Global Natural Logarithm Formula

[sandokhan]

The common way to calculate logarithm involves tables of logarithm, the mantissa and the characteristic. What is needed is a global, explicit formula for the logarithm.

LN V = 2ⁿ x (V^1/2n+1 - 1/V^1/2n+1)

This is the first explicit global formula for the natural logarithm, which can be used immediately to find LN V without resorting to logarithm tables, or calculators which feature the logarithm key: all we need is a calculator which has the four basic operations and the square root key. It links algebraic functions with elementary and higher transcendental functions.

For a first approximation,

LN V = 2ⁿ x (V^1/2n - 1)

First results appear for n = 8 to 12, all the remaining digits for n = 19 and higher...

Example: x = 100,000        LN x = 11.5129255

with n = 20 the first approximation is LN x = 11.512445 (e^11.512445 = 100,001.958)

 

Proof

I first derive the global cosine formula, then the global hyperbolic cosine formula, then one can easily find the global natural logarithm formula as well.

 

The hypothenuse is labeled as c (which unites points A and C), side a is located on the x axis (which unites points A and B), and we also have side b. Angle θ is located between sides c and a (cos θ = a/c).

Point D will be the intersection of the circle with the positive x axis.

We first calculate the value of segment CD, in terms of a, b and c: (2c² - 2ac)^1/2

We then succesively bisect the chord CD, and each hypothenuse thus obtained (if we divide CD in half the midpoint will be E, and the intersection of the segment AE with the circle will be labeled as F; then we calculate this new hypothenuse CF in terms of the values obtained earlier, and so on, aiming to get as close to the value of s [arc length of CD] as possible], into smaller and smaller equal segments, calculating each succesive value in terms of a, b and c, in order to obtain a very close approximation of s (the arc length between points C and D); since s = rθ, where r = c = 1, by acquiring an exceptional figure for s, we correspondingly then get the value of θ.

Letting c = 1, we finally obtain:


COS θ =  1/2 x (({ [( (2 - θ²/2^N)² - 2)²...]² - 2}² - 2))    (n/2 + 1 evaluations)

COS^-1 θ =  2ⁿ x {2 - ((2 + {2 + [2 + (2 + 2θ)^1/2]^1/2}^1/2...))}^1/2   (n + 1 evaluations)


The cosine formula is a GLOBAL formula; by contrast the Maclaurin cosine series is a local formula:

 

My global cosine formula is the SUM of the Maclaurin cosine expansion.


We know that the Maclaurin hyperbolic cosine expansion is:

cosh x = 1 + x²/2! + x⁴/4! + ...

Therefore, by just changing the sign in the global cosine formula, we obtain immediately the GLOBAL hyperbolic cosine formula:

COSH V =  1/2 x (([(({[(2 + V²/2ⁿ)² - 2]²} - 2))²...-2]² - 2))   (n/2 + 1 evalutions)

This is the global hyperbolic cosine formula which is the sum of the corresponding local Maclaurin power series expansion.


We then immediately obtain the GLOBAL natural logarithm formula:

LN V =  2ⁿ x ((-2 + {2 + [2 + (2 + 1/V + V)^1/2]^1/2...}^1/2))^1/2   (n+1 evaluations)


By summing the nested continued square root function, we finally obtain:


LN V = 2ⁿ x (V^1/2n+1 - 1/V^1/2n+1)

We also can get the corresponding arctangent formula:

ARCTAN V =  2ⁿ x ((2- {2+ [2+ (2+ 2{1/(1+ V²)}^1/2)^1/2]...^1/2}))^1/2 (n+1 parentheses to be evaluated)


ERROR ANALYSIS

Here is the Maclaurin expansion for e^x:

Let us obtain a remainder form for the Maclaurin expansion for e^x (Lagrange remainder):

R_n(x) = f⁽ⁿ⁺¹⁾(c)[xⁿ⁺¹]/(n+1)! , where c is between 0 and x

f⁽ⁿ⁺¹⁾(c) = e^c


An approximation is said to be accurate to n decimal places if the magnitude of the error is less than 0.5 x 10^-n.

e¹ to four decimal place accuracy:

R_n = e^c/(n+1)!

since c<1, then e^c < e

since e<3, then R_n < 3/(n+1)!

then, for n=8 we will obtain four decimal place accuracy.


Local formulas are difficult to use because of their very slow convergence.


By contrast, my formula is a GLOBAL formula, which rapidly converges to the result, even for large x.


Cosh v = (e^v + e^-v)/2 =~ 1/2e^v = 1/2 x (({[( ( (2 + v²/2ⁿ)²) -2)²] -2)² ...-2}² -2))       (n/2 +1 evaluations)

We can turn this formula into an exact formula for e^v by simply substituting y for e^v, and then solve the quadratic equation for y.

One might ask, could you not use Taylor expansions to obtain cosh 10 (as an example)? No, because you would need some other value to start with, cosh 9.5 or cosh 9.8 or cosh 10.3, to apply Taylor series.

With my global formula, no such approximations are needed, we can start directly with the value v = 10.

My formula also has a built-in remainder approximation estimation: the term v²/2ⁿ.

That is, we can estimate the accuracy from the very start: this is the power of a global formula.

The higher the value of n, the better the approximation that we will obtain.


Example:

COSH 10 = 11013.233

10²/2²⁰ = 0.00009536

Using the global hyperbolic cosine formula with n = 20, we get: 11012.762

 

Second Proof

One of the readers of my messages has sent this proof for my global logarithm formula which however uses mathematical analysis:

[studiot]

1 hour ago, sandokhan said:

For a first approximation,

LN V = 2ⁿ x (V^1/2n - 1)

First results appear for n = 8 to 12, all the remaining digits for n = 19 and higher...

Example: x = 100,000        LN x = 11.5129255

with n = 20 the first approximation is LN x = 11.512445 (e^11.512445 = 100,001.958)

If I put n = 10 and use your V = 100000 , as well as a calculator, I get

[math]\ln 100000 = {2^{10}}\left\{ {{{100000}^{\frac{1}{{{2^{10}}}}}} - 1} \right\} = 11.578[/math]

So the formula is not bad even for n = 10.

 

However how am I supposed to evaluate

[math]{{{100000}^{\frac{1}{{{2^{10}}}}}}}[/math]

?

I used the y^x button, but how is that better than using the lnx button directly?

 

Edit I see the formula also works for decimal fractions eg

n = 10, V = 0.01 yields -4.5945

ln(0.01) yields -4.6

Edited March 20, 2019 by studiot

[sandokhan]

We use only the square root function to evaluate V^1/2n

In fact, using a continued fraction algorithm to calculate the sequence of square roots, we'd only need a very simple calculator which features the four basic arithmetic operations. What this formula also does is to show that the logarithm is actually a sequence of nested square roots.

Edited March 21, 2019 by sandokhan

[John Cuthber]

It's an interesting identity, but I'm not sure it's useful.

How fast can you calculate it and how fast is the conventional method?

[sandokhan]

Each of the global formulas is the sum of the corresponding Maclaurin expansion. We now have a global, explicit formula for the natural logarithm: it makes any and all logarithm tables obsolete. 

What you are referring to is the case where you have a hand held calculator (or an online calculator) and you need to find the value of lnx or e^x. Even in that case, this direct and explicit formula can replace the algorithms currently used to calculate this function, which are quite complex. A continued (nested) square root function provides a much faster algorithm to calculate the needed logarithm expression. I can calculate any logarithm using only a hand held calculator which features the four basic arithmetic operations, using this global formula. 

[studiot]

4 hours ago, sandokhan said:

We use only the square root function to evaluate V^1/2n

In fact, using a continued fraction algorithm to calculate the sequence of square roots, we'd only need a very simple calculator which features the four basic arithmetic operations. What this formula also does is to show that the logarithm is actually a sequence of nested square roots.

Thank you for the answer.

I hadn't appreciated you were actually taking repeated square roots.

What method woud you use for taking the 10 to 20 square roots of a 10 digit number on a four function calculator?

[sandokhan]

I would use either the Newtonian iteration formula (if tools from mathematical analysis are permitted) or a continued fraction algorithm.

[studiot]

21 minutes ago, sandokhan said:

I would use either the Newtonian iteration formula (if tools from mathematical analysis are permitted) or a continued fraction algorithm.

So the proceedure has many more steps than meets the eye at first glance.

[sandokhan]

I always try to obtain formulas without the use of mathematical analysis. In fact, the humble tools of basic arithmetic are needed to resolve the Riemann hypothesis, a fact discovered recently:

It is my belief that RH is a genuinely arithmetic question that likely will not succumb to methods of analysis. Number theorists are on the right track to an eventual proof of RH, but we are still lacking many of the tools. 

J. Brian Conrey

And there are much more difficult problems than the basic RH: what to these zeros of the zeta function actually represent? Is there a hidden pattern to them, unobserved to this day?

My formula simply tells us that the logarithm function is actually a sequence of continued square roots, and that it can be used to replace the currently used algorithms (for finding the logarithm) in hand held calculators/online calculators. We no longer need the tedious logarithm tables, a direct global formula where we can see at a glance everything we need to know. The global arctangent formula also might have applications as it is a continued function and provides a faster to compute the arctan function than using the extended series from calculus. I obtained these formulas without using mathematical analysis, just simple computational procedures.

[studiot]

1 hour ago, sandokhan said:

I always try to obtain formulas without the use of mathematical analysis. In fact, the humble tools of basic arithmetic are needed to resolve the Riemann hypothesis, a fact discovered recently:

It is my belief that RH is a genuinely arithmetic question that likely will not succumb to methods of analysis. Number theorists are on the right track to an eventual proof of RH, but we are still lacking many of the tools. 

J. Brian Conrey

And there are much more difficult problems than the basic RH: what to these zeros of the zeta function actually represent? Is there a hidden pattern to them, unobserved to this day?

My formula simply tells us that the logarithm function is actually a sequence of continued square roots, and that it can be used to replace the currently used algorithms (for finding the logarithm) in hand held calculators/online calculators. We no longer need the tedious logarithm tables, a direct global formula where we can see at a glance everything we need to know. The global arctangent formula also might have applications as it is a continued function and provides a faster to compute the arctan function than using the extended series from calculus. I obtained these formulas without using mathematical analysis, just simple computational procedures.

 

I am not suprised that an arithmetic connection can be made since powers and roots are linked arithmetically already both between different powers and the base of those powers.

 

[John Cuthber]

If you ask most people how they would calculate square roots (without a calculator ) they would say they use logarithms...

 

Is this new method of finding ln(x) quicker or less demanding of memory than the usual methods (I think they used chebyshev polynomials last time I looked into it)?

If not, it's interesting, but not really useful.

[sandokhan]

At the present time, a global/explicit formula for the logarithm or the summing of the Maclaurin expansions for the cos/arccos/arctan/cosh functions are not envisioned to be possible. Yet, these formulas prove that it is possible. As for the computational power, I think it would be best to have this formula compared to the usual algorithms (which are very complex) utilized in hand held/online calculators.

The Riemann hypothesis and the factorization of large semiprimes require methods, new ideas, from arithmetic in order to be finally solved.

Let us briefly discuss the factorization of large semiprimes.

For a 200 digit number (semiprime), the required computational time (1990) for the methods then used in integer factorization will take 4 x 10¹⁵ years.

For a 300 digit number, we would need 5 x 10²¹ years

For a 500 digit number, the figure would rise to 4.2 x 10³² years.

b₁, a₁ and c₁ are the three sides of a right triangle

b₁² + a₁² = c₁²

b₁ = d₁ x d₂ (divisors of b₁)

a₁ = (d₁² - d₂²)/2

c₁ = (d₁² +d₂²/2

If b₁ is prime, then b₁² + a₂² = c₂² (where c₂ = (b₁² +1)/2 )

Modern geometry/trigonometry tells us that Pythagoras' theorem is the only known relationship relating the three sides of a right triangle, in a single equation.

But there is another equation relating the three sides of right triangle:

b₁^2sc + a₁^2sc =~ [(b₁ + a₁ + c₁)/2]^2sc + ...

sc = either 0.63662... (the sacred cubit) or 0.618034... (phi), this matter needs to be resolved.

Since a₁ + c₁ = d₁², with a reasonable estimate for a₁, we can obtain a very good approximation for d₁.

The Fibonacci numbers are actually sacred cubit numbers.

1,618034 = 4sc² (1sc = 0.636009827, in this case)

Then F_n = 1/(8sc² -1) x 2²ⁿ x sc²ⁿ

The first formula proves to be enough to completely solve the large integer factorization involving a semiprime having 10 or less digits. The right side of the equation is an asymptotic expansion, I was able to obtain the main term; of course, adding more terms of this expansion (a very difficult endeavor), would mean we can factorize semiprimes which have more than 10 digits, the accuracy depending on the number of terms in the expansion.

Of course, to attempt to solve the large semiprime factorization problem beyond the case where b₁ has more than 10-20 digits, would mean we need a more precise algorithm

List of Fibonacci numbers (F_n) (sacred cubit sequences):

http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibtable.html


b₁ has less than 10 digits

How to obtain a reasonable estimate for a₁


b₁ = 8141 x 131071 = 1073602561

1073602561 = 28657² + 252378911 = 28657 x 46368 - 255165215  

F₂₃ = 28657 

F₂₄ = 46368 

If b₁<a₁, then the a₁ term will be of the form F₂₄² - ..., F₂₄ x F₂₅ - ..., F₂₅ x F₂₆ - ..., or F₂₅² - ...

if b₁>a₁, then a₁ will equal F₂₃ x F₂₄ - ..., F₂₂² - ... , that is, only 4-6 possible choices.


In order to get a very good estimate for d₁, we will use the first remainder (and a few subsequent remainders if needed, more explanation below) obtained from the b₁ for each of the above choices .

For the a₁ =  F₂₅ x F₂₆ - ... choice, using a₁₀ = 255165215, and substituting in the first formula, we get: 

d₁ = 132578.957, an excellent approximation.

Actually, a₁ = 8556257280 = 75025² + 2927506655 = 75025 x 121393 - 551252545


b₁ = 65537 x 131071 = 8590000127 = 75025² + 2961249502 = 75025 x 121393 - 517509698

F₂₅ = 75025

Using the same reasoning and the same formula, we get a first estimate for d₁, d₁ = 130095.707

The sacred cubit hidden pattern of the natural number system can be used to obtain as much information as possible out of the b₁ semiprime.

b₁ = 821 x 941 = 772561

772561 = 610 x 987 + 170491 = 987² - 201608

170491 = 377² + 28362 = 377 x 610 - 59479

F₁₅ = 610
F₁₆ = 987

We use each and every remainder obtained by dividing b₁ by Fibonacci numbers, in a similar sequence: each subsequent remainder expressed as in the classic division formula (a = qd + r, where q and d are Fibonacci numbers, while r is the remainder to be used in the next division process)

201608 = 377 x 610 - 28362 = 377² + 59479

F₁₄ = 377

28362 = 144 x 233 - 5190 = 144² + 7899

59479 = 233² + 5190 = 233 x 377 - 28362

7899 = 89² - 22 = 89 x 55 + 3004

5190 = 89 x 55 + 295 = 89² - 2731


3004 = 55² - 21 = 55 x 34 + 1134

2731 = 55² - 294 = 55 x 34 + 861


1134 = 34² - 22 = 34 x 21 + 420

861 = 34 x 21 + 147 = 34² - 295

420 = 21² - 20 = 21 x 13 + 147

294 = 21 x 13 + 21 = 21² - 147 ; 147 + 21 = 168

147 = 21 x 8 - 21 = 13 x 8 + 43

43 = 8 x 5 + 3 = 8² - 21

21 = 5 x 3 + 6 = 5² - 4


Interestingly, we can immediately obtain a first approximation for d₁, d₁ = 918; by summing the remainders of b₁ in their corresponding order, for a 3 digit d₁ divisor. Several such sums can be obtained (where d₁ can be assumed to have 3, 4, 5 digits) and one of them will actually represent a nice estimate of d₁.


The crucial observation is that we can actually get the remainders of the a₁ term either by noticing that 6 and 4 (remainders obtained by dividing 21 by F₅ and F₄) can be used to initiate the a₁sequence of remainders starting from the bottom up, or by using a very interesting shortcut involving b₁^sc, where this can be applied.

Actually, a₁ = 105720 = 377² - 36409 = 377 x 233 + 17879

Using the same scheme as above for the a₁ term (same division by Fibonacci numbers algorithm as was utilized for the b₁ term) we finally get:

40 = 8² - 24 = 8 x 3 + 16

16 = 5² - 9 = 3² + 7

9 = 3 x 5 - 6 = 2 x 3 - 3


65 = 8² + 1 = 8 x 13 - 39

39 = 8 x 5 - 1 = 5² + 14

14 = 3 x 5 -1 = 2 x 5 + 4


Knowing that 6 and 4 are the remainders of a₁, we can see that from the possible choices we eventually get (11, 19, 9, and 14) only 9 and 14 will make any sense, given the fact that the remainders at each stage of the calculation have to be expressed as in the classic division formula (a = qd + r, where q and d are Fibonacci numbers, while r is the remainder).


One of the remainders of a₁ will be 2857.


3004 - 2857 = 147



772561^sc = 5530


5530 - 5063 = 2 x 233

5530 - 2857 = 89 x 30

(5063, another a₁ remainder)

That is, there is a certain symmetry and relationship between b₁^sc and some of the a₁ remainders.



Another example.

b₁ = 1000009

For 1000009 = 3413 x 293, we get a first estimate of 3486, and by summing the remainders of b₁(576230 + 204130 + 62001 + 25840 + 5104 + 2817 + 947 ...) we get an estimate of 3400, which is amazing, because we only use the remainders from b₁ and very simple approximations.


For 1000009, b₁^sc = 6515.72

9368 - 6515.72 = 610 x 4.66 = 987 x 2.88  (4.66 = 2 x 2.33 , and 2.88 = 2 x 1.44, both 233 and 144 are Fibonacci numbers)

9368 is one of the a₁ remainders

Another a₁ remainder is 3448

6515.72 - 3448 =~ 55² = 233 x 13



Thus, the factorization of semiprimes is related to the sacred cubit, and I believe the above algorithm is a start in studying further this new approach to solving this problem, based on the power of the sacred cubit.

For moderately large b₁ such as:

2³¹ - 1 = 2147483647

2⁶¹ - 1 = 2.3059 x 10¹⁸

b₁ = (2³¹ - 1) x (2⁶¹ - 1) = 4.951760152 x 10²⁷

a computer which can handle the full/entire number of digits could be used to verify the algorithm proposed above, and to see if the same relationship exists between the sequence of remainders obtained for both b₁ and a₁.

In fact, with an a₁ trial function 4 x 10³⁵, we get an estimate for d₁ = 2.353 x 10¹⁸.

Of course for the exact answer, we would need to verify the correctness of the above algorithm, involving the sequence of remainders obtained upon division by the corresponding Fibonacci numbers.
 

Edited March 22, 2019 by sandokhan