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Apply the cosine double-angle identity twice:This finite trigonometric sum is already the Fourier series. In increasing harmonic order, including the constant term, the first two nonzero terms are

With the real Fourier series convention,where the Fourier coefficients areFor a square-integrable periodic function, Parseval's identity isThe Fourier coefficients require integrability, and the stated Parseval identity requires square integrability. The expansion agrees with $f$ in the L2 space sense; for piecewise smooth data it converges pointwise to the average of the two one-sided limits. Without such qualifications an arbitrary periodic function need not have a pointwise-convergent Fourier series.

At a fixed radius, Parseval's identity applied to the preceding Fourier series givesIn polar coordinates, $dA=rdrd\theta$. Integrating over radius and interchanging the nonnegative sum and integral givesThe interchange follows from the monotone convergence theorem. This is the special case of the area norm of the Poisson extension in the unit disk for the odd step boundary data.

For period two, the Fourier series convention isSince $f$ is an even function, every $b_n$ is zero. Also $a_0={\int }_{-1}^1\cosh xdx=2\sinh 1$. For $n\ge 1$, an antiderivative useful for the cosine coefficient isEvaluation at $\pm 1$ gives $a_n=2(-1{)}^n\sinh 1/(1+n^2{\pi }^2)$. ThusThe periodic extension is continuous at $x=1$, so its Fourier series converges to $\cosh 1$ there. Since $\cos (n\pi )=(-1{)}^n$,Therefore $\boxed{p=2,q=\pi }$, with $q=-\pi$ giving the same expression because only $q^2$ occurs. This is the $\beta =1$ case of Fourier series of periodic hyperbolic sine and cosine.

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.

For the period-two Fourier series of a square-integrable function $w$, Parseval's identity statesFor $w=f-g$, write $s=\sinh 1$. Its coefficients are $a_0=2s$, $a_n=2(-1{)}^{n}s/(1+n^2{\pi }^2)$ and $b_n=n\pi a_n$, because $g$ is subtracted. ThereforeThe sum was obtained in part (a). As a separate check, $f-g=e^{-x}$ in the interval, whose square has integral ${\int }_{-1}^1{e}^{-2x}dx=\sinh 2$. Endpoint jumps have no effect on this integral.

Use the product-to-sum formula and the double-angle trigonometric identities:By Fourier orthogonality, these are already its only nonzero Fourier series coefficients:

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.

Set $x=0$ in the convergent Fourier series from part (b). Since $f(0)=0$ and every cosine equals one,Rearranging givesConvergence at the evaluation point is essential when extracting a numerical series identity from a Fourier series. Here the function is continuous there and the coefficients are absolutely summable.

The standard Fourier series iswith Fourier coefficientsThese follow by taking inner products with the individual basis vectors and using their orthogonality and squared norms.

A standard complete orthogonal family is the real Fourier basisCompleteness gives expansion in the L2 space for a square-integrable function. Mere integrability alone does not guarantee pointwise convergence of a Fourier series; the explicit function in part (b) is continuous and piecewise smooth, so this issue does not obstruct its expansion.

Separate the already finite sine terms from the even part $g(x)=x^2-\pi |x|$. Since $g$ is an even function, it has no sine coefficients. Its constant coefficient isFor $n\ge 1$, integration by parts givesTherefore the cosine Fourier coefficients areOnly even frequencies survive: $a_{2m}=1/m^2$. The sine Fourier coefficients are $b_1=1$, $b_2=-2$, with all others zero. HenceThe periodic function is continuous at zero and at its endpoint joins, and its coefficients are absolutely summable. Thus this Fourier series represents it everywhere, including those joins.

Evaluate at $x=\pi /2$, where the original expression gives $f(\pi /2)=1-{\pi }^2/4$. The Fourier series givesRearranging provesThe convergence at this point is justified by the continuity properties noted in part (b).

The product-to-sum formula gives the finite Fourier seriesThus the only nonzero Fourier coefficients are $b_2=b_4=1/2$.

Use the Maclaurin series $\sin 3x=3x-\frac{9}{2}{x}^3+O(x^5)$ and $\cos x=1-\frac{1}{2}{x}^2+O(x^4)$. Multiplication givesThe same terms follow from the two sine harmonics in the Fourier series found above.

Termwise differentiation of a Fourier series, applied to the continuous periodic extension in part (c), gives for $-1<x<1$Hence the required Fourier sine series isThe endpoint deserves care: at $x=1$ every term is zero, so the Fourier series sums to $0$, the average of the periodic one-sided limits $1$ and $-1$. It therefore does not reproduce the specified endpoint value $g(1)=1$. Changing a value at one point does not change any Fourier coefficient, so this is still the Fourier series of the stated function on $(-1,1]$.

The trigonometric identity gives $\sin x\cos x=\frac{1}{2}\sin 2x$, and the complementary-angle trigonometric identity gives $\cos (\pi /2-x)=\sin x$. Thus the Fourier series is already the finite expressionThe only nonzero Fourier coefficients are $b_1=1$ and $b_2=1/2$.

The analogous orthogonal projection onto the cosine terms givesFor $n=0$, this means $a_0=L^{-1}{\int }_{-L}^{L}f(x)dx$; the actual constant in the Fourier series is $a_0/2$ because the constant function has squared norm $2L$.

Take the inner product of the Fourier series with $\sin (n\pi x/L)$. Fourier orthogonality eliminates the constant, all cosine terms, and all sine terms except the $n$th. The remaining squared norm issoFor a square-integrable function, the calculation is justified by taking inner products of finite Fourier partial sums and passing to their mean-square limit.

The function is an odd function, so $a_0=a_n=0$. Here $L=1$, and the Fourier coefficients areWrite $k=n\pi$. One integration by parts has zero endpoint terms because $x(1-x)$ vanishes at both endpoints:A second integration by parts givesTherefore $b_n=4(1-(-1{)}^n)/(n^3{\pi }^3)$, zero for even $n$ and $8/(n^3{\pi }^3)$ for odd $n$. The Fourier series isThe coefficients are absolutely summable, and the periodic function is continuous even at the joins $x=\pm 1$, so the displayed Fourier series represents it everywhere.