Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability / 5

The chain rule givesThe exponential function never vanishes, so all critical points areIn particular $f^{\prime }(0)=0$.

Use the Taylor series $e^u=1+u+u^2/2+\cdots$ with $u=x^3-x^5$. Since $u$ begins at cubic order and $u^2$ begins at sixth order, through cubic terms the Taylor series isThe linear and quadratic coefficients are zero. The stronger remainder $O(x^5)$, in big O notation, follows because the fourth-order term is also zero; the next term is $-x^5$.

Near zero, $x^3-x^5=x^3(1-x^2)$ has the sign of $x$, so $f(x)<1$ just to the left and $f(x)>1$ just to the right. Thus the critical point is neither a local minimum nor a local maximum. Moreover the Taylor series gives $f^{\prime \prime }(x)=6x+O(x^3)$, which changes sign at zero. ThereforeThis is the point called a “saddle point” in the question; stationary inflection point specifies the one-variable meaning.

Solving the linear ordinary differential equation and imposing the initial condition givesAs $\lambda >0$, the requested limit is

Let $h=1/N$, $x_j=jh$, and let $y_j$ approximate $y(x_j)$. The forward Euler method gives the difference equationThis geometric sequence has solutionFor fixed $\lambda$, the limit as $N\to \infty$ is $2e^{-\lambda }$, as required by the exact solution.

The amplification factor is $R=1-\lambda /N$. To reproduce decay, the forward Euler method requires $|R|<1$, equivalentlyAt $N=\lambda /2$, $R=-1$ and the approximations alternate with constant magnitude $2$; for $N<\lambda /2$, $|R|>1$ and the magnitude grows. This is the condition of absolute stability for this decaying differential equation.

Stability is only a minimum requirement. Values just inside the bound can still give a poor endpoint approximation; resolving the decay accurately generally requires a smaller step size. If positivity at every step is desired, the stronger condition $N\ge \lambda$ ensures $R\ge 0$. The figure shows the difference between decay of the exact solution and numerical oscillation.

For positive edge lengths the volume is $V=abc$ and the surface area constraint is $2(ab+ac+bc)=S$. Introduce a Lagrange multiplier $\mu$:The stationary point equations areMultiplying these by $a,b,c$, respectively, and equating their left sides gives $\mu c(a-b)=0$ and $\mu b(a-c)=0$. Because all edge lengths are positive, the equations imply $\mu >0$; hence $a=b=c$. The constraint gives $6a^2=S$.

To establish a global maximum, rather than only a stationary point, apply the arithmetic-geometric mean inequality to $ab,ac,bc$:Equality holds exactly when $ab=ac=bc$, or $a=b=c$. Therefore