-1)}{T}_{p^{a+b-2j}}$; combining the prime factors gives the full relation. Hence all Hecke operators commute, $T_1=1$, and the Hecke algebra of modular forms is generated by the $T_p$ together with scalars.

The Petersson inner product on the finite-dimensional cusp form space isThe integrand is invariant and the integral converges because cusp forms decay exponentially at the cusp. The Hecke operators are self-adjoint for the Petersson inner product at level one. To see the underlying adjoint calculation, unfold the finite correspondence for a determinant-$n$ matrix and change variable $\tau \mapsto \alpha \tau$. The invariant hyperbolic measure and the determinant-normalized slash factors transfer $\alpha$ from one side of the inner product to ${\alpha }^{-1}$. Scalar matrices act trivially, so one can replace ${\alpha }^{-1}$ by the integral adjugate $n{\alpha }^{-1}$, which again has determinant $n$. This reverses the correspondence and permutes the same collection of terms, with the same real prefactor $n^{k/2-1}$. Summing givesCommuting self-adjoint operators on a finite-dimensional inner-product space admit a common orthonormal basis of eigenvectors. Thus $S_k(\Gamma )$ has a basis of simultaneous Hecke eigenforms, with real eigenvalues.

A nonzero simultaneous cuspidal Hecke eigenform has $a_1\ne 0$: if $T_{n}f={\lambda }_{n}f$ and $a_1=0$, comparison of the first coefficients gives $a_n={\lambda }_n{a}_1=0$ for every $n\ge 1$, so $f=0$. Normalize $a_1=1$. ThenConsequently the full Fourier series is determined by the eigenvalues, and a simultaneous eigenvalue system has a one-dimensional eigenspace. The Hecke multiplication relations translate intoIn particular the coefficients are multiplicative at coprime indices and are determined by the prime coefficients. This is the elementary multiplicity-one statement for normalized level-one Hecke eigenforms.

The Hecke eigenvalues are algebraic integers. One way to explain the arithmetic input is to use the rational structure supplied by the graded ring $M_{\ast }(\Gamma ,\mathbb{Q})=\mathbb{Q}[E_4,E_6]$. The rational cusp form space has a basis with integral Fourier coefficients, for example from appropriate monomials in $E_4,E_6,\Delta$. Its subgroup of forms whose entire expansions are integral is a full lattice: a finite initial coefficient map is injective by the valence formula for the modular group, so it embeds this subgroup into a finite-rank integer module. The coefficient formula shows that every $T_n$ preserves this lattice, and therefore has a monic integral characteristic polynomial. Its eigenvalues are algebraic integers. Simultaneous diagonalization also makes their number field finite over $\mathbb{Q}$; conjugating the rational coefficient data gives another eigenform, whose eigenvalues are again real by self-adjointness. Thus the coefficient field is totally real.

For a normalized cuspidal Hecke eigenform, define its L-function of a cusp form by $L(f,s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$. The coefficient relations yield the Euler product of a Hecke eigenformIndeed the prime-power recurrence gives ${\sum }_{r\ge 0}{a}_{p^r}{X}^r=(1-a_{p}X+p^{k-1}{X}^2{)}^{-1}$, and multiplicativity assembles these factors. These identities hold analytically in an initial right half-plane, as well as formally. For instance boundedness of $y^{k/2}|f(\tau )|$ on the full complex upper half-plane gives $a_n=O(n^{k/2})$, so $\mathrm{Re}s>k/2+1$ suffices for absolute convergence.

The Mellin transform of a cusp-form L-function relates this arithmetic to the modular transformation:Splitting at one and using $f(i/y)=i^k{y}^{k}f(iy)$ givesExponential cusp decay makes the last integral entire in $s$. This proves the analytic continuation and functional equation of the L-function of a cusp form, complementing its Euler product. The root number is $i^k=(-1{)}^{k/2}$.

Two examples show the scope of the theory. For even $k\ge 4$, the Eisenstein series $E_k$ is an eigenform with $T_n{E}_k={\sigma }_{k-1}(n)E_k$. This follows from its divisor-sum expansion and the same multiplicativity relations; its associated positive-index Dirichlet series is proportional to $\zeta (s)\zeta (s-k+1)$. The constant coefficient gives $M_k(\Gamma )=\mathbb{C}{E}_k\oplus S_k(\Gamma )$, so an eigenbasis of the cusp space together with $E_k$ gives an eigenbasis of the whole modular-form space. The ordinary Petersson inner product above is only used on the cusp space, where it converges. In weight twelve the valence formula for the modular group gives $\dim S_{12}=1$: cancellation of the first coefficients of two independent cusp forms would give a nonzero form with order at infinity at least two, exceeding $12/12$. Thus the normalized modular discriminant is a Hecke eigenform; writing $\Delta ={\sum }_{n\ge 1}\tau (n)q^n$ gives the Ramanujan tau function relations $\tau (mn)=\tau (m)\tau (n)$ for coprime $m,n$ and $\tau (p^{r+1})=\tau (p)\tau (p^r)-p^{11}\tau (p^{r-1})$.

For a full-rank Euclidean lattice $\Lambda \subset {\mathbb{R}}^n$, its dual lattice isIf $\Lambda =B{\mathbb{Z}}^n$, then ${\Lambda }^{\prime }=B^{-\mathsf{T}}{\mathbb{Z}}^n$, $m(\Lambda )=|\det B|$ and $m({\Lambda }^{\prime })=m(\Lambda {)}^{-1}$. The characters of a real torus identify the additive dual lattice with the multiplicative character group byThe map is well defined precisely because $u\in {\Lambda }^{\prime }$, and is injective. For surjectivity, a continuous group homomorphism from the compact torus into ${\mathbb{C}}^{\times }$ has compact image. Its modulus has logarithm a homomorphism into $(\mathbb{R},+)$ with compact image, hence is zero, so the image lies in the unit circle. Pull the character back to ${\mathbb{R}}^n$. Its continuous real lift under $t\mapsto e^{2\pi it}$, normalized to zero at the origin, is additive: its additive defect is an integer-valued continuous function and vanishes at the origin. A continuous additive function is $⟨u,x⟩$ for a unique $u$. Triviality on $\Lambda$ says $u\in {\Lambda }^{\prime }$, proving surjectivity.

Use the Fourier transform convention $\hat{f}(u)={\int }_{{\mathbb{R}}^n}f(x)e^{-2\pi i⟨u,x⟩}dx$, and take $f$ to be a Schwartz function. The periodization of a Schwartz function $P_f(x)={\sum }_{\lambda \in \Lambda }f(x+\lambda )$ is smooth and $\Lambda$-periodic. On a fundamental cell $F$, the coefficient of the torus character ${\chi }_u$ isThe equality follows by translating each cell and using $e^{2\pi i⟨u,\lambda ⟩}=1$. The rapidly convergent Fourier series can be evaluated at zero, yielding the Poisson summation formula for a Euclidean latticeThis argument keeps track of the covolume factor rather than tacitly assuming a unit-volume lattice.

For $\tau$ in the complex upper half-plane, let $g_{\tau }(x)=e^{\pi i\tau |x{|}^2}$. Scaling the self-dual real Gaussian function gives its Fourier transform at $\tau =it$, $t>0$, and holomorphic continuation in $\tau$ gives the complex Gaussian Fourier transformThe branch is $\exp [-(n/2)\mathrm{Log}(-i\tau )]$ with the logarithm on the right half-plane; it is positive for $\tau =it$. Both the integrals and the lattice sums are locally normally convergent on the complex upper half-plane. Applying the Poisson summation formula proves the lattice theta functional equationNo integrality or self-duality hypothesis on the lattice is needed.

Put $a=n/2$, $m=m(\Lambda )$ and ${\theta }_{\Lambda }(t)={\Theta }_{\Lambda }(it)$. The Epstein zeta function converges absolutely for $\mathrm{Re}s>a$, and its Mellin transform representation isAt infinity, ${\theta }_{\Lambda }(t)-1$ decays exponentially. Define the entire functionOn $(0,1)$, substitute ${\theta }_{\Lambda }(t)=m^{-1}{t}^{-a}{\theta }_{{\Lambda }^{\prime }}(1/t)$ and then $u=1/t$. Isolate the two elementary terms before integrating; this gives the pole-subtracted theta integral for an Epstein zeta functionThe formula initially holds for $\mathrm{Re}s>a$ and continues the completed Epstein zeta function meromorphically to all $\mathbb{C}$. Apply the same formula to the dual lattice at $a-s$, using $m({\Lambda }^{\prime })=m^{-1}$ and ${\Lambda }^{\prime \prime }=\Lambda$. The entire terms and the two rational terms match, provingFinally $E_{\Lambda }(s)={\pi }^s{\mathcal{E}}_{\Lambda }(s)/\Gamma (s)$ also continues meromorphically. Its only pole is a simple one at $s=n/2$, with residue ${\pi }^{n/2}/[m(\Lambda )\Gamma (n/2)]$; the pole of the completed function at zero is cancelled by $1/\Gamma (s)$, and $E_{\Lambda }(0)=-1$. The residues and this cancellation make explicit why subtracting the constant theta term was necessary.

Oddness gives zero cosine coefficients. Integration by parts gives the sine Fourier coefficientsThe Fourier series of the piecewise smooth periodic extension converges to its value at interior continuity points. Termwise integration from 0 to $x$ is valid: on the interval between these points, which stays inside $(-\pi ,\pi )$, the partial sums of $\sum (-1{)}^{n+1}\sin (nt)$ are uniformly bounded by the geometric-sum formula, since $|1+e^{it}|$ is bounded away from zero. The Dirichlet test therefore gives uniform convergence there. Integrating givesThe integrated series is absolutely and uniformly convergent. ConsequentlyIndependently, the constant Fourier coefficient is $a_0={\pi }^{-1}{\int }_{-\pi }^{\pi }{x}^{2}dx=2{\pi }^2/3$. Comparison yieldsThe original sine series is not uniformly convergent on the whole open interval; that stronger assertion is unnecessary for the justified integration.

Use the Fourier transform normalization $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-2\pi ix\xi }dx$. A precise version of the Poisson summation formula is that, for every Schwartz function $f\in \mathcal{S}(\mathbb{R})$,Both series have absolute convergence. Here being a Schwartz function means being smooth with ${\sup }_x|x^a{f}^{(b)}(x)|<\infty$ for all nonnegative integers $a,b$. This hypothesis makes the sums and the termwise operations below legitimate; the formula is not being asserted for arbitrary integrable functions.

Form the periodization of a Schwartz function $P_f(x)={\sum }_{n}f(x+n)$. Every differentiated series has uniform convergence on $[0,1]$, by rapid decay, so $P_f$ is a smooth periodic function of period one. Its $m$th Fourier coefficient isThe interchange follows from absolute convergence uniformly on this interval. Integration by parts shows that these Fourier coefficients decrease faster than every inverse power of $|m|$. Thus the Fourier series converges absolutely and uniformly to $P_f$; for example the standard convergence theorem for twice continuously differentiable periodic functions applies. Evaluating at zero proves the Poisson summation formula.

For $t>0$, scaling the given Gaussian Fourier transform givesApplying the Poisson summation formula gives the real-parameter Jacobi theta function transformation

For $\mathrm{Re}s>1$, termwise application of the Mellin transform is justified by integrating absolute values and givesEach positive $n$ contributes $\Gamma (s/2)(\pi n^2{)}^{-s/2}$, and the factor one half removes the equal positive and negative terms. This is the Mellin representation of the completed Riemann zeta function.

Split the integral at one. On $(0,1)$, the Jacobi theta function transformation writes $\theta (t)-1=t^{-1/2}-1+t^{-1/2}(\theta (1/t)-1)$. Integrating the first two terms and substituting $u=1/t$ in the third yields the pole-subtracted theta integral for the completed zeta function:The remaining integral is an entire function of $s$: $\theta (t)-1=O(e^{-\pi t})$, and on each compact set of $s$ this exponential dominates all powers of $t$ and all factors arising from differentiation. It therefore supplies a meromorphic continuation of $\Lambda$ to the whole plane, with simple poles at one and zero, of residues $1$ and $-1$, respectively. Its expression is unchanged by $s\mapsto 1-s$.

The reciprocal Gamma function is entire, with simple zeros at its nonpositive integer arguments. Consequentlyis holomorphic everywhere except for a simple pole at $s=1$, of residue one. The apparent pole at $s=0$ cancels; in fact $\Gamma (s/2{)}^{-1}\sim s/2$ gives $\zeta (0)=-1/2$. This proves the analytic continuation of the Riemann zeta function. The completed functional equation of the Riemann zeta function isThese are equalities of meromorphic functions; at apparent singularities they are interpreted by continuation. Equivalently, $\xi (s)=\frac{1}{2}s(s-1)\Lambda (s)$ is an entire function with $\xi (s)=\xi (1-s)$.

Apply the method of multiple scales with independent fast time $t$ and slow time $T={ϵ}^{2}t$. A preliminary slow time $ϵt$ produces no resonant forcing at first order, so its leading amplitudes are constant; the first nontrivial modulation is at $T$. Write $x=x_0+ϵx_1+\cdots$, withAt order $ϵ$, the inhomogeneous linear differential equation is $({\partial }_t^2+1)x_1=-\cos t{x}_0$. Absorbing its homogeneous part into the amplitudes givesAt order ${ϵ}^2$, the forcing is $-2{\partial }_t{\partial }_T{x}_0-k{x}_0-\cos t{x}_1$. Its resonant Fourier series coefficients areThe solvability condition in the method of multiple scales sets both coefficients to zero, giving the amplitude equations. DefineFor bounded initial conditions $x(0)=x_{\ast }$, $\dot{x}(0)=v_{\ast }$, the requested leading approximation isThis holds on bounded intervals of $T$, hence $t=O({ϵ}^{-2})$, with the constant in the error allowed to depend on the bounded $T$ interval. If desired, retain the displayed $ϵx_1$ and adjust $A(0),B(0)$ by $O(ϵ)$ to match initial conditions at that order. No particular initial conditions were supplied.

Since $M^2=abI$, the matrix exponential is $I\cosh (\sigma T)+(M/\sigma )\sinh (\sigma T)$ when ${\sigma }^2=ab>0$. When $ab<0$, put $\omega =\sqrt{-ab}$ and replace this with $I\cos (\omega T)+(M/\omega )\sin (\omega T)$. At $ab=0$ it is $I+TM$. Thus the leading second instability tongue of a weak Mathieu oscillator has exponentially growing solutions for $-1/12<k<5/12$, and the leading modulation is bounded for every initial condition precisely whenAt either displayed endpoint, $M$ is a nonzero nilpotent operator; some initial conditions give a secular term proportional to $T$. This leading calculation alone must not be mistaken for an exact finite-$ϵ$ stability decision at the endpoints.

For an exact small-$ϵ$ interpretation, put $t=2s$. The standard Mathieu equation parameters are $a_{\mathrm{M}}=4+4k{ϵ}^2$ and $q=-2ϵ$. Its two relevant Mathieu characteristic values are obtained by projecting $-d^2/d{s}^2+2q\cos 2s$ onto even and odd Fourier series modes. Near the unperturbed value $4$, the even modes $1,\cos 2s,\cos 4s,\cos 6s$ have diagonal entries $0,4,16,36$ and adjacent couplings $\sqrt{2}q,q,q$ in an orthonormal basis; the odd modes $\sin 2s,\sin 4s,\sin 6s$ have diagonal entries $4,16,36$ and couplings $q,q$. Substituting $4+c_2{q}^2+c_4{q}^4$ into their characteristic polynomials givesHigher modes of the Fourier series first contribute at higher order. Therefore the fourth-order edges of the second Mathieu instability tongue areFloquet theory gives boundedness of every solution outside the closed gap, $k<k_{-}(ϵ)$ or $k>k_{+}(ϵ)$, in this fixed-$k=O(1)$ neighborhood. At the actual gap edges the second independent solution is generically unbounded. In particular, if $k$ is fixed exactly at $-1/12$ or $5/12$, it lies just outside the true gap and is stable for each sufficiently small positive $ϵ$. The strict boxed criterion is the robust leading-order, uniformly separated criterion; the endpoints require this higher-order refinement, and their amplification bounds need not remain uniform as $ϵ\to 0$.

Use the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-ix\xi }dx$. An orthonormal multiresolution analysis is a family of closed vector subspaces $V_j\subset L^2(\mathbb{R})$, indexed by integers, with $V_j\subset V_{j+1}$, $\overline{{\bigcup }_j{V}_j}=L^2(\mathbb{R})$, and ${\bigcap }_j{V}_j=\{0\}$. Its dilation condition is $f\in V_j$ if and only if $f(2\cdot )\in V_{j+1}$. The space $V_0$ is invariant under integer function translations and admits a scaling function $\phi$ whose integer function translations form an orthonormal basis. Consequently ${\phi }_{j,k}=2^{j/2}\phi (2^j\cdot -k)$ form an orthonormal basis of $V_j$.

Since $\phi \in V_1$, expansion in that orthonormal basis gives the scaling refinement equationThe series converges in the L2 norm. The associated low-pass filter of a multiresolution analysis is the periodic Fourier series symbolInitially the symbol is defined almost everywhere. Orthonormal integer function translations give the quadrature mirror filter identity $|m(\xi ){|}^2+|m(\xi +\pi ){|}^2=1$ almost everywhere. For a Lebesgue integrable scaling function, its Fourier transform is continuous and $|\hat{\phi }(0)|=1$; choosing a constant phase makes $\hat{\phi }(0)=1$ and the continuous representative near zero has $m(0)=1$. Different conventions absorb $\sqrt{2}$ into the refinement coefficients; the displayed convention fixes that ambiguity.

The homogeneous solutions $(C_1+C_{2}x)e^{-x}$ are not $2\pi$-periodic unless $C_1=C_2=0$, so there is a unique periodic solution. For a forcing term $a_n\cos (nx)$, a trial solution $A_n\cos (nx)+B_n\sin (nx)$ givesUsing the Fourier series from part (a), for which only odd $n$ occur and $a_n=-4/(\pi n^2)$, yieldsThe coefficients decay fast enough to differentiate the series twice term by term, verifying the equation and periodicity.

For a function smooth on finitely many closed pieces with finite one-sided derivatives, integration by parts gives its Fourier series coefficients, for $n\ne 0$,where $J_r=f(x_r+)-f(x_r-)$ includes the jump of the periodic extension at the endpoint when necessary. In particular $|c_n|\le C/|n|$. Keeping frequencies $|n|\le K$, with $N\asymp K$, gives squared error $O({\sum }_{n>K}{n}^{-2})=O(N^{-1})$.

A best N-term approximation cannot improve this worst-case rate. For example, $f={\chi }_{[0,1/2)}$ has $|c_n|=1/(\pi |n|)$ on odd nonzero $n$ and zero on even nonzero $n$. Even the optimal selection therefore leaves a squared tail comparable to $N^{-1}$. Thus for the class with jumps,or $N^{-1/2}$ in the $L^2$ norm. Individual functions with no jumps, or additional cancellation, can converge faster. “Smooth except at finitely many points” must include controlled one-sided smoothness; smoothness merely on open pieces permits pathological behavior near their endpoints.

Diagonalize $T$ with eigenvalues ${\lambda }_a$ and use the periodic Fourier seriesEach Berezin integral contributes its quadratic coefficient, soPairing $k$ with $-k$ and using the infinite product for the hyperbolic sine gives, up to the local regularization factor,The product of the prefactors is $e^{\mathrm{tr}T/2}=1$ because $T$ is traceless. ThusThis agrees with the direct identity that the supertrace of an induced linear map on an exterior algebra is its characteristic determinant.

Count independent sinusoidal ripples by their spatial Fourier series indices. A mode with $n$ cycles across $D$ has $\Lambda =D/n$, so the given half-wave index is $j=2n$. The Nyquist spatial frequency permits $n\le N/2$. In two dimensions, a circular cutoff therefore contains approximately $\pi (N/2{)}^2$ integer wavevectors.

For a real wavefront error, the wavevectors $\mathbf{n}$ and $-\mathbf{n}$ describe the same ripple with conjugate coefficients, so count each pair once. The continuum mode-counting approximation givesThis counts one ripple with amplitude and phase per conjugate pair, not one real coefficient. Exact finite-grid counts are integers and have boundary corrections; the formula is the circular area estimate implicit in the question, with piston excluded.

Expand the unknown in Fourier modes:By Termwise differentiation of a Fourier series,Also,Therefore the coefficient of $e^{i\pi mx}$ on the left-hand side isThe right-hand side has coefficientsbecause the coefficient of each of $e^{\pm i\pi nx}$ is $1/(n^2+1)$ and the constant coefficient is one. Hence the Fourier spectral system for a cosine-modulated second derivative isIts coefficient array is an infinite tridiagonal matrix.