## Notes on Logarithmic Computations## Gaussian Logarithms
Tables of Gaussian Logarithms can be used to calculate the addition or subtraction
of number values for which (e.g. due to a certain calculation scheme) only the logarithmic values
are available.
Using Gaussian Logarithms for this purpose, avoids converting the logarithmic values
to the natural number representation and back to the logarithmic value after
performing the requested addition or subtraction. +-----------+ +----------+----------+ | | | A | B | | x | | log(x) |log(1+1/x)| +-----------+ +----------+----------+ | 1.000 | | 0.00000 | 0.30103 | | 1.002 | | 0.00100 | 0.30053 | | 1.005 | | 0.00200 | 0.30003 | | 1.007 | | 0.00300 | 0.29953 | | 1.009 | | 0.00400 | 0.29903 | .. .. .. | 50118.723 | | 4.70000 | 0.00001 | | 63095.734 | | 4.80000 | 0.00001 | | 79432.823 | | 4.90000 | 0.00001 | |100000.000 | | 5.00000 | 0.00000 | +-----------+ +----------+----------+
Suppose that log(a) and log(b) are given, with log(b) being smaller than log(a),
and that the result for log(a+b) is to be found. A = log(a) - log(b) = log(a/b)For the argument A=log(a)-log(b), or x=(a/b), the function value "B" from the Gaussian Logarithms Table is: B = log(1+b/a) = log((a+b)/a)If the value of log(a) is added to this result: log(a) + log((a+b)/a) = log(a+b)the requested result for log(a+b) is obtained! Notice that the argument x=(a/b) is not explicitly required to obtain this result.
Hence for the logarithm of addition, the following rule is derived: For logarithm of subtraction, the Table of Gaussian Logarithms must be extended with the function C=log(1+x), which is now used as table entry. It is still assumed that a>b so the difference (a-b) is a positive number. Hence if the argument "C" is log(1+x) = log(a) - log(b) or (1+x)=(a/b), the function "B" will give log(1+1/(a/b-1)) = log(a/(a-b)), which subtracted from log(a) will give log(a-b). +-----------+ +----------+----------+----------+ | | | A | B | C | | x | | log(x) |log(1+1/x)| log(1+x) | +-----------+ +----------+----------+----------+ | 1.000 | | 0.00000 | 0.30103 | 0.30103 | | 1.002 | | 0.00100 | 0.30053 | 0.30153 | | 1.005 | | 0.00200 | 0.30003 | 0.30203 | | 1.007 | | 0.00300 | 0.29953 | 0.30253 | | 1.009 | | 0.00400 | 0.29903 | 0.30304 | .. .. .. .. | 50118.723 | | 4.70000 | 0.00001 | 4.70001 | | 63095.734 | | 4.80000 | 0.00001 | 4.80001 | | 79432.823 | | 4.90000 | 0.00001 | 4.90001 | |100000.000 | | 5.00000 | 0.00000 | 5.00000 | +-----------+ +----------+----------+----------+ In case log(a)-log(b) is smaller than log(2)=0.30103, the column "C" cannot be used as table entry. Instead column "B" is used as entry and column "C" as function output. This gives the following relationships: B=log(a)-log(b) or (1+1/x)=(a/b), and thus the function "C" will give log(1+b/(a-b)) = log(a/(a-b)), which subtracted from log(a) will give log(a-b).
Hence for the logarithm of subtraction, the following rule is derived: ## Examples
Given log(a)=1.2300 and log(b)=0.6700, get the logarithm of the addition: log(a+b).
First calculate log(a)-log(b)=0.5600, and enter column "A" with this value to obtain
the result from column "B" (with appropriate interpolation): B=0.1056. ## Sources1. "Logarithmic Computations" by Professor H.A. Howe (University of Denver, Colorado), published in the "Sidereal Messenger", Vol 2, April 1883, pp. 45-50.2. "Gaussian Logarithms and Navigation" by Charles H. Cotter, in "The Journal of Navigation", Volume 24 - Issue 4, October 1971, pp. 569-572 3. "Gauß'sche Additionslogarithmen feiern dieses Jahr ihren 200. Geburtstag" by Herman Kremer in "de.sci.mathematik", August 2002 |

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