Seed values for estimating square roots.

[aarciga]

This is taken from Wikipedia:

Many of the methods for calculating square roots of a positive real number S require an initial seed value. If the initial value is too far from the actual square root, the calculation will be slowed down. It is therefore useful to have a rough estimate, which may be very inaccurate but easy to calculate. If S ≥ 1, let D be the number of digits to the left of the decimal point. If S < 1, let D be the negative of the number of zeros to the immediate right of the decimal point. Then the rough estimation is this:

If D is odd, D = 2n + 1, then use \sqrt{S}\approx2\cdot10^{n}
If D is even, D = 2n + 2, then use \sqrt{S}\approx6\cdot10^{n}

Two and six are used because

\sqrt{\sqrt{1\cdot10}}=\sqrt[4]{10}\approx2 and \sqrt{\sqrt{10\cdot100}}=\sqrt[4]{1000}\approx6

Im just wondering if anyone could elaborate further as to why 2 and 6 are used.

I see why i works in that it gives you an estimate with the same number of digits as \sqrt{S}

but I am confused as to what 4th root of 1*10 and 10*100 represent. where did those numbers come from?

any clarification is appreciated.

 

[LCKurtz]

This is taken from Wikipedia:

Many of the methods for calculating square roots of a positive real number S require an initial seed value. If the initial value is too far from the actual square root, the calculation will be slowed down. It is therefore useful to have a rough estimate, which may be very inaccurate but easy to calculate. If S ≥ 1, let D be the number of digits to the left of the decimal point. If S < 1, let D be the negative of the number of zeros to the immediate right of the decimal point. Then the rough estimation is this:

If D is odd, D = 2n + 1, then use \sqrt{S}\approx2\cdot10^{n}
If D is even, D = 2n + 2, then use \sqrt{S}\approx6\cdot10^{n}

Two and six are used because

\sqrt{\sqrt{1\cdot10}}=\sqrt[4]{10}\approx2 and \sqrt{\sqrt{10\cdot100}}=\sqrt[4]{1000}\approx6

Im just wondering if anyone could elaborate further as to why 2 and 6 are used.

I see why i works in that it gives you an estimate with the same number of digits as \sqrt{S}

but I am confused as to what 4th root of 1*10 and 10*100 represent. where did those numbers come from?

any clarification is appreciated.

I'm just guessing here, but it might come from something like this for D even:

10^{2n+2} = 10^210^{2n}\approx 10^{\frac 3 2}10^{2n}=(10^{\frac 3 4}10^n)^2\approx (6\cdot 10^n)^2

which gives a nice round number to start the iteration.