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#21 |
The Doctor
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Makes sense heh
I think I found the problem with my formula - My Nth root of .5 * 5 formula only works for picking values 0 - 5, I don't know how to factor in starting at 1. What it's multiplied by determines the asymptote and I can't Add a constant, or the result will go over 5. Edit:// Ok I was close I think it's (xth-root of .5)*4 +1 Last edited by TheSaxRunner05; 07-26-2017 at 08:25 PM.. |
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#22 | |
Spun a twirly fruitcake,
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![]() that doesn't work
using 1 to 5, doing it twice before taking highest average: 95 / 25 = 3.8 using your formula: 0.5^(1/2)*4+1 = 3.8284~~~
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Last edited by SKG_Scintill; 07-26-2017 at 08:46 PM.. |
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#23 | |
The Doctor
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Thanks for going back and forth on this, I'm going to toy with the equation you posted a bit more. |
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#24 |
Kawaii Desu Ne?
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![]() I came up with this: http://puu.sh/wU5Jx/8ad6c883ab.pdf
Got the equation (5n+1)/(n+1) If there's any errors or questions about my work, let me know. Or if someone finds a more elegant solution, I'd also be up for seeing it. EDIT: Here's a graph ![]() Last edited by reuben_tate; 07-26-2017 at 09:13 PM.. |
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#25 | |
Constantly Improving!
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#26 |
The Doctor
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Plugging 2 into this equation gives 11/3 or 3.666....
Is there an error in what Scintill showed earlier to show an answer of exactly 3.8? |
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#27 |
Kawaii Desu Ne?
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It looks like that is the case if you're looking at the discrete probability (the only options are 1, 2, 3, 4, 5) as opposed to the continuous probability (you can pick any real number between 1 and 5). So I'd expect it to be close but not exactly the same.
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#28 |
Kawaii Desu Ne?
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It seems that might be the case...wolframalpha won't do the integral in the general case for me...so I'll have to compute it by hand. I can double check.
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#29 |
The Doctor
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Sorry if I mistook what you said earlier, I thought you were essentially recapping what I said in my first paragraph of the OP, but you were much more on the right track than I was.
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#30 | |
The Doctor
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I really need to take a calculus course sometime to better understand integrals |
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#31 | ||
Constantly Improving!
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BUT, I KNEW there is a connection somehow, and thanks to ben's formula, I think I found it! I hope he double checks to make sure I'm right, and if not, I'm sure he'll think of something more solid! :P Thanks nonetheless! EDIT: I JUST started taking AP Calculus AB, (essentially first year college calculus in high school) and I am learning how to use limits correctly and master that skill. After THAT, I will be learning how to use integrals correctly, so it should be fun!
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Last edited by MarcusHawkins; 07-26-2017 at 09:49 PM.. |
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#32 | |
Spun a twirly fruitcake,
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![]() graphs are cute
for 1-5 ![]() for 1-k ![]() courtesy of wolframalpha, wonderful site edit: by the way, are we just neglecting my formula because it isn't elegant enough? :P edit v2: rewrote it this way, is it elegant enough?
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Last edited by SKG_Scintill; 07-26-2017 at 10:07 PM.. |
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#33 | |
Kawaii Desu Ne?
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The formula isn't quite as nice...at least it doesn't look like it simplifies to what we thought it might be: ![]() EDIT: Made it a tad bit nicer: ![]() Last edited by reuben_tate; 07-26-2017 at 10:21 PM.. |
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#34 | |
Kawaii Desu Ne?
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![]() Like I said before, I didn't mean to neglect your formula...I'm just trying to solve the "continuous" case instead of the discrete case :P |
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#35 | |
Kawaii Desu Ne?
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#36 | |
Spun a twirly fruitcake,
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![]() as a bit of visualization for my approach
![]() in case I: you take a number from 1-5 twice, so I put the 2 numbers as seperate axes the squares show how many times the number is the highest you can see that the occurrences of number N are N squared minus (N-1) squared aka: n^2 - (n-1)^2 same in case III where I put it as 3 axes and the occurrences of N are N cubed minus (N-1) cubed aka: n^3 - (n-1)^3 case II is where I multiplied the occurrences of case I by the weight of the number obviously I can't visualize the same for case III as it would be a 4d shape at this point I've done n(n^2-(n-1)^2) for case II so I did a summation for all the numbers N to get the sum of all amounts with that sum I now have to divide it by the amount of combinations, which is just the size of case I I tried "re-ordering the blocks" so they'd fit neatly inside eachother, but that doesn't seem to have a simple way (pretty obvious since the averages are decimal numbers) for the latest rewritten formula I approached it like this: take case I, make it one dimension greater, so it's [the highest number]^3, then subtract the sum of the smaller numbers squared (*1 to make it little cubes, but that's redundant) so... summarized average is ( [the highest number] to the power of ([amount of times you take a number] + 1) minus the sum of [all smaller numbers] to the power of [amount of times you take a number] ) divided by [the highest number] to the power of [amount of times you take a number] average = ( [k] ^ ([m]+1) - sigma [from n=0 to n = k-1] of n ^ [m]) / k^m also to be written as ... no idea if that helps, just having fun you can see my steps, maybe go a different route
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Last edited by SKG_Scintill; 07-27-2017 at 10:33 AM.. |
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