mersenneforum.org Calculating parabolical curve problem
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 2006-01-20, 04:00 #1 jasong     "Jason Goatcher" Mar 2005 350510 Posts Calculating parabolical curve problem Because of the fact I have an IQ of 130, but an education cut short by mental illness, there are a lot of math problems where I should know what to do, but don't. This is an example. Mind you, I'm giving the simplistic version because I don't think I'm smart enough to include all the variables. Here's the thing: In Riesel Sieve, we're LLRing numbers around n=2,038,000. On average from one n-value to the next there's a difference about 4. Points are approximately proportionate to the time any computer takes to finish a test, relative to the computer itself. The formula is 7.5^(ln(n/524288)*100. The average score per day over the last thirty days is about 1.04 million for everyone involved(total score of ALL tests for thirty days divided by thirty). Basically, if the pattern stays the same, and no n-values are removed by sieving, nor primes found, how long would it take to get to n=2^22, or n=2097152. I know the first step is to graph the parabolical curve which is created by the 7.5^... formula for n=2097152, and remove the part that has to do with n=1 to n=2,038,000. Take the area and divide by 4(since not every value is represented). Unfortunately, I don't have the education to understand the formula given in the calculator, which by the way is for a double parabola(I'm assuming that's the right formula). Assuming it's not too involved(I just started a Plane Trig. class and am doing fairly well, if that helps), could someone dredge up the formula and tell me what the values mean. If someone confirms what type of formula it is, I'd be happy to attempt(!) to print it on these forums, although it wouldn't be in that special language that was installed a while back. Thanks. Last fiddled with by jasong on 2006-01-20 at 04:02
 2006-01-20, 11:59 #2 alpertron     Aug 2002 Buenos Aires, Argentina 32·149 Posts I wrote the following program in UBASIC using the data you present above: Code: 10 Sum=0 20 for N=2038000 to 2097152 step 4 30 Sum=Sum+7.5^(log(N/524288))*100 40 next N 50 print Sum/1040000 The answer is 22.57 days. Does it make sense to you? Last fiddled with by alpertron on 2006-01-20 at 12:00
 2006-01-20, 12:31 #3 alpertron     Aug 2002 Buenos Aires, Argentina 53D16 Posts Since this is the math forum and not the programming forum, this is another response with the same answer, of course: $f(x) = 7.5^{\log (n/524288)}*100$ $\log f(x) = \log (n/524288)* \log \,7.5 + \log \,100$ $f(x) = (n/524288)^{\log \,7.5} * 100$ $f(x) = (n/524288)^{2.014903} * 100$ $f(x) = n^{2.014903} * 2.989775 * 10^{-10}$ The antiderivative of f(x) is: $g(x) = n^{3.014903} * \frac {2.989775 * 10^{-10}}{3.014903}$ $g(x) = n^{3.014903} * 9.91665 * 10^{-11}$ The number of days is: $D = \frac{g(2097152) - g(2038000)} { 4 * 1040000}$ $D = \frac {9.91665 * 10^{-11}} { 4 * 1040000} * (2097152^{3.014903} - 2038000^{3.014903})$ $D = 2.38381 * 10^{-17} * (114.57793 * 10^{17} - 105.10868 * 10^{17})$ $D = 2.38381 * 9.46925 = 22.572$
2006-01-20, 17:11   #4
jasong

"Jason Goatcher"
Mar 2005

5×701 Posts

Quote:
 Originally Posted by alpertron I wrote the following program in UBASIC using the data you present above: Code: 10 Sum=0 20 for N=2038000 to 2097152 step 4 30 Sum=Sum+7.5^(log(N/524288))*100 40 next N 50 print Sum/1040000 The answer is 22.57 days. Does it make sense to you?
It makes perfect sense. Thank you.

 2006-01-20, 17:14 #5 jasong     "Jason Goatcher" Mar 2005 5×701 Posts Question: Was my parabolical curve idea feasible? And if it is, which would take more computation, the program above, or graphing on a calculator?

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