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Ln(x) using repeated square root extraction
03-21-2016, 12:03 AM (This post was last modified: 04-11-2020 12:00 AM by Gerson W. Barbosa.)
Post: #1
Ln(x) using repeated square root extraction
Algorithm:

1. Enter x;
2. Take the square root 6 times;
3. Multiply by 2;
4. Subtract 1;
5. Take the square root;
6. Subtract 1;
7. Multiply by 64.

For x = 2, on an 8-digit calculator like the Canon LC-37, you should get ln(2) ~ 0.693152. For 10-digit calculators, in step 2 take the square root 9 times and multiply by 512 in the last step. Thus, on the HP-12C, which has two guard digits, we get ln(2) ~ 0.693147648. Ordinary 10-digit calculator might give less approximate results.

A 4-step less accurate algorithm is available in this old thread:

http://www.hpmuseum.org/cgi-sys/cgiwrap/...ead=145192

The accuracy range should be tested for your particular calculator. Use these at your own risk.

Gerson.


Method:


\[\ln (x)=\int_{1}^{x}\frac{du}{u}=\int_{1}^{x}u^{-1}du=\lim_{v\rightarrow 0}\left | \frac{u^{v}}{v} \right |_{1}^{x}=\lim_{v\rightarrow 0} \left ( \frac{x^{v}}{v}-\frac{1^{v}}{v} \right )=\lim_{v\rightarrow 0} \left ( \frac{x^{v}-1}{v} \right )\]

By testing the limit with a few values for v close to zero, an empirical second term has been able to be added to the expression:

\[\ln (x)=\lim_{v\rightarrow 0} \left ( \frac{x^{v}-1}{v}-\frac{v\cdot \ln^{2}(x)}{2} \right )\]

After turning the limit into an equality and solving the resulting quadratic equation for ln(x), we get

\[\ln (x)=\lim_{v\rightarrow 0} \left ( \frac{\sqrt{2\cdot x^{v}-1}-1}{v} \right )\]


Example:

Let x = 2 and v = 0.001

Then

\[\ln (2)\approx \frac{\sqrt{2\cdot 2^{0.001}-1}-1}{0.001}\approx 0.69314723\]

which is good to 6 decimal places ( ln(2) = 0.69314718 ).

P.S.: Perhaps 8 square root extractions and final multiplication by 256 (2^8) is a better compromise for the range 2..100 on the Canon LC-37. Contrary to what I thought its SILVA-CELL 189 battery is still working after all these years, so I was able to make some more tests. :-)


-----------------------------------

Good Friday 2020 Update


The following expression allows us to go further:

\[\ln (x)=\lim_{v\rightarrow 0} \left ( \frac{x^{v}-1}{v}-\frac{v\cdot \ln^{2}(x)}{2!} -\frac{v^{2}\cdot \ln^{3}(x)}{3!} -\frac{v^{3}\cdot \ln^{4}(x)}{4!} - \cdots \right )\]

By going up to the third power of ln(x) and solving the corrisponding cubic equation, we get

\[\ln (x)=\lim_{v\rightarrow 0}\frac{1}{v}\left ( \sqrt[3]{\sqrt{9x^{2v}-6x^{v}+2}+3x^{v}-1} - \frac{1}{\sqrt[3]{\sqrt{9x^{2v}-6x^{v}+2}+3x^{v}-1}} - 1 \right )\]

For example, if x = 2 and v = 0.01 then ln(2) ~ 0.69314719

This approximation is rather complicated for manual calculations. Also, it requires one cubic root extraction. It can be used, however, as an alternative method for implementing the ln(x) function to 7 significant digits on calculators that lack it, like the HP-16C. The e^x function has also been implemented. Since it's based on the first formula, obtained by solving a quadratic equation, and uses a constant number of loops, it gives less significant digits. The cubic root algorithm is optimized for the narrow range required by the approximation formula and was based one provided by Albert Chan here.

Code:

001- LBL A; LN
002- 1
003- EEX
004- CHS
005- 2
006- CF 1
007- 1
008- +
009- STO I
010- CLx
011- LASTx
012- x⇆y
013- x≤y
014- GTO 2
015- LBL 0
016- √x
017- RCL I
018- x>y
019- GSB 1
020- R↓
021- x⇆y
022- ENTER 
023- +
024- x⇆y
025- GTO 0
026- RTN
027- LBL 1
028- R↓
029- STO 0
030- R↓
031- 2
032- ×
033- STO 1
034- RCL 0
035- ENTER 
036- ×
037- 9
038- ×
039- 6
040- RCL 0
041- ×
042- -
043- 2
044- +
045- √x
046- 3
047- RCL 0
048- ×
049- +
050- 1
051- -
052- GSB 3
053- ENTER 
054- 1/x
055- -
056- 1
057- -
058- RCL 1
059- ×
060- F?1
061- CHS
062- RTN
063- LBL 2
064- SF1
065- 1/x
066- RTN 
067- LBL 3
068- 2
069- STO I
070- ENTER 
071- +
072- x⇆y
073- ×
074- STO 0
075- LASTx
076- 2
077- -
078- 6
079- /
080- √x
081- 1
082- +
083- LBL 4
084- ENTER 
085- ENTER 
086- RCL 0
087- x⇆y
088- /
089- LASTx
090- ENTER 
091- ×
092- -
093- 3
094- /
095- √x
096- +
097- 2
098- /
099- DSZ
100- GTO 4
101- RTN
102- LBL B; eᵡ
103- 1
104- 3
105- STO I
106- R↓
107- 8
108- 1
109- 9
110- 2
111- +
112- LASTx
113- /
114- ENTER
115- ×
116- 1
117- +
118- 2
119- /
120- LBL 5
121- ENTER 
122- ×
123- DSZ
124- GTO 5
125- RTN

Examples:

2 GSB A -> 0.6931471(360)
GSB B -> 2.0000(13165)

12345 GSB A -> 9.421006(848)
GSB B -> 12345.0(1957)

6.789 EEX 79 GSB A -> 183.8195(343) GSB B -> 6.(686887E79)

0.12345 GSB A -> -2.091919(104)
GSB B -> 0.123450(119)

230 GSB B -> 7.496895E99
GSB A -> 229.97041(15)

1 GSB A -> 0.000000000
GSB B -> 1.000000000
GSB B -> 2.7182(92170)
GSB B -> 15.1544(4181)
GSB A -> 2.71829(4016)
GSB A -> 1.000004(736)
GSB A -> 0.000004736

0.693147181 CHS GSB B -> 0.500000(16)

10 GSB A -> 2.302585(344) STO 2

2 GSB A RCL 2 / -> 0.3010299(435)

0 GSB A -> Error 0

2 CHS GSB A -> Error 0
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03-21-2016, 05:09 AM (This post was last modified: 03-21-2016 05:12 AM by Gerson W. Barbosa.)
Post: #2
RE: Ln(x) using repeated square root extraction
For the sake of completeness here is the recursive limit definition of ln(x), even though this is not quite necessary for our purpose:

\[\ln (x)=\lim_{n\rightarrow \infty} \left [ n\cdot \left ( x^{\frac{1}{n}}-1 \right )-\sum_{k=2}^{\infty }\frac{\ln ^{k}(x)}{k!\cdot n^{k-1}} \right ]\]

If the limit is removed and n is set to 1 then we'll have the following recursive series representation:

\[\ln (x)= \ x-1 -\sum_{k=2}^{\infty }\frac{\ln ^{k}(x)}{k!}\]
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03-21-2016, 07:11 AM
Post: #3
RE: Ln(x) using repeated square root extraction
I've got a feeling I tried something along these lines for the LN function of the 34S. The LN function has always been very slow on the 34S and I've tried several implementations to try to get both performance and accuracy. The current implementation is much faster than the original but still far slower than I'd like. I ended up with:

Code:
/* Natural logarithm.
 *
 * Take advantage of the fact that we store our numbers in the form: m * 10^e
 * so log(m * 10^e) = log(m) + e * log(10)
 * do this so that m is always in the range 0.1 <= m < 2.  However if the number
 * is already in the range 0.5 .. 1.5, this step is skipped.
 *
 * Then use the fact that ln(x^2) = 2 * ln(x) to range reduce the mantissa
 * into 1/sqrt(2) <= m < 2.
 *
 * Finally, apply the series expansion:
 *   ln(x) = 2(a+a^3/3+a^5/5+...) where a=(x-1)/(x+1)
 * which converges quickly for an argument near unity.
 */


- Pauli
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03-03-2022, 11:20 PM
Post: #4
RE: Ln(x) using repeated square root extraction
Formula

We can use the following identity:

\(\log(x) = n \log\left(x^{\frac{1}{n}}\right) = n \log\left(\sqrt[n]{x}\right)\)

For say \(n = 2^{10} = 1024\) and \(1 \leqslant x \leqslant 100 \) the value of \(\sqrt[n]{x}\) is close to \(1\).

Thus we can use the Taylor series to calculate the logarithm:

\(\log(1 + \varepsilon) = \varepsilon - \frac{\varepsilon^2}{2} + \frac{\varepsilon^3}{3} - \frac{\varepsilon^4}{4} + \frac{\varepsilon^5}{5} + \mathcal{O}(\varepsilon^6)\)

Program

Here's a program for the HP-42S that calculates both the logarithm and its approximation:
Code:
LN  LASTX
SQRT  SQRT  SQRT  SQRT  SQRT
SQRT  SQRT  SQRT  SQRT  SQRT
1  -
4  1/X  RCL× ST Y
3  1/X  X<>Y  -  RCL× ST Y
2  1/X  X<>Y  -  RCL× ST Y
1  X<>Y  -  ×
1024  ×

It's easy to extend if you want to use more terms.

Example

x = 2

0.69314718056
0.69314718056


Comparison

We can compare this to your solution:

\(\sqrt{2(1 + \varepsilon) - 1} - 1\)

The Taylor series agrees for the first two terms:

\(\varepsilon - \frac{\varepsilon^2}{2} + \frac{\varepsilon^3}{2} - \frac{5 \varepsilon^4}{8} + \frac{7 \varepsilon^5}{8} + \mathcal{O}(\varepsilon^6)\)
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03-05-2022, 05:32 PM
Post: #5
RE: Ln(x) using repeated square root extraction
I just tried this on my CASIO fx-4000P, and the result was 0.693147648
I did the square root 9 times, and multiplied with 512 in the last step.
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03-05-2022, 05:53 PM
Post: #6
RE: Ln(x) using repeated square root extraction
(03-05-2022 05:32 PM)Dan C Wrote:  I just tried this on my CASIO fx-4000P, and the result was 0.693147648
I did the square root 9 times, and multiplied with 512 in the last step.

I just had to try this on my CASIO fx-4500P also, and the result was the same,
0.693147648.

Now i just have to try this on my FX-602P also Smile
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03-05-2022, 08:02 PM
Post: #7
RE: Ln(x) using repeated square root extraction
On my FX-602P this gives 0.693147371
hmm, different from my fx-4000p and fx-4500p.
What does this tell of the FX-602P?
Less accurate?
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03-05-2022, 08:29 PM
Post: #8
RE: Ln(x) using repeated square root extraction
The extended limit for ln(x) does very well with values of v = 0.001 or less. Very impressive!

Namir
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03-05-2022, 08:39 PM
Post: #9
RE: Ln(x) using repeated square root extraction
We experience cancellation when calculating \(\varepsilon = \sqrt[n]{x} - 1\).
The program was executed on free42 with 34 decimal digits of precision.
Thus this effect is not noticed.

It is a trade-off between coming close to \(1\) and how many terms to use in the Taylor series.

For a calculator like the HP-15C with 10 digits using \(n = 2^6\) might be the best choice.

For \(x = 2\) we end up with the following sequence of iterated square roots:

2.000000000
1.414213562
1.189207115
1.090507733
1.044273783
1.021897149
1.010889286

This leaves us with \(\varepsilon = 0.010889286\) and thus 5 terms of the Taylor series should be enough.
We notice the cancellation since we have now only 8 significant digits left.

Here's the program for the HP-15C or similar calculators:
Code:
   001 {          11 } √x̅
   002 {          11 } √x̅
   003 {          11 } √x̅
   004 {          11 } √x̅
   005 {          11 } √x̅
   006 {          11 } √x̅
   007 {           1 } 1
   008 {          30 } −
   009 {          36 } ENTER
   010 {          36 } ENTER
   011 {          36 } ENTER
   012 {           5 } 5
   013 {          10 } ÷
   014 {           4 } 4
   015 {          15 } 1/x
   016 {          34 } x↔y
   017 {          30 } −
   018 {          20 } ×
   019 {           3 } 3
   020 {          15 } 1/x
   021 {          34 } x↔y
   022 {          30 } −
   023 {          20 } ×
   024 {           2 } 2
   025 {          15 } 1/x
   026 {          34 } x↔y
   027 {          30 } −
   028 {          20 } ×
   029 {           1 } 1
   030 {          34 } x↔y
   031 {          30 } −
   032 {          20 } ×
   033 {           6 } 6
   034 {           4 } 4
   035 {          20 } ×

For \(x = 2\) we get:

0.6931471776

While for \(\ln(2)\) we get:

0.6931471806

This leaves us with a difference of:

0.0000000030
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03-06-2022, 12:04 AM
Post: #10
RE: Ln(x) using repeated square root extraction
(03-03-2022 11:20 PM)Thomas Klemm Wrote:  We can compare this to your solution:

\(\sqrt{2(1 + \varepsilon) - 1} - 1\)

The Taylor series agrees for the first two terms:

\(\varepsilon - \frac{\varepsilon^2}{2} + \frac{\varepsilon^3}{2} - \frac{5 \varepsilon^4}{8} + \frac{7 \varepsilon^5}{8} + \mathcal{O}(\varepsilon^6)\)

log1p(ε) = 2*atanh(y) = 2*(y+y^3/3+y^5/5+...) ≈ 2y, where y=ε/(2+ε)

2ε/(2+ε) is simpler than √(1+2ε)-1, and twice as accurate, when ε is tiny.

CAS> series(2*ε/(2+ε),ε) // estimate for log1p(ε)

ε - 1/2*ε^2 + 1/4*ε^3 - 1/8*ε^4 + 1/16*ε^5 + ε^6*order_size(ε)

CAS> (1/3-1/2)/(1/3-1/4)

-2.

Is is funny sqrt approximation formula give better estimate for log1p, than sqrt itself.

CAS> pade(√(1+2*x)-1,x,2,2)

2*x/(x+2)
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03-06-2022, 12:50 AM
Post: #11
RE: Ln(x) using repeated square root extraction
(03-05-2022 08:39 PM)Thomas Klemm Wrote:  For \(x = 2\) we end up with the following sequence of iterated square roots:

2.000000000
1.414213562
1.189207115
1.090507733
1.044273783
1.021897149
1.010889286

This leaves us with \(\varepsilon = 0.010889286\) and thus 5 terms of the Taylor series should be enough.
We notice the cancellation since we have now only 8 significant digits left.

It is more work, but we could improve accuracy by "pulling out" 1.

CAS> f(x) := x/(sqrt(1+x)+1) // == sqrt(1+x)-1
CAS> 1. // x-1 = 2-1 = 1
CAS> f(Ans)
0.414213562373
0.189207115003
9.05077326653e−2
4.42737824274e−2
2.18971486541e−2
1.08892860517e−2

The recursive formula, log1p(x) = 2*log1p(x/(sqrt(1+x)+1)), is similar to atan formula
(05-31-2021 09:51 PM)Albert Chan Wrote:  \(\displaystyle\arctan(x) = 2\arctan\left( {x \over \sqrt{1+x^2}+1} \right)\)
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03-06-2022, 09:33 AM
Post: #12
RE: Ln(x) using repeated square root extraction
I was curious to try Pauli's idea and use:

\(\log\left(\frac{1+\varepsilon}{1-\varepsilon}\right) = 2 \left( \varepsilon + \frac{\varepsilon^3}{3} + \frac{\varepsilon^5}{5} + \mathcal{O}(\varepsilon^7) \right) \)

We set:

\(\frac{1+\varepsilon}{1 - \varepsilon} = x\)


This leads to:

\(\varepsilon = \frac{x - 1}{x + 1}\)


Here's the corresponding 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 {          36 } ENTER
   008 {          36 } ENTER
   009 {           1 } 1
   010 {          30 } −
   011 {          34 } x↔y
   012 {           1 } 1
   013 {          40 } +
   014 {          10 } ÷
   015 {          36 } ENTER
   016 {          36 } ENTER
   017 {          36 } ENTER
   018 {           5 } 5
   019 {          10 } ÷
   020 {          20 } ×
   021 {           3 } 3
   022 {          15 } 1/x
   023 {          40 } +
   024 {          20 } ×
   025 {          20 } ×
   026 {           1 } 1
   027 {          40 } +
   028 {          20 } ×
   029 {           1 } 1
   030 {           2 } 2
   031 {           8 } 8
   032 {          20 } ×

The computation of \(\varepsilon\) is now a bit more complicated, but we have fewer terms to compute.
In the end, the program is shorter.

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

0.6931471773


Compared to the previous result, it is only slightly off in the last digit.
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03-09-2022, 07:47 PM
Post: #13
RE: Ln(x) using repeated square root extraction
Instead of reducing argument, √...√x, we can blow it up, and get log from AGM

From Series[EllipticK[1-(4x)^2], {x,0,4}], and ignored imaginery parts:

K(m=1-(4/x)^2) = log(x) + (log(x)-1)*(4/x^2) + O(1/x^4)

lua> x = 2^14
lua> a, b = 1, 4/x
lua> c = b/x
lua> for i=1,6 do a,b = (a+b)/2, sqrt(a*b); print(a,b) end
Code:
0.5001220703125         0.015625
0.25787353515625        0.08839913658307309
0.17313633586966154     0.15098277337311447
0.162059554621388       0.16168056210089243
0.1618700583611402      0.16186994744240293
0.16187000290177156     0.16187000290176207

AGM have quadratic convergence.
With 6 sqrt, (a,b) already close enough so that (a+b)/2 ≈ AGM

lua> k = pi/(a+b)
lua> k / 14
0.6931471898242749
lua> (k+c)*(1-c) / 14
0.6931471805599455
lua> log(x) / 14 -- = log(2)
0.6931471805599453

Note: if c=4/x^2 small enough, below machine epsilon, log(x) = k, no need for correction.
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03-10-2022, 05:18 AM
Post: #14
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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03-10-2022, 03:17 PM (This post was last modified: 03-12-2022 03:49 PM by Albert Chan.)
Post: #15
RE: Ln(x) using repeated square root extraction
(03-10-2022 05:18 AM)Thomas Klemm Wrote:  The first 3 terms of the Taylor series of this expression agree with those of \(\log\left(\frac{1+\varepsilon}{1-\varepsilon}\right) \):

\( \frac{2\varepsilon}{9} \left( 4 + \frac{5}{1 - \frac{3\varepsilon^2}{5}} \right) = 2 \varepsilon + \frac{2\varepsilon^3}{3} + \frac{2\varepsilon^5}{5} + \frac{6\varepsilon^7}{25} + \frac{18\varepsilon^9}{125} + \mathcal{O}(\varepsilon^{11}) \)
FYI, approximation formula based from atanh(ε) pade approximation:

atanh(ε) = ε + ε^3/3 + ε^5/5 + ... ≥ ε + (ε^3/3) / (1 - 3/5*ε^2)

Let d = 1 - 3/5*ε^2, we have ε^2/3 = 5/9*(1-d)

ε + (ε^3/3)/d = ε*(1 + 5/9*(1-d)/d) = ε/9*(9 + 5*(1/d-1)) = ε/9*(4 + 5/d)

ln((1+ε)/(1-ε)) = 2*atanh(ε) ≥ 2ε/9*(4 + 5/(1-3/5*ε^2))

(10-22-2021 02:15 PM)Albert Chan Wrote:  ln(n) ≈ g - g/(2.7 + 24/g^2)       , where g = (n-1)/√n (*)

We doubled accuracy if we use above formula instead (Bonus, less code steps)
Again, taylor series of ln((1+ε)/(1-ε)), using above formula for ln:

XCAS> g := (x-1)/sqrt(x)
XCAS> f := g - g/(27/10+24/g^2)
XCAS> series(f(x=(1+ε)/(1-ε)),ε,0,10)

2*ε + 2/3*ε^3 + 2/5*ε^5 + 123/400*ε^7 + 3217/12000*ε^9 + O(ε^11)

XCAS> (c - 6/25) / (c - 123/400) | c=2/7.

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