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The complex modulus $|z-(-1-i)|$ is the Euclidean distance from $z$ to $-1-i$. Thus the locus isa circle with centre $(-1,-1)$ and radius $8$. Its extreme horizontal coordinates are $-9$ and $7$, and its extreme vertical coordinates are also $-9$ and $7$.

For real cube roots, set $u=a^{1/3}$ and $v=b^{1/3}$, so $(ab{)}^{1/3}=uv$. The sum of two cubes identity gives

The difference of two cubes identity gives the real polynomial factorizationExpanding the right side cancels the two mixed terms of each degree. The quadratic factor satisfies $a^2+ab+b^2=(a+b/2{)}^2+3b^2/4$.

With homogeneous Dirichlet boundary condition, the eigenfunctions are $\sin (n\pi x/L)$ and all modes decay. WriteReflection about the midpoint makes every even-$n$ coefficient zero. For odd $n$, putting $u=\pi x/L$ and using the product-to-sum identity givesIn particular $b_1=8T_0/(3\pi )$ and $b_3=-8T_0/(15\pi )$. ThusThis uses the Fourier sine series of a squared sine. The coefficients are $O(n^{-3})$, so the initial series converges uniformly; for $t>0$ the exponential factors also justify termwise differentiation to check the heat equation. Every mode vanishes at both endpoints, and the initial series gives the specified profile. Unlike part (a), there is no constant mode: the temperature tends to zero, with leading long-time term

For the heat equation ${\Theta }_t=\kappa {\Theta }_{xx}$, separation of variables with homogeneous Neumann boundary condition at both ends gives spatial modes $\cos (n\pi x/L)$ and decay factors $e^{-\kappa (n\pi /L{)}^{2}t}$. The $n=0$ mode is constant and does not decay.

The initial profile already contains only two such modes:Evolving each mode therefore yieldsThe endpoint derivatives vanish, and substitution checks both the equation and the initial profile. The spatial integral remains $L{T}_0/2$: no heat escapes through insulated ends, so the limiting uniform temperature is $T_0/2$.

For a right prism with equilateral triangular ends, the triangle area is $\sqrt{3}{\alpha }^2/4$. Thus its volume and external area areApply the Lagrange multiplier method to $V-\lambda (A-A_0)$ with positive lengths. The stationarity equations areThe second gives $\lambda =\sqrt{3}\alpha /12$. Substitution in the first then gives $6\beta =\sqrt{3}\alpha +3\beta$, or $\beta =\alpha /\sqrt{3}$. The area constraint becomesConsequentlyThe resulting volume is $V_{\max }=3^{3/4}/4{\mathrm{m}}^3$. As a check on the maximum, eliminating $\beta$ gives $V(\alpha )=\sqrt{3}{A}_0\alpha /12-{\alpha }^3/8$, whose second derivative is negative for every admissible positive $\alpha$; volume also tends to zero at both degenerate boundary shapes.

Here $f(x{)}^2=x^2$, soSubstituting the coefficients from part (b) into Parseval's identity givesHence the sum over all positive odd integers is ${\pi }^4/96$. The requested series starts at $n=1$, whose first denominator is $3^4$, so it excludes the contribution $1^{-4}$:

For a real $2L$-periodic square-integrable function with the convention $f\sim a_0/2+{\sum }_{n\ge 1}(a_n\cos (n\pi x/L)+b_n\sin (n\pi x/L))$, Parseval's identity isIt states that the squared L2 norm equals the sum of the squared components in the orthogonal Fourier basis, with the normalization factors shown. In the present even-function case every $b_n$ vanishes.

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.

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.

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.