POV-Ray : Newsgroups : povray.binaries.images : It's been a while! : Re: It's been a while! Server Time
18 Aug 2024 00:21:56 EDT (-0400)
  Re: It's been a while!  
From: Oldstench
Date: 14 Jul 2001 03:36:21
Message: <7580lt033mpl2t3hhb6v5qvvqohfknqbbi@4ax.com>
"Mike Metheny" <lon### [at] vtedu> wrote:

><<
>I don't get it.  Wouldn't a 2d function (real+imaginary) have a
>tangent-plane, not a tangent line?  And the intersection of that with
>another plane is a line.  You'd want a point...  explain this to me.
>>>
>
>Okay; take the function that generates this image:
>
>we want the roots of z^4 + 4.  so we need z^4 = -4
>
>it's to the 4th power, so it has 4 roots.  we can directly solve for the
>roots by doing (-4)^(1/4) and using the formula:
>
>r^(1/n) * [cos( (theta + 2*pi*k) / n ) + i * sin( (theta + 2*pi*k) / n ) ]
>
>where r is the radius of the complex number (here we're using -4 as our
>complex number; it has no imaginary part so the radius is just 4) and theta
>is the angle from the +x axis around to that point in the complex plane( a
>point at 1 + i would have a theta of pi/4, for our example, our point -4 +
>0i we have a theta of pi).
>
>n is the number of roots, 4, and k ranges from 0 to n-1.  this gives us all

>
>You're right in that newton's method is a slope method, but don't get too
>caught up on the visualizations.  what we do with NM is guess a root, and
>then subtract from that guess the function evaluated at that guess divided
>by the derivative of the function evaluated at that guess, or
>
>z - f(z) / f'(z) gives us our new guess.
>
>this will work just fine to find complex roots.  lets say we guess 2 + i as
>a root, for instance.
>
>f(z) = z^4 + 4 = (2+i)^4 + 4 = -3 + 24i
>f'(z) = 4*z^3 = 8 + 4i
>
>(2+i) - (-3+24i)/(8+4i) = 1.1 - 1.55i
>
>Making this our new guess, our next iteration gives us approximately 0.965 -
>1.20i, our third iteration gives 0.968 - 1.02i, etc etc.  You see that we
>are converging to one of the roots, namely 1 - i.
>
>However, if you were to start with something like 2 as your guess, your new
>approximation would diverge; and bounce around all over the place like crazy
>with this function; since there are no real roots.  So NM works just fine to
>find complex roots, but you have to start with a complex number in the first
>place, otherwise you will not converge to a complex number.
>
>I think you're just getting a bit hung up on the graphical description of
>newton's method (pick a point, draw a tangent line from that point to the
>axis, use this as your new x value, and repeat).

Uh...yer like smart and stuff...  jeez my head hurts just looking at this.
:)

Oldstench................


Post a reply to this message

Copyright 2003-2023 Persistence of Vision Raytracer Pty. Ltd.