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Started by Gath, December 05, 2011, 07:24:41 PM
Quote from: Ertxiem on December 06, 2011, 06:49:02 PMDid you knew that Gauss, when he was aged 10, discovered by himself that to add the numbers between 1 and 60 all he had to do was to multiply 30 by 61? And he explained it's because you have 30 pairs that add 61 (1+60, 2+59, 3+58,...). Pretty amazing for a 10 year old!After this brief drift only slightly off-topic...Thanks flack. I forgot to mention that to add the numbers between 1 and n is equivalent to compute n*(n+1)/2.You can read this expression in 2 ways:- If n is even, you have n/2 pairs equal to n+1 (1 + n, 2 + n-1, ...);- If n is odd, you have (n+1)/2 terms equal to n (1 + n-1, 2 + n-2, ..., and the lonely n).
Quote from: flack on December 06, 2011, 06:28:49 PMLet s = starting level, t= level you want to train to. Where t>s. We are given that when s = 1, the cost starts at 25. Hence, the total cost is the following sum:25s + 25(s+1) + ... +25(t-1) = 25(s + (s+1) + ... + (t-1)) = 25 ([1+ 2 +... + (t-1)] – [1+ 2+ ...(s-1)]) = 25(t(t-1)/2 – s(s-1)/2) = 25(t-s)(t+s-1)/2This is how you derive the formula.
Quote from: Ertxiem on December 06, 2011, 04:11:17 AMI prefer the formula:[ n(n-1) - c(c-1) ] * 25 / 2 = (n - c)(n + c - 1) * 25 / 2where n is the new level and c is the current level.I think that the 2nd expression is easier: the difference between levels times the sum minus 1 times 25 over 2 (actually I divide by 2 the factor that is even before multiplying everything else).
Quote from: Duckling on December 11, 2011, 12:38:26 AMHow did you know?; I'm highly interested in this formula. Personally, I prefer:C = (n2 - c2 + n - c) × 25 ÷ 2because it's easy to remember. ... For me. [spoiler=Xkcd math]But I, too, got your formula, but derived it differently:[n (n - 1) - c (c - 1)] * 25 / 2= [[(n - c) (c - 1)] § (n + c (- 1))] * 25 / 2 1= [(n - c) [(c - 1) § (n + c - 1)]] * 25 / 2 2= [(n - c) (n + c - 1)] * 25 / 2 3Note 1: This operation takes the sum of the quotients of the factoring taking place between steps one and two, while the two negative ones are grouped into one bigger negative one. In exchange, the binomial factors are made equivalent.Note 2: Here, the associativity of the operation is shown, while the summed term swaps itself with the c-1, and closing the metaspace portal created in step one, forever banishing the useless thing into another realm. Meanwhile, the negative one has lost weight.Note 3: The completed formula. The entire process was mathematically sound; look at the results![/spoiler]But honestly, I'm wondering how best to describe it such that it's easy to compute quickly and efficiently.Perhaps it had best be described as... 25 [c2(-(c - n)2 + (3n - c)) + n2((c - n)2 - (3c + n))] = C 2 (c - n)2I think that that's easy enough to remember! (It actually works, if anyone wants to test it out) ;)Who could ever want to punch numbers in a calculator when they had a lovely, elegant equation like that to work with?(But really, the c-n symmetry of the equation astounds me. It's beautiful, and yet baffling at the same time.)