Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 72 / 1

For $ϵ>0$, dominant balance for algebraic roots gives two roots on the scale $\sqrt{ϵ}$ and a third on the scale ${ϵ}^{2/3}$. On the first scale put $t=\sqrt{ϵ}s$. The equation becomes $s^5-s^3+\sqrt{ϵ}=0$. The derivative of $s^5-s^3$ at either $s=1$ or $s=-1$ is two, so the first correction is $-\sqrt{ϵ}/2$. On the second scale put $t={ϵ}^{2/3}v$; then $1-v^3+{ϵ}^{1/3}{v}^5=0$. Writing $v=1+a{ϵ}^{1/3}+\cdots$ gives $a=1/3$. Thus all three real-root expansions for positive $ϵ$ areThese are fractional-power root expansions. To check that none is missing, the derivative of the polynomial is $t^2(5t^2-3ϵ)$. Its nonzero critical points are $\pm \sqrt{3ϵ/5}$; for sufficiently small positive $ϵ$ the polynomial is positive at the negative one and negative at the positive one. Its three monotone ranges therefore contain exactly these three real roots. The remaining two small roots correspond to the nonreal cube roots of unity in the second balance.

The printed limit does not specify the sign of $ϵ$. If a two-sided real limit is intended, also put $ϵ=-\eta$ with $\eta >0$. The polynomial $t^5+\eta t^3-{\eta }^3$ is strictly increasing and has just one real root. The same dominant balance for algebraic roots gives the negative-parameter branch

Use the principal complex logarithm in the upper half-plane. Deform the interval upward into a half-strip: the vertical side at zero is traversed upward, and the side at one downward. The top side vanishes exponentially; a small indentation at zero contributes $O(r|\log r|)$ and vanishes. Thus the contour deformation gives the exact representation, for $x>0$,Since $\log (is)=\log s+i\pi /2$, scaling $u=xs$ and using the integral defining Euler's constant givesFor the second endpoint, $\log (1+is)=is+s^2/2+O(s^3)$. Watson's lemma, or integration of these powers against $e^{-xs}$ with an exponentially small tail, yieldsIn particular the required first terms areThe two different endpoint scales come from the logarithmic singularity at zero and the simple zero of the amplitude at one.

The stationary points of the exponent are $t=1/3$ and $t=1$, because ${\varphi }^{\prime }(t)=(3t-1)(t-1)$. The first is a local maximum with $\varphi (1/3)=4/27$ and ${\varphi }^{\prime \prime }(1/3)=-2$; the second is a local minimum. Alsoso the endpoint overtakes the interior maximum only when $b>4/3$.

For fixed $0<b<1/3$, the upper endpoint is the unique maximum. Linearizing the exponent there gives the Laplace endpoint estimate $e^{\lambda \varphi (b)}/(\lambda {\varphi }^{\prime }(b))$. At $b=1/3$, only half of the local Gaussian integral is included. For $1/3<b\le 4/3$, the whole Gaussian near $1/3$ contributes. At $b=4/3$ an endpoint has the same exponential height, but its prefactor is ${\lambda }^{-1}$, smaller than the interior ${\lambda }^{-1/2}$ prefactor. For $b>4/3$, the endpoint has strictly greater height and dominates.

Consequently all fixed-$b$ regimes areThe estimate follows from Laplace's method; the tail toward negative infinity is exponentially harmless since the cubic tends to negative infinity.