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Substitute the power series $y={\sum }_{n\ge 0}{a}_n{x}^n$ and compare the coefficient of $x^n$. This givesorThe two arbitrary coefficients $a_0,a_1$ determine all solutions. The condition $y^{\prime }(0)=0$ sets $a_1=0$, leaving the even series. It terminates precisely when ${\lambda }^2=(2k{)}^2$ for some nonnegative integer $k$. Therefore the permitted real values are $\lambda =0$ or $\lambda =\pm 2k$ for $k\ge 1$; these are the even Chebyshev polynomials up to normalization.

The degrees below six occur for $\lambda =0,\pm 2,\pm 4$. Taking $a_0=1$, the corresponding polynomials are

For the forced equation, try $y=a_0+a_2{x}^2+a_4{x}^4$. Comparing the coefficients of $x^4,x^2,1$ givesThus one polynomial satisfying $y^{\prime }(0)=0$ is

Independence of random variables and the two unit-rate exponential distributions giveThe linear transformationhas determinant $2$ and inverseThe inequalities $x,y\ge 0$ become $8v/3\le u\le 3v$. By the change-of-variables formula for probability densities,The variables $U$ and $V$ are not independent: the support of their joint density is a wedge rather than a Cartesian product of one-dimensional supports. For example, conditioning on a positive value of $V$ constrains $U$ to a finite interval depending on that value.

Substituting the proposed form $\mathbf{v}(t)=e^{\lambda t}\mathbf{u}$ into the linear system of ordinary differential equations givesIf $\lambda$ is not an eigenvalue of $B$, the matrix $\lambda I-B$ is invertible, soprovides the required particular solution.

In the stated two-dimensional case,and therefore $\mathbf{u}=(3,2{)}^T$. The eigenvalues of $B$ are $\pm \sqrt{3}$, with corresponding eigenvectors $(\sqrt{3},1{)}^T$ and $(-\sqrt{3},1{)}^T$. Adding the general solution of the homogeneous system yields

The characteristic polynomial of the homogeneous linear ordinary differential equation isso its complementary solution is $A{e}^{3x}+B{e}^{-x}$. Because the forcing contains the resonant factor $e^{-x}$, set $y=e^{-x}v$. Substitution givesWriting $w=v^{\prime }$, a polynomial particular solution of $w^{\prime }-4w=-16x$ is $w=4x+1$, and hence $v=2x^2+x$. ThusBoundedness as $x\to \infty$ forces $A=0$, while the initial condition $y(0)=1$ gives $B=1$. Therefore

Let $X_i=1_{A_i}$ be the indicator random variable of $A_i$. Then $X={\sum }_i{X}_i$, so linearity of expectation givesUsing variance of a sum andwe obtain

For the circular array, let $X_i$ indicate that bulbs $i$ and $i+1$ are both on, with indices read modulo $n$. Then $\mathbb{E}{X}_i=p^2$. Non-neighbouring indicators are independent, whileThere are $n$ individual terms and $2n$ ordered neighbouring pairs in the double covariance sum, so

The event $B$ is exactly $\{X\ge 1\}$. If $p=n^{-0.6}$, Markov inequality givesIf $p=n^{-0.4}$, then $B^c=\{X=0\}$ implies $|X-\mathbb{E}X|\ge \mathbb{E}X$. By Chebyshev inequality,Therefore $\boxed{\mathbb{P}(B)\to 1}$ in the second regime.

Let $h_k$ be the probability that the simple random walk started at $k$ hits $n$ before zero. The first-step analysis recurrence and boundary conditions areIf $p\ne q$, the characteristic equation of a linear recurrence has roots $1$ and $q/p$, so the boundary conditions giveIf $p=q=1/2$, the repeated root gives the limiting formula $\boxed{h_m=m/n}$. This is the gambler's ruin probability.

For a fixed integer stake $a\le N$, the fact that $a$ divides $N!$ lets us measure Patricia's wealth in units of £$a$. She starts at $M=N!/a$ and succeeds upon reaching $2M$. The formula above factors asFor $p=18/37$, $r=19/18>1$, so this probability increases as $M$ decreases. She should therefore choose the largest stake, $\boxed{a=N}$. For $p=19/37$, $r=18/19<1$, so the probability increases as $M$ increases; she should choose the smallest stake, $\boxed{a=1}$.

The Cauchy functional equation first gives $g(0)=0$ and $g(-x)=-g(x)$. If $h\to 0$, then continuity at the given point $z$ givesso $g$ is continuous at zero. For every $x$,so $g$ is continuous everywhere.

Put $c=g(1)$. Additivity gives $g(n)=nc$ for integers $n$ and then $g(m/n)=(m/n)c$ for rational numbers. For any real $x$, choose rationals $q_j\to x$; continuity gives

For the multiplicative equation, $h(0)=h(0{)}^2$. If $h(0)=0$, then $h(x)=h(x)h(0)=0$ for every $x$. Otherwise $h(0)=1$. In that caseand $h(x)$ cannot vanish because $1=h(0)=h(x)h(-x)$. Hence $h$ is everywhere positive. The continuous function $g=\log h$ satisfies the Cauchy functional equation, so $g(x)=cx$. Therefore the complete family is