How did Napier round his logarithms?By repeated subtractions Napier calculated (1−10^(−7))^L for L ranging from 1 to 100. The result for L=100 is approximately 0.99999=1−10^(−5). Napier then calculated the products of these numbers with 10^7(1 − 10^(−5))^L for L from 1 to 50, and did similarly with 0.9998~~(1−10^(−5))^(20) and 0.9~~0.99520.What was his rule for rounding? May one say that he was computing in the ring ZZ[x]/(10x−1,x^8)?

Scamuzzig2

Scamuzzig2

Answered question

2022-10-02

How did Napier rounded his logarithms?
How did Napier round his logarithms? Wikipedia says:
By repeated subtractions Napier calculated ( 1 10 7 ) L for L ranging from 1 to 100. The result for L = 100 is approximately 0.99999 = 1 10 5 . Napier then calculated the products of these numbers with 10 7 ( 1 10 5 ) L from 1 to 50, and did similarly with 0.9998 ( 1 10 5 ) 20 and 0.9 0.99520
What was his rule for rounding? May one say that he was computing in the ring
Z [ x ] / ( 10 x 1 , x 8 )?

Answer & Explanation

Derick Ortiz

Derick Ortiz

Beginner2022-10-03Added 11 answers

This is not "Napier's rule" but to round ( 1 10 7 ) L you can use the binomial theorem and only keep the first (or two if L is large) elements in the expansion :
( 1 10 7 ) L 1 L 10 7 + L 2 2 10 14
Hence only using one term: ( 1 10 7 ) 100 1 100 10 7 = 1 10 5 = 0.99999
This also explains ( 1 10 5 ) 20 1 20 0.00001 = 1 0.0002 = 0.9998
Generally, taking only the first term is enough unless L is very large.
Note that what you are doing is calculating ( ( 1 10 7 ) 100 ) 20 = ( 1 10 7 ) 2000 0.9998 always using the same rule.

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