Earlier this month, I received a request from Jason Foose. These are six graphs of y = x * rand^rand: two with the TI84+, two with the Casio Prizm (fxCG 10, and two with the HP 39gii.
On the Prizm, I had to use the catalog to find the random number function (labeled Ran#).
Reediting the function will cause the function to be redrawn.
Enjoy!
Eddie
A blog is that is all about mathematics and calculators, two of my passions in life.
Sunday, November 25, 2012
f(x) = x * random ^ random
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That is VERY COOL!!!!! Thanks for the thought and effort Eddie. I am impressed by the screen resolution of the Casio GC10 and the
ReplyDeleteHP 39II. Hhmmmm, what if we throw in a powerball and use the reciprocal value of Hp's normal random (<1 >0) generated value?
Okay. I've had some time to digest this. Why is the model still relatively predictable? There's unpredictable spikes but the there is still a smooth overlying upward trend. My initial thoughts were that this would graph maniacally. Any thoughts or insight???
ReplyDeleteJason,
ReplyDeleteIn a typical random number function, the number returned, let's say r, is between 0 and 1. Typically, 0 ≤ r < 1.
For r^r, we have 0^r ≤ r^r < 1^r
The lower bound is either undefined, in the case of 0^0. If r ≠ 0, 0^r = 0.
The upper bound is 1, for any r.
Now we multiply 0^r ≤ r^r < 1^r by x to get the following bounds:
0^r * x ≤ x * r^r < x
r^r is a "pseudocoefficient" of x. While we are seeing a seemingly random pattern, the growth is expected to be linear.
Eddie
Hello Eddie,
ReplyDeleteI think of the "quality" behind the random. The average of random must be near of 0.5 and the standard deviation 1/(sqrt(12)).
On my hp35s, with 1000 rand and a seed of 0.5, found 0.4967 and 0,2883 (good in 1:30 minute)
HP 15c : 0,5149 and 0,2891 (average a little high, in 17 minutes and the same seed)
With r^r : i think the average near 1/sqrt(2) but have no idea for the standard deviation.
Then f(x) = x * r ^ r is in average near x/sqrt(2).
What do you think about that ?
It's time to sleep here, good night !
Caloubugs,
DeleteYou are on track with the average.

To get an idea for the sample standard deviation, I am going to generate nine lists, three each on a TI84, HP calculator, and Excel, of 100 data points of random^random. This is no where near a proof of course, just to get an idea of what the deviation would be.
Sample Standard Deviation for each list were:
(TI)
0.2547639802
0.2575450285
0.2551388739
(HP)
0.262557143565
0.266684230048
0.265612924289
(Excel)
0.269713444
0.254391699
0.241800699
Average of the sample standard deviation from the nine lists is approximately 0.25868978, not too far away from 1/sqrt(15).
Eddie
Thanks Eddie !
ReplyDeleteIncluding random in fonctions was a discover for me and behind that, i think there's a mathematic theory (High numbers laws or something like that).
My maths studies are 20 years ago... English too :D
It wakes my brain !
Pascal
OK, thinking of f(x,y)=x^y doesn't let me sleep.
ReplyDeleteThen as the average of f(x)=1 is 0.5 (intégration of f(x) between 0 and 1), the average of f(x,y)=x^y is the double intégration of f(x,y) dx dy (not really easy to write math here ;)
We find ln 2 = 0,69315 which is correct with average of random^random.
I find the result with wiris.net
The standard deviation is then :
Sqrt( intg[0,1] intg[0,1] x^(2y) dx dy  (ln 2)^2 ) = 0,2624
Correct with your random^random standard deviation.
Calculate with www.wiris.net too...
Have fun !
Pascal
I was in the range with the nine list samples. Thank you posting the integration  and yes it is not easy to type math  especially using only text.
DeleteI use my iPad to type my blogs and customized the keyboard using the Settings and the Unicode Pad app. I make the shortcut "intgl" for the integral symbol, "pi" for pi, etc.
Eddie