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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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