## 170 – Homework 2

This set is due Friday, March 8, at the beginning of lecture.

• Solve problem 2 from Chapter 7. In each case, use the bisection method to approximate within $0.01$ a value of $x$ for which we have $f(x)=0$. Recall that in the bisection method, at each stage we have an interval ${}[a,b]$ and we know that $a and $f(a)<0 (or $f(b)<0). We let $c$ be the midpoint of the interval. If $f(c)=0$, we let $x=c$ and we are done. More likely, either $f(c)<0$, and we have $c, our new interval is ${}[c,b]$, and we iterate the process, or $0, and we have $a, our new interval is ${}[a,c]$, and we iterate the process.
• Solve problem 3 from Chapter 7. As before, approximate $x$ within $0.01$.
• Find the first few convergents to $\sqrt5$, and use them to find $\sqrt5$ within $0.0001$. Recall that the convergents of $\sqrt5$ are obtained by the following process:

Define $a_0,a_1,a_2,\dots$ so that: $a_0$ is the largest integer below $\sqrt5$. Let $t_1=1/(\sqrt5-a_0)$, and let $a_1$ be the largest integer below $t_1$. Let $t_2=1/(t_1-a_1)$, and let $a_2$ be the largest integer below $t_2$. Let $t_3=1/(t_2-a_2)$, and let $a_3$ be the largest integer below $t_3$, etc.

The first convergent to $\sqrt5$ is the number $a_0$. The second convergent is $\displaystyle a_0+\frac1{a_1}$. The third convergent is $\displaystyle a_0+\frac1{a_1+\frac1{a_2}}$. Etc.

The number $\sqrt5$ is sandwiched between the convergents, in the sense that it is larger than the first, smaller than the second, larger than the third, smaller than the fourth, etc.

• Approximate $\sqrt2$ following the following algorithm: Let $x_0$ be an arbitrary number that you choose, presumably not too far from $\sqrt2$. Given $x_n$, we define a new approximation $x_{n+1}$ by the formula $\displaystyle x_{n+1}=\frac{1}{x_n}+\frac{x_n}{2}.$ Check with the help a calculator that these numbers approach $\sqrt2$ very quickly. Use this to find the first $10$ digits of $\sqrt2$.
• Extra credit problem: Why does the algorithm of the last problem work?