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Define interaction-picture states by $|{\psi }_I(t)⟩=e^{i{H}_{0}t/\hslash }|{\psi }_S(t)⟩$ and $V_I(t)=e^{i{H}_{0}t/\hslash }V(t)e^{-i{H}_{0}t/\hslash }$. Differentiation cancels the free Hamiltonian operator and yieldsAssume the interaction is switched off sufficiently fast in the remote past for the following integral to converge. Starting in $|n⟩$, the first iteration of the integral equation givesfor $m\ne n$. The picture change only multiplies the measured amplitude by a phase. Squaring therefore givesFor the harmonic oscillator, $|1⟩=a^{†}|0⟩$ is normalized and $⟨1|(a+a^{†})|0⟩=1$. The energy difference is $\hslash \omega$, andHence the lowest nonzero transition probability isThe stronger remainder here follows from parity: every insertion of $a+a^{†}$ reverses oscillator parity, so the amplitude between $|0⟩$ and $|1⟩$ has only odd powers of $\lambda$. The general displayed $O({\lambda }^3)$ estimate remains valid.

Put $q={\mu }^2-1/4$ and $S=\log y$. The differential equation becomes the Riccati equation $S^{\prime \prime }+(S^{\prime }{)}^2=1/4+q/x^2$. The inverse-power exponential hierarchy begins withso choose $S_0=\epsilon x/2$, $S_1=0$ with $\epsilon =\pm 1$; additive constants specify the overall normalization. At order $x^{-2}$, $2S_0^{\prime }{S}_2^{\prime }=q/x^2$, and at subsequent ordersThese equations are formally equivalent coefficient by coefficient; the order-zero transport term happens to be constant, so its chosen zero value is not interpreted as a literal denominator in a successive-ratio hypothesis. The ansatz $S_j=c_j{x}^{-(j-1)}$ is consistent for $j\ge 2$, withEmpty sums are zero. In particular $c_3=-q$, which agrees with direct expansion of the Riccati equation.

Exponentiating the inverse-power series gives $y_{\epsilon }\sim e^{\epsilon x/2}{\sum }_{j\ge 0}{A}_j^{\epsilon }{x}^{-j}$ with $A_0=1$. Substitute this amplitude series into $u^{\prime \prime }+\epsilon u^{\prime }=q{x}^{-2}u$ to obtainFormal matching alone does not prove existence of actual solutions with these asymptotics. Here actual solutions are obtained from the Modified Bessel differential equation and the large-argument asymptotic expansion of a modified Bessel function: substituting $y=\sqrt{x}w(x/2)$ yields solutions $\sqrt{\pi }\sqrt{x}{I}_{\mu }(x/2)$ and $\sqrt{x/\pi }{K}_{\mu }(x/2)$, using the Modified Bessel function of the first kind and Modified Bessel function of the second kind, whose positive-real-axis asymptotic expansions have precisely these normalized growing and decaying series. They are linearly independent. The series need not converge; it is an asymptotic expansion with a remainder after each fixed truncation. When $q=j(j+1)$ the amplitude recurrence terminates.

For $x>0$, $N(x)=\lfloor 1/x\rfloor$ and $T(x)=1/x-\lfloor 1/x\rfloor$. Iterating this Gauss continued-fraction map on an irrational number never reaches zero, and its continued fraction isThe branches $T:(1/(n+1),1/n)\to (0,1)$ have inverse $x_n(y)=1/(n+y)$, for $n\ge 1$. The change of variables formula for a probability density function gives, almost everywhere,The endpoints of these branches are a countable set of probability zero. Thus $T(X)$ and $X$ have the same distribution; this Gauss measure is an invariant measure for the Gauss continued-fraction map.

Assume a finite optimum with $X>0$ and continuous price near its completion time. With costate $p^{\ast }$ for the remaining-file state, the minimization Hamiltonian of an optimal-control problem is $H=u(p(t)-p^{\ast })$. It is independent of $x$, so $p^{\ast }$ is constant. By the Pontryagin maximum principle, pointwise minimization over $[0,1]$ givesThis is a bang-bang control away from equality. At equality partial transmission or either endpoint rate is allowed. This also follows by exchanging a small amount of transmission from a more expensive time to a cheaper unused time. The free-terminal-time transversality condition is ${\min }_{u}H(T,u)+2\gamma T=0$. Since $T>0$, it forces $p^{\ast }>p(T)$ and terminal transmission at rate one, yieldingWithout regularity/existence hypotheses an arbitrary known price need not have a finite differentiable optimum; the equality is the interior terminal-time condition used here.

For $p(t)=t+1/t$, $X=1$, feasible completion requires $T>1$ for finite cost. The price decreases to its minimum at one and then increases. Its cheapest unrestricted interval of length one has endpoints $a,a+1$ of equal price, giving $a(a+1)=1$ and $a+1=\phi =(1+\sqrt{5})/2$. For $1<T\le \phi$, the cheapest length-one subset of $[0,T]$ is the terminal interval $[T-1,T]$: its left endpoint has at least the price of its right endpoint, and the earlier times are still more expensive. For $T\ge \phi$ the transmission minimum is unchanged from that unrestricted interval, while the delay cost increases, so no optimum lies beyond $\phi$.

For the terminal-interval policy the total cost isIts derivative is strictly increasing on $T>1$, since $J^{\prime \prime }(T)=(2T-1)/[T^2(T-1{)}^2]+2\gamma >0$. It is negative near one and positive at $\phi$, so the unique global optimum isThere is exactly one positive root: none lies in $(0,1)$ because the left side is negative, and the strictly monotone comparison gives exactly one in $(1,\phi )$. Its threshold is $p(T-1)=p(T)+2\gamma T$.