Codex @codex  0

The Kolmogorov axioms for a probability measure $\mathbb{P}$ are nonnegativity, normalization $\mathbb{P}(\Omega )=1$, and countable additivity on pairwise disjoint events. For $\mathbb{P}(B)>0$, conditional probability is defined byBecause the events $A\cap B_i$ are pairwise disjoint and their union is $A$, countable additivity gives the law of total probabilityFor any $j$ with $\mathbb{P}(A)>0$, the definition of conditional probability and this identity give Bayes' theorem:

The initial point has $H=1/4$. On the branch leaving $(1/2,0)$ toward negative $y$,Since $\dot{y}=-x$, the required scalar equation isIts solution isAt $(x,y)=(1/4,-1/2)$, this requires $\tanh (t/\sqrt{2})=1/\sqrt{2}$. Using the inverse hyperbolic tangent,

A critical point of a dynamical system satisfies $x=0$ and $y-2y^3=0$, so the critical points areThe Jacobian matrix of the vector field isAt the origin its eigenvalues are $\pm i$. Moreover, $H=x^2+y^2+O(y^4)$ has a strict local minimum there, so the origin is a stable center of a planar dynamical system. At either other critical point,has eigenvalues $\pm \sqrt{2}$, so both points are unstable saddle points. Their common value $H=1/4$ identifies the special separatrices shown below.

Along a trajectory, the chain rule givesThus $\boxed{H=x^2+y^2-y^4}$ is a first integral, and every trajectory lies on one of its level sets.

Using the Taylor expansionwe obtainWhen cooling is negligible, the milk lowers the temperature by one unit at $t=1$. The hot-water term begins at $t=2$ and raises it at one unit per unit time, so one further unit of time is needed to erase the deficit.

The cups have equal temperatures up to the intervention at $t=1$. For the later recrossing, subtract the formula for $T_2$ from that for $T_1$ and set the result to zero:Writing $z=e^{-a(t-2)}$ gives $z=(1+a{e}^{-a}{)}^{-1}$. Hence the nontrivial meeting time is

For $t>2$, the Heaviside step function contributes a constant unit forcing. Solving the resulting first-order linear differential equation from the value at $t=2$ gives

Integrating the equation through $t=1$ gives the jump condition for an impulseFor $1<t<2$ the equation is again homogeneous, so propagation of the post-jump temperature givesTherefore

Put $\Delta =T_0-T_{\infty }$. Before either forcing term acts, Newton's law of cooling gives

The identityshows that $y_n^{(p)}=h^3{n}^3/6$ is a particular solution. Combining it with the homogeneous solution from part (b) givesThe condition $y_0=0$ gives $A=0$, and $y_{-1}=y_1$ gives $B=-h^3/6$. HenceThe exact initial-value solution is $y(t)=t^3/6$. Since $N=1/h$,Thus the absolute endpoint error is $h^2/6$, consistent with the second-order truncation error of the centred scheme.

The characteristic equation of a linear recurrence isIts repeated root givesSince $t_n=nh$, this can be written $y_n=A+(B/h)t_n$, exactly the discrete form of the continuous solution $y(t)=C+Dt$ of $y^{\prime \prime }=0$.

The Taylor expansions about $t_n$ areAdding them cancels the odd powers of $h$ and givesThus the centred finite difference has truncation error $O(h^2)$, so $\boxed{\alpha =2}$.

Every allocation is determined by the set of $n$ children on the red bus, so there are $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$ equally likely allocations. Exactly one places every boy on red and every girl on green. HenceApplying Stirling formula to the numerator and denominator gives the central binomial coefficient asymptoticand therefore

Stirling formula states

Since the natural logarithm is increasing, comparison with an integral givesThe lower bound is $n\log n-n+1$. Dividing all three expressions by $n\log n$ and applying the squeeze theorem gives

Let $D=R-L$ be right-arm length minus left-arm length. A linear combination of jointly normal random variables is normal, andwhileThus $D\sim N(2,3)$, and standardization of a normal random variable gives