The event $T\ge 5$ is the same as $2^T\ge 32$. Apply Markov's inequality to the nonnegative random variable $2^T$ and use part (iv):The strict numerical comparison is enough even though the probabilistic inequality itself is non-strict.

By the definition of the probability generating function,The evaluation at two is valid because two is inside the power-series radius $14/5$.

Markov's inequality states that for a nonnegative random variable $Z$ and $a>0$,It follows immediately from $Z\ge a{1}_{\{Z\ge a\}}$ and monotonicity of expectation. A finite expectation makes the bound informative.

For a probability generating function, $\mathbb{E}T=g^{\prime }(1)$ and $\mathbb{E}[T(T-1)]=g^{\prime \prime }(1)$. HereThus $\mathbb{E}T=5/54$ and $\mathbb{E}[T(T-1)]=25/324$. Adding the first moment gives $\mathbb{E}{T}^2=55/324$, hence

Expand the analytic factor around zero by the binomial series:Every coefficient is nonnegative. Adding $2/3$ to the constant coefficient gives probabilities $p_j$ with sum $g(1)=2/3+1/3=1$. Define $\mathrm{Pr}(T=j)=p_j$ for $j\ge 0$. This is a nonnegative integer-valued random variable and its probability generating function is exactly $g$. The power series has radius $14/5>1$, so the evaluations and derivatives used below are finite.

A characteristic function is useful for proving convergence in distribution, including for laws without a finite moment-generating function: the continuity theorem says that pointwise convergence of characteristic functions to a characteristic function continuous at zero implies convergence in distribution to its law. For example, the Cauchy distribution has a characteristic function although its moment-generating function is not finite on any neighbourhood of zero.

A moment-generating function is especially useful for exponential tail bounds: for $t>0$ where it is finite, Markov's inequality givesOptimizing over $t$ yields a Chernoff bound. The positive real exponential in this argument permits the inequality; a unit-modulus complex exponential does not give the corresponding tail bound.

The moment-generating function is $M_X(t)=\mathbb{E}[e^{tX}]$ for real $t$ where this expectation is finite. Usually one requires finiteness on a neighbourhood of zero when using it to characterize a distribution. The characteristic function isIt always exists because the integrand has absolute value one. In both definitions the expectation is taken with respect to the distribution of $X$.

Because $2|x|\le 1+x^2\le Q$, we have $|f_x(x,y)|=2|x|/Q^2\le 1/Q\le 1$; similarly $|f_y|\le 1$. Apply the one-variable mean value theorem on the horizontal and vertical segments joining $(x,y)$ to $(u,y)$ to $(u,v)$. The triangle inequality givesThus the function is globally one-Lipschitz for the ${\ell }^1$ metric on the plane.

The derivative of the curve isIts image lies on the unit circle, so $f(g(t))=1/2$ and $\boxed{D(f\circ g)(t)=0}$. Meanwhileand multiplication by $J_g(t)$ gives $-\sin t\cos t/2+\cos t\sin t/2=0$. This verifies the chain rule explicitly.

At $(-1,0)$ the gradient is $(1/2,0{)}^T$. The total derivative is therefore the linear functional

Put $Q=1+x^2+y^2>0$. Differentiation of $Q^{-1}$ givesThe positive denominator also shows that the function and these derivatives are smooth throughout ${\mathbb{R}}^2$.

The chain rule at the specified point isIn standard coordinates this becomesThe matrix dimensions are respectively $k\times n$, $k\times m$, and $m\times n$, matching the composition of the two linear derivative maps.

For a scalar-valued function, the gradient is the column vectorWhen $f$ is differentiable, its total derivative is the linear functionalThus the Jacobian is the transpose of the gradient in standard Euclidean coordinates.

The $j$-th partial derivative is the vector limitwhen it exists. The total derivative is a linear map $Df(x):{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe Jacobian matrix is $J_f(x)=({\partial }_j{f}_i(x){)}_{i,j}$. When the total derivative exists, all partial derivatives exist and $J_f(x)$ is its matrix in the standard bases; its $j$-th column is $Df(x)e_j$. Existence of partial derivatives alone does not guarantee total differentiability.

Use the nonnegative parts $h^{+}=\max (h,0)$ and $h^{-}=\max (-h,0)$, so $h=h^{+}-h^{-}$. The zero-mean assumption gives $H:={\int }_0^{\pi }{h}^{+}={\int }_0^{\pi }{h}^{-}<\infty$. On the $k$-th period, each of the positive and negative weighted integrals lies between $H/((k+1)\pi )$ and $H/(k\pi )$. Their difference therefore satisfies the stronger estimateThe sum of the absolute values of these block integrals converges, proving the existence of the finite limit of the integrals from $\pi$ to $n\pi$.

For genuine Lebesgue integrability we instead need the integral of the absolute value. Periodicity givesIf $H>0$, summation diverges by the harmonic series. If $H=0$, then $h=0$ almost everywhere and the function is integrable. ThusThe finite signed block limit therefore need not be a Lebesgue integral.

Let $H_0={\int }_0^{\pi }|h(u)|du<\infty$. On each period, substitute $x=j\pi +u$ and use periodicity:Summing over $j\ge 0$ gives the convergent geometric boundMeasurability is preserved under multiplication by the exponential, so this proves Lebesgue integrability.

False. Let $k$ be the indicator of the positive integers. It is measurable, bounded by one and zero almost everywhere, so its Lebesgue integral on $(0,\infty )$ is zero. Yet $k(n)=1$ for every positive integer $n$, so it does not tend to zero at infinity. Lebesgue integrability is unchanged by modifications on a null set, whereas a pointwise limit can change.

False. Take $k(x)=1/(1+|x|)$. This is bounded and measurable on $\mathbb{R}$, and tends to zero as $x\to \infty$, butThus decay to zero alone does not imply Lebesgue integrability.

An integer square is a square modulo every prime, with the zero class allowed at primes dividing the integer. Conversely, suppose an odd integer $a$ is not an integer square. If every prime exponent in $|a|$ is even, the nonsquare condition forces $a=-b^2$, handled by part (i). Otherwise some odd prime has an odd exponent, so $a=\pm q^{t}b$ as in part (ii). In either case there is a positive odd $k$ with $(a/k)=-1$.

Factor $k$ into odd primes. A product of Legendre symbols equalling minus one has at least one factor equal to minus one. Thus $a$ fails to be a square modulo one of those primes, contradicting the hypothesis. HenceThe solvability formulation makes explicit the zero-residue convention needed by the printed final assertion. The prime two causes no extra issue because every odd integer is one modulo two.

Choose a nonzero quadratic nonresidue $c$ modulo $q$. By the Chinese remainder theorem, choose a positive integer $k$ satisfyingThe moduli are pairwise coprime, and $k$ is coprime to $qb$. Jacobi reciprocity, with $k\equiv 1(\mathrm{mod}4)$, givesSince $t$ is odd, multiplying these symbols yieldsWhen $b=1$ its congruence imposes no restriction and its Jacobi factor equals one.