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TakeThenWith $F(0)=1$, separation givesand henceon its maximal interval of existence.

The floor-half generating function isSumming $u_n-2u_{n-1}=\lfloor n/2\rfloor$ for $n\ge 1$ givesso

Sincea particular solution isAdding the homogeneous term $C{2}^n$ and imposing $u_0=1$ gives $C=5/3$. Therefore

The homogeneous characteristic equation isA linear particular solution $u_n=an+b$ givesso $a=-1/4$ and $b=-11/16$. ThusThe initial data yield $A=1$ and $B=-5/16$, hence

For the ordinary generating function of a recurrence, multiply by $x^n$ and sum for $n\ge 2$. Using $u_0=0,u_1=2$ givesEquivalently,Expanding each geometric series gives exactly the displayed formula for $u_n$, verifying consistency.

Rewrite the equation asOn $[2,6]$, one has $q(x)\le 1/3$. If a nontrivial solution had two zeros, choose two consecutive ones $\alpha <\beta$ in that interval. Then $\beta -\alpha \le 4<\pi \sqrt{3}$.

The solutionof ${\phi }^{\prime \prime }+\frac{1}{3}\phi =0$ is strictly positive on $[\alpha ,\beta ]$. Since $q\le 1/3$ and the coefficients are not identical on this interval, the Sturm comparison theorem requires $\phi$ to have a zero between $\alpha$ and $\beta$, a contradiction. Hence every nontrivial solution has

TakeIts consecutive zeros are $-1$ and $\pi -1$. For the stated equation,and $q_2$ is not identically equal to $1$ on the interval. The Sturm comparison theorem therefore shows that every nontrivial solution has at least one zero in

Suppose ${\phi }_2$ has no zero in $(x_1,x_2)$. It has a constant sign there, so replace it by its negative if needed and apply part (i). The conclusion is $q_1\equiv q_2$ on that interval. Taking the contrapositive proves the Sturm comparison theorem: unless the coefficients agree identically, ${\phi }_2$ has a zero between the consecutive zeros of ${\phi }_1$.

After multiplying either solution by $-1$ if necessary, suppose both are positive on $(x_1,x_2)$. SetThe equations giveSince ${\phi }_1$ is positive between consecutive zeros,ThereforeBut $W$ is nondecreasing, so both endpoint values coincide at zero and $W^{\prime }\equiv 0$. Since both solutions are strictly positive inside,This is the equality case in the wronskian proof of Sturm comparison.

LetThey are jointly normal because they are linear combinations of a Gaussian vector. When ${\sigma }_1={\sigma }_2=\sigma$,Hence uncorrelated jointly normal variables are independent. Their means and variances areandThereforeindependently.

Completing the square in $x_1$ in the joint density shows that the conditional distribution of a bivariate normal variable isThus its conditional density is

The bivariate normal distribution has density

The Markov inequality states that for a nonnegative random variable $Y$ and $a>0$,For $t>0$, the event $X\ge x$ is the same as $e^{tX}\ge e^{tx}$. Applying Markov's inequality to the nonnegative variable $e^{tX}$ gives the exponential Markov bound

Sinceintegration givesPart (i) gives $D=0$, so the tangent addition functional equation has the solutionsfor constants $C$ such that no pole lies in $(-a,a)$. Conversely, the tangent addition formula verifies the functional equation. If $f^{\prime }(0)=2$, then $C=2$, andon any such interval, necessarily with $a\le \pi /4$.

Using the addition law with $y=h$,Differentiability makes $f$ continuous, so $f(h)\to f(0)=0$, andWriting $C=f^{\prime }(0)$ and taking the limit gives

Set $x=y=0$ in the functional equation:The denominator is defined. If $f(0)\ne 0$, cancellation would give $f(0{)}^2=-1$, impossible for a real-valued function. Hence

Multiplyingby $y^{-n}$ and setting $z=y^{1-n}$ givesTherefore the Bernoulli differential equation becomes

For the stated equation,Set $z=\sqrt{x}$. ThenThe integrating factor is $e^{t^3}$, soHenceThe condition $x(1)=4$ gives $C=2e-\frac{1}{2}$. Consequently

The zero function is a solution. On an interval where $y\ne 0$, putThenThe differential equation becomesHence $z=Cx$, and integration givesThus all nonzero solutions aretogether with $\boxed{y\equiv 0}$.

Write $N={\sum }_i{I}_i$ as in part (ii). For $i\ne j$, the two vertices are both isolated exactly when their combined $2n-3$ incident edges are absent, soThereforeLet $A=\mathbb{E}N=n(1-p{)}^{n-1}$. The displayed formula gives

Now take $p=c\log n/n$ with $c<1$, and choose $\alpha >1$ such that $\alpha c<1$. For all sufficiently large $n$, the supplied inequality gives $1-p\ge e^{-\alpha p}$, whenceAlso $p\to 0$, so both terms in the variance ratio tend to zero. Part (iv) now applies and proves the lower side of the isolated-vertex threshold:

If $N=0$, then $|N-\mathbb{E}N|=\mathbb{E}N$. Hence the Chebyshev inequality givesThe assumed limit therefore provesThis is the second moment method in its simplest form.