Erlang BDate: 10/22/2002 at 14:11:22 From: Jamie Bell Subject: Logarithms I need to know how to calculate the addition of numbers using logarithms: 1 + 2 + 3 I have a formula in an Excel file that produces numbers large enough to exceed Excel's calculation limits; e.g.: 200! and 10^1000. I have converted all of the factorial and expotential expressions to logs, but there is a step that requires adding numbers that exceed Excel's limits. How can I use logs to add the numbers? The formula is Erlang B: P = (A^N/N!) / Sum 0 through N of A^n/n! Thanks. Date: 11/14/2002 at 17:00:01 From: Doctor Peterson Subject: Re: Logarithms Hi, Jamie. There are some useful things to say that have nothing to do with logarithms. Here is one page I found that will be useful to you: A Robust and Efficient Algorithm for Evaluating Erlang's Formula - Erlang B - Dave Albert, Jun Zhu http://www.dcss.mcmaster.ca/~qiao/publications/erlang/newerlang.html This gives a modified version of the formula that is easier to use. I'll explain and derive it in what I think is a clearer form. We can avoid calculating factorials when we are able to cancel, as is usually done in calculating combinations. We can do that in this case by distributing the numerator; that is, we can divide the numerator and denominator by A^N/N!: A^N/N! P = -------------------- Sum[k=0 to N] A^k/k! 1 = ------------------------------ Sum[k=0 to N] (A^k/k!)(N!/A^N) 1 = ------------------------------------ Sum[k=0 to N] (N!/k!)/A^(N-k) 1 = ------------------ Sum[k=0 to N] a[k] where N!/k! a[k] = -------, for k=0 to N A^(N-k) Look at the sequence a[k]: N(N-1)...(3)(2)(1) a[0] = ------------------ A^N N(N-1)...(3)(2) a[1] = --------------- A^(N-1) N(N-1)...(3) a[2] = ------------ A^(N-2) ... N a[N-1] = --- A^1 1 a[N] = --- A^0 We can calculate the sequence a[k] recursively in reverse (starting from a[N] and working back to a[0]), as a[N] = 1 a[k-1] = a[k] * k/A for k=N to 1 You will never get large numbers when you calculate this way, and will have as much precision as you can get. You can do it in a simple loop, adding in each a[k] as you calculate it, and then taking the reciprocal of the result. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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