For convenience we write \(a\) instead of \(1.37\). Then \(1 / a\) is the root of the equation

\[

f(x)=0 \text { where } f(x)=a-\frac{1}{x} .

\]

We have \(f^{\prime}(x)=1 / x^{2}\), and therefore the Newton Method yields the iteration

\[

x_{n+1}=x_{n}-\frac{a-1 / x_{n}}{1 / x_{n}^{2}}=x_{n}-x_{n}^{2}\left(a-1 / x_{n}\right)=x_{n}\left(2-a x_{n}\right) .

\]

Note that the expression \(x_{n}\left(2-a x_{n}\right)\) can be evaluated on our defective calculator, since it only involves multiplication and subtraction.

Pick \(x_{0}\) reasonably close to \(1 / 1.37\). The choice \(x_{0}=1\) would work out fine, but I will start off a little closer, maybe by noting that \(1.37\) is about \(4 / 3\) so its reciprocal is about \(3 / 4\). Choose \(x_{0}=0.75\). We will report answers as they come out of the calculator.

We get \(x_{1}=x_{0}\left(2-1.37 x_{0}\right)=0.729375\). Thus \(x_{2}=0.729926589\), and \(x_{3}=0.729927007\). And it turns out that \(x_{4}=x_{3}\) to the 9 decimal places that my calculator shows. So we can be reasonably confident that \(1 / 1.37\) is equal to \(0.72992701\) to 8 decimal places.

I went out and spent almost \(\$ 9\) on a calculator with a " \(1 / x\) " button. It tells me that \(1 / 1.37\) is indeed equal to \(x_{3}\) to 9 decimal places. But it was not necessary to spend all that money. To check that \(0.72992701\) is correct to 8 decimal places, it is enough to check by multiplication that \((1.37)(0.729927005)<1\) and \((1.37)(0.729927015)>1\).

Note. In the early days of computing, the technique for finding \(1 / a\) described above was of great practical importance. Computers had addition, subtraction, and multiplication "hard-wired." But division was not hard-wired, and had to be done by software. Note that \(x / y=\) \(x(1 / y)\), so if multiplication is hard-wired, we can do division if we can find reciprocals. And Newton's Method was used to do that.