NetNews Usenet Archive 1992 #27/NN_1992_27.iso/spool/sci/math/15321

Text File | 1992-11-20 | 1.8 KB | 39 lines

Newsgroups: sci.math
Path: sparky!uunet!cs.utexas.edu!sun-barr!ames!sgi!wdl1!wdl39!mab
From: [e-mail] (Mark A Biggar)
Subject: Re: Square root
Message-ID: <[e-mail]>
Sender: [e-mail]
Organization: Loral Western Development Labs
References: <n9146070.722056383@waldorf> <[e-mail]>
Date: Fri, 20 Nov 1992 22:59:52 GMT
Lines: 27

In article <[e-mail]> [e-mail] (John C. Baez) writes:
>In article <n9146070.722056383@waldorf> [e-mail] (Jacob C Kesinger) writes:
>>There was a previous post about ways to find the square root of a number...
>>Here's my contribution:
>>Find the Taylor series about a (I used a=1, for simplicity)
>>(x-1)^.5=1+ (x-1)/2- (x-1)^2/4 +3(x-1)^3/8 -......
>>This is nasty, messy, ugly, and very likely wrong.  However, it IS
>>a method (*grin*)
>It'll work, assuming you got the right coefficients in your Taylor
>series, as long as |x-1| < 1.  It's pretty good if you just want a few
>digits and x is near 1.
>The best use of this formula is to impress undergraduates by working out
>in your head, for example, that the square root of 1.23 is about 1.115.

This is also not bad for generating the mantissa of an initial guess
when computing square roots of binary floating point numbers.  A binary
floating point number (e.g. IEEE) has a mantissa that is .5<=m<1 which meets
the above inequality.  After renormalizing to make the floating point exponent
even you now have a mantissa that is .25<=m<1 and the inequaltity still
holds, so use the above 3 or 4 terms to get the mantissa of the guess,
then run 2 to 4 interations of the Newton-Raston formula to firm up the
answer and your done.

--
Mark Biggar
[e-mail]