Newton's division-free reciprocal

MediumCalculus~5m

Some hardware has fast multipliers but no divider, so $1/a$ is computed by Newton's method on a function whose root is $1/a$ - and the resulting iteration must contain no division. Derive that iteration, then compute one step for $a=3$ starting from the guess $x_0 = 0.3$ . Give $x_1$ (to 4 decimals).

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Use $f(x)=1/x - a$ so the root is $1/a$ ; its derivative is $-1/x^2$ .
The Newton step simplifies to $x_{n+1}=x_n(2-a x_n)$ - no division. Plug $a=3$ , $x_0=0.3$ .

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0.33 (± 0.0001)

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Newton's division-free reciprocal

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