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Re: Square root of six
Posted:
Aug 27, 2012 11:00 PM
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On Aug 27, 2012, at 1:11 PM, Peter Duveen <pduveen@yahoo.com> wrote:
> I think Newton's method requires calculus to strictly justify it, at least according to a wikipedia article on it.
Actually, the newton method for square roots can be justified without calculus, or algebra, especially the Babylonian version of it. It traps the square root as well. It can be shown that the iterations always straddle the sqrt and yet always fall within bounds of the previous iterations. It requires arithmetic and a number line.
Find the square root of N...
1. Start with a guess of sqrt(N), call it x_0.
2. If x_0 < sqrt(N) then N/x_0 > sqrt(N), or if x_0 > sqrt(N) then N/x_0 < sqrt(N). In other words, sqrt(N) lies between x_0 and N/x_0.
3. Calculate the average of x_0 and N/x_0. Call this x_1.
4. if x_1 < sqrt(N) then N/x_1 > sqrt(N), or if x_1 > sqrt(N) then N/x_1 < sqrt(N). In other words, sqrt(N) lies between x_1 and N/x_1.
5. Also, x_1 and N/x_1 lies between x_0 and N/x_0.
6. And so on...
Essentially, each pair of x_n and N/x_n enclose the sqrt and are themselves enclosed by the previous pair of x_n and N/x_n.
Note: I generally consider expressions involving radicals to be algebraic. Not technically, but pedagogically. The above method is very easy to see on a number line, as long as you understand that N / sqrt(N) = sqrt(N). Square roots (other than perfect squares) are generally not dealt with till algebra, but for the sake of this discussion we can assume that we are doing a deeper dive than usual during arithmetic. Which is very possible.
Bob Hansen
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