Using the probability mass function of the Poisson distribution directly givesFor each fixed positive integer $n$, this series converges: the ratio of successive nonzero terms is $(1+1/j{)}^n/(j+1)$, which tends to zero. Equivalently, using the polynomial identity in (ii) and the factorial moments from (a)(iv) gives $\mathbb{E}[X^n]={\sum }_{k=1}^{n}A(n,k)$; the extension of that identity to $X=0,1$ is justified in (iv).

Group the elements of a colouring by equality of colour. This creates a unique partition of a set into nonempty unlabelled blocks, each of one colour, with distinct blocks assigned distinct colours.

For a partition of a set with $k$ blocks, assigning distinct colours to the blocks gives the falling factorial $x(x-1)\cdots (x-k+1)$ possibilities. To see that no extra factor $k!$ is needed, order the blocks canonically by their smallest elements, then assign colours in that order. There are $A(n,k)$ possible set partitions with $k$ blocks; these numbers are the Stirling numbers of the second kind.

Every colouring is counted exactly once in exactly one value of $k$, soFor $k>x$, no colouring of the blocks is possible and the displayed falling factorial is zero, because one factor is $x-x$.

Each of the $n$ distinguishable elements has $x$ independent choices of colour. By the multiplication principle, the number of functions from the $n$-element set to the $x$-element colour set is

For the Poisson distribution with parameter $1$, the exponential series givesEvery derivative is $G^{(k)}(s)=e^{s-1}$, so the factorial moments areIn particular $G^{\prime }(1)=G^{\prime \prime }(1)=1$, and the variance is

Since $X^2=X(X-1)+X$, the probability generating function identities giveSubtracting the squared mean yields the variance

By hypothesis the power series converges at every positive real argument, so its radius of convergence is infinite. Termwise differentiation of a power series is therefore valid in a neighborhood of $s=1$. For $k\ge 1$,At $s=1$, the right side is the expectation of the falling factorial of $X$:The special case $k=1$ givesFor $k=0$, take the empty product as $1$; the equality becomes $G(1)=1$. These are the factorial moments.

The probability generating function of a nonnegative integer-valued random variable isUse $s^0=1$, including at $s=0$. The power series converges absolutely for every real $s\in [-1,1]$, since ${\sum }_{j}p(j)|s{|}^j\le 1$. In particular it is always defined on the usual probability-generating interval $[0,1]$. It may also be defined for $|s|>1$, depending on the radius of convergence and convergence at its endpoints; the precise real domain of finite expectation is $\{s\in \mathbb{R}:{\sum }_{j}p(j)|s{|}^j<\infty \}$.

Each $X_i$ has the same probability distribution as $-X_i$. By independence, the entire vector and its negative have the same joint law, so $S_m$ has a symmetric distribution about zero. The two disjoint tail events consequently have equal probabilities:Apply Chebyshev's inequality with the mean and variance from (i):Using only containment in the two-sided event would give $8/(3m)$; the symmetry supplies the necessary extra factor $1/2$.

For each random variable, the probability mass function givesLet $S_m={\sum }_{i=1}^m{X}_i$. By linearity of expectation and variance additivity for independent random variables,The independence is needed for the variance calculation, but not for the mean.

The absorbing states give $e_0=e_n=0$. Multiplying the difference equation in (i) by $3$ givesThe second finite difference of $k^2$ is $2$, so a particular solution is $-3k^2/2$; the homogeneous solutions are $A+Bk$. Imposing the two boundary conditions gives $A=0,B=3n/2$. ThereforeUniqueness follows because the difference of any two solutions has zero second finite difference, hence is affine function, and vanishes at both endpoints. This is the mean absorption time of a lazy symmetric random walk with holding probability $1/3$.

Let $\tau$ be the absorption time. Its expectation is finite: from every interior state, a string of at most $n$ successive left moves reaches $0$, and the probability of such a string is at least $p=3^{-n}$. Hence the Markov property gives ${\mathbb{P}}_k(\tau >jn)\le (1-p{)}^j$, a geometric tail bound, and ${\mathbb{E}}_k[\tau ]<\infty$.

For $1\le k\le n-1$, one step is always spent before absorption or the remaining walk. Apply first-step analysis and the law of total expectation to the three possible first moves. The conditional expectations of the remaining absorption time are $e_{k+1},e_k,e_{k-1}$, respectively; independence of future moves gives the same law as a fresh random walk from the new position. Thuswhich rearranges toThe term $1$ also appears when the walk stays at $k$, because staying still consumes a step.

For an integrable random variable $X$ and a partition of a sample space $(B_i)$, the law of total expectation statesIt follows by writing $X={\sum }_{i}X{1}_{B_i}$ and using $\mathbb{E}[X{1}_{B_i}]=\mathbb{P}(B_i)\mathbb{E}[X\mid B_i]$. Integrability justifies the countable sum. The same equality is valid with nonnegative, possibly infinite expectations when $X\ge 0$.

A partition of a sample space is a finite or countable family of nonempty events such thatThus exactly one event in the family occurs in each outcome. To use ordinary event-conditioned conditional expectations, one usually assumes $\mathbb{P}(B_i)>0$; any zero-probability blocks can instead be omitted from the expectation formula.

The heat equation and its homogeneous Dirichlet boundary conditions are linear, so superposition of the separated modes givesThe Fourier sine basis represents the initial profile on the interior of the interval. At $t=0$ its Fourier sine series must equal $x(1-x)$. Multiplying by $\sin (m\pi x/L)$, integrating from $0$ to $L$, and using orthogonality givesTherefore the required Fourier sine coefficients areFor every $t>0$, the exponential factors ensure convergence of the differentiated series, and every term satisfies the partial differential equation and the boundary conditions. The Fourier sine series recovers the smooth initial profile at every $0<x<L$ as $t\downarrow 0$.

The scan explicitly prints $x(1-x)$ on $(0,L)$; it must not silently become $x(L-x)$. For $L\ne 1$, the interior profile tends to $L(1-L)\ne 0$ at the right endpoint, whereas the prescribed boundary condition is zero for $t>0$. The solution is therefore not continuous at the corner $(L,0)$, although it satisfies every condition on the printed open ranges. A solution continuous on the closed space-time domain with these data would require $L=1$.^[ref]

For the spatial mode with eigenvalue ${\kappa }_n=(n\pi /L{)}^2$, the temporal ordinary differential equation isIts general solution isThe printed reference to “each solution $X(x)$ ... in part (b)” refers to (b)(i). The zero spatial mode gives $u=0$ irrespective of $T$ and contributes no additional nonzero solution.^[ref]

For a nonzero separable solution $u=XT$, separation of variables givesThese Dirichlet boundary conditions follow because $T$ is not identically zero. If $\kappa =0$, the affine function $X=A+Bx$ obeys both conditions only when $A=B=0$. If $\kappa =-{\rho }^2<0$, $X=A\cosh (\rho x)+B\sinh (\rho x)$ also has only the zero solution under both conditions.

For $\kappa ={\omega }^2>0$, $X=A\cos (\omega x)+B\sin (\omega x)$. The first boundary condition forces $A=0$, and the second forces $\omega L=n\pi$ for a nonzero solution. Hence all nonzero spatial modes areThe identically zero function is also a solution. Allowing $B_n=0$ includes it in the displayed family. Here $L>0$ is the interval length and $n$ labels the eigenvalues of the operator $-d^2/d{x}^2$ with these boundary conditions.

Write the sum in the question as the right Riemann sumRelative to the composite trapezoidal rule, the only change is in the two endpoint weights:The integrand $f(x)=1/(1+x^2)$ is continuous on $[0,1]$, so $T_N\to I$. Since an antiderivative is $\arctan x$,Thus

With step size $h=1/N$ and nodes $x_j=j/N$, the composite trapezoidal rule givesThe endpoint weights are half the interior weights, and the endpoint values are $1$ and $1/2$.

The amplification factor is $R=1-\lambda /N$. To reproduce decay, the forward Euler method requires $|R|<1$, equivalentlyAt $N=\lambda /2$, $R=-1$ and the approximations alternate with constant magnitude $2$; for $N<\lambda /2$, $|R|>1$ and the magnitude grows. This is the condition of absolute stability for this decaying differential equation.

Stability is only a minimum requirement. Values just inside the bound can still give a poor endpoint approximation; resolving the decay accurately generally requires a smaller step size. If positivity at every step is desired, the stronger condition $N\ge \lambda$ ensures $R\ge 0$. The figure shows the difference between decay of the exact solution and numerical oscillation.

Let $h=1/N$, $x_j=jh$, and let $y_j$ approximate $y(x_j)$. The forward Euler method gives the difference equationThis geometric sequence has solutionFor fixed $\lambda$, the limit as $N\to \infty$ is $2e^{-\lambda }$, as required by the exact solution.