|
Senuka Harbinger
A-Life, one bit at a time
Join date: 24 Oct 2005
Posts: 491
|
04-12-2007 02:52
I'm using some numbers from a fractal generated by hand and I'm trying to get the recursive function for the fractal, but I'm being thrown for a loop as to how to construct the equation. t is my iteration number (time) in this case it allows for both negative and positive values X=2^t value chart for Xn and Yn X_____Y 4.000|189.12500 2.000|187.62500 1.000|188.37500 0.500|188.00000 0.250|188.18750 0.125|188.09375
I know that Yn - Y(n-1) follows this pattern: -1.5 .75 -.375 .1875 -.09375 that is, the difference between two y values is the negative half the previous y value (.75 is -1.5/-2, -.375 is .75/-2) I'm just at a blank as to how to construct this into an equation which I can then feed into my script.
_____________________
My SLExchange shopTypos are forgiven; desecrating the english language with reckless abandon and necrophilic acts is not. The function is working perfectly fine. It's just not working the way you wanted it to work.
|
|
Ged Larsen
thwarted by quaternions
Join date: 4 Dec 2006
Posts: 294
|
04-12-2007 06:49
Sorry I can't help you -- don't know anything about fractals -- but the puzzle seems interesting. For a non-fractal knowledgeable person, could you explain just what you are trying to solve for?
Your value chart seems to be for values of t ranging from 2 to -3, for the X values anyhow. Do you mean Xt? Or is n something different? What are the Y values coming from?
Are you trying to come up with an equationthat will give you what you call Yn - Y(n-1)?
If so, your alternating + and - signs will mean that you'll want to start that equation with:
(-1)^n times something
The "something" seems to be exponentially decaying by a factor of 2, so: divide by 2^n
So, if I even understand what you're trying to get, would it be:
3.0 * [ (-1)^n ] / [ 2^n ]
That would seem to give you your Yn - Y(n-1), if ranging n from 1 incrementing upwards by 1 each time.
Or am I lost?
_____________________
- LoopRez, flexi prim skirt generating tool - LinkRez, a necklace chain generator
|
|
Senuka Harbinger
A-Life, one bit at a time
Join date: 24 Oct 2005
Posts: 491
|
04-12-2007 07:03
From: Ged Larsen Sorry I can't help you -- don't know anything about fractals -- but the puzzle seems interesting. For a non-fractal knowledgeable person, could you explain just what you are trying to solve for? The raw numbers for Y are the position along the X axis for each piece in the iteration, and the values for X is a multiplier for the size of the prim I'm using. both the X and Z coordinates for each prim follow this rule of behaviour (the way the Y value changes), with only a constant difference between them. The resulting fractal resembles a butterlfly of sorts. From: someone Your value chart seems to be for values of t ranging from 2 to -3, for the X values anyhow. Do you mean Xt? Or is n something different? What are the Y values coming from?
t is actually ranging from 2 to -3. X will always be a power of 2 (negative powers are decimals, positive powers will resemble the familiar 1,2,4,8,16,32,64 ... sequence, negative values for t produces x being 1/2, 1/4, 1/8, 1/16, etc...). The Y values is the x coordinate for each piece. From: someone Are you trying to come up with an equation that will give you what you call Yn - Y(n-1)?
no, this is just one of the steps I recall from my classes in solving a geometric formula. I could be wrong since it's been years since my last mathematics class of any kind. From: someone If so, your alternating + and - signs will mean that you'll want to start that equation with:
(-1)^n times something
The "something" seems to be exponentially decaying by a factor of 2, so: divide by 2^n
So, if I even understand what you're trying to get, would it be:
3.0 * [ (-1)^n ] / [ 2^n ]
That would seem to give you your Yn - Y(n-1), if ranging n from 1 incrementing upwards by 1 each time.
Or am I lost?
You're essentially on the right track. ultimately I need an equation that when I put in the X value for the T chart it spits out the Y value. I've a feeling it's going to be something like: Y=a*-2^(b*2^t)+c which I don't remember how to set up the series of equations to solve for a,b, and c respectively since it's a nonlinear (exponential) equation; a simple matrix solution will not render a correct result
_____________________
My SLExchange shopTypos are forgiven; desecrating the english language with reckless abandon and necrophilic acts is not. The function is working perfectly fine. It's just not working the way you wanted it to work.
|
|
Ged Larsen
thwarted by quaternions
Join date: 4 Dec 2006
Posts: 294
|
04-12-2007 07:44
I see, and nope, this is beyond my abilities  Good luck!
_____________________
- LoopRez, flexi prim skirt generating tool - LinkRez, a necklace chain generator
|
|
Ee Maculate
Owner of Fourmile Castle
Join date: 11 Jan 2007
Posts: 919
|
04-12-2007 08:12
OK.. think I've transferred this from back of envelope to forum correctly... treating your first term as term 0 (i.e. y(0) ) then what you have is y(n) = 188.125 + (-1)^n * (1/2)^n and x(n) = 4 * (1/2)^n giving y(n) = 188.125 + (-1)^n * x(n)/4 This still involves that (-1)^n term which is a pain, but you can retrieve n from x(n) by taking logs: n = log( x(n)/4)) / log (1/2) but since you only care if n is even or odd you can use log(x(n)) / log (1/2) So finally your equation is y = 188.125 + (-1)^[log(x)/log(0.5)] * 0.25* x Plug in your x value and it should correctly give you your y value (unless I've made a mistake somewhere). I'll let you test it to make sure it works! 
|
|
Ee Maculate
Owner of Fourmile Castle
Join date: 11 Jan 2007
Posts: 919
|
04-12-2007 08:20
P.S. Technically the problem you were solving was a linear one....
2*y(n) - y(n-1) - y(n-2) = 0
x(n) -0.5 x(n-1) =0
|
|
Senuka Harbinger
A-Life, one bit at a time
Join date: 24 Oct 2005
Posts: 491
|
04-12-2007 09:26
From: Ee Maculate OK.. think I've transferred this from back of envelope to forum correctly... treating your first term as term 0 (i.e. y(0) ) then what you have is y(n) = 188.125 + (-1)^n * (1/2)^n and x(n) = 4 * (1/2)^n giving y(n) = 188.125 + (-1)^n * x(n)/4 This still involves that (-1)^n term which is a pain, but you can retrieve n from x(n) by taking logs: n = log( x(n)/4)) / log (1/2) but since you only care if n is even or odd you can use log(x(n)) / log (1/2) So finally your equation is y = 188.125 + (-1)^[log(x)/log(0.5)] * 0.25* x Plug in your x value and it should correctly give you your y value (unless I've made a mistake somewhere). I'll let you test it to make sure it works!  I don't have acess to SL right now, but it looks like it's working from some quick numbers I plugged in. thanks a heap 
_____________________
My SLExchange shopTypos are forgiven; desecrating the english language with reckless abandon and necrophilic acts is not. The function is working perfectly fine. It's just not working the way you wanted it to work.
|
|
Ee Maculate
Owner of Fourmile Castle
Join date: 11 Jan 2007
Posts: 919
|
04-12-2007 11:22
No probs. 
|