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Galandros
Active Member
Joined: 29 Aug 2008 Posts: 565

Posted: 19 Aug 2009 05:55:44 pm Post subject: 


Data:
a b
0 10000
1 20000
2 40000
3 75000
4 140000
5 255000
6 470000
7 865000
8 1590000
9 2920000
10 5355000
11 9820000
12 18005000
13 33005000
14 60510000
15 110925000
16 203350000
17 372785000
18 683385000
19 1252785000
20 2296600000
EDIT: correct values thanks do DarkerLine
I want to find the formula that gives this. Dependent variable is a. (of course)
I have tried the most usual regressions. Linear (no hopes on it at start though), Power, exponential, polynom, etc..
Hint: a big *maybe*: it looks something like: b = 100000 + 50000 * (RoundUp(1.6^a)1)
I worked on trying to figure if the formula is like that one with a spreadsheet a this online tool: http://www.xuru.org/rt/TOC.asp
No big results. I made everything I could... Maybe I need more serious tools.
Let the fun begin!
PS: why a need this formula? It's a non TIproject of mine.
PS2: I learned a bit with this So I am happy even if we don't find it.
Last edited by Guest on 20 Aug 2009 01:46:40 pm; edited 1 time in total 

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DarkerLine ceci n'est pas une 
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Joined: 04 Nov 2003 Posts: 8328

Posted: 19 Aug 2009 06:11:53 pm Post subject: 


That data looks pretty clearly exponential to me. Is there any reason you don't like that model?
Edit: oh, and the fit becomes much better if you ignore the first few terms (which are a bit off the model, presumably because they have less significant digits).
Last edited by Guest on 19 Aug 2009 06:18:59 pm; edited 1 time in total 

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Galandros
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Joined: 29 Aug 2008 Posts: 565

Posted: 19 Aug 2009 06:29:23 pm Post subject: 


DarkerLine wrote: That data looks pretty clearly exponential to me. Is there any reason you don't like that model?
Edit: oh, and the fit becomes much better if you ignore the first few terms (which are a bit off the model, presumably because they have less significant digits).
The idea is a formula that gives perfect results. I think I can get other values behind 21.
I will begin to cut always the first values then. Sometimes I tried without them. 

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DarkerLine ceci n'est pas une 
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Joined: 04 Nov 2003 Posts: 8328

Posted: 19 Aug 2009 06:36:28 pm Post subject: 


Perfect results are generally not the idea. For actual data, perfect results are impossible since the data depends on a bunch of other factors including measurement error. In your case, it looks like every value of B is rounded to the nearest 5K, so any formula that gives a correct result up to that tolerance is as good as perfect.
Of course, knowing where the data comes from generally helps make such a judgment. 

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Galandros
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Joined: 29 Aug 2008 Posts: 565

Posted: 19 Aug 2009 06:51:38 pm Post subject: 


DarkerLine wrote: Perfect results are generally not the idea. For actual data, perfect results are impossible since the data depends on a bunch of other factors including measurement error. In your case, it looks like every value of B is rounded to the nearest 5K, so any formula that gives a correct result up to that tolerance is as good as perfect.
Of course, knowing where the data comes from generally helps make such a judgment.
Well the formula comes from something made by man. It has some formula behind. I am quite sure it has a formula. ;)
Sorry about not clarifying about this before, it is a important information. 

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Weregoose Authentic INTJ
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Joined: 25 Nov 2004 Posts: 3976

Posted: 19 Aug 2009 07:10:54 pm Post subject: 


Verify that b=1,297,589,000 is correct in your list; it's a little strange.
5000int(2.5(11/6)^a[font="verdana"]³√([font="times new roman"]e^(2a61/3 is good for 3[font="verdana"]<a[font="verdana"]<19. I will look more into this.
[EDIT] – Now that the table in the first post has been altered, it turns out that 1 through 3 from this expression were correct.
Last edited by Guest on 05 Jul 2010 08:11:43 am; edited 1 time in total 

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DarkerLine ceci n'est pas une 
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Posted: 19 Aug 2009 09:48:26 pm Post subject: 


That's a little too complex compared to the other ones.
It's definitely exponential, plus some bit of rounding that got thrown in there. You'd have to experiment to figure out the exact form of the formula (the rounding could happen anywhere), but this tells us that the exponent is around 1.84 (the formula is probably fairly simple, so I'm guessing it's either 1.8 or, less likely, 1.85). Then again, the values might have been hardcoded  maybe they were originally calculated by a formula, but then altered. 

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Galandros
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Joined: 29 Aug 2008 Posts: 565

Posted: 20 Aug 2009 04:38:04 am Post subject: 


Weregoose wrote: Verify that b=1,297,589,000 is correct in your list; it's a little strange.
5000int(2.5(11/6)^a[font="verdana"]³√([font="times new roman"]e^(2a61/3 is good for 3[font="verdana"]<a[font="verdana"]<19. I will look more into this.
I noticed that too. Every of the others are multiple of 5000 except that one.
Looks too much complex than the others.
DarkerLine wrote: That's a little too complex compared to the other ones.
It's definitely exponential, plus some bit of rounding that got thrown in there. You'd have to experiment to figure out the exact form of the formula (the rounding could happen anywhere), but this tells us that the exponent is around 1.84 (the formula is probably fairly simple, so I'm guessing it's either 1.8 or, less likely, 1.85). Then again, the values might have been hardcoded  maybe they were originally calculated by a formula, but then altered.
Definitely the exponent is somewhere between 1,8 and 1,92. An exponential regression works very well with the old formula of the data (100000 + 50000 * (RoundUp(1.6^a)1))
Yesterday I miss telling many things... Here they are:
A hint of mine for the formula is something like:
10000 + numberA * SOMEROUND( MAX(0;(numberB ^ level)  2 )) )
But there is somethind odd going on there... That max(0 thing maybe is not the way to go. And with numberB = 1,8, numberA = 10000 and ROUND to 0 decimals you get this promising values:
10000
10000
20000
50000
110000
210000
390000
730000
1360000
2530000
4690000
8680000
16060000
29720000
55000000
101750000
188250000
348260000
644290000
1191950000
2205120000
If it were to a model of a real situation, it would be good. This or other I have found.
This exponential is very strange, although it seems between 1,8 and 1,92 from this values the results are:
 or it gets very accurate for the first values in a
 or for the high values in a
Is there expression something related to exponential that solves this? DarkerLine your link gave idea. Maybe the formula looks like: numberA * a * numberB ^ a. Maybe it uses "a" more than once.
Maybe instead of the max(0; we could just accept the 1 and later the formula gets care of that.
When I analyse the data, I subtract 10k... Right now I start to loose little trust on this subtract.
I am curious to see the formula.
Last edited by Guest on 05 Jul 2010 08:11:02 am; edited 1 time in total 

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DarkerLine ceci n'est pas une 
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Posted: 20 Aug 2009 12:24:06 pm Post subject: 


My guess is that the formula is simply iterated, and the rounding is done at each step. In that case, the formula is more or less 11/6 (or maybe an approximation to it) times the previous value, rounded to the nearest 5000. There's some trickiness with the rounding here, but this is much closer to the truth than anything else.
Also, here are several corrections to your values:
1: 20000 not 10000
2: 40000 not 20000
3: 75000 not 40000
19: 1252785000 not 1297589000 

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Galandros
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Joined: 29 Aug 2008 Posts: 565

Posted: 20 Aug 2009 12:32:14 pm Post subject: 


DarkerLine wrote: My guess is that the formula is simply iterated, and the rounding is done at each step. In that case, the formula is more or less 11/6 (or maybe an approximation to it) times the previous value, rounded to the nearest 5000. There's some trickiness with the rounding here, but this is much closer to the truth than anything else.
Also, here are several corrections to your values:
1: 20000 not 10000
2: 40000 not 20000
3: 75000 not 40000
19: 1252785000 not 1297589000
That makes things easier. I think I got some formulas that throw that.
Minina tricked me with a wrong table...(or things got changed) I was waiting to check level 19. Thanks anyway.
Where did you find the level 19 at this point? Thanks, I will see if now I can find it.
Last edited by Guest on 20 Aug 2009 01:28:22 pm; edited 1 time in total 

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DarkerLine ceci n'est pas une 
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Posted: 20 Aug 2009 12:39:17 pm Post subject: 


The value of 1297589000 was a typo that one person made and everyone else copied. I searched Google for 683.385.000 but not 1.297.589.000 and found a table with a better number. 

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Galandros
Active Member
Joined: 29 Aug 2008 Posts: 565

Posted: 20 Aug 2009 03:11:20 pm Post subject: 


DarkerLine words to coding:
10000
FOR(1, 20)
store = CEILING( (11/6) * ans ;0,5)
END
The problem is that the step is around 11/6 like this:
1 2
2 2
3 1,875
4 1,866666667
5 1,821428571
6 1,843137255
7 1,840425532
8 1,838150289
9 1,836477987
10 1,83390411
11 1,833800187
12 1,833503055
13 1,833101916
14 1,833358582
15 1,833168071
16 1,833220645
17 1,833218589
18 1,833188031
19 1,833205294
20 1,83319564
I tried some roundings (up, down, normal before the ceiling but no perfect results).
This steps make me remember some iterative algorithms like with the Greek method to find a square root (it gets close of the square root with fractions).
Last edited by Guest on 20 Aug 2009 03:13:29 pm; edited 1 time in total 

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DarkerLine ceci n'est pas une 
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Joined: 04 Nov 2003 Posts: 8328

Posted: 20 Aug 2009 06:04:57 pm Post subject: 


The reason for that convergence is the rounding. As the numbers get larger, the rounding stops mattering as much, and therefore the ratio converges to the true value of around 1.8332. 

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