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.

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.

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 $\mu =-1$, ${\mathbf{c}}_2=-{\mathbf{c}}_1$, while ${\mathbf{c}}_1$ and ${\mathbf{c}}_3$ are independent. Their span is a plane. A solution requires $\mathbf{b}$ to lie in that plane, so every scalar triple product formed from two columns and $\mathbf{b}$ must vanish. The product with ${\mathbf{c}}_1,{\mathbf{c}}_2$ vanishes identically; the other two give the plane condition.

Explicitly, ${\mathbf{c}}_1\times {\mathbf{c}}_3=(-7,1,3)$, soThus $\lambda =7$ is necessary and, since the two columns span the plane, sufficient. The equations then give $x-y=1$ and $z=2$, soThe solution is an affine line parallel to the homogeneous null space from part (c), obtained by intersecting the compatible constraint planes. Every point on it is nonzero because $z=2$.

Multiplication of a matrix by a column vector forms the linear combination of its columns with coefficients from that vector. Hereso $A\mathbf{r}=x{\mathbf{c}}_1+y{\mathbf{c}}_2+z{\mathbf{c}}_3$. Equating this expression with $\mathbf{b}$ reproduces the original three linear equations.

Reading off coefficients gives the matrix equation

Under $y=xu$, the product rule gives $y^{\prime }=u+x{u}^{\prime }$ and $y^{\prime \prime }=2u^{\prime }+x{u}^{\prime \prime }$. Substitute these into the original equation:Since $x>0$, this is precisely $x{u}^{\prime \prime }+u^{\prime }=0$, or $(x{u}^{\prime }{)}^{\prime }=0$. Integrating twice gives $x{u}^{\prime }=C_1$ and $u=C_1\ln x+C_2$. ThereforeThe two independent solutions of a differential equation supply the general solution of this second-order linear ordinary differential equation.