Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 19

Continuity. The function is continuous at $x_0$ when ${\lim }_{x\to x_0}f(x)=f(x_0)$. Equivalently, for every $\epsilon >0$ there exists $\delta >0$ such that $|x-x_0|<\delta$ within the domain of a function implies $|f(x)-f(x_0)|<\epsilon$; this is the epsilon-delta definition of continuity.

Differentiability. At an interior point $x_0$, the difference quotient must have a finite two-sided limit of a function:This defines differentiability and the derivative. At an endpoint one may instead state a one-sided definition relative to the interval. Differentiability implies continuity, because the increment is $h$ times a quotient with finite limit, but continuity alone does not guarantee differentiability.

As $x\to 0$ through nonzero values, $1/x^2\to +\infty$ and $e^{-1/x^2}\to 0=f(0)$. Thus $f$ is continuous at zero.

For the derivative at zero use its definition, rather than substituting into a formula obtained away from zero:Indeed, putting $t=1/|h|$ makes the absolute value $t{e}^{-t^2}$, which tends to zero since Gaussian decay dominates every power. Therefore $f$ is continuous and differentiable at zero, with $f^{\prime }(0)=0$. Away from zero, the chain rule gives $f^{\prime }(x)=2x^{-3}{e}^{-1/x^2}$.

For $x\ne 0$, induction giveswhere each $P_n$ is a polynomial. This recurrence follows from the product rule and $d(1/x)/dx=-1/x^2$. Since Gaussian decay dominates every power,These limits also describe the derivatives of the extension at zero. If $f^{(n)}(0)=0$, its next difference quotient is $P_n(1/h)e^{-1/h^2}/h$, which again tends to zero. Induction proves existence and continuity of every derivative, all zero there. This is a flat function, covered more generally by polynomial multiples of a Gaussian flat function.

Insert $u=-1/x^2$ into the exponential series $e^u={\sum }_{n=0}^{\infty }{u}^n/n!$. For every fixed $x\ne 0$, this gives the expansion with absolute convergenceThis is a Laurent series in inverse powers of $x$, as requested. It is not a Maclaurin series about zero: that series has every coefficient zero and cannot equal $f$ on a punctured neighborhood of zero. There is no contradiction, since the inverse-power expansion does not apply at $x=0$.

The flat function $f$ is an even function, has its unique zero and global minimum at the origin, and approaches the horizontal asymptote $1$ as $x\to \pm \infty$. It decreases for negative $x$ and increases for positive $x$, meeting the origin with a horizontal tangent line.

Its derivative is an odd function: $f^{\prime }(x)=2x^{-3}{e}^{-1/x^2}$ off zero, with $f^{\prime }(0)=0$. It is negative to the left and positive to the right, tends to zero both at the origin and at infinity, and has one negative trough and one positive crest. Indeed $f^{\prime \prime }(x)=2e^{-1/x^2}(2-3x^2)/x^6$ away from zero, so the change in slope of $f^{\prime }$ occurs once on each side. The exact crest and trough coordinates are not needed for the sketch.

A constant sequence must satisfy the fixed point equation $y_{\ast }=\sqrt{2+y_{\ast }}$. The square root forces $y_{\ast }\ge 0$. Squaring gives $(y_{\ast }-2)(y_{\ast }+1)=0$, so the negative algebraic root is inadmissible. HenceStarting at $y_0=2$ produces the constant sequence $2,2,\dots$.

The cobweb plot uses the curve $v=\sqrt{2+u}$ and the diagonal $v=u$. From a point on the diagonal, a vertical step to the curve computes the next iterate; a horizontal step returns its value to the diagonal.

For $0\le u<2$, $(2+u)-u^2=(2-u)(1+u)>0$, so $u<\sqrt{2+u}<2$. Thus a start below $2$ gives a strictly increasing sequence bounded above by $2$. For $u>2$ the inequalities reverse to $2<\sqrt{2+u}<u$, giving a strictly decreasing sequence bounded below by $2$. Both types of staircase approach the same intersection.

The bounded monotone sequence theorem proves existence of a limit, and continuity then makes it a nonnegative fixed point. Part (i) givesThis proves the global convergence of a square-root iteration; the graph below illustrates both cases without relying on a finite numerical sample as proof.