Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 19

The Newton root-finding iteration follows the current tangent line to its horizontal-axis intersection. Since $f^{\prime }(x)=2x$, it givesA positive initial guess is appropriate for the positive square root; negative starting values instead target the negative root.

Write $r=\sqrt{a}>0$ and subtract it from the iteration:This is an Exact Newton error identity for a quadratic. If $x_n>r$, then $x_n>0$ and $e_n>0$, so $e_{n+1}>0$. Induction from $x_0>r$ proves $x_n>r$ for every $n$. Also $x_{n+1}-x_n=(a-x_n^2)/(2x_n)<0$, so the iterates decrease while remaining above the root. Their limit must satisfy $L=(L+a/L)/2$, hence $L=r$. The squared error numerator explains the ensuing quadratic convergence.

The supplied bracket gives $0<e_0=1.5-\sqrt{2}<0.1$. Part (b) ensures $x_n>\sqrt{2}>1.4$, so each error satisfiesWriting $d_n=e_n/2.8$ gives $d_{n+1}<d_n^2$ and $d_0<0.1/2.8=1/28$. After three steps the exponent has doubled three times:This rigorous error bound uses only the given bracket, not advance knowledge of the digits of $\sqrt{2}$.

The actual rational iterates are $x_1=17/12$, $x_2=577/408$, and $x_3=665857/470832\approx 1.414213562374690$. They illustrate the bound, but the proof comes from the exact error identity and the positive lower bound on the denominators.

The function $g(x)=1/(x\ln x)$ is positive and decreasing for $x\ge 2$, since $g^{\prime }(x)=-(\ln x+1)/(x^2{\ln }^{2}x)<0$. The integral test for convergence therefore applies. Its comparison improper integral isConsequently $\sum _{n=2}^{\infty }1/(n\ln n)$ diverges.

The function $g(x)=1/(x{\ln }^{2}x)$ is likewise positive and decreasing on $[2,\infty )$, with $g^{\prime }(x)=-(\ln x+2)/(x^2{\ln }^{3}x)<0$. NowThe integral test for convergence shows that $\sum _{n=2}^{\infty }1/(n{\ln }^{2}n)$ converges. This comparison integral determines convergence, not the numerical value of the series itself.