>> What part of
>>
>> | U x V | = |U| * |V| * cos a
>>
>> do you *not* understand? :-P

> Would be a lot simpler if the damn stuff you have to use it in
> "understood" all that shit, natively. The problem I always run into is
> that you can find a perfectly comprehensible form of something some
> place, but it is only applicable is you a) do it by hand, or b) know how
> to derive some completely bloody different set of equations, that the
> damn computer will understand. Its like knowing, sort of, how to speak
> some obscure Chinese dialect, but then finding out that you need to
> *write* the information down in German, which for which the only work
> you know is the one applying to yourself, Dummkopf.
>
> Well, not exactly the same case, but if you don't have all the other
> stuff in between the two concepts, understanding what the math is doing
> in the "human" version won't get you any closer to understanding how the
> hell the computer needs to deal with it.

I'm having difficulty following what you're trying to say here.

> The original post in this, describing deriving the two equations needed
> for Mandelbrot, from the original non-computer usable one, is a perfect
> example. My reaction is, "Show the math, step by step, because WTF?" lol

Yeah. All the textbooks advanced enough to talk about complex dynamical 
systems assume that you *already* know everything there is to know about 
complex numbers (a much less advanced topic, comparatively speaking).

The first equation *is* useable in some software packages. But usually 
you need to expand out the real and imaginary parts seperately - a step 
which isn't mentioned anywhere and isn't described in any detail because 
it's presumed to be "obvious".



I assume that was a rhetorical question, but I'll answer anyway...

Start with Z^2 + C. Assume that Z = X + Y i, and C = A + B i. For that 
is what complex numbers are: a real part, and an imaginary part. 
(Explaining *that* is a whole /other/ essay...)

Substituting into the original formula, we have

   (X + Y i)^2 + (A + B i)

We can open the second pair of brackets without changing anything:

   (X + Y i)^2 + A + B i

Now, the first pair of brackets. The binomial theorem tells us the 
answer in canned form, but it's perfectly easy to do it all by hand. 
Since K^2 means the same thing as K * K, we can say

   (X + Y i) (X + Y i) + A + B i

Opening the first pair of brackets, we have

   X (X + Y i) + Y i (X + Y i) + A + B i

Opening the first pair of brackets, we get

   X X + X Y i + Y i (X + Y i) + A + B i

Obviously X X becomes X^2:

   X^2 + X Y i + Y i (X + Y i) + A + B i

Do the same sort of thing with the last pair of brackets:

   X^2 + X Y i + Y i X + Y i Y i + A + B i

The Y i X term is of course the same thing as X Y i:

   X^2 + X Y i + X Y i + Y i Y i + A + B i

Since there are two X Y i terms, we can gather them together:

   X^2 + 2 X Y i + Y i Y i + A + B i

The Y i Y i term is the same thing as Y Y i i. And that's obviously the 
same as Y^2 i^2. Now, BY DEFINITION, i^2 = -1. And Y^2 * -1 = -Y^2. So 
we have

   X^2 + 2 X Y i - Y^2 + A + B i

If we now put all the terms containing "i" into one pair of brackets, 
and everybody else in other pair of brackets, we get

   (X^2 - Y^2 + A) + (2 X Y i + B i)

This is still the same formula, we've just added some redundant brackets 
and rearranged the order a bit. Now, if we take everything in the left 
set of brackets and divide through i, we get

   (X^2 - Y^2 + A) + (2 X Y + B) i

This is still the same formula. (If we open the brackets, we get back 
what we had a few lines ago.) But now we have two purely real 
subexpressions, which a computer can calculate:

   X := X^2 - Y^2 + A
   Y := 2 X Y + B

QED.



I saw pictures in books of the "cubic Mandelbrot", and tried all sorts 
of modifications of the formula to make it cubic instead of quadratic. 
Of course, I didn't know about the binomial theorem yet. In order to get 
a cubic, you need to do this:

   (X + Yi)^3 + (A + Bi)
   (X + Yi)(X + Yi)(X + Yi) + A + Bi
   X(X + Yi)(X + Yi) + Yi(X + Yi)(X + Yi) + A + Bi
   X^2(X + Yi) + XYi(X + Yi) + Yi(X + Yi)(X + Yi) + A + Bi
   X^3 + X^2 Yi + XYi(X + Yi) + Yi(X + Yi)(X + Yi) + A + Bi
   X^3 + X^2 Yi + X^2 Yi + X Y^2 i^2 + Yi(X + Yi)(X + Yi) + A + Bi
   X^3 + X^2 Yi + X^2 Yi - X Y^2 + Yi(X + Yi)(X + Yi) + A + Bi
   X^3 + 2X^2 Yi - X Y^2 + Yi(X + Yi)(X + Yi) + A + Bi
   X^3 + 2X^2 Yi - X Y^2 + XYi(X + Yi) + Y^2 i^2(X + Yi) + A + Bi
   X^3 + 2X^2 Yi - X Y^2 + XYi(X + Yi) - Y^2(X + Yi) + A + Bi
   X^3 + 2X^2 Yi - X Y^2 + X^2 Yi + X Y^2 i^2 - Y^2(X + Yi) + A + Bi
   X^3 + 2X^2 Yi - X Y^2 + X^2 Yi - X Y^2 - Y^2(X + Yi) + A + Bi
   X^3 + 2X^2 Yi - X Y^2 + X^2 Yi - X Y^2 - X Y^2 + Y^2 i + A + Bi
   X^3 + 2X^2 Yi - X Y^2 + X^2 Yi - 2 X Y^2 + Y^2 i + A + Bi
   X^3 + 2X^2 Yi - 3 X Y^2 + X^2 Yi + Y^2 i + A + Bi
   X^3 + 3X^2 Yi - 3 X Y^2 + Y^2 i + A + Bi
   (X^3 - 3 X Y^2 + A) + (3 X^2 Y i + + Y^2 i + Bi)
   (X^3 - 3 X Y^2 + A) + (3 X^2 Y + Y^2 + B)i

   X := X^3 - 3 X Y^2 + A
   Y := 3 X^2 Y + Y^2 + B

As you can see, the result looks *nothing like* the quadratic case. If 
you didn't know what the derivation procedure was, you would have no 
idea why this is so.

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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