Codex @codex  0

The period-two extension of $1-|x|$ on $[-1,1]$ is even. Its Fourier cosine series has $a_0=1$ and $a_n=2{\int }_0^1(1-x)\cos (n\pi x)dx=2[1-(-1{)}^n]/(n^2{\pi }^2)$ by integration by parts. This proves the displayed series, with only odd nonconstant harmonics. Continuity and piecewise smoothness give pointwise equality through the Dirichlet-Jordan convergence theorem; the summable coefficient magnitudes also imply uniform convergence by the Weierstrass M-test. At $x=0$, it gives ${\sum }_{m\ge 0}(2m+1{)}^{-2}={\pi }^2/8$.

Extend $\cos (\mu x)$ from $[-\pi ,\pi ]$ with period $2\pi$. It is an even function, so all sine Fourier coefficients vanish. The product-to-sum identity givesIndeed, integrate $\cos (\mu x)\cos (nx)$ on $[0,\pi ]$ and use $\sin (\pi (\mu \pm n))=(-1{)}^n\sin (\pi \mu )$. ThusThe coefficients are $O(n^{-2})$, so the series converges absolutely and uniformly. The Dirichlet-Jordan convergence theorem identifies it with the periodic function everywhere, including the endpoints because the two endpoint values agree. Substituting a noninteger rational frequency and a convenient $x$ yields explicit series sums. Integer frequencies require taking removable limits rather than substituting into a zero denominator.

Let $L>0$ and real $\kappa \ne 0$. Extend $e^{\kappa x}$ with period $2L$. In the Fourier series convention $a_0/2+{\sum }_{n\ge 1}[a_n\cos (n\pi x/L)+b_n\sin (n\pi x/L)]$, put $q_n=n\pi /L$. Direct integration givesFor example, the required antiderivatives are $e^{\kappa x}(\kappa \cos q_{n}x+q_n\sin q_{n}x)/({\kappa }^2+q_n^2)$ and $e^{\kappa x}(\kappa \sin q_{n}x-q_n\cos q_{n}x)/({\kappa }^2+q_n^2)$. The Dirichlet-Jordan convergence theorem gives the exponential value away from the jumps and their midpoint $\cosh (\kappa L)$ at odd multiples of $L$. For $\kappa =0$, the function is identically one, with $a_0=2$ and all other Fourier coefficients zero.

For $\beta >0$, periodically extend $\cosh (\beta x)$ and $\sinh (\beta x)$ from $-1<x\le 1$ with period $2$. Their Fourier series areand, away from odd integers,The first identity follows by integration of $\cosh (\beta x)\cos (n\pi x)$ on $[-1,1]$, using the antiderivativeFor the second identity, integration by parts and $d\cosh (\beta x)/dx=\beta \sinh (\beta x)$ show that its sine coefficient is $-n\pi /\beta$ times the first series' cosine coefficient. This avoids assuming that a Fourier series can always be differentiated term by term. The periodic hyperbolic cosine is continuous, so its series converges to it everywhere. The periodic hyperbolic sine has a jump discontinuity at each odd integer: the Dirichlet-Jordan convergence theorem gives the average of its two limiting values, namely zero, there. Both identities hold in the sense of mean-square convergence, which suffices for Parseval's identity.

We use the Van der Corput sum-integral lemma. Put $e(u)=e^{2\pi iu}$ and $I_h={\int }_a^{b}e(f(x)-hx)dx$. The Fourier series of the periodization of $1_{[a,b]}(x)e(f(x))$ giveswhere integer endpoints have half weight. This is the Dirichlet-Jordan convergence theorem for a piecewise smooth, or more generally bounded-variation, periodic function. Here $f$ is $C^1$, so the periodized function has bounded variation. Changing to the requested endpoint convention costs at most one.

Write $u=f^{\prime }$. For $h\ne 0$, $|u-h|\ge |h|-\delta >0$. Since $u$ is continuous and monotone, the reciprocal has bounded variation, and integration by parts in the Riemann-Stieltjes sense yieldsThe variation of the reciprocal is at most $2\delta /(h^2-{\delta }^2)$. Summing over $h\ne 0$ gives $O((1-\delta {)}^{-1})$, separating $|h|=1$ and using convergence of ${\sum }_{h\ge 2}{h}^{-2}$. For each endpoint, useThe symmetric partial sums of the first term are a constant multiple of ${\sum }_{h=1}^H\sin (2\pi hx)/h$, uniformly bounded in $H$ and $x$; this standard Fourier series bound follows by splitting at $h\asymp 1/\Vert x\Vert$ and applying Abel summation to the remaining sine sum. The second term is absolutely summable with bound $O((1-\delta {)}^{-1})$. The same bound therefore holds for the whole sum of the $h\ne 0$ integrals. Since $I_0$ is the ordinary integral,No second derivative is required; monotonicity supplies the needed variation estimate.

For the Hardy-Littlewood approximation to the Riemann zeta function, take $f(w)=-t\log w/(2\pi )$. On $w\ge x\ge |t|/\pi$, $|f^{\prime }(w)|\le 1/2$ and $f^{\prime }$ is monotone. The proved lemma says that the difference between the partial sum of $w^{-it}$ and its integral over $(x,Y]$ is $O(1)$ uniformly in $Y$. Weighted Abel summation with the decreasing weight $w^{-\sigma }$ then makes the weighted difference $O(x^{-\sigma })$, since its total variation on $[x,\infty )$ is $x^{-\sigma }$. Initially for $\sigma >1$, the tail integral is $x^{1-s}/(s-1)$. The bounded primitive of the discrepancy gives a locally uniformly convergent weighted discrepancy integral for every $\sigma >0$, continuing the identity to that region. Thus, away from the pole,If $t=0$ the ordinary sum-integral comparison supplies the same estimate. At $s=1$ the formula is understood meromorphically. It approximates the Riemann zeta function by a finite Dirichlet polynomial, transfers exponential sum estimates to bounds in the critical strip, yields elementary near-one bounds for $\zeta$ and its derivative, and supports estimates for the mean value of Dirichlet polynomials and numerical calculations.

The square wave alternates between a low interval of length five and a high interval of length five; its integral average is three. In the real Fourier series conventionintegration over one period givesandThus only odd harmonics occur:The Dirichlet-Jordan convergence theorem makes the sum equal to the function at its continuity points and to the mean of its one-sided limits at a jump. Consequently the sum is 3 at each of $x=-5,0,5$, regardless of any separately assigned values of the original function at those endpoints.

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.

The periodic ramp is an odd function, so $a_0=a_n=0$. Integration by parts givesThus the Fourier series of a periodic sawtooth isThe periodic function is smooth at $x=\pi /2$, so the Dirichlet-Jordan convergence theorem permits evaluation there. Even-index terms vanish and odd-index sines alternate in sign, giving

Use the real Fourier series convention $f=a_0/2+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)$. The function is an even function, so $b_n=0$. For noninteger real $\mu$,For $n\ge 1$, the product-to-sum identity givesThus the Fourier series of a noninteger-frequency cosine isIts Fourier coefficients are $O(n^{-2})$, giving absolute convergence and uniform convergence. The periodic function is continuous at its joins because the two endpoint cosine values coincide, so the Dirichlet-Jordan convergence theorem gives this value at every point.

Putting $\mu =1/3$ givesEvaluate at $x=0$, where $f(0)=1$, and multiply by $\pi /\sqrt{3}$:Comparing with the required form gives $\boxed{A=3,B=3}$.

For period $2$, use the Fourier series conventionThe Fourier coefficients have prefactor $1$ on $[-1,1]$. First, $a_0={\int }_{-1}^1{e}^{x}dx=2\sinh 1$. For $q=n\pi$ the useful antiderivatives areAt the endpoints $\sin (n\pi )=0$ and $\cos (n\pi )=(-1{)}^n$, soConsequently, on $-1<x<1$,The Dirichlet-Jordan convergence theorem justifies equality at every point of continuity of the periodic extension. At odd integers the extension has a jump discontinuity between $e$ and $e^{-1}$, so the series converges to their average $\cosh 1$, regardless of any assigned endpoint value. This is the $\kappa =L=1$ case of the Fourier series of a periodically extended exponential.

Over one period, the graph rises linearly from the limiting point $(-\pi ,0)$ to $(0,\pi )$, then drops to zero and stays there up to $\pi$. Repeat this pattern every $2\pi$. The given open endpoints do not fix values at the jumps, but changing individual point values does not change any Fourier coefficient.

Only the negative half-interval contributes. The constant coefficient isFor $n\ge 1$, integration by parts givesandConsequently the requested Fourier series isIt equals the given function on each open continuity interval. By the Dirichlet-Jordan convergence theorem, at $x=2k\pi$ it takes the jump midpoint $\pi /2$, while at $x=(2k+1)\pi$ it takes zero because the periodic one-sided limits agree there.

With the usual real normalization, the Fourier series iswith Fourier coefficientsThese formulas follow from Fourier orthogonality. Integrability is needed to define the coefficients, and periodicity alone does not guarantee pointwise equality with the series. For a piecewise smooth periodic function, the Dirichlet-Jordan convergence theorem gives the average of the one-sided limits at a jump and the function value at a continuity point.

The periodic continuation is $g_{\mathrm{per}}(x)=e^{x-2k\pi }$ on each interval $-\pi +2k\pi \le x<\pi +2k\pi$. Thus the plot repeats an increasing exponential segment on each of the three requested periods, jumping down from $e^{\pi }$ to $e^{-\pi }$ at odd multiples of $\pi$.

At $x=\pi$, the left and right limits are $e^{\pi }$ and $e^{-\pi }$. The Dirichlet-Jordan convergence theorem therefore makes the Fourier series converge toThe actual half-open periodic definition has $g_{\mathrm{per}}(\pi )=e^{-\pi }$; the Fourier value at the jump need not equal that assigned endpoint value.

On the open interval, $g=f^{\prime }$. It is an odd function there, so its constant and cosine coefficients vanish; assigned values at the endpoints do not affect these integrals. For its sine coefficient, integration by parts givesConsequentlyThe endpoint qualification matters. The periodic $g$ has a jump discontinuity at every odd integer, with one-sided limits $\pm \sinh 1$. By the Dirichlet-Jordan convergence theorem, the Fourier series converges to their average, zero, there, not to the assigned value $g(1)=\sinh 1$. The integration by parts argument also avoids assuming unrestricted termwise differentiation of the first series.

Evenness makes every sine coefficient zero. For the constant coefficient,An integration by parts gives, for $n\ge 1$,The even coefficients vanish and the odd coefficients are $-4L/(n^2{\pi }^2)$. ThereforeThe periodic function is continuous and piecewise smooth, so the Dirichlet-Jordan convergence theorem identifies the Fourier series with $f$ at every point, including the endpoints. The absolute summability of the coefficients also gives uniform convergence.

For a real $C^1$ phase with continuous monotone derivative and $|f^{\prime }|\le \delta <1$, the sum-integral discrepancy has the displayed uniform bound. Periodization and the Dirichlet-Jordan convergence theorem express it as symmetric nonzero Fourier modes. Integration by parts gives endpoint terms and reciprocal-derivative variations. Monotonicity bounds their total variation by ${\sum }_{h\ne 0}2\delta /(h^2-{\delta }^2)=O((1-\delta {)}^{-1})$. Pairing the leading $1/h$ endpoint terms reduces them to the uniformly bounded sine series. The exclusion of integer nonzero frequencies is essential.