Modern Electronics (was Re: List charter mods & headcount... ;

So, you use Newton-Raphson; he uses binomial. Nothing wrong with either one. (Though I'm not familiar with how to use binomial expansions to find square roots.) Today, it is, when a computer can cough up several thousand digits in the time it takes you to work out three by hand. Knowing how to do it, though, involves learning enough of the theory behind it to stand you in good stead in other cases, though. (Assuming you actually understand it. Learning how to do it by rote, without understanding why, _is_ pretty pointless.) There is, but it is significantly more elaborate to perform. Longhand square root extraction is based on (X+d)^2 = X^2 + 2Xd + d^2. For example, to work out sqrt(15.5) (in decimal), we start with the first digit. (If you can't get the first digit right without assistance, you have no business trying something this elaborate! :-) ___3.___________ ) 15.50 00 00 00 9 6 -- 6 50 Now, we have written 15.5 = (3^2 + 2?3?d + d^2), with d to be determined. The 6.50 is 2?3?d + d^2. The 6 sitting off at the right is what I saw described as the memorandum column; it's twice the root-so-far. Now, we look for a digit X such that 6X * X <= 650 (this is 2?3?d+d^2). We want the largest digit that fits; 9 works, so we fill it in, and write in and subtract off 69*9: ___3._9_________ ) 15.50 00 00 00 9 6 -- 6 50 6 21 78 ---- 29 00 This gives us 15.5 = 3.9^2 + .29, and we need to write .29 as 2*3.9*d+d^2, for some value of d in [0,.1). The mechanics of this is that we find X such that 78X * X < 2900. 783*3=2349, but 784*4=3136, so we use 3 (that is, we've decided d is in [0.3,0.4)): ___3._9__3______ ) 15.50 00 00 00 9 6 -- 6 50 6 21 78 ---- 29 00 23 49 786 ----- 5 51 00 Et cetera. Und so weiter. Og s? videre. Usually, you can get the next digit right by looking at just the high digit or two of the memorandum column; only when it's close to the edge may you have to try multiplying out different choices in full. In the next position, for example, I'd look and say "55/7 = 7+, let's try 7", and indeed, 7 works (barely - the remainder is so low the next two digits of the root are zero). You can do the same thing for cube roots, writing (N+d)^3=N^3+3N^2d+3Nd^2+d^3, but the arithmetic with the memorandum column becomes a good deal messier. Well...for the first few digits, I'd probably do something ad-hoc, bisection and/or cut-and-try. If I needed more precision, I'd likely use Newton-Raphson, but I'd truncate the quotient to about the number of places that I know are good, since I also know that N-R is fully capable of recovering from a more or less arbitrary guess, so the lost digits will, at most, slow convergence - and I care less about the number of N-R steps required than I do about the amount of work I need to do during them. If I needed a high-precision root, of course the last steps would be N-R. But under no-tools circumstances, it's unlikely the root is needed to more than about three significant digits anyway. /~\ The ASCII der Mouse \ / Ribbon Campaign X Against HTML mouse(a)rodents.montreal.qc.ca / \ Email! 7D C8 61 52 5D E7 2D 39 4E F1 31 3E E8 B3 27 4B