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Math and Science => Technology & Calculator Open Topic
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Galandros

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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 TI-project 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
DarkerLine
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Super Elite (Last Title)

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
Galandros

Active Member

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.
DarkerLine
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Super Elite (Last Title)

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 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.
Galandros

Active Member

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.
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^(2a-61/3 is good for 3[font="verdana"]
DarkerLine
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Super Elite (Last Title)

Joined: 04 Nov 2003
Posts: 8328

 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 hard-coded -- maybe they were originally calculated by a formula, but then altered.
Galandros

Active Member

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^(2a-61/3 is good for 3[font="verdana"]
DarkerLine
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Super Elite (Last Title)

Joined: 04 Nov 2003
Posts: 8328

 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
Galandros

Active Member

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
DarkerLine
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Super Elite (Last Title)

Joined: 04 Nov 2003
Posts: 8328

 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.
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
DarkerLine
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Super Elite (Last Title)

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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