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Square root formulas with initial control of the correct decimals

precision output.

Mones Kasem Jaafar

Independent Researcher

monesjaafar@gmail.com

P.O. Box 11231 Damascus Syria

Abstract

All the square root calculation methods and algorithms can not replace the square root

√ function in equations, they requires an initial guess or starting value and / or many

calculations steps or iterations to reach the square root at an arbitrarily precision

which mean they need an algorithm to perform the calculation, until now it is

impossible to deduce the square root of positive real number from a function.

I have successfully build and tested many formulas to calculate the square root of all

the positive real number in one step of calculation, based on a discovery of a pattern,

this is not a calculation method only it is a mathematic law and formulas that illustrate

the incontestable relation between the square root and the trigonometric function

according to a specific pattern, where the output precision could be controlled and

predetermined by determining the desired correct numbers of decimals initially that

do not necessitate any guess or starting values. These formulas can replace the square

root function in equations for many purposes according to a predetermined

precisions.

Keywords

Square root formulae; Root; controlled precision; starting value; guess value; Square

root calculation method

1. Introduction

The square root calculation methods such as Newton method [5,7,8], continued

fraction expansion [1] , reciprocal square roots [1] and many algorithms [6], are based

on guess and / or iterations and many calculations steps [1,6,7] to reach a desired

precision of the square root, with a developed algorithm we can certainly calculate

exactly the square root of any number at an arbitrarily and non predictable precision,

but there is no formula that can substitute the square root function √ in any equation

to calculate the square root in one step with a chosen and predictable precision.

I have developed many formulas to calculate the square root of any positive real

number without starting or guess value and without iteration or many calculations

steps or algorithm, instead I Introduced a precision factor input to control the desired

correct numbers of decimals initially according to the need of calculation, because in

practice the number of decimals is always fixed and limited, whatever the purpose of

the calculation even if we know exactly all the correct decimals that may extent to the

infinity, this is completely different from an approximate calculations since the

accuracy is controlled and known initially without limitation through this precision

factor , the formula is calculated using Pythagorean theorem [2]. which is not a

calculation method only it is mathematic law that illustrate the incontestable relation

between the square root and the trigonometric function according to a specific pattern.

These formulas can replace the square root √ function in mathematic and scientific

calculation and analysis for many purposes.

2. Methods

Consider the right triangle such as:

10n is the precision factor that determine initially the output precision of the square

root calculation:

n=1,2,3…n

m is the number of digits of x

n ≥ m to ensure the accuracy

γ = π/2

By Pythagoras we have that :

b2 = (√x)2 + (10n x)2

b = √((√x)2 + (10n x)2)

b = √(x + 102n x2) (1)

n = 1

n = 2

n = 3

n = 4

x

b = √(x + 102x2)

b = √(x + 104x2)

b = √(x + 106x2)

b = √(x + 108x2)

1

10.0498756211209

100.0049998750060

1000.0004999998800

10000.0000499999998

2

20.0499376557634

200.0049999375020

2000.0004999999400

20000.0000499999999

3

30.0499584026334

300.0049999583340

3000.0004999999600

30000.0000499999999

4

40.0499687890016

400.0049999687500

4000.0004999999700

40000.0000499999999

5

50.0499750249688

500.0049999750000

5000.0004999999700

50000.0000499999999

6

60.0499791840097

600.0049999791670

6000.0004999999800

60000.0000499999999

7

70.0499821556009

700.0049999821430

7000.0004999999800

70000.0000499999999

8

80.0499843847580

800.0049999843750

8000.0004999999800

80000.0000499999999

9

90.0499861188218

900.0049999861110

9000.0004999999900

90000.0000499999999

10

100.049987506246

1000.004999987500

10000.000499999990

100000.000049999999

11

110.049988641526

1100.004999988640

11000.000499999990

110000.000049999999

12

120.049989587671

1200.004999989580

12000.000499999990

120000.000049999999

Table (1) b calculation using formulae (1) with different n values.

b

10n x

√ x

α

β

γ

3. Results

Through numerical analysis and as shown in table (1) I found that there is a pattern

concerning b value proportional to n and 10n which is limited to (n+1) decimals

precision and correct decimals of b so that we can write the formula (1) according to

m digits input of x and for (n+1) correct number of decimals precision for b

calculation:

b = 10n x + 0.5. 10-n (2)

The mechanism of the new formula (2) is that the rules of n and 10n is to emphasis

the pattern by extremely decreasing α and increasing β and pushing the "mess" of

numbers away and "replacing" them by a controllable and predetermined known

number of correct decimals and precision by simply increasing n value.

The rule of the precision factor 10n is to control the precision output of the

calculation and reorganize the formula, this factor has a contradictive effect at the

same time on formula (2) it increase greatly the term 10n x and decrease greatly the

term 0.5. 10-n, this very small term must not be rounded to "0" or discarded in any

case, these two terms must stay together to perform the calculations.

Since this formula is correct up to (n+1) correct decimals output of b calculation, it

is not important anymore and we will not consider decimals after that, instead we can

expand indefinitely the precision of b calculation and the corrects decimals by simply

increasing n value.

Since that the formula (2) is not an approximate formula but a controlled and

predictable precision formula of correct decimals, so the term 0.5. 10-n must not be

rounded in any case specially for big value of n

We can conclude then the following formulas:

β = asin(10n x/b)

β = asin( x / (x + 0.5. 10-2n )) (3)

α = acos(10n x/b)

α = acos(x / (x + 0.5. 10-2n ) ) (4)

tan (β) = (10n x/√x) = 10n √x

√x = tan (β)/ 10n

√x = tan (asin( x / (x + 0.5. 10-2n ))) / 10n (5)

tan (α) = √x /( 10n x) = 1/( 10n √x)

√x = 1/( 10n tan (α) )

√x = 1/( 10n tan(acos(x / (x + 0.5. 10-2n )))) (6)

sin(α) = √x/b

√x = b sin(α)

√x = (10n x + 0.5. 10-n) sin(acos(x / (x + 0.5. 10-2n ))) (7)

cos(β) = √x /b

√x = b cos(β)

√x = (10n x + 0.5. 10-n) cos(asin( x / (x + 0.5. 10-2n ))) (8)

From numerical analysis, and according to the precision chosen, we can consider the

angle α as equal to tan(α ) = 1/(10n √x), according to the following formulas:

√x = 1/(10n acos( x / (x + 0.5. 10-2n ))) (9)

We can write also

√x = 10n x acos( x / (x + 0.5. 10-2n )) (10)

3.1. Example of √x calculation using formula (5)

√x = tan (asin( x / (x + 0.5. 10-2n ))) / 10n

n = 1

n = 2

n = 3

n = 4

x

√x

Formula (5)

Formula (5)

Formula (5)

Formula (5)

1

1

0.9987523389

0.9999875002

0.9999998750

0.9999999987

2

1.4142135624

1.4133305067

1.4142047236

1.4142134739

1.4142135640

3

1.7320508076

1.7313295705

1.7320435907

1.7320507352

1.7320508068

4

2

1.9993752928

1.9999937500

1.9999999385

1.9999999993

5

2.2360679775

2.2355091700

2.2360623873

2.2360679209

2.2360679769

6

2.4494897428

2.4489795918

2.4494846397

2.4494896923

2.4494897422

7

2.6457513111

2.6452789820

2.6457465865

2.6457512626

2.6457513105

8

2.8284271247

2.8279852866

2.8284227054

2.8284270786

2.8284271243

9

3

2.9995834201

2.9999958333

2.9999999607

2.9999999995

10

3.1622776602

3.1618824496

3.1622737073

3.1622776249

3.1622776613

11

3.3166247904

3.3162479654

3.3166210215

3.3166247540

3.3166247914

12

3.4641016151

3.4637408276

3.4640980067

3.4641015851

3.4641016147

Table (2) Square root calculation using formula (5) with different n values.

3.2. High precision example

let us choose n=18 and calculate√2 using formula (5) with high precision program [9]:

√2 = tan (asin( 2 / (2 + 0.5. 10-36 )) / 1018

The result is :

1.4142135623730950488016887242096980784812835277286296

2 n = 36 correct decimals

Since the correct square root of 2 is given as follow:

1.414213562373095048801688724209698078569671875376948073

3.3 Derivative formulas:

Through testing and numerical analysis I found that we can derivate the following

formulas:

Since we have:

x = (√x)2

then we can write also:

x = tan2 (asin( x / (x + 0.5. 10-2n )) / 102n (11)

this can be done for all the above formulas

Through testing and numerical computing analysis I found that:

tan (asin((x + 0.5. 10-2n ) / x )) / 10n = -√x i (12)

tan (asin( x / (x + 0.5. 10-2n )±π/2)).10n = - 1/√x (13)

tan ( k. asin( x / (x + 0.5. 10-2n )))/ 10n = √x / k (14)

for k odd and integer and n > 3

tan ( k. asin( x / (x + 0.5. 10-2n ))). 10n = - k / √x (15)

for k even and integer and n > 3

10n/(tan(k . acos(x / (x + 0.5. 10-2n )) ±π/2) = -k/√x (16)

k ≤ 10 n/2 real number

(x + 0.5. 10-2n) sin(acos(x / (x + 0.5. 10-2n )) ±π/2) = ± x (17)

(x + 0.5. 10-2n) cos(asin( x / (x + 0.5. 10-2n ))±π/2) = x (18)

d√x /dx = d(tan (asin( x / (x + 0.5. 10-2n ))) / 10n)/dx (19)

∫ √x dx = ∫ (tan (asin( x / (x + 0.5. 10-2n ))) / 10n) dx (20)

The equation (17,18) are numerically equal and can be done for all the formula

(5,6,7,8,9,10) according to a chosen precision of calculation

There are many minor results for this section…

3.4. Example of square root function replacement:

In physique and engineering the kinetic energy is defined as below:

Ek = 1/2 m v2

v = √( 2 Ek / m)

v = tan (asin(2 Ek / (2 Ek + m 0.5. 10-2n ))) / 10n (21)

we can use the formulas (6,7,8) for the same purpose.

In mathematics using Pythagorean theorem we can write:

d = √(a2 + c2)

d = tan (asin((a2 + c2) / ((a2 + c2) + 0.5. 10-2n ))) / 10n (22)

We can chose the number of correct decimals precision using an appropriate

value of n.

In general we can replace x by any function f(x) and if the function f(x) contain also

square root we can replace it also by one of the equation of √x

√ (f(x))= 1/(10n acos(f(x) / (f(x) + 0.5. 10-2n ))) (23)

4. Discussion

The formulas (5,6,7,8,9,10) are correct up to 2n decimals output of the square root

calculation √x, as shown in table (2) and paragraph (3.2.), so 10n as an input value

will control the precision of the output value of √x, the trigonometric function should

be calculated with high precision, which mean just correctly, with higher correct

decimals calculation precision [3], because the numerical accuracy of the functions

"asin" near −π/2 and π/2 and "acos" near 0 and π is ill-conditioned and will thus

calculate the angle with reduced accuracy in a computer implementation (due to the

limited number of decimals) [4]

The formulas (9,10) are the simplest formula for calculation because they use one

function which reduce calculation time and increase efficiency.

Our square root √x calculation is limited to 2n precision so it is not important to

consider number after that. And there is no limit to expand the precision calculation

by just increasing n value and thus 10n value.

Many commercial programs limits the accuracy to 15 correct decimals and round or

discard the small part 0.5. 10-2n and 0.5. 10-n especially for big value of n which will

yield in a wrong results using the above formulas for n > 3, and for big numbers, so it

is very important to keep the terms 0.5. 10-2n and 0.5. 10-n without rounding

especially for n > 3 and use a high precision calculation programs for the

trigonometric functions to ensure the correct results of the square root formulas with

2n number of correct decimals precision.

It is possible to use another number as a precision factor that lead to different pattern

which may be instable, but I found that 10n is the simplest number for calculation that

gives always a stable and accurate results.

These formulas are dynamic and realistic and practical because in practice the number

of correct decimals are always fixed and limited whatever the method chosen to

calculate the square root and even if we know exactly all the correct decimals.

It is possible to divide √x by a precision factor with or without multiplying x by the

precision factor, but I found from numerical analysis that the above method is the best

and most stable and the simplest which is an important feature to consider.

It is possible to replace √x function by one of the formulas (5,6,7,8,9,10) above as

shown by an example in paragraph (3.4.) according to a chosen accuracy and correct

decimals output and perform calculation and mathematical operations

5. Conclusion

The formulas of the square root (5,6,7,8,9,10) are not a calculation method only they

are mathematic law that illustrate the incontestable relation between the square root

and the trigonometric function according to a specific pattern, they can replace the

square root function in any equation for a chosen precision for any application, and

can be used to calculate precisely and efficiently the square root of any real positive

number at any chosen precision and number of correct decimals, which is certainly an

excellent and more efficient substitute to the existing methods and algorithms of

square root calculations, for an initial fixed and limited number of correct decimals,

that could be expanded indefinitely, the small term 0.5. 10-2n in the above formulas,

which is usually discarded by the most scientific minds, is very important and should

not anymore rounded or discarded now in any case to perform the correct

calculations, the limited use of correct decimals is a calculation method adopted in all

scientific and engineering fields for calculations. These formulas are a new tool for

mathematicians, engineers and scientists that I suggest to use in future and study the

applications of these formulas and explore how they are useful and practical and

conclude advantages and features since the applications of the square root is very vast,

I hope that it will make a progress for calculations and mathematical analysis.

Acknowledgement

I am an independent researcher and a professional solution provider,

Funding

This research did not receive any specific grant from funding agencies in the public,

commercial, or not-for-profit sectors.

Declarations of interest: none

References

[1] Methods of computing square roots.

https://en.wikipedia.org/wiki/Methods_of_computing_square_roots

access date December 3 – 2018

[2] Pythagorean theorem. https://en.wikipedia.org/wiki/Pythagorean_theorem

access date December 3 – 2018

[3] List of trigonometric identities.

https://en.wikipedia.org/wiki/List_of_trigonometric_identities

access date December 3 – 2018

[4] Inverse trigonometric functions.

https://en.wikipedia.org/wiki/Inverse_trigonometric_functions

access date December 3 – 2018

[5] Root-finding. http://mathworld.wolfram.com/topics/Root-Finding.html

access date December 3 – 2018

[6] Root-finding algorithm. https://en.wikipedia.org/wiki/Root-finding_algorithm

access date December 3 – 2018

[7] Newton method. https://en.wikipedia.org/wiki/Newton%27s_method

access date December 3 – 2018

[8] Newton method. https://www.encyclopediaofmath.org/index.php/Newton_method

access date December 3 – 2018

[9] wolframalpha. https://www.wolframalpha.com/

access date December 3 – 2018