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Author Topic: An old formula revised  (Read 14937 times)
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laser blaster
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« Reply #105 on: March 28, 2014, 08:36:26 PM »

I've been playing around with the fragmentarium version, and it's a very beautiful fractal indeed! Definitely the closest thing to the holy grail yet (visually). There are still plenty of stretched areas, but it's a much nicer looking, more orderly stretch than the whipped cream effect of other formulas. Did you specifically construct the formula to create grail-like features, or did you just stumble onto it by chance? Do you happen to have any insight as to how the formula creates the detailed bulb-on-bulb structure for power 2, when other formulas fail to do so? I'm very curious about this.
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M Benesi
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« Reply #106 on: April 11, 2014, 11:03:17 PM »

  Hi LB,

  I don't know if I believe in chance, but I certainly don't have the mathematical skills to construct the formula with a preconceived purpose in mind. 

  I was messing around with various ideas, and they all came to fruition with that formula.  Basically, with "luck" (a greater hand behind my efforts) and perseverance I arrived at the formula. 

  Jesse (and I) did something similar with the Amazing Box fractal a while back by blending the AB fractal with the MandelBulb and got some interesting results.  I had created the non-blended version of this fractal (look towards the beginning of the thread), then attempted to combine the fractal with the MandelBulb and didn't get spectacular results. 

  Because this specific fractal was aligned with the magnitude axis (-1,-1,-1) to (1,1,1), and was very distorted when mixed with the MandelBulb, I eventually came to the idea that the fractal should be rotated in between iterations to successfully combine it with the MandelBulb. 

  Eventually, trying different things lead to the very simple version of the formula-  I was attempting a power 1 version of the Mag vs. XYZ formula from the beginning of this thread and came up with the simplified |x| -1, |y|-1, |z|-1 version.  Which makes me think that the Mag vs. XYZ formula might have a more logically derived version, but I haven't bothered to investigate it yet.

  In other news, I am going to try and implement a new type of periodic function, similar to the exponential function, in Fragmentarium.
   
  A little while back, I had an idea to create a new type of exponential function (e^n) based on my desire to develop a way to divide by zero.  I messed around and found something similar to the exponential function that had a corresponding logarithmic function.  The function allows the input of more than one angle, and creates greater variations over the whole.  If the first angle is pi, the second angle's sin and cosine vary from -1 to 1 and 1 to 3 respectively.  If the first angle is pi/2, the second angle's sin and cosine vary from -2 to 0 and 0 to 2 respectively.  If the first angle is 0, the second angle's sin and cosine are normal (vary the same way).  If the first angle is 3/2 pi, the second angle's sin and cosine both vary for 0 to 2. 

  I've got to check if it's a direct correspondence-  anyways, gotta go. 
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jehovajah
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« Reply #107 on: April 21, 2014, 08:53:00 PM »

Interesting. Something in your description reminds me of DeBroglies analogy of corpuscle and associated or corresponding wave.

I would be interested to see the results. grin
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May a trochoid of ¥h¶h iteratively entrain your Logos Response transforming into iridescent fractals of orgasmic delight and joy, with kindness, peace and gratitude at all scales within your experience. I beg of you to enrich others as you have been enriched, in vorticose pulsations of extravagance!
M Benesi
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« Reply #108 on: May 31, 2014, 08:20:55 AM »

  Hey JJ, just checked back in today.  It ended up not working well whatsoever.  hehe...  Don't know if I mentioned it in the other thread I posted about it here, but it ended up being e^(a+b) - e^a.  Nothing spectacular.  This explained the variation of the sin and cosine functions.... sin (a+b) - sin (a) and cos (a+b)-cos(a). 

  Didn't really work all that well at all.
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kram1032
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« Reply #109 on: May 31, 2014, 09:53:31 AM »

dividing by zero is already done by people all the time. It's called taking the derivative cheesy
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M Benesi
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« Reply #110 on: May 31, 2014, 10:24:44 PM »

hey Kram1032... I totally don't get the humor, and that is frustrating....
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kram1032
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« Reply #111 on: June 01, 2014, 01:49:37 PM »

Taking a derivative is essentially the correct way to divide by zero - based on taking the limit.
If you can take the limit, you can divide. If there is no unique limit, you can't.
The notation \frac{d^nf}{dx^n} essentially divides by 0 n times.

Of course, that's not the only case in which you actually can divide by 0.

In case of \frac{\sin{x}}{x}, you can look at the taylor polynomial of \sin{x} which is \sin{x}=\sum_{k=0}^\infty \frac{{\left(-1\right)}^k x^{1+2k}}{{\left(1+2k\right)}!} which, in turn, you can easily divide by x to obtain the series:  \frac{\sin{x}}{x}=\sum_{k=0}^\infty \frac{{\left(-1\right)}^k x^{2k}}{{\left(1+2k\right)}!}.
In doing so, you'll find, that, for x=0, the function returns 1.

Another method to figure out a division by zero is via L'Hôpital's rule:

Take two completely generic functions f\left(x\right) and g\left(x\right).
Let's assume, at a point x_0, g\left(x\right)=0.
Thus \frac{f\left(x_0\right)}{g\left(x_0\right)} would be a division by 0.
However, if this is possible, a value can be found at some point by looking at \frac{f^'\left(x_0\right)}{g^'\left(x_0\right)}.
If that still results in a division by 0 (i.e. g^'\left(x_0\right)=0), you just have to keep going. At some point, the n-th derivative  g^{\left(n\right)} \left( x_0 \right) \neq 0. and the division \frac{f^{\left(n\right)}\left(x_0\right)}{g^{\left(n\right)}\left(x_0\right)} can be done.
The value you get in that case will be precisely the value you'll get for the very first division  \frac{f\left(x_0\right)}{g\left(x_0\right)}.
Technically, to be more exact, you'd need to use limits.
But to cut a long story short, it already is known how to properly divide by 0. You just need to keep track of extra information, namely the behavior of the two functions you divide as the denominator approaches 0. In case of two constant functions, with, say, f\left(x\right)=1 and g\left(x\right)=0, this will just result in a directed infinity.

So really, when people say, dividing by 0 is undefined, what they really mean is, dividing two mere numbers, not functions, by 0, is underdefined. More information can (but doesn't have to) resolve the problems. If the problems remain, you get a singularity. Most singularities, however, can be dealt with pretty well. (Polynomial singularities, namely)
Exponential singularities are rarer but also more problematic.
« Last Edit: June 01, 2014, 07:35:56 PM by kram1032 » Logged
M Benesi
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« Reply #112 on: June 01, 2014, 07:12:21 PM »

Nicely written, easy to understand.  Always nice to read something written like that.
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