Use the measurable partition $B_n=\{N=n\}$ for $n\ge 0$, omitting any null cells. For $0\le s\le 1$, the random variable $s^X$ is bounded, so the law of total expectation applies without moment assumptions. Conditional on $N=n$, the sum contains exactly $n$ terms. Independence of $N$ from the whole sequence preserves their joint law under this conditioning; independence of the $Y_k$ then givesFor $n=0$ the empty sum is zero and the empty product is one, in agreement with the formula. Therefore the probability generating function of an independent random sum is

Summing a geometric series gives the probability generating function of the zero-based geometric distribution:The series converges for $|(1-p)s|<1$, in particular on $[0,1]$. At $p=1/2$, $G_N(s)=1/(2-s)$, $G_N^{\prime }(s)=(2-s{)}^{-2}$, and $G_N^{\prime \prime }(s)=2(2-s{)}^{-3}$. The factorial moment identities give $\mathbb{E}(N)=G_N^{\prime }(1)=1$ and $\mathbb{E}[N(N-1)]=G_N^{\prime \prime }(1)=2$. Thus

For a nonnegative integer-valued random variable $X$, its probability generating function isAt $s=0$, use $0^0=1$ so $G_X(0)=\mathbb{P}(X=0)$. The series also converges absolutely for complex $|s|\le 1$; for $s$ near zero, its coefficients recover the probability mass function.

For an integrable random variable $X$, the law of total expectation on the finite or countable measurable partition isHere $\mathbb{E}(X\mid B_i)=\mathbb{E}[X{1}_{B_i}]/\mathbb{P}(B_i)$ for positive-probability cells; null cells contribute zero and need no conditional value. The series is absolutely convergent since its absolute terms sum to at most $\mathbb{E}|X|$. For a nonnegative $X$, the same identity is valid with possible value $+\infty$.

A partition of a set $\Omega$ is a collection of nonempty subsets $B_i$ such that $B_i\cap B_j=\varnothing$ for $i\ne j$ and ${\bigcup }_{i\in I}{B}_i=\Omega$. Thus every outcome lies in exactly one cell. For a probability-space partition the cells must also be measurable events. Nonempty cells may nevertheless have probability zero.

The weak law of large numbers gives $S_n/n\to p$ in probability. If $\alpha <p$, the event $S_n/n\le \alpha$ implies $|S_n/n-p|\ge p-\alpha$, whose probability tends to zero. If $\alpha >p$, its complement $S_n/n>\alpha$ implies $|S_n/n-p|>\alpha -p$, whose probability also tends to zero. ThusFor an explicit independent bound, Chebyshev's inequality bounds either unwanted probability by $p(1-p)/[n(\alpha -p{)}^2]$. The boundary $\alpha =p$ is outside the two requested cases and is not decided by the weak law of large numbers alone.

Summing the probability mass function of the binomial distribution up to the largest permitted integer givesThe floor is implicit in the original integer-indexed sum.

By linearity of expectation, $\mathbb{E}(S_n)={\sum }_{i=1}^{n}p=np$. Part (ii) gives $\mathrm{Var}(X_i)=\mathbb{E}(X_i^2)-[\mathbb{E}(X_i){]}^2=p-p^2$. Independent random variables have zero pairwise covariance, so the variance of their sum is the sum of their variances. Therefore

An indicator random variable equals $1$ with probability $p$ and $0$ with probability $1-p$, so $\mathbb{E}(X_i)=1\cdot p+0\cdot (1-p)=p$. Also $X_i^2=X_i$ pointwise, whence

The $X_i$ are independent Bernoulli random variables with success parameter $p$. Their sum has the binomial distribution $S_n\sim \mathrm{Bin}(n,p)$, whose probability mass function isChoose the $k$ successful events in $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ ways; independence gives the same product probability for each choice. At $p=0$ or $p=1$, interpret the formula by its degenerate distribution or by continuity.

For $f^{\prime }=\Phi (x,f)$, the standard explicit midpoint method isThe implicit midpoint rule instead uses $y_{i+1}=y_i+h\Phi (x_i+h/2,(y_i+y_{i+1})/2)$. When the derivative depends only on $x$, both give $y_{i+1}=y_i+h{f}^{\prime }(x_i+h/2)$.

The endpoint-only request in (e) suggests another possible convention: averaging the endpoint slopes, $y_{i+1}=y_i+\frac{h}{2}[\Phi (x_i,y_i)+\Phi (x_{i+1},y_{i+1})]$. This is the trapezoidal rule, and generally differs from the implicit midpoint rule. A further use of “midpoint” is the two-step explicit midpoint linear multistep method, $y_{i+1}=y_{i-1}+2h\Phi (x_i,y_i)$, which needs a second starting value. The printed question supplies no convention to distinguish these meanings; (e) spells out the resulting alternatives.

For $f^{\prime }=\Phi (x,f)$, the Backward Euler method uses the slope at the end of the step:This generally requires solving an implicit equation for $y_{i+1}$. For a derivative depending only on $x$, it reduces to $y_{i+1}=y_i+h{f}^{\prime }(x_{i+1})$.

Write the derivative equation as $f^{\prime }(x)=\Phi (x,f(x))$ with initial approximation $y_0$. The forward Euler method uses the slope at the start of the step:If the derivative is a known function of $x$ alone, this is $y_{i+1}=y_i+h{f}^{\prime }(x_i)$.

With the same three cosine values, Simpson's rule givesUsing the table's $n=12$ row gives $0.2618+4(0.1851)=1.0022$, with the same final rounding. That table row refers to the prefactor $\pi /12$; it does not change the specified number of strips.

Two strips give $h=\pi /4$, with cosine values $1,1/\sqrt{2},0$. The trapezoidal rule therefore givesThe exact integral is $1$, so the estimate is low, consistent with the concavity of cosine on this interval. The table can be used as $T_2=\frac{1}{2}(0.7854)+0.5554$ before rounding.

For the same uniform grid and even $n$, the composite Simpson's rule integrates a quadratic interpolant over each successive pair of strips:The weights are $1,4,2,4,\dots ,2,4,1$, which explains the even number of strips.

Set $h=(b-a)/n$ and $x_i=a+ih$. Summing the areas of the $n$ endpoint trapezia gives the composite trapezoidal ruleEvery interior value is shared by two adjacent strips; the two endpoint values occur only once with half weight. Evenness of $n$ is not needed for this rule.

Apply integration by parts with $u=x$ and $dv=\sin (\mu x)dx$. Part (i) gives $v=-\cos (\mu x)/\mu$, so

The chain rule gives $\frac{d}{dx}\cos (\mu x)=-\mu \sin (\mu x)$. Dividing by the nonzero constant $-\mu$ gives the antiderivative

We prove the stronger explicit bound $n!\le n^n/2^{n-1}$ by mathematical induction. The case $n=1$ is equality. The binomial theorem gives $(1+1/n{)}^n\ge 1+n/n=2$, so $(n+1{)}^n\ge 2n^n$. Consequently the induction hypothesis yieldsTherefore $n!=O(n^n/2^n)$, with constant $2$ and threshold $1$.

The second bound is asymptotically tighter. Using $n\le 2^n$,Thus its bounding function is negligible relative to the first in little o notation; this compares the bounds without needing Stirling's formula.