Ln(x) using repeated square root extraction

[Thomas Klemm]

In the initial thread An old logarithm algorithm the "pages 33 and 34 of the manual for the Texas Instruments SR-10" are quoted.

However in this manual for the Texas Instruments electronic slide rule calculator SR-10 we find on page 30:

Quote:Logarithmic and Exponential Function

\(\ln a = \left[ \left( - \frac{3}{5} b^2 + 1 \right)^{-1} \times 5 + 4 \right] \frac{2b}{9} \)


\( 0.7 < a < 1.6 \)

where \( b = \frac{a - 1}{a + 1} = \left( a + 1 \right)^{-1} \times (-2) + 1 \)

This expression yields values with an error less than 0.0003% over the range of a from 0.7 to 1.6.

The first 3 terms of the Taylor series of this expression agree with those of \(\log\left(\frac{1+\varepsilon}{1-\varepsilon}\right) \):

\( \frac{2\varepsilon}{9} \left( 4 + \frac{5}{1 - \frac{3\varepsilon^2}{5}} \right) = 2 \varepsilon + \frac{2\varepsilon^3}{3} + \frac{2\varepsilon^5}{5} + \frac{6\varepsilon^7}{25} + \frac{18\varepsilon^9}{125} + \mathcal{O}(\varepsilon^{11}) \)

Again a program for the HP-15C:

Code:

   001 {          11 } √x̅
   002 {          11 } √x̅
   003 {          11 } √x̅
   004 {          11 } √x̅
   005 {          11 } √x̅
   006 {          11 } √x̅
   007 {           1 } 1
   008 {          40 } +
   009 {           2 } 2
   010 {          34 } x↔y
   011 {          10 } ÷
   012 {           1 } 1
   013 {          34 } x↔y
   014 {          30 } −
   015 {          36 } ENTER
   016 {       43 11 } g x²
   017 {           3 } 3
   018 {          20 } ×
   019 {           5 } 5
   020 {          10 } ÷
   021 {           1 } 1
   022 {          34 } x↔y
   023 {          30 } −
   024 {           5 } 5
   025 {          34 } x↔y
   026 {          10 } ÷
   027 {           4 } 4
   028 {          40 } +
   029 {          20 } ×
   030 {           1 } 1
   031 {           2 } 2
   032 {           8 } 8
   033 {          20 } ×
   034 {           9 } 9
   035 {          10 } ÷

The result for \(x = 2\) is:

0.6931471786