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For a smooth real phase with no stationary point, integration by parts using $e^{ixf}=(ix{f}^{\prime }{)}^{-1}(e^{ixf}{)}^{\prime }$ gives an $O(x^{-1})$ contribution, provided $f^{\prime }$ stays away from zero and the differentiated amplitude is integrable. Near an interior nondegenerate stationary point $u_0$, write $f(u)=f(u_0)+f^{\prime \prime }(u_0)(u-u_0{)}^2/2+O((u-u_0{)}^3)$ and scale $u-u_0=x^{-1/2}s$. The Fresnel integral then gives the stationary phase contributionOne sums these contributions over the stationary points and adds endpoint terms where needed. Mere differentiability is insufficient to guarantee this particular nondegenerate formula; it requires the stated local smoothness and nonzero second derivative. Degenerate or endpoint stationary points need their own scaling.

For the gamma integral, rotate the positive real contour toward the positive imaginary axis, using the logarithm in the first quadrant. The limiting imaginary-axis integral is understood in the Abel sense, with a damping factor removed after integration; it is not an ordinarily convergent integral at infinity. With $u=iv$ and then $v=ns$ this givesThe rotation can first be made to an angle below $\pi /2$, where the exponential decays, and then continued to the boundary. There is one stationary point at $s=1$, with phase $-1$ and second derivative $-1$. Thus its contribution is $e^{-in-i\pi /4}\sqrt{2\pi /n}$. To check that the endpoints do not compete, use a cutoff around one. On the remaining region integrate by parts with $1/(1/s-1)$: it vanishes at zero, its derivative is integrable away from one, and Abel damping controls infinity, giving $O(n^{-1})$. This is smaller than the saddle contribution.

Multiplication by the prefactor therefore yieldsIn particular the modulus behaves as $\sqrt{2\pi n}{e}^{-\pi n/2}$; retaining the contour-rotation factor is essential to this decay and to the phase $\pi /4$.

The polynomial form of Runge theorem says that a function holomorphic on a neighbourhood of a compact set $K\subset \mathbb{C}$ with connected complement can be approximated uniformly on $K$ by polynomials.

For $n\ge 3$, takeThe complement is connected: the missing angular sector connects the inner disk to the exterior, and removing the single point zero from the inner disk does not disconnect it. On a neighbourhood of the annular sector choose an analytic logarithm with arguments between $-1/(2n)$ and $2\pi -1/(2n)$. Its exponential $\exp (3\log z/2)$ agrees with the prescribed function on the sector. On a disjoint small neighbourhood of zero prescribe the analytic function zero. These two neighbourhoods define a holomorphic function $g_n$ near $K_n$ with $g_n=f$ on $K_n$.

By Runge theorem, choose a polynomial $p_n$ with ${\sup }_{K_n}|p_n-g_n|<1/n$. Every fixed nonzero point of the disk eventually belongs to all $K_n$, including points on the positive real axis; zero belongs to every $K_n$. ConsequentlyEach polynomial restricts to an analytic function on $D$. The excluded sector lies immediately below the positive axis and shrinks with $n$, so the construction correctly preserves the chosen value on the branch cut. It gives pointwise convergence, not local uniform convergence across that cut. This is pointwise approximation by a Runge exhaustion.

The heat kernel is $H_t(x)=(4\pi t{)}^{-1/2}{e}^{-x^2/(4t)}$, so the homogeneous solution for smooth bounded data is $u(t,x)=(H_t\ast g)(x)$. Differentiating under the integral gives the heat equation, and the approximate-identity property gives the initial value.

For a tempered distribution, its derivative is defined by $⟨T^{\prime },\psi ⟩=-⟨T,{\psi }^{\prime }⟩$ for every Schwartz test function. A fundamental solution of a constant-coefficient operator $P$ is a distribution $E$ satisfying $PE={\delta }_0$. Put $E(x)=e^{-|x|}/2$. Away from zero, $E^{\prime \prime }=E$; its first derivative has jump $-1$ at zero. Distributional differentiation therefore gives $E^{\prime \prime }=E-{\delta }_0$, andFor an ansatz $u=e^{t}v$, the forced equation becomes $(1-D^2)v=\varphi$. With Fourier convention $\hat{v}(k)=\int e^{-ikx}v(x)dx$, its unique solution in the Schwartz space isThe smooth multiplier and its derivatives have polynomial bounds, so this quotient is Schwartz. For uniqueness, multiplication of the difference's Fourier transform by $1+k^2$ gives zero, forcing the difference to vanish.

Subtract this particular solution and evolve the remaining initial data by the heat kernel:For smooth bounded $f$, the second term remains uniformly bounded, so $e^{-t}u(t,\cdot )\to v$ uniformly. If additionally $f-v$ is integrable, its uniform norm is $O(t^{-1/2})$, so the remainder itself decays. General bounded data need not have a decaying heat remainder, and no such stronger claim follows without a condition on the data. The homogeneous uniqueness assertion is understood in the bounded or usual tempered-growth solution class; the uniqueness of the particular Schwartz ansatz is unconditional within that stated class.

A Markov control chooses its action from the current time and state, $a=u(k,x)$, rather than the full past. It gives transition probabilities $P_k^u(x,y)=P(y\mid k,x,u(k,x))$, hence a time-inhomogeneous Markov chain.

For any admissible control, the given inequality and conditional expectation implyTake expectations and sum from zero to $n-1$. The value terms telescope, and $V(n,X_n)=C(X_n)$ gives $V(0,x)\le {\mathbb{E}}_x[{\sum }_{k<n}c(k,X_k,A_k)+C(X_n)]$. For $u^{\ast }$ every inequality is equality, so its cost is exactly $V(0,x)$ and it achieves the lower bound. Thus $u^{\ast }$ is optimal. Boundedness of $V$ and the usual integrability of the costs justify these expectations; the argument even compares it with history-dependent controls.

For the cards, let $r,b$ be the remaining red and black counts, $m=r+b\ge 2$. Betting now succeeds with probability $q(r,b)=r(r-1)/[m(m-1)]$. For $m\ge 3$, elementary simplification giveswith zero-probability transitions omitted when a colour count vanishes. Thus continuing one reveal has exactly the same conditional winning probability as betting now. The backward value function for maximal success probability is $q$ at $m=2$, and induction makes it $q$ at every earlier state: the stop and continue values coincide. Equivalently the conditional success probabilities are a bounded martingale, and every permitted stopping time is bounded by the time at which two cards remain.

Therefore any adapted betting rule that makes exactly one bet while at least two cards remain is optimal; there is no strictly advantageous moment. Betting immediately is one simple optimal rule, with winning probability $26\cdot 25/(52\cdot 51)=25/102$. The stake size only rescales a fixed payoff, and does not alter this comparison. This is unchanging betting probability under sampling without replacement; allowing the player never to bet would be a different problem.