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Use the positive-support convention for the geometric distribution and put $q=1-p$. Differentiating the convergent geometric series ${\sum }_{k\ge 0}{q}^k=(1-q{)}^{-1}$ givesConsequentlySubtracting the square of the expectation from the second moment yields

For the die, this is the coupon collector problem. When $r$ scores are still missing, let $W_r$ count rolls up to and including the next newly observed score. Every fresh roll has success probability $r/6$, so $W_r$ has a geometric distribution of parameter $r/6$. These six successive waiting times are independent: conditional on the entire past at the start of a stage, its waiting-time distribution depends only on the number $r$, not on the identities of the missing scores. Iterated conditioning therefore factors their joint distribution. The total number of rolls is $N={\sum }_{r=1}^6{W}_r$.

Using linearity of expectation givesUsing variance additivity for independent random variables, or the coupon collector waiting-time variance formula just derived by this decomposition, givesHence the requested standard deviation isThe hint rounds the sum of squares from $53.69$ to $53.7$; using that rounded value gives $\sqrt{39}\approx 6.245$, consistent at the precision of the hint. The exact computation preserves the slightly more precise value above.

Write $s=\sqrt{1-{\rho }^2}>0$. The inverse change of variables is $y=\rho x+sz$, whose Jacobian determinant has absolute value $s$. Completing the square givesThe change of variables formula therefore transforms the joint probability density function toIt factors into two normalized standard normal distribution densities. Thus $X$ and $Z$ are independent $N(0,1)$ variables; factorization proves the requested independence, rather than simply zero covariance.

The standard Gaussian random vector $(X,Z)$ has a rotation-invariant density. In polar coordinates, $X=r\cos \theta$, $Z=r\sin \theta$, its density with the area factor is $(2\pi {)}^{-1}r{e}^{-r^2/2}$, so the angular coordinate is uniform on an interval of length $2\pi$.

Put $\alpha =\arcsin \rho \in (-\pi /2,\pi /2)$, so $s=\cos \alpha$ andIn the right half-plane, choose $-\pi /2<\theta <\pi /2$. The additional condition $Y>0$ then requires $-\alpha <\theta <\pi /2$. This wedge has angle $\pi /2+\alpha$; its boundary has probability zero. ConsequentlyAt $\rho =0$ this is $1/4$, as required by independence. The limits as $\rho \to 1$ and $\rho \to -1$ are $1/2$ and zero, respectively. The angular calculation is also the geometric basis of the Gaussian sign-correlation identity.

Let $H$ be a Hermitian matrix and $He=\lambda e$ with $e\ne 0$. The inner product $e^{†}He$ is real because its conjugate equals $e^{†}{H}^{†}e=e^{†}He$. But it also equals $\lambda e^{†}e$, with $e^{†}e>0$. Thus every eigenvalue is real. For two eigenvectors,When ${\lambda }_i\ne {\lambda }_j$, this proves $e_i^{†}{e}_j=0$, so the eigenvectors are orthogonal.

For a real antisymmetric matrix, $(iA{)}^{†}=-i{A}^T=iA$, so $iA$ is a Hermitian matrix. Its real eigenvalues imply that the eigenvalues of $A$ are purely imaginary. To ensure a nonzero one without assuming it, writeExpansion of the characteristic polynomial gives $\det (tI-A)=t(t^2+|w{|}^2)$. Therefore the nonzero eigenvalue $-i\theta$ exists with $\theta =|w|>0$, and a nonzero complex eigenvector $e_1=u+iv$ satisfiesSeparating real and imaginary parts givesNeither real vector can vanish, since either vanishing would force the other to vanish. Antisymmetry yields $0=u\cdot Au=\theta u\cdot v$ andThus $u,v$ are orthogonal and have equal length.

In odd dimension $\det A=\det A^T=\det (-A)=-\det A$, so $\det A=0$. Its real homogeneous system therefore has a nonzero real eigenvector $e_3$ with $A{e}_3=0$. It is perpendicular to $u,v$, because $e_3\cdot Au=-(A{e}_3)\cdot u=0$ and similarly for $Av$. Normalize it and choose its sign so $(u/|u|,v/|v|,e_3)$ is positively oriented; this is an orthonormal basis.

The matrix exponential series converges absolutely, allowing even and odd terms to be grouped. Since $A^{2}u=-{\theta }^{2}u$ and $A^{2}v=-{\theta }^{2}v$, it acts asConsequently the skew-symmetric exponential as an axial rotation has, in the displayed orthonormal basis, the matrixIt is orthogonal with determinant one, fixes its axis and rotates the perpendicular plane. Hence $R=e^A$ is a rotation matrix, with angle $\theta$ about the chosen oriented axis.

If $e_1,\dots ,e_n$ are orthonormal, any relation ${\sum }_i{c}_i{e}_i=0$ gives $c_j=0$ on taking the dot product with $e_j$. Thus the vectors are linearly independent. A linearly independent set of $n$ vectors in the $n$-dimensional vector space ${\mathbb{R}}^n$ is a basis, proving the claim.

Expand both $x$ and $f$ in this orthonormal basis, writing $f_i=e_i\cdot f$. Since $A{e}_i={\lambda }_i{e}_i$, the linear equation becomesComparing the unique basis coefficients gives, when $\mu$ is not an eigenvalue,If $\mu ={\lambda }_1$, the equation is solvable exactly when $e_i\cdot f=0$ for every $i$ with ${\lambda }_i=\mu$. For those indices $a_i$ is arbitrary; for all other indices the displayed formula still applies. Thus repeated eigenvalues require orthogonality to the entire corresponding eigenspace, not merely to one chosen eigenvector. This is the finite-dimensional Fredholm solvability condition for a self-adjoint operator.

For the specific real symmetric matrix, chooseTheir dot products with $f$ are $-1/\sqrt{2}$, $3/\sqrt{2}$ and $3$. At $\mu =2$ this givesOne can check $(A-2I)x=f$ directly. At $\mu =1$, the first coefficient equation would require $0\cdot a_1=-1/\sqrt{2}$, which is impossible. Hence there is no solution when $\mu =1$; the nonzero projection onto its eigenspace is the obstruction.

For two upper triangular matrices and $i>j$,Indeed a potentially nonzero factor $A_{ik}$ requires $k\ge i$, while a potentially nonzero $B_{kj}$ requires $k\le j$. These requirements cannot both hold when $i>j$. Thus their matrix product is upper triangular.

Direct matrix multiplication for the given $A$ givesAdding $A^3+A^2-A$ yields $I$, so $A(A^2+A-I)=(A^2+A-I)A=I$. HenceEvery nonnegative matrix power is in $\mathrm{span}\{I,A,A^2\}$: reduce powers of degree at least three using $A^3=-A^2+A+I$. More formally, repeated reduction shows that products of any two polynomials of degree at most two in $A$ again lie in this span. Since $A^{-1}$ also lies in it, positive powers of $A^{-1}$ prove the same assertion for all negative integers. This is an instance of the integer powers for an annihilating polynomial with roots one and minus one, because $A^3+A^2-A-I=(A-I)(A+I{)}^2=0$.

To find the required entries for every integer, use the block upper-triangular structure. The top-left entry of $A^n$ is $1$ for positive and negative powers. Its lower-right block isThe identity $(I_2-N{)}^n=I_2-nN$ holds for all integers: multiplication adds the coefficients of $N$, and $(I_2-N{)}^{-1}=I_2+N$ proves the negative cases. ThereforeWrite $s=(-1{)}^n$ and $A^n={\alpha }_n{A}^2+{\beta }_{n}A+{\gamma }_{n}I$. Comparing those three entries givesSubtracting the first two equations finds ${\beta }_n$; the third then finds ${\alpha }_n$, and the first finds ${\gamma }_n$. The unique solution isFor clarity, substituting also gives the full matrix power

Fix $x\in (0,1)$. For any sufficiently small nonzero $h$, the additivity of the Riemann integral givesTaking absolute values, for either sign of $h$,Because $f$ is continuous at $x$, the right side tends to zero. Thus the fundamental theorem of calculus here follows directly from the difference quotient:Only continuity at the point under consideration was used.

Without continuity, the assertion is false. Take the Riemann integrable step function $f(t)=0$ for $t<1/2$ and $f(t)=1$ for $t\ge 1/2$. Its primitive is $F(x)=0$ for $x\le 1/2$ and $F(x)=x-1/2$ for $x\ge 1/2$. The left and right difference quotients at $1/2$ are respectively zero and one, so the primitive need not be differentiable at every interior point. A function with this single jump is Riemann integrable; changing its value at the jump does not change the integral or the counterexample.

The ratio test states that, for a series $\sum a_n$ with nonzero terms eventually, if $|a_{n+1}/a_n|\to L<1$, the series converges absolutely; if $L>1$, including $L=\infty$, it diverges because its terms do not tend to zero. At $L=1$ the test is inconclusive: the harmonic series diverges while $\sum n^{-2}$ converges.

For $|x|<1$, the ratio test applied to $x^n/n$ gives absolute convergence when $x\ne 0$, and at $x=0$ every such term vanishes. ThusAt $x=1$ all terms are zero, so the series converges to zero. At $x=-1$, the even terms vanish and the odd terms are $-2/n$, whose partial sums tend to $-\infty$. Finally, if $|x|>1$, then $|x^n-1|/n\ge (|x{|}^n-1)/n\to \infty$, so the term test for divergence applies. Consequently

De Moivre's theorem says that, for a real angle $\theta$ and an integer $m$,For a complex number $z=\rho (\cos \theta +i\sin \theta )$, the corresponding formula is $z^m={\rho }^m[\cos (m\theta )+i\sin (m\theta )]$, with $z\ne 0$ required for negative powers.

Put $q=e^{i\theta }$. If $\theta \notin 2\pi \mathbb{Z}$, the finite geometric series givesTo extract the real part, multiply numerator and denominator by $1-q^{-1}$. The denominator becomes $2(1-\cos \theta )$, and the numerator is $q-1-q^{n+1}+q^n$. Thus this finite trigonometric sum isAt $\theta \in 2\pi \mathbb{Z}$, the sum equals $n$; the printed quotient is undefined there, though its continuous limiting value is $n$.

Now take $\theta =2p\pi /(n+1)$ with $1\le p\le n$. This is not a multiple of $2\pi$, while $(n+1)\theta =2p\pi$ and $\cos (n\theta )=\cos (2p\pi -\theta )=\cos \theta$. Substitution gives