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A calculator is defective: it can only add, subtract, and multiply. Use the equation $1 / x=1.37$, the Newton Method, and the defective calculator to find $1 / 1.37$ correct to 8 decimal places.
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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.
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