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By De Moivre's theorem, $(e^{i\alpha }{)}^j=e^{ij\alpha }$. The complex sum whose real part is wanted is therefore a finite geometric series:Use $1-e^{i\beta }=-2i{e}^{i\beta /2}\sin (\beta /2)$ in the numerator and denominator. This yieldsTaking its real part proves the finite trigonometric sumWhen $\alpha \in 2\pi \mathbb{Z}$, the printed quotient is undefined; the original sum is directly $(n+1)\cos \theta$, also the limiting value of the displayed identity. The denominator restriction cannot be omitted from a literal use of the quotient.

For $r\ne 1$, differentiate the finite geometric series and multiply by $r$:The required terms are $-n{r}^n$ with $r=-1/3$. The finite weighted geometric sum therefore givesSince $N/3^N\to 0$, the correction tends to zero, proving convergence and yieldingIn fact the ratio test also proves absolute convergence, since the ratio of consecutive absolute terms tends to $1/3$.

Use the common eigenvectors of $A$ and $A^{\mathsf{T}}$. For each $k$,Thus, without constructing $S$ explicitly, its eigenvalues areThe eigenvalue $1$ has direction $(1,1,1{)}^{\mathsf{T}}$. The repeated eigenvalue $-1/2$ has a real eigenspace consisting of vectors whose coordinates sum to zero. This is the general rule for the symmetric part of an orthogonal matrix. Adding unrelated lists of eigenvalues would not suffice; here the shared eigenvectors justify the pairing.

The symmetric part of a matrix isIt satisfies $S^{\mathsf{T}}=S$. The complementary term $(A-A^{\mathsf{T}})/2$ is a skew-symmetric matrix.

Since $A^{\mathsf{T}}=A^{-1}$, each eigenvector ${\mathbf{v}}_k$ from part (d) remains an eigenvector, with reciprocal eigenvalue:The set of eigenvalues is again $\{1,e^{2\pi i/3},e^{4\pi i/3}\}$, with the two nonreal values exchanged relative to the same eigenvectors. Equivalently, for $\mu =e^{2\pi ik/3}$ one can choose the eigenvector $(1,\mu ,{\mu }^2{)}^{\mathsf{T}}$. Pairing each eigenvalue with its eigenvector matters in the next part.

For an eigenvector $\mathbf{v}=(v_1,v_2,v_3{)}^{\mathsf{T}}$, the equation $A\mathbf{v}=\lambda \mathbf{v}$ readsA nonzero solution requires ${\lambda }^3=1$. Write the three roots of unity in the requested polar form of a complex number:Indeed ${\mathbf{v}}_k=(1,{\lambda }_k^2,{\lambda }_k{)}^{\mathsf{T}}$ satisfies all three equations. These eigenvectors are unnormalized; any nonzero scalar multiples are equally valid.

The preceding cycles show the action on a full basis, not only on one direction: $A$ sends ${\mathbf{e}}_1\to {\mathbf{e}}_2\to {\mathbf{e}}_3\to {\mathbf{e}}_1$, and $A^{\mathsf{T}}$ reverses each arrow. Therefore either composition fixes every basis vector, and by linearity fixes every vector. ThusIn particular $A$ is an orthogonal matrix, $A^{\mathsf{T}}=A^{-1}=A^2$, and $A^3=I$.

Interchanging the rows and columns gives the matrix transposeIt cycles the three basis vectors in the opposite direction. Hence

Let ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$ be the Cartesian basis vectors, with the given vector $\mathbf{x}={\mathbf{e}}_1$. The permutation matrix cycles them forward:Thus three applications return the original vector.

First apply integration by parts with $u=x^k$ and $dv=\sin xdx$:For $k>1$, integrate the remaining integral by parts once more. Its boundary term $[x^{k-1}\sin x{]}_0^{\pi }$ is zero, soAt $k=1$, direct integration gives $F(1)=\pi$, agreeing with the same formula; the coefficient of $F(-1)$ vanishes and that integral is finite. Now follow the odd chain:This is the sine-weighted power integral recurrence.

For $b>0$, split the interval into $n$ equal widths $\Delta x=b/n$. Since $x^2$ increases on this interval, the left and right rectangles give lower and upper Riemann sums:Using the finite sum of squares,The rectangle bounds differ by $U_n-L_n=b^3/n\to 0$. The squeeze theorem therefore gives, directly from elementary areas,For $b=0$ the result is zero. If negative $b$ is allowed, reverse the integration limits and use the reflection $x=-u$; the same signed formula holds. No antiderivative is needed for this first-principles calculation.

Distinguish the $N$ balls by labels. For a fixed position $n$, each label appears there in exactly $(N-1)!$ of the $N!$ equally likely permutations. Since $N_B$ labels are black,This is the same marginal law at every position, even the last one. The argument averages over earlier draw outcomes; it does not assert that a conditional probability after a specified history remains $N_B/N$. The original PDF uses $B_n$, correcting the transcription's uppercase subscript $B_N$.