NetNews Usenet Archive 1992 #16/NN_1992_16.iso/spool/comp/programm/2048

Text File | 1992-07-21 | 1.9 KB | 54 lines

Newsgroups: comp.programming
Path: sparky!uunet!mcsun!sun4nl!alchemy!jeroen
From: [e-mail] (Jeroen Fokker)
Subject: Re: Calculating a cube root
Message-ID: <[e-mail]>
Date: Wed, 22 Jul 1992 08:16:58 GMT
References: <[e-mail]> <[e-mail]> <[e-mail]> <[e-mail]> <[e-mail]> <[e-mail]>
Organization: Utrecht University, Dept. of Computer Science
Lines: 43

This thread is floating away from the original subject a bit,
but let me add just one posting to it.

I suggested the following code for "approximate-equal-to":

>>>> return( a-b<EPSILON && b-a<EPSILON );

Dave Schaumann ([e-mail]) first claimed that this
was incorrect, then retracts his claim, but remarks:

> After careful observation, I see that your code is correct
> (tho not at all clear, IMHO).  I would've greatly preferred
>
>       #define abs(x) ((x) > 0 ? (x) : (-x))
>       return abs(a-b) < EPSILON ;

This may be more clear, but alas, it is not correct.
A classic pitfall with macroexpansion: abs(1-1) yields
-2 this way.

Now for some obligatory programming:
you can apply Newton's method to find the inverse of _any_
function f by iterating:

     b - (f(b)-x)/f'(b)

where f' is the derivative of f.
The starting value of b depends on the function at hand.
For monotonic f, x will do most times.
Of course, f should be differentiable.
If f is not monotonic, the iteration may not terminate.

Formulas for sqareroot and cuberoot are special cases:

For sqrt,   iterate  b - (b^2-x)/2b,   that is (b+x/b)/2
For cubert, iterate  b - (b^3-x)/3b^2, that is (2b+x/b^2)/3
For arcsin, iterate  b - (sin(b)-x)/cos(b).

--
Jeroen Fokker                                 |  [e-mail]
dept.of Computer Science, Utrecht University  |  tel.[phone]
PObox 80089, 3508TB Utrecht, the Netherlands  |  fax.[phone]