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Joined: 19 May 2009
Posts: 520

Posted: 30 Mar 2011 11:04:39 pm    Post subject:

Okee dokey then, so I have been working for the past two days on forming equations for finite sets of data. For example, if I have {0,-33,i,27,6}, I would want to find an equation where if x=0, then the result is 0, and if x=1, the result is -33, et cetera. I have a method that accomplishes this and I have explored it much in the past few days, but I get the feeling there is either an easier way or this method is just something I haven't gotten to, yet. Anyway, here is what I do...
First, to make things simpler, I want to make two sets (one real and the other imaginary), so I end up with:
{0,-33,0,27,6}+i{0,0,1,0,0}. I want to tackle one at a time, so I arrange the first list into a column like so:

Now from there, you make a second column that is the difference of Un+1 and Un:

-33   -33-0=-33
  0   0--33=33
 27    27-0=27
  6    6-27=-21

And from then you add another column using the same idea until you reach one element in the column:
[attachment=3458:Example 1, Step 1.gif]
Now from there, fill the last column space with 30. From there, you can work your way backwards to fill in the rest of the columns:
[attachment=3457:Example 1, Step 2.gif]
At this point, we should denote the column spaces in this manner:
a-First column space (the original numbers)
b-Second column space
c-Third column space
d-Fourth column space
e-Last column space

So, looking at the matrix, e follows the equation y=30. Now we move to d which must be -162+sum(e,n,0,x) or -162+30x. c is going to be 300+sum(d,x,0,n) which can be represented as 300-162x+30(x2+x)/2). This can be simplified, but for the sake of making the concept clear, I will not simplify it. As we continue with the last two, we get:

And that is the equation for the real part of the original set of numbers. Applying the same method, we get for the imaginary part:

And now we simply need to combine the two equations to get the ridiculously large equation:

Anyway, that is the method I have used and explored and that is the long, tedious way of obtaining results. The faster way will require the use or knowledge of Pascal's Triangle and by knowledge... hehehe, this ties into some of my other explorations that I haven't gotten into here...
For a quick way to obtain the diagonals, let us represent those as {a,z,y,x,w,...} and we can look at it in this form:


[[a l m n o ...
  b z
  c   y
  d     x
  e       w
...         ]]

If you notice those coefficients, you will note the relationship to Pascal's Triangle. There is another relationship you may not have noticed in the first example. Firstly, it should be pretty clear that taking the values in row 1, we have the coefficients required for the end equation. Those coefficients go along with some polynomial and those polynomials are the equations to the diagonals of Pascal's Triangle. So, if the first diagonal (all 1s) is represented as P0, then the equations are as follows:
The pattern is fairly easy:
So P4 should be read as the sum of the sum of the sum of the sum of 1. SO using the above notation (in the code box), the equation that hits {a,b,c,d,e,...} would be a*P0+l*P1+m*P2+n*P3+o*P4...

So it would be nice to have {a,l,m,n,o,...}, right? Well that, too, makes use of Px as well as an alternating negative sign. As examples:
If you notice the coefficients there, you will note that:
a= uses 0 for its coefficients
l= uses P0 for its coefficients
m= uses P1 for its coefficients
n= uses P2 for its coefficients
o= uses P3 for its coefficients

So now with that in mind, we can model this simple set of data: {π,-3,6,2}



a=a         = π        = π
z=b-a       = -3-π     = -3-π
y=c-2b+a    = 6+6+π    = 12+π
x=d-3c+3b-a = 2-18-9-π = -25-π


a=a         = π                  = π
l=z-y+x     = -3-π-12-π-25-π   = -40-3π
m=y-2x      = 12+π+50+2π         = 62+3π
n=x         = -25-π              = -25-π

And finally the equation:
Which turns to:

So anywho, there are probably a few things I missed and maybe a few mistakes, but is there an easier approach to this?
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