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In the wiki page of square root calculation, I see an additional term $4Nd$ in Bakhshali approximation than in the Taylor series with 3 terms. I do not see how Bakhshali approximation is "equivalent to two iterations of the Babylonian method beginning with N." To me, "two iterations of the Babylonian method beginning with N." will give exactly the Taylor series with 3 terms.

Can anybody please help clarify this? How to understand the term $4Nd$ added in Bakhshali approximation and why does it make the convergence faster than 3-term Taylor series?

Thanks!

EDIT

OK. I got it. Just being really stupid earlier. Actually quite simple.

Applying once Babylonian method beginning with $N$, one can get $$\sqrt{N^2+d} \approx \frac{1}{2}(N+\frac{N^2+d}{N})=N+\frac{d}{2N}$$

Then apply again Babylonian method beginning with this new result $N+\frac{d}{2N}$, one can get $$\sqrt{N^2+d} \approx \frac{1}{2}((N+\frac{d}{2N})+\frac{N^2+d}{(N+\frac{d}{2N})})=N+\frac{d}{2N}-\frac{d^2}{8N^3+4Nd}$$

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I do not see how Bakhshali approximation is "equivalent to two iterations of the Babylonian method beginning with N."

This is what two applications of the Babylonian method for $\sqrt{s}$ with seed $n$ looks like:

$$\frac12\left(\frac12\left(n+\frac{s}{n}\right)+\frac{s}{\frac12\left(n+\frac{s}{n}\right)}\right)$$

This is what the Bakhshali method looks like, after expanding out the intermediate quantities given in the wiki article:

$$\left(n+\frac{s-n^2}{2n}\right)-\frac{\left(\frac{s-n^2}{2n}\right)^2}{2\left(n+\frac{s-n^2}{2n}\right)}$$

Showing that these two simplify to the exact same rational expression is left as an exercise in algebraic manipulation.

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