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The periodically extended $x^2$ is even, so every sine coefficient is zero. With the convention $f=a_0/2+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)$, the Fourier coefficients areFor the second integral, two applications of integration by parts give ${\int }_0^{\pi }{x}^2\cos nxdx=2\pi (-1{)}^n/n^2$. ConsequentlyThe periodic extension is continuous even at the joined endpoints, where both one-sided values equal ${\pi }^2$. The pointwise convergence of a piecewise smooth Fourier series therefore gives the displayed values throughout the interval; the coefficient bound also gives uniform convergence by the Weierstrass M-test. Setting $x=\pi$ and then $x=0$ yields

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 $\pm \sin \pi a$.

Because the function is even, all sine Fourier coefficients vanish. Its average is zero, so $a_0=0$. The Fourier cosine coefficients of a sine mode follow from the product-to-sum identity:For $n=2$, the integral is directly zero rather than an undefined substitution into a denominator. ThusThe Fourier series equals the continuous function at every point, including the corners. Its coefficients decay quadratically, so the series is also absolutely convergent.

Use the real Fourier series convention $f(x)\sim a_0/2+{\sum }_{n\ge 1}[a_n\cos (n\pi x)+b_n\sin (n\pi x)]$ for period two. The Fourier coefficients areThe last two integrals follow by integration by parts. Thus the Fourier series of a half-wave rectified ramp isOn $(-1,1)$, $f(x)=|x|+x$. Its even part is $f_e(x)=|x|$ and its odd part is $f_o(x)=x$, so the cosine and sine terms separate:At a jump, the Fourier series converges to the average of the one-sided limits. Thus the first series for $f$ takes the value $1$ at odd integers, while the Fourier series of a periodic sawtooth for $f_o$ takes the value $0$ there. The even function has a continuous periodic extension.

At $x=0$, the formula for $f_e$ gives $0=1/2-(4/{\pi }^2){\sum }_{r\ge 0}(2r+1{)}^{-2}$. At $x=1/2$, the formula for $f_o$ has only odd-index terms and gives $1/2=(2/\pi ){\sum }_{r\ge 0}(-1{)}^r/(2r+1)$. Hence

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 $\boxed{A=3,B=3}$.

In the real Fourier series convention for period $2\pi$,where the Fourier coefficients areThese formulas apply to integrable periodic functions. Writing equality at each point requires appropriate convergence hypotheses; the periodic solution sought below is smooth, so its series causes no such difficulty.

Fourier orthogonality isolates a coefficient. Multiplication by $\sin (mx)$ and integration over $[0,2\pi ]$ annihilates the constant, all cosine terms and all sine terms except the one with $n=m$, whose squared integral is $\pi$. Thus the corresponding coefficients on the two sides must agree for each positive integer $m$.

Multiplication by $\cos (mx)$ works in the same way, while integration against the constant mode extracts $a_0$. This is the uniqueness of the Fourier coefficients; it does not follow from comparing a few point values of the series.

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 required property is odd symmetry, $\boxed{f(-x)=-f(x)}$, up to irrelevant changes on a set of measure zero when discussing Fourier coefficients. An odd function has zero mean, and its products with cosine are odd, so every cosine coefficient vanishes. Its sine coefficients may remain nonzero.

For pointwise representation, the Fourier series takes the average of the one-sided limits at a jump. In particular a pure Fourier sine series has value zero at the origin and at the endpoints; arbitrary assigned values there need not equal the series sum.

If $\ell =mL$ with integer $m>0$, then $k_n\ell =2\pi nm$. All sine phase factors vanish and all cosine phase factors equal one. Thereforefor the periodic extension, and every Fourier coefficient is unchanged.

Apart from its endpoint convention, the ramp is odd, so its constant and cosine Fourier coefficients vanish. For the sine coefficient, evenness of the integrand and integration by parts giveHenceThis is the Fourier series of a periodic sawtooth. At the periodic jump, the series sums to zero rather than the assigned endpoint value $K/2$; changing a value at one endpoint does not change the coefficients.

For a real $2\pi$-periodic square-integrable function, use the Fourier series conventionwith Fourier coefficientsThe trigonometric orthogonality relations are ${\int }_{-\pi }^{\pi }\cos nx\cos mxdx=\pi {\delta }_{nm}$ for positive $n,m$, with the same formula for sines; sine-cosine products integrate to zero. The constant has squared integral $2\pi$ and is orthogonal to every positive harmonic. Thus the squared norm of the $N$th partial sum isTaking the limit in the L2 space gives the Fourier-series Parseval identity:Square integrability is the natural hypothesis for this finite norm identity; additional smoothness can be used when pointwise convergence is required.

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.

For fixed $r<1$, $\sum r^n\cos (n\theta )$ converges absolutely and uniformly. Thus for integrable $f$, one may integrate the kernel series term by term. The cosine addition formula givesSubstitute this into the defining integral. The constant kernel term gives $\frac{1}{2\pi }\int f=a_0/2$; the remaining terms givewhereThese are the same Fourier coefficients as in part (a). The kernel $2K$ is the Poisson kernel for the unit disk; the factors $r^n$ suppress the higher angular frequencies inside the disk.

The boundary data are odd except at isolated endpoint choices, which do not change integrals. Thus every cosine Fourier coefficient, including $a_0$, vanishes. For the sine Fourier coefficients,Consequently the extension isIt converges absolutely for every $r<1$; its limiting boundary values at jumps are the midpoint of the two one-sided limits.

Assume $f\in L^2[-L,L]$, so the squared-error integral is finite. WriteAt a minimum, differentiating with respect to each coefficient givesBy Fourier orthogonality, the constant has squared norm $2L$, each nonconstant cosine has squared norm $L$, and all distinct modes are orthogonal. ThusThese are exactly the Fourier coefficients of the Fourier cosine series of the even function. To check that the stationary choice is a unique global minimum, perturb the coefficients by $\delta a_n$. The residual is orthogonal to every fitted mode, so the cross terms vanish andwith equality only when every perturbation is zero. The truncated series is the orthogonal projection onto a finite-dimensional subspace spanned by the constant and the first $N$ cosine modes.

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

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.