Mathematical SeriesDate: 04/19/2005 at 14:43:52 From: James Subject: Mathmatical series I have a question relating to the transposition of a formula. The formula relates to reliability: mtbf = 1/a + 1/2a + 1/3a + .... + 1/na I can see that this is a mathematical series--I know mtbf is 23998 and I also know a = 0.0008, but I am having difficulty finding n since it changes at each stage and therefore I cannot put a direct value to it. I could possibly single out n, but this is proving difficult because what value do you start with? Date: 04/20/2005 at 21:38:57 From: Doctor Vogler Subject: Re: Mathmatical series Hi James, Thanks for writing to Dr. Math. If you multiply both sides of your equation mtbf = 1/a + 1/2a + 1/3a + .... + 1/na by the number a, then it becomes mtbf*a = 1/1 + 1/2 + 1/3 + ... + 1/n. This is much easier to work with, if you are trying to find n. Now, the simple way to find n is to multiply mtbf by a, and then subtract off 1/1, and then subtract off 1/2, and then 1/3, and 1/4, and so on, until you get something that is zero or less. The number of fractions you had to subtract off is your n. That's the simple way. But that's not the fast way. In fact, if mtbf*a is bigger than 19, then you'll need to subtract off over a hundred million fractions before you get negative. That's because 1/1 + 1/2 + 1/3 + ... + 1/n is approximately the natural log of n, ln n. And so if you want ln n > 19 then you need n > e^19 > 100,000,000. Of course, you should also notice that if you only have four or five digits of precision on mtbf*a (and you actually only showed one digit of precision on the number a), then you can't get the number n exactly, since adding one more term 1/n will only change the sum by less than 0.00000001. And it is certain that you will never get EXACTLY 23998*0.0008, so the best you can hope for is a number n that gives you close to the right amount. So what is the fast way to find n? The fast way uses Euler's constant gamma (that's a Greek letter that looks rather like a y with a loop on the tail) and some approximations for the sum. gamma = 0.5772156649015328606065120900824024310421593359399235988... (You'll probably only need a few digits of this, like five or six, and no more than ten.) A good estimate that will certainly suffice for your purposes is ln (n+1) + gamma - 1/(2n) < 1/1 + 1/2 + 1/3 + ... + 1/n < ln (n+1) + gamma - 1/(2n+2). For example, if you know what you want the sum to equal M = 1/1 + 1/2 + 1/3 + ... + 1/n, then n will be approximately n0 = e^(M - gamma), and you can get better estimates by changing n0 to e^(M - gamma + 1/(2*n0)). For example, if you take M = 23998*0.0008 = 19.1984 then you get n0 = 122201800.06 and another iteration gives 122201800.56 So we take n = 122,201,800, and we find that the sum 1/1 + 1/2 + 1/3 + ... + 1/n is between 19.1984000035373101405886978465851071841052388010642051815130 and 19.1984000035373101740709601813505258621221779810226505580255 But if we use one less term (n = 122,201,799), then the sum is between 19.1983999953541246228630992399870771093374538183247805496923 and 19.1983999953541246563453621227356287486853252543122747286441 For your information, I did these computations using the free program GNU Pari, available for download at http://pari.math.u-bordeaux.fr/ In Pari, the constant gamma is called Euler(), and so I asked it to compute things like n = 122201800 log(n+1) + Euler() - 1/(2*n) By the way, if you need a better estimate for the sum (which is not likely), then you can use ln (n+1) + gamma - 1/(2n+2) - 1/12n(n+1) < 1/1 + 1/2 + 1/3 + ... + 1/n < ln (n+1) + gamma - 1/(2n+2) - 1/12(n+1)(n+2). If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 04/21/2005 at 16:31:13 From: James Subject: Thank you (Mathmatical series) Thanks very much for helping--it is very much appreciated. It's nice to know that there are people out there willing to help others. James |
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