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Mathematical Series

Date: 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
Associated Topics:
High School Sequences, Series

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