Ln(x) using repeated square root extraction
03-10-2022, 05:18 AM
Post: #14
 Thomas Klemm Senior Member Posts: 1,769 Joined: Dec 2013
RE: Ln(x) using repeated square root extraction
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
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 Messages In This Thread Ln(x) using repeated square root extraction - Gerson W. Barbosa - 03-21-2016, 12:03 AM RE: Ln(x) using repeated square root extraction - Gerson W. Barbosa - 03-21-2016, 05:09 AM RE: Ln(x) using repeated square root extraction - Paul Dale - 03-21-2016, 07:11 AM RE: Ln(x) using repeated square root extraction - Thomas Klemm - 03-03-2022, 11:20 PM RE: Ln(x) using repeated square root extraction - Albert Chan - 03-06-2022, 12:04 AM RE: Ln(x) using repeated square root extraction - Dan C - 03-05-2022, 05:32 PM RE: Ln(x) using repeated square root extraction - Dan C - 03-05-2022, 05:53 PM RE: Ln(x) using repeated square root extraction - Dan C - 03-05-2022, 08:02 PM RE: Ln(x) using repeated square root extraction - Namir - 03-05-2022, 08:29 PM RE: Ln(x) using repeated square root extraction - Thomas Klemm - 03-05-2022, 08:39 PM RE: Ln(x) using repeated square root extraction - Albert Chan - 03-06-2022, 12:50 AM RE: Ln(x) using repeated square root extraction - Thomas Klemm - 03-06-2022, 09:33 AM RE: Ln(x) using repeated square root extraction - Albert Chan - 03-09-2022, 07:47 PM RE: Ln(x) using repeated square root extraction - Thomas Klemm - 03-10-2022 05:18 AM RE: Ln(x) using repeated square root extraction - Albert Chan - 03-10-2022, 03:17 PM

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