Wonder Relation In Mathematics

Where do you reside?”

“In front of Post office.”

“Where is then post office?”

“It is in front of my house.” The answer has baffled all at one time or the other as the reference is related and is not fixed. In fact, both locations are the reference for one another. That makes the location not only unlocatable but puts one into never ending loop of hopping from one location to the other. Such loops are called closed and never ending.

But where is mathematics involved in it? Per Se there appears no mathematics in it but in mathematics, there are abundant such situations where there are such unending loops. I recollect vividly as a teacher of computer programming way back in eighties, I used to make programme involving loops based on mathematics questions, language at that time was simple ‘basic’ and commands were go to that line and from that back to original to puzzle the calculating machine. But machines are beyond nervousness and confusion. To my amusement, Computer when asked to display the status used to indicate, “Still calculating.” Hardly the machine knew, it had been put in never ending task.

Coming to my initial question of location of John’s house, if John says, his house is situated in front of Post Office in Peter Lane, then reference as Post Office is fixed but one when goes to Post Office, there one finds, Post Office is comprised of a huge building and there are a number of houses in front of that building and John’s house can not be found. Again John will be contacted and he informs, it is a grey coloured house in front of telegraph section of Post Office. But there are a number of grey houses in front of that section. On again inquiring, John replies, it is in front of cashier’s window. In situation highlighted here, there is a loop but it is localising on each cycle thus reaching nearby its destination. Such loops in mathematics are called convergent loops. Relationship of Post Office and house is called recursive relation.

Hope, I could clarify the concept of recursive relation. Such recursive relations have charismatic effects and can land mathematicians in new unexplored land of mathematics. Continued fractions, continued Ramanujan nested radicals, continued products are gifts given by such relations.

I take up some examples to bring home further the concept of recursive relation. Let us consider a right angled triangle ABC with base BC, perpendicular AC and right angle at C, ratio of perpendicular AC to hypotenuse BC is called sin B where B is angle at point B. This angle B can be halved by formula,

sin B = ½.sin B/2.cos B/2. ………………………………………………………….(1)

Examination of this formula reveals that sin of an angle appears in both left hand side LHS and right hand side RHS. This sine on right hand side RHS will reflect it back to the formula of sine written in left hand side LHS. This reflection of sine formula from LHS to RHS and then from RHS to LHS continues indefinitely. But in the process angle B goes on halving from B to B/2, from B/2 to B/4 and so on infinitely. This is analogous to reflection from Post Office to opposite building and opposite building to post office and continuous halving of angle is analogous to localising the house to smaller area.

Equation (1) is recursive relation and applying this formula successively, the angle B will be reduced to infinitesimal value B/2^n. How this angle will be reduced by application of recursive relation is explained in steps below.

sin B = ½.sin B/2.cos B/2.

Therefore, sin B/2 on RHS can be written as 2.sin B/4.cos B/4. That makes

sin B = 2.cos B/2. 2.sin B/4.cos B/4 = 2^2. cos B/2.cos B/4.sin B/4.

Again, sin B/4 on RHS can be written as 2.sin B/8.cos B/8. That makes

sin B = 2^3. cos B/2.cos B/4.2.sin B/8.cos B/8 = 2^3 cos B/2.cos B/4.cos B/8.sin B/8.

In this way, it can written,

sin B = 2^n cos B/2.cos B/4.cos B/8…………………………….sin B/2^n. …………………………….(2)

Angle B will go on reducing but for derivation of formula, it has been aborted at n terms.

Let us see what happens to this equation if a little mathematics is applied to it. Sin A = A is true when angle A is very very small tending to zero. In equation (2), B/2^n also tends to zero on successive halving particularly when n tends to infinitely large, therefore, sin B/2^n equals B/2^n. Applying it to equation (2), it transforms to

sin B = 2^n cos B/2.cos B/4.cos B/8…………………………….B/2^n.

On transposing B/2^n to RHS,

(sin B)/(B/2^n = 2^n cos B/2.cos B/4.cos B/8……………………………

Or (sin B)/(B/2^n = 2^n cos B/2.cos B/4.cos B/8……………………………

Or 2^n.(sin B)/(B) = 2^n cos B/2.cos B/4.cos B/8…………………………

On cancelling 2^n on LHS and RHS,

sin B/B = cos B/2.cos B/4.cos B/8…………………………………. (3)

This is a beautiful formula, first derived by great mathematician Euler and also indirectly used by another great mathematician François Viète. The formula derived here is the outcome of application of recursive relation and was widely used particularly for determining the value of π. Euler used it to factorise sin B in terms of angle whereas François Viète used it as it is.

Another example of recursive relation is angle halving formula for Cosine of an angle. For that I refer to right angled triangle as mentioned before wherein cos B is the ratio of base BC to hypotenuse AC and it can be written as

cos 2.B = 2.cos^2 B – 1 and it can be rearranged as

cos B = {(1+ cos 2.B)/2}^2 = ½.(2+ 2.cos 2.B)^2. …………………………………………………(4)

Examination of equation (2) reveals that it contains cosine of an angle in LHS and also in RHS, therefore, it can form a never ending loop. Cos 2.B in right hand side can be replaced by ½.(2+ 2.cos 4.B)^2 and again cos 4.B again replaced by ½.(2+ 2.cos 8.B)^2 and so on. Therefore, cos B can be written as

I take another example of angle tripling formula whereby sin 3.B is written as

sin 3.B = 3.sin B – 4.sin^3 B = sin B.(3- 4.sin^2 B) or

sin B = sin 3.B/(3- 4.sin^2 B) …………………………………………………………………………………..(6)

Inspection of equation (6) reveals that its RHS has sine of angles in numerator and denominator and LHS also has sine of an angle. In RHS, sin 3.B in numerator will be considered for recursive relation and will be replaced by sin 9.B/(3- 4.sin^2 3.B) and sin 9.B by sin 3^3.B/(3- 4.sin^2 3^2.B) and so on till sin 3^n.B = sin B. At that stage, sin B on LHS and RHS cancels and denominator which is product of terms (3- 4.sin^2 B).(3- 4.sin^2 3.B), …………………{3- 4.sin^2 3^(n-1) B} equals 1, giving rise to an identity.

For illustration, let us take the example of sin π/10.

Here angle B = π/10 which can be written as

sin π/10 = sin 3π/10/(3- 4.sin^2 π/10) = sin 9 π/10/{(3- 4.sin^2 π/10).(3- 4.sin^2 3π/10)}. That is

But sin 9.π/10 = sin π/10, therefore,

Here recursive relation stops when sin^n B equals sin B or – sin B. From this, it can be concluded


Π(3-4.sin^2 3^(j-1) = 1 …………………………………………………………………………..(7)


when sin (3^k B)= sin B and


Π(3-4.sin^2 3^(j-1) = – 1 …………………………………………………………………………(8)


when sin (3^k B)= – sin B.

In this way, a new trigonometric identity is established using recursive relation.

Coming to equation (5) which is an equation that contains roots of root, such continuous nested radicals were first used by François Viète when he derived formula for pi first time from entire Europe. Recently, these have been abundantly used by Indian mathematician Ramanujan. As such, use of recursive relation gave unexpected and astonishing results that mesmerised English mathematician Hardy.

I also experimented on recursive relations and results were encouraging. If you feel a bit interest in it, don’t shy away. Take a note book and jot down a formula. Find all possibilities for its recursive nature, try differently and look for different result. You will find one I am sure and that would be blissful. Do you feel interested in recursive relation, you may go through this article at http://www.ijma.info/index.php/ijma/article/download/5227/3074



1. A part of this article is excerpts from my paper written on “MORE TRIGONOMETRIC IDENTITIES ,” available at http://www.ijma.info/index.php/ijma/article/download/5227/3074

2. Title image courtesy brewbooks at https://www.flickr.com/photos/brewbooks/184343329/

About author

Writer is an Electronics and Electrical Communication Engineering graduate and was earlier Scientist, then Instrument Maintenance Engineer, then Civil Servant in Indian Administrative Service (IAS). After retirement, he writes on subjects, Astronomy, Mathematics, Yoga, Humanity etc.


Mathematicians Think Differently

Yes, it is correct mathematicians think differently and extraordinarily. It is expected of them as traditional approach will lead to traditional results and halt the process of evolution of new results and formulae. It reminds me of Johann Carl Friedrich Gauss, sometimes referred to as Princeps mathematicorum. He was a German mathematician and also contributed to many fields including number theory, algebra, statistics, analysis, differential geometry, geodesy, geophysics, mechanics, electrostatics, magnetic fields, astronomy, matrix theory and optics.When Gauss was studying in school, to inflict punishment upon him for misbehaving, his school teacher gave him a cumbersome and time consuming problem of addition of all integers from 1 to 100. But to the astonishment of the teacher, he added the quantities with in few seconds and showed the result. The result was correct and it earned appreciation for young Gauss in the heart of the teacher.

What Gauss did even baffle now to most of us. He first wrote the sum as

S = 1 + 2 + 3 + ………………………………………up to 100 and again wrote it but in a different way

S = 100 + 99 + 98 +………………………………………………up to 1 and added both as

2S = 101 + 101 + 101 + ……………………up to 100 times and then

2S = 101 x 100 = 10100 and 

S = 5050.

That produced the result wonderfully in a neat and clean manner. Had there been some other student, he would have adopted the normal procedure of adding 1 with two then with 3 and then with four until he would have reached the integer 100. Even the teacher expected it so from Gauss but Gauss was not a simple student, he was extraordinary.

In the same way, another student at the university of Copenhagen was asked in examination how the height of a skyscraper can be measured with barometer. Naively, he replied by tying the barometer with a rope and then hanging it from top of the tower touching the ground and measuring the length of rope plus the height of barometer that will give the height of the skyscraper. Not impressed with the answer, the examiner awarded zero marks to this answer. 

On appeal by the student, it was held that answer is correct but there is no physics involved and the student was called for oral examination. When the student was asked to give answer to the same problem with in five minutes, he spent most of the time thinking for answer. Ultimately, he was warned the time was running out and he should put forth his answer immediately. However, the student replied, he has many answers to the problem and was thinking which answer was the best. When he was asked to give all the answers, he politely replied,” Height of skyscraper can be known by throwing the barometer from top and noting the time taken by it to touch the ground. Since it may cost barometer, the height can alternatively be known by measuring the length of barometer and also its shadow and then measuring the shadow of the skyscraper, ratio of barometer to its shadow will equal ratio of height of building to its shadow and simple arithmetics will give height of skyscraper. He continued, there is yet another way, difference of barometer reading at top and bottom of the sky scrapper will equal to atmospheric pressure equivalent to height of skyscrapers. He still added, If one is too scientific, one can note down the time period of a pendulum at the base of the skyscraper and and also at the top of skyscraper and difference in time period will be a measure of height.”
That student was genius mathematician and physicist Neil Bohr who propounded quantum theory later in his life.  

It is also said of Bohr that during a foot ball match where Bohr was a goalkeeper, the ball remained in the opponent half but when the attacker brought the ball to Bohr’s half, a spectator shouted, ” Be aware Bohr, the ball is in your half,” and at that time, Bohr was found solving mathematical problem on the ground.

Mathematicians think differently from an ordinary person. I am certain, I will be correct if I say, most of inventions have been made when unconventional route was adopted.

Let us take the case of determination of value of pi written as π which is a mathematical constant and is defined as a quantity when multiplied with diameter of a round object gives its circumstance. Normal thinking leads us to determine its value by measuring circumference of a round figure and then dividing it by diameter.

Method is correct but error in measurement of circumference and diameter will result in error in value of pi. And pi so calculated will culminate in large error in the area or circumference of a round object if that happens to be of large diameter. That led mathematicians to determine pi independent of measurement of its circumference or its area. This is, thinking differently.

Without going into the history of pi, I will attempt to explain how earlier mathematician and physicist thought of pi. Pi, according to mathematician and physicist Archimedes, can be calculated by averaging the area of a polygon circumscribed and inscribed in a circle. His thought points to the fact that circumference of a circle is equal to half the circumference of polygon just outside plus half the the circumference just inside the same circle. That was an apt thinking and gave great approximation of value of pi provided sides of regular polygon are quite large. Probably, such thought would generally not come to ordinary person.

Idea that occupied to most of the mathematicians in the subject of determination of pi was approximating regular polygon of extremely large number of sides with circle. Extremely large number here is that number which is greater than the greatest.

Let us attempt finding value of pi by above said approximation.

 I submit, a point forms an angle of 2. Pi around it and let us take polygon or Triangle of three equal sides AB, BC, CA. That will make, its three corners or points as equidistant from each other. Let there be a point O inside the triangle such that these points are also equidistant from point O ie OA = OB = OC = radius r. That means a circle with centre O , circumscribes triangle ABC. Let us find out perimeter of triangle ABC which is equal to sum of AB, BC and CA.

Side BC = twice length BD and BD = OB x sine of angle BOD.

Since ABC is a polygon of three sides, therefore angle BOC = 2.pi/3.

If regular polygon has n sides then angle BOC = 2.pi/n.

 For n =3, angle BOD = ½. 2.pi/3 = pi/3.

BC = 2.BD = 2r.sin pi/3. For n sided regular polygon, it would equal 2r.sin pi/n.

AB + BC + CA = 3. BC = 6.r. Sin pi/3. For n sided regular polygon, perimeter would be 2.n.r sin pi/n.

First approximation is made by equating perimeter of three sided regular polygon with circumference of circle (2.pi.r). That is

6.r. Sin pi/3= 2.pi.r.

Or pi = 3. under root 3/2 = 3. (.866)= 2.698.

Next approximation would be by taking n = 4,

 pi = 4.(sin pi/4)= 4./2^1/2 = 2.828.

Next approximation when n= 6,

 pi = 6. Sin pi/6 = 3.

It is clear from above as n increases, value of pi approximates better to actual value of pi.

For n sided regular polygon,

 pi = n.sin pi/n……………………………………………………………(1)

If we put x = pi/n, pi equals pi.sin x/x when x is very small.

It has been the endeavour of mathematician to find value of sin x/x as it contains the value of pi in it.

In Europe, first attempt to determine value of pi was made by François Viète*, a French mathematician. His work on new algebra was an important step towards modern algebra, due to its innovative use of letters as parameters in equations. He was a lawyer by trade, and served as a privy councillor to both Henry 3 and Henry 4 of France. He attempted equating polygon with circle by Geometry and finally gave a formula for pi in nested radicals. What he practically solved was determination of value sin x/x. Sin x is a trigonometric ratio of perpendicular to hypotenuse in a right angled triangle with angle x facing perpendicular.

He geometrically reached the formula for sin x as

 sin x = x.cos x/2.cos x/4. Cos x/8. Cos x/4…………………………………up to infinity with last term as 1.

When x equals pi/4, equation takes the form,

(½)^1/2 = pi. cos pi/8. cos pi/16. cos pi/32………………………

Where cos pi/8. cos pi/16. cos pi/32 are known from double angle formula,

 cos pi/8 = ½.(2+ 2^1/2)^1/2),

 cos pi/16 = ½{2+(2+ 2^1/2)^1/2}^1/2,

cos pi/32 = ½[2+ {2+(2+ 2^1/2)^1/2}^1/2]^1/2,


On putting these values,

1/pi = 2^1/2. ½.(2+ 2^1/2)^1/2). ½{2+(2+ 2^1/2)^1/2}^1/2. ½[2+ {2+(2+ 2^1/2)^1/2}^1/2]^1/2…….(2)

There came then most eminent mathematician of 18th century Leonhard Euler. He thought differently to determine the value of pi from sin x/x. He factorised sin x in infinite product.

 sin x/x = {1- (x/pi)^2}.{1- (x/2pi)^2}.{1- (x/3pi)^2}…………. up to infinity………………..(3)

Sin x can also be written in series expansion as x – x^3! + x^5!-…………. up to infinity.

Or sin x/x = 1- x^2/3! + x^3/5!-…………………………………………………….(4)

Comparing coefficient of x^2 of equations (3) and (4),

1/pi^2.[1/1 + ½^2 + 1/3^2+………]= 1/3!

Or pi^2= 6.[1/1 + ½^2 + 1/3^2+………]

 Another English mathematician John Wallis thought differently from Euler and Viete, he, in stead, straight way considered equation (3) of sin x/x and put x = pi/2. That gave the value of pi as

2/pi = (1-1/4).(1- 1/16).(1-1/36).(1-1/64)……………

Or pi = 2/1.4/3. 16/15. 36/35. 64/63……………..

Thus it can be summarised as equation of a circle in trigonometric ratios is pi = n.sin pi/n where n is number of sides of regular polygon. Also sin x/x where x = pi/n, when equals 1 then n tends to infinity. If it does not equal 1, it means, n has finite value and sin x/x has also value other than 1. Thus (sin x)/x is an equation that contains value of pi. Or in other words pi is determinable from (sin x)/x.

Different ways have been adopted by mathematician to calculate pi from (sin x)/x and that had given different formulae of pi. If an attempt is made to find (sin x)/x, new formula will originate. Is it not then worth trying to find some other way to express (sin x)/x?

I submit, mathematicians are not restricted to sin x divided by x to determine pi, statistical methods, probability theory and many more have already been introduced and pi freshly calculated. Still more methods will be put forth because mathematicians think differently.


1. Viète* For further reading, ‘When Law Met Mathematics’ at https://narinderkw.wordpress.com/2017/08/07/when-law-met-mathematics/

2. https://en.m.wikipedia.org/wiki/Carl_Friedrich_Gauss

3. http://felix.physics.sunysb.edu/~allen/Jokes/bohr.html

4. https://en.m.wikipedia.org/wiki/François_Viète

5. http://www2.mae.ufl.edu/~uhk/EULER-SINE.pdf

6. Figure courtesy https://www.quora.com/Find-the-perimeter-of-an-equilateral-triangle-of-side-l-cm-is-inscribed-in-a-circle-with-radius-r-cm

7 Cover photo courtesy artist Domenico Fetti, Italian Painter, current location Gemaldegalerie Alte Meister, Source/Photographer http://archimedes2.mpiwg-berlin.mpg.de/archimedes_templates/popup.htm 


About author:

Writer is an Electronics and Electrical Communication Engineering graduate and was earlier Scientist, then Instrument Maintenance Engineer, then Civil Servant in Indian Administrative Service (IAS). After retirement, he writes on subjects, Astronomy, Mathematics, Yoga, Humanity etc.      

It Is Surprising But True, Trigonometric Ratios Of Sine, Cosine Of An Angle Can Be Larger Than One

In basic plane trigonometry, trigonometric ratios of Sine or Cosine of an angle can have any value between -1 and + 1 and these can not be more than 1 or less than -1. To define sine of an angle, it is ratio of perpendicular to hypotenuse and cosine of an angle is ratio of base to hypotenuse in a right angled Triangle. From these definitions also and since hypotenuse is always larger than any of other side, these ratios are natural to be less than one. A number of problems have been solved by all assuming sin (n.x) and cos (n.x) to be equal to any value between -1 and + 1 when angle n.x tends to infinite with n approaching infinity. Assumption of (sin x)^n or (cos x)^n as approaching zero when these ratios are not equal to one with n tending to Infinity has always been made. All these assumptions are based on fact that magnitude of trigonometric ratios of sine and cosine are always less than or equal to one.

Are such assumptions then incorrect in view of the statement that these ratios can be larger than one? It is unmathematical to have answer in “yes,” and “no” to same question. This, therefore, needs analysis to reach logical result about the magnitude of these ratios. For that, it will be examined what deters to equate these trigonometric ratios to quantities larger than one.

Let us assume sin x equal to a quantity larger than one and let it be square root 5. Cos x can be calculated from identity (1- sin^2 x)^1/2 and that makes cos x equal to (-4)^1/2 or 2i where i is (-1)^1/2 and is called imaginary number. A question then arises whether imaginary quantity is practical and realisable. Simple answer to this query is “yes” if it is assumed that imagination is the result of real quantity. If a real quantity can give birth to imaginary then reverse should also be true and imagination should also give birth to reality.

Friedrich August Kekulé, a German chemist saw in his dream that snakes are entangled eating tails of each other and that led him to proclaim valency of carbon four and the structure of benzene. Apart from this, imaginary number raised to power imaginary number* ie i^i equals e^-pi/2 Here base is imaginary and power is also imaginary but the result is real and astonishing as it relates to pi which is the worlds most important mathematical constant.

Imaginary quantity can, therefore, not be dubbed as insignificant as it is analogous to operator 90 degree. That is it rotates the real quantity by 90 degree, for example 2i means, a quantity of magnitude 2 rotated by 90 degree, in other words, it is a quantity of magnitude 2 that lies on y axis. Similarly, i.i or -1 means rotation by 180 degree or – x axis,

i.i.i or -i means rotation by 270 degree or -y axis and

i.i.i.i or 1 means rotation by 360 degree or x axis.

This analogy of imaginary number is utilised in electrical engineering where current leads or lags according to capacitive or inductive nature of load and is represented by C.i or -C.i where C is the current flowing in the load. It can then be summed up, imaginary quantity is as important, significant and realisable as real quantity.

Coming to cos x = 2i calculated from Pythagorean equality, it is quite apparent that angle x must be complex so as to make cos x as imaginary quantity. A complex quantity is that which has real and imaginary part. If ‘a’ and ‘b’ are real then a + i.b is a complex quantity where a can be zero also. Let us assume angle 

x = a+ i.b, then

cos (a+ i.b) = 2.i

or cos a. cos i.b – sin a. sin i.b

= cos a.cosh b – i sin a.sinh b,

since cos ia = cosh a and sin ia = i.sinh b.


cos a.cosh b – i sin a.sinh b = 2.i.

Equating real and imaginary parts,

cos a.cosh b= 0,

i sin a.sinh b = 2i

or sin a.sinh b = 2.

That means a =(2k + 1). pi/2 or (2k- 1).pi/2.,Substituting this value,

1.sinh b = 2 or b = sinh inverse 2.

Sinh b can also be written as ½.(e^b – e^-b) = 2.

On putting e^b= y, we get

y – 1/y = 4 or y^2- 4y – 1= 0.

That gives

y = e^b = 2 + 5^1/2 or 2 – 5^1/2.

This can also be written as

y = 2Φ+ 1 or 3- 2Φ

where Φ is golden ratio equal to

½.(1+ 5^1/2).

From this, b can be written

= log (2 + 5^1/2) or log (2 – 5^1/2) or log (2Φ+ 1) or log (3- 2Φ)

Therefore complex angle whose cosine is 2i and sine is square root 5 is

= Pi/2 +i.log (2 + 5^1/2) or Pi/2 +i.log (2Φ+ 1).

On taking the principal value of a and positive value of b, we can also write

Sin [Pi/2 +i.log (2 + 5^1/2)] = 5^1/2,

Or sin [Pi/2 +i.log (2Φ+ 1)]= 5^1/2

Or cosh [log (2Φ+ 1)] = 2Φ-1,

cosh [log (2 + 5^1/2)] = 5^1/2,

Also sinh (log (2 + 5^1/2) = 2

or sinh [log (2Φ+ 1)] = 2.

From above analysis, we can say, cosine or sine of an angle can be larger than one provided the angle is complex containing imaginary part. Imaginary part is not an imagination but is realisable by way of cosh and sinh of imaginary angle as given below.

Cos ix = cosh x = ½.(e^x + e^-x),

sin ix = i.sinh x = ½.i.(e^x – e^-x)

and this leads to Pythagorean identities

cos^2 x + sin^2 x = cosh^2 x – sinh^2 x = 1.

Next question that arises is how to draw a complex or imaginary angle. In Plane Trigonometry, real angle can be drawn in x-y Plane and trigonometric ratios can be found out by drawing a right angled triangle in that plane. When an angle is drawn in a plane, it always has real value.That means, it can never be complex and resultantly cosine and sine of an angle as ratios of base to hypotenuse and perpendicular to hypotenuse can never be larger than one as hypotenuse in a Plane trigonometry is always larger than its sides. Therefore sine and cosine of an angle is always equal or less than one in plane trigonometry.

Coming to drawing of an imaginary angle, I submit, such a case must satisfy identity cosh^2 x – sinh^2 = 1 and in Plane trigonometry, imaginary angle would be such that the straight lines forming it, never join. When the lines do not join, it can never be defined as an angle. Therefore, in a Plane x-y where real angle can be drawn, imaginary angle can never be drawn.

That necessitates there should be another plane an if that Plane is considered perpendicular to the plane that contains real angle, then imaginary part of the angle can be drawn in that perpendicular plane. Let us see how. It is well known to those who have basic knowledge of plane trigonometry that as the angle increases the length rotates in a circle. Or we can say that if length starts from point (0,0) and ends at a point (a cos x, a sin x), then with increase in angle x, the line rotates in a circle of radius a in counter clockwise directin. The locus of the point in Plane trigonometry with change in angle is circle. Why locus is a circle can be proved from the identity

cos^2 x + sin^2 = 1

On multiplying both sides by a^2, we get

a^2.cos^2 x + a^2. Sin^2 x = a^2,

and that is an equation of circle. It is reiterated that when angle is real in plane trigonometry, a point can be defined as a point on a circle.

It is explicit from the explanation already given that when angle changes to imaginary angle say i.x then cosine of imaginary angle takes the form cos i x and that is equal to cosh x and cosh x is equal to ½(e^x + e^-x). When angle is imaginary, sine of imaginary angle takes the form sin ix and that is equal to i sinh x and i sinh x is equal to 1/2.i(e^x -e^-x). Therefore, it leads to identity,

cosh^2 x – sinh^2 x = 1.

On multiplying both sides by a^2, we get

a^2.cosh^2 x –a^2. sinh^2 x = a^2.

Let a^2.cosh^2 x be x and sinh^2 x = y, then locus of point x, y represents

x^2 – y^2 = 1

and that is an equation of hyperbola. That is why cosh x and sinh x are called hyperbolic functions.

It follows from this, change in imaginary angle drawn on perpendicular plane moves the point a.cosh x, a sinh x on hyperbola. We can, therefore, conclude that whenever there is trigonometric ratio and if that pertains to complex angle, real part of the angle will decide the length on the circumference of a circle in a plane and its imaginary part the length on the hyperbola in a plane perpendicular to the plane of real angle.

Based on this analysis let me answer the questions I raised at the start of this paper. Sine and cosine of a real angle that lies in a plane is always either equal to or less than one. It is true in plane trigonometry what is taught in schools. Length of the line rotates in a circle as the real angle is changed if we keep magnitude of length constant.

When the angle is complex (that is it has imaginary part also), sine and cosine of complex angle can be larger than one and in that case, imaginary part of the angle lies in a plane perpendicular to the in which real angle lies and the point moves upon hyperbola drawn in perpendicular plane with the change in imaginary angle. Finally, sine or cosine of an angle is Plane trigonometry is equal or less than one but if it pertains to complex angle containing imaginary part, it can be greater than one.




1. Title Image courtesy Kan8eDie at https://upload.wikimedia.org/wikipedia/commons/e/eb/Principle_branch_of_arg_on_Riemann_%28small%29.png

 2.Kekulé Dream by Michael Verderese

Professor Heinz D. Roth at https://web.chemdoodle.com/kekules-dream/

3.* Eternal Bonds Of Togetherness Between Natural Number, Pi, Tangent And Logarithm Of A Quantity at https://narinderkw.wordpress.com/2017/08/12/eternal-bonds-of-togetherness-between-natural-number-pi-tangent-and-logarithm-of-a-quantity/

4.Image of complex angle courtesy Richard Hammack at A Geometric View of Complex Trigonometric Functions http://www.people.vcu.edu/~rhammack/reprints/cmj210-217.pdf

About Author:

Writer is an Electronics and Electrical Communication Engineering graduate and was earlier Scientist, then Instrument Maintenance Engineer, then Civil Servant in Indian Administrative Service (IAS). After retirement, he writes on subjects, Astronomy, Mathematics, Yoga, Humanity etc.      



Eternal Bonds Of Togetherness Between Natural Number, Pi, Tangent And Logarithm Of A Quantity

Eternal bonds of love between the characters are not found only in plays of Shakespeare or literature but these are profound in mathematical functions. I am presenting this mathematical affinity amongst Natural number e, Mathematical constant pi, Tangent and logarithm of a quantity and a bit more. I will be using power series expansion, calculus and complex numbers. These mathematical functions also exhibit strong bonds for each other. I submit wherever there is a logarithm of a quantity, presence of natural number can not be denied. Wherever a natural number comes into being, mathematical constant pi appears from oblivion.Tangent of the quantity is also not far behind, it is hidden in the logarithm as the baby concealed in belly pouch of its mother Kangaroo. The efflorescence in which these quantities are born from each other urges me to state that there is some unseen eternal bonds between these. Cosmic cycles will revolve but there bonds not diminishing even by an iota.  


Before I proceed further let there be brief introduction. For tangent, I refer you to triangle ABC with right angle at corner B, base AB, perpendicular BC and hypotenuse CA that

makes an angle x radians at A. Ratio of perpendicular to base ie BC/AB is the tangent of angle x. 


                                        Graph Of Tan Inverse x and Cot inverse x.

If value of tangent of an angle is given, angle can be determined by taking inverse of tangent.
Natural number abbreviated as e* is defined as (1+ 1/n) multiplied with itself infinite times where n tends to infinity. 


Mathematically, it is written as  

e = (1 + 1/n)^n where n tends to infinity and sign ^ denotes raised to power. Numerically,
           e = 1+ 1 + ½! + 1/3! + ¼! + 1/5! +….up to Infinity.                                                                                      And this sums up as 2.71828.  

Image of circle showing Pi as ratio of circumference to diameter, Courtesy author Klonjee.

Pi**abbreviated as π is defined as ratio of circumference of a circle to its diameter. Numerically, it equals 3.14159.  

Logarithm*** of a quantity is that power of natural number which equals the quantity. If the quantity is e^x then natural logarithm is x. 

Graph of function of logarithm courtesy Krisnavedala

Theory And Proof

Having introduced these quantities, I take up function

F(x) = 1/(1+ x^2)

      = (1+ x^2)^-1.

Above function can be expanded as binomial expansion,

F(x) = (1+ x^2)^-1 = 1- x^2 + x^4 – x^6 + x^8- …………..

On integrating above function with respect to x, we get

tan inverse x =c + x – (x^3)/3 + (x^5)/5– (x^7)+ x^8- ………up to Infinity

where c is a constant of proportionality.

If x is 0, then tan inverse is either zero or is integral multiple of pi and is written as k.pi. On putting x equal to zero, right hand side of above equation becomes c.

That is c equals k.pi where k is any number 0,1,2, 3 ……. Substituting this value of c in above equation, we get

tan inverse x = k.pi + x – (x^3)/3 + (x^5)/5– (x^7)+ x^8- …………..

Taking principal value by putting k equal to zero, value of c also equals 0. The equation then can be rewritten as

tan inverse x = x – (x^3)/3 + (x^5)/5– (x^7)+ x^8- ….up to Infinity                                                      (1)

Now we take another function and will try to make it equal to tan inverse x expansion. Let this be

f(x) = 1/(1- x^2).

It is obvious from this function that when x would be substituted by i x this function will be indistinguishable from 1/(1+ x^2) which is derivative of tan inverse x. On integration with respect to x, this will result in tan inverse x and 1/(1- x^2) on integration will yield logarithmic function in x. These on substitution with x as i x will have close relations.

Based on this theory, let us decompose this function into partial fraction as

f(x) = 1/(1- x^2) = 1/(1+ x).(1-x) = A/(1+ x) + B/(1- x).

Or 1 = B.(1+ x) + A.(1- x).

On comparing constant term and coefficient of x of left hand side LHS with right hand side RHS, we get

A = ½ = B.

Then fx) = 1/(1- x^2) = (1/2)/(1+ x) + (1/2)/(1- x).

However 1/(1- x^2) can be expanded according to Binomial Theorem as

 1/(1- x^2) = 1+ x^2 + x^4 + x^6 + x^8- …………..up to infinity.


F(x) = 1/(1- x^2) = (1/2)/(1+ x) + (1/2)/(1- x) =1+ x^2 + x^4 +x^6 + x^8+ …………..up to Infinity.

On integrating with respect to x, we get

½. log (1+ x) – ½.log (1-x) = c1+ x + (x^3)/3 + (x^5)/5 +(x^7)/7 + (x^9)/9+ …………..up to infinity

where c1 is constant of proportionality.

At x = 0, 0 = c1.


½. log (1+ x) – ½.log (1-x) = x + (x^3)/3 + (x^5)/5 +(x^7)/7 + (x^9)/9+ …………..up to infinity.

Or ½.log (1+x)/(1-x) = x + (x^3)/3 + (x^5)/5 +(x^7)/7 + (x^9)/9+ ………….. up to infinity                    (2)

On putting x as i x in equation (1), it transforms to

tan inverse ix = ix + i(x^3)/3 + i(x^5)/5+ i(x^7)+ ix^8- ….up to Infinity.

Or (1/i) tan inverse ix = x + (x^3)/3 + (x^5)/5+ (x^7)+ x^8- ….up to Infinity……………………………………………………….                                                                                  (2/1)

This equation is same as equation (2), therefore equating these two we get,

1/i. tan inverse ix = ½.log (1+x)/(1-x) or

tan inverse ix = ½.i.log (1+x)/(1-x)                                                                                                            (3)

Let (1+x)/(1-x) = z, then x = (z-1)/(z+1) and

½.i log z = tan inverse i.(z- 1)/(1+z)                                                                                                            (4)

It is explicit from above, wherever the term logarithm of a quantity comes, tan inverse is a part and parcel of it and is given by above equations. Logarithm as it was defined earlier denotes the power of natural number e that makes it equal to given quantity and for tangent, that power points to an angle the tangent makes so as to equal the quantity. It is submitted, in other words, logarithm of a quantity is joint to the angle that the tangent makes, by an unbreakable bond. The bond is stated in relations as defined by equations (3) and (4).

If x is substituted by – i in equation (3), it takes the form

tan inverse 1 = ½. i.log (1-i)/(1+ i) or

k pi + pi/4 = ½. i.log {(1-i)(1+i)}/{1+ i)(1+i)}

                  = ½. i.log 2/2i = ½. i.log1/i = -1/2.i log i where i is (-1)^1/2 and k is any number 0, 1, 2, 3………….

Taking principal value by putting k = 0,

pi/4 = – ½.i log i or

– pi/2 = i. log i                                                                                                                                       (5)

That means logarithm of pure imaginary number multiplied by itself that is, pure imaginary number always equals minus half pi. Imaginary number i that equals (-1)^1/2 is impractical and unachievable but when its logarithm is multiplied with imaginary number, gives result as minus half pi. This again is un breakable bond between pi and logarithm of imaginary number.

If we combine equations (3) with (5), it transforms to

tan inverse ix – pi/4 = 1/2.i log (1+x)/(1-x) + ½ i log i

                                = ½.i log i.(1+x)/(1- x).                                                                                            (6)

Again this is an equation where logarithm, tangent, pi and natural number are all all connected with each other.

We revert to equation (5),

-pi/2 = log i^i

 e^-pi/2 = i^i                                                                                                                                                 (7).

This is another equation which joins natural number e with mathematical constant pi through imaginary number iota i.

Again coming to equation (3), tan inverse ix = ½.i.log (1+x)/(1-x), by rearranging I can write this as

Tan [½.i.log (1+x)/(1-x)] = i x.

From this, I can get

 sec [½.i.log (1+x)/(1-x)] = [1+ tan^2 {1/2i.log (1+x)/(1-x)}]^1/2

                                     = (1- x^2)^1/2.

By its reciprocal, value of cosine can be found as

Cos [½.i.log (1+x)/(1-x)] = (1- x^2)^-1/2.

This can also be written as

Cosh [½.log (1+x)/(1-x)] = (1- x^2)^-1/2.

Similarly value of sinh can also be found out as

Sinh ½.i.log (1+x)/(1-x)] = x/(1- x^2)^1/2

If Cosh [½.log (1+x)/(1-x)] = (1- x^2)^-1/2 is integrated with variable x from range 0 to 1, it equals sin inverse x with range 0 to 1 and that equals pi/4. Or

Integral Cosh [½.log (1+x)/(1-x)] range 0 to 1 = integral (1- x^2)^-1/2 range 0 to 1 = sin inverse x range 0 to 1 = pi/2. Logarithm of (1+x)/(1-x) is also connected by equation

Integral Cosh [½.log (1+x)/(1-x)] range 0 to 1 = pi/2


Natural Number, Pi, Tangent And Logarithm Of A Quantity are all connected with each other as has been proved from equations,

tan inverse ix = ½.i.log (1+x)/(1- x)                                                                                              (3)

½.i log z = tan inverse i.(z- 1)/(1+z)                                                                                               (4)

k pi + pi/4 = -1/2.i log i

where i is (-1)^1/2 and k is any number 0, 1, 2, 3………….

and for principal value on putting k = 0,

– pi/2 = i. log i                                                                                                                                        (5)

tan inverse ix – pi/4= ½.i log i.(1+x)/(1- x)                                                                                       (6).

e^-pi/2 = i^I,                                                                                                                                           (7)

sin [½.i.log (1+x)/(1-x)] = (1- x^2)^-1/2,

cos ½.i.log (1+x)/(1-x)] = x/(1- x^2)^1/2 and

cosh ½.log (1+x)/(1-x)] = x/(1- x^2)^1/2.

These bonds are not affected by physical matters as unfortunately human bonds are. Their relationships are intrinsic present and will last till eternity. Nothing in the universe can break these.

Keywords: Natural Number, Logarithm, Tangent, Pi, Eternal Bond, Binomial Expansion, Integration, Imaginary Number Iota, Tangent Inverse, Complex Number.


1) Graph of tan Inverse x courtesy Geek3 at https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions

2) Graph of natural number courtesy http://www.mathsisfun.com/numbers/e-eulers-number.html

3) Image of circle showing Pi as ratio of circumference to diameter by author Kjoonlee at https://en.m.wikipedia.org/wiki/Pi

4) Graph of function of logarithm courtesy Krisnavedala at https://en.m.wikipedia.org/wiki/Logarithm

5) * Natural number details can be read at https://narinderkw.wordpress.com/2017/07/05/natural-number-is-not-much-natural/

6) ** Pi details can be read at https://narinderkw.wordpress.com/2017/04/10/pi-is-sweeter-than-pie/

7) *** Logarithm details can be read at https://narinderkw.wordpress.com/2017/06/16/goodbye-to-log-table-and-calculator-determining-logarithm-is-easy-now/

8. Title Image English: Sunrise over the bay, Little Gasparilla Island, Florida, Source courtesy  Author Mmacbeth at https://commons.m.wikimedia.org/wiki/File:Little_Gasparilla_sunrise.jpg#mw-jump-to-license 


About Author

Writer is an Electronics and Electrical Communication Engineering graduate and was earlier Scientist, then Instrument Maintenance Engineer, then Civil Servant in Indian Administrative Service (IAS). After retirement, he writes on subjects, Astronomy, Mathematics, Yoga, Humanity etc.      





























































When Law Met Mathematics

He was a lawyer, a qualified lawyer and had studied law at Poitiers, graduating in 1559. He began his career as an attorney at a quite high level, with cases involving the widow of King Francis I of France and also Mary, Queen of Scots. He generally pleaded important cases of royal families and won also. He was always logical in his dealings and practice. He argued his cases on logics and never hid any facts irrespective these went against him.

But law was not his cup of tea. He was always curious solving mathematical problems. What makes a circle round, a rectangle cornered and a line straight? Can a polygon be a circle? He used to ponder over such problems sitting on his desk with elbow on the study table, taking food with other hand and constantly concentrating on the same problem even for as many as three days. He was François Viète of France, a Lawyer and Mathematician, a combination that is uncommon and he was uncommon.

Nonetheless, he was highly successful in law. By 1590 he was working for King Henry IV. The king admired his mathematical talents, and Viète soon confirmed his worth by cracking a Spanish cipher, thus allowing the French to read all the Spanish communications they were able to obtain.

He did not abandon his passion for mathematics and In 1591, François Viète came out with an important book, introducing what is called the new algebra: a symbolic method for dealing with polynomial equations.

How I came to know of this great Mathematician, is also worth telling. A few months back, it was Pie (written as π) day and there was a great hustle and bustle and magazines full of articles on Pi. This inspired me though late to contribute something to Pi which is considered the great mathematical constant.

For a person like me who had studied mathematics from the point of view of its applicability to Engineering, writing on Pi by me with contents noticeable among mathematics fraternity was analogous to passing post graduation by a school primer. With passing of each day, celebrations of Pi were dimming but my determination remained bright. Specially during morning walk when neurones were upbeat, different description of Pi focussed on my mental screen and finally I took this shining and glassy device in my hand and typed my notions about pie with title,”Pi Is Sweeter Than Pie.”

It was my whole hearted effort aimed at revealing something new which was not known earlier about Pi.I had already gone through various description evaluating Pi by means of infinite series and infinite products. Why I can’t give something alike of these, was the voice of my inner self.

Writing sine of an angle, I started sub dividing it infinitely into sines and cosines of an angle. Equating sin x as x where x diminished to very small quantity, Lo, there appeared an identity not seen by me. And I substituted x as Pi/2 and there appeared the value of Pi in nested form.

Immediately, I checked the identity on google search engine, yes it was correct, no one had done it the way I did. But still there was a lurking doubt in my mind, if I could do, others would have also done it. It were Eureka moments till my further search led me to Viete Formula for Pi. I looked at it intently, final formula of Viete had more or less the resemblance with my formula but his method was altogether different. There was a geometrical drawing, certain perpendiculars, triangles, everything was different but the result had the resemblance. Derivation done by me was based upon the trigonometry and calculus whereas great Mathematician Viete has derived it from geometry. This assuaged my feelings a bit, after all my approach was independent unguided by any help. An idea had cropped up in my mind and that led me to this derivation. “There are different ways to reach a destination and journey by each route had its own pleasures. Those who discover new ways, enjoys the beauty of virgin paths,” said I to myself.

Lifting my morales, I completed, “Pie is sweeter than pie,” which have many formulae to ponder and there one can find a bit of mathematician in oneself. Later on I found, it indeed proved sweeter than pie to many readers to whom I am indebted for showing interest in my works. Now whenever, I attempt any mathematical derivation, similarity of the result with Viete formula keeps me on tenter hooks. I am writing a paper on golden ratio and there too appears a nested function (similar to Viete Pi formula) to determine value of  φ^1/k where k = 2^n, φ = (1+ 5^1/2)/2 and n is 0, 1, 2, 3, ……. any number. But great Mathematician Viete has not done, according to me, work on golden ratio  φ and I feel secure.

Besides this, I find, attempting a mathematical derivation without having any knowledge of previous works in that field, gives one great satisfaction on knowing later on that ones approach was unique and undefiled. I am benefitted in this regard, there is an ocean of mathematics to attempt and derive independently. Do you endorse, books bias ones thinking? Whatsoever your answer may be, I am certain, it is one of the most difficult question to answer.

All said and done, Francoise Viete will remain great Mathematician in the annals of history. He was the first one from Europe to give formula in nested radicals of Pi to the world.

(Viete Beautiful Formula For Pi, see only integer 2 is used.)

If you happen to see this formula, it uses only one integer 2 to calculate Pi. One wonders if 2 was sufficient to evaluate Pi, what are then other integers meant for? Such great men will always be remembered like Benjamin Jonson said in poem, “In Short Measure Life May Perfect Be.” It is virtue that never dies, it is like a wood that is seasoned to last for ever.


1. Image Pi courtesy Pi visualized. Saw it over at /r/dataisbeautiful and had to make it. – View on Imgur: https://m.imgur.com/r/wallpapers/ZCUW7js

2. Pi and Golden ratio by John Baez at https://johncarlosbaez.wordpress.com/2017/03/07/pi-and-the-golden-ratio/amp/

3. “Pi Is Sweeter Than Pie,” at https://narinderkw.wordpress.com/2017/04/10/pi-is-sweeter-than-pie/

4. Viete’s Formula for Pi at World Of Pi http://www.pi314.net/eng/viete.php


About Author

Writer is an Electronics and Electrical Communication Engineering graduate and was earlier Scientist, then Instrument Maintenance Engineer, then Civil Servant in Indian Administrative Service (IAS). After retirement, he writes on subjects, Astronomy, Mathematics, Yoga, Humanity etc.

Solution Of Cubic And Higher Degree Equation With Combination Of Continued Fractions

 Solution of cubic and higher degree equations has always engaged attention of mathematicians from ancient times. Different approaches have been adopted to solve cubic equation, some transformed this equation to depressed form, some eliminated all the coefficients of this equation except cubic and constant term. By and large, all solutions seem tedious and requires constant concentration in their derivation. With the purpose of finding a solution which is easily doable, use of combined two continued fractions have been adopted and this has given inspiring results.

A fraction as we generally consider, is a part of whole quantity. Extending it, it can also be said, a fraction is a part plus complete quantity also and can be expressed as complete quantity plus fraction but certainly, it is never a whole in itself. A fraction thus can be written as ‘p’ divided by ‘q’ or p/q where p and q are whole number and q does not equal 1.

Let us take a fraction p/q = 71/137 and 71 being smaller can not be divided by 137 but 137 can be divided by 71. One can write it as 1 divided by1/(137/71) or (1)/1/(71/137). It is based on the principle that reciprocal of a reciprocal is same quantity.

Also (1)/1/(71/137) can be written as 1/137/71.

137 on being divided by 71 equals 1+ 66/71, therefore,

71/137 = 1/(1+ 66/71).

Again 66 can not be divided by 71 as 66 is smaller than 71. Again taking reciprocal of reciprocal of 66/71 as was done above, it can be written as 1/(71/66) and the procedure can be repeated.

Fraction 71/66 equals 1+ 5/66, thus

Fraction 71/137 = 1/{1+ 1/(1+ 5/66)}

5/66 can be written as 1/(66/5) or 1/(13+ 1/5), therefore,

71/137 = 0 + (1)/[1+1/{1+ 1/(13+ 1/5)}] or the fractions

0 + 1/[1+ 1/{1+ 1/(13+ 1/5)}] is called continued fractions of 71/137.

0 is added to this continued fraction as 71 could not be divided by 137 or it can be said that it is divided 0 time.

It is evident from above that procedure adopted above is same as finding the greatest common divisor (g c d) of 71 and 137. That is why great mathematician Euler described continued fractions as determining g c d.

By adopting the same procedure, fraction

15/11 can be written as 1+ 1/{2+ 1/(1+ 1/3)}.

Continued fraction can be written in generalised form as

pn/qn = a0 + b0/[a1 + b1/{a2 + b2/{a3+……an + bn/1)}}]………(1) 

where coefficients a0, a1, a3………. are called partial quotients.

When b0, b1,b2,b3 all are equal to 1, continued fraction can be written as

pn/qn = a0 + 1/[a1 + 1/{a2 + 1/{a3+…………an-1 +1/an)}}]……(2).

Such a fraction is also abbreviated as (a1; a2, a3, a4, ……an).

Continued fractions which terminate at a point as we have seen, 71/137 terminated at 1/5, 15/11 at 1/3 and a0…….., b0 at bn/1, are called close continued fractions. But there are continued fractions which do not terminate but continue infinitely. These are called infinitely continued fractions or periodic continued fractions. Some examples are given below.

y = 1/[2+1/{2+ 1/(2+ 1 /2………… up to Infinity……………….(3),

y = 2+ 1/[2+ 1/{2+ 1/(2+ 1………up to Infinity………………….(4).

Partial fractions (3) and (4) are infinitely continued but on examination, it is observed that fraction (3) repeats all its terms even after first quotient whereas equation (4) repeats all its terms after two quotients, therefore equation (3) has period one and equation (4) has period 2. Period is number of quotients or terms after which the fraction continues all its terms.

It is easily discernible from equation (3) that it can be written as

y = 1/(2+ y) as it is infinitely continued after first quotient.

Or y^2 + 2y -1 = 0 which is a quadratic in y and has two roots as

y = -1+ 2^1/2 or -1- 2^1/2.

Since fraction (3) only contains positive signs, its sum must be positive, therefore fraction (3) equals -1+ 2^1/2.

Similarly fraction (4) has value

y = 2+ 1/y

y^2- 2y – 1 = 0.

This is also a quadratic in y and has value equal to

(1+ 2^1/2) ignoring the negative value.

We can thus write fractions (3) as

(2^1/2- 1) = 1/[2+1/{2+ 1/(2+ 1 /2………………. up to infinity or 2^1/2 =1+ 1/[2+1/{2+ 1/(2+ 1 /2………………. up to infinity……(5)

and equation (4) as

(1+ 2^1/2) = 2+ 1/[2+ 1/{2+ 1/(2+ 1………up to Infinity. 

Mathematician Lagrange on the basis of this nature of infinitely continued fraction stated that all algebraic irrationalities can be expressed only and only in quadratic equations. Conversely, only and only quadratic equations express the algebraic irrationalities of infinitely continued fractions.

This property of infinitely continued fraction is utilised in solving quadratic equations and also finding square roots of quantities.

Let us say square root of quantity 2 is to be calculated. First step is to find that value which approximates 2. If I assume approximate value of square root of 2 as 1, on squaring, it becomes equal to 1 but if I assume its value as 2, on squaring, it becomes 4, therefore, 1 is in better proximity of square root of 2 than 2. Let x is that quantity when added to 1 gives the exact value of square root of 2. That means

1+ x = 2^1/2.

Or x^2 + 2x + 1 = 2

Or x^2 + 2x = 1

Or x = 1/(2 + x)

Here we find x in the denominator term x+2 but x = 1/(2 + x) and on substituting this value in the denominator, we get

x = 1/{2 + 1/(2+ x)}.

On successively substituting value of x = 1/(2 + x), we get never ending fractions as we substitute the value of x, another x further appears in subsequent denominator. Therefore x can be written as

x = 1/{2 + 1/(2+ 1/2+…………. up to infinity.

Therefore 2^1/2 = 1+ x = 1 + 1/{2 + 1/(2+ 1/2+…………. up to infinity.

Let us generalise the formula for determining square root of a quantity n.

Let m be any integer which approximates square root of n and x is an added quantity that exacts it to square root of n.

Then (x + m) = n^1/2

Or x^2 + 2.m.x + m^2 = n

Or x = (n- m^2)/(2m + x).

Substituting the value of x successively, we get

x = (n- m^2)/[2m + (n- m^2)/{2m+(n- m^2)/(2m+……………….up to infinity.

To determine convergence of a continued fraction, let us inspect fraction (5), sum of first row

p0/q0= 1,

sum of first plus second row, p1/q1 = 1+1/2 = 3/2.

Similarly, p2/q2= 7/5, p3/q3 =17/12 , p4/q4 =41/29 , p5/q5 = 99/70……….so on.

Therefore, p0/q0- p1/q1= 1- 3/2 = – ½,

p1/q1 – P2/q2 = 3/2- 7/5 = 1/10,

p2/q2- p3/q3 = 7/5- 17/12 = -1/84, 

p3/q3- p4/q4=17/12- 41/29 = 1/348,

p4/q4- p5/q5 = 41/29 – 99/70 = -1/2030,



So on.

From above, it is clear that as sum of more quotients is taken up, difference between consecutive sum decreases and that means fraction is convergent. Pn/qn is also called nth convergent. It also reveals a pattern where difference alternates from negative to positive like that of swing span of a physical balance. This will be explained in detail in example section.  

By mathematical induction, we can make a generalised statement from fraction (2) that

po /qo = a0/1, p1/q1=(a1a0+1)/a1, p2/q2 = (a2a1a0 +a2 +a0)/(a2a1 + 1) p3/q3 = (a3a2a1a0 +a3a2 +a3a0 +a1a0 +1)/ (a3a2a1 + a3 + a1) 

 ………………………………………………………………….so on.

Convergence of this fraction can also be checked the way we did for fraction (5).

Graph of a cubic function with 3 real roots (where the curve crosses the horizontal axis—where y = 0). Here the function is f(x) = (x3 + 3×2 − 6x − 8)/4.

Determining Roots Of a Cubic Equation

With this background, we proceed to determine Roots of a cubic equation which can in general be written as

x^3 + px^2 + qx + r =0 ………………………(6),

where p, q and are coefficients of x^2, x and constant term.

Values of x that satisfy this equation are called its roots and this being a cubic equation has three roots as number of roots equals degree of equation. If one of its roots is determined, the equation can be reduced to a quadratic and its remaining two roots can easily be found out.

Approximation Of A Root

A root can be approximated, if x is assigned some value and that value is put in the cubic equation, it will either have positive value or negative value or even zero value. If the equation, on substitution, has zero value, then we are lucky enough as it would be its roots and the matter to find a root ends. But in the rare cases, such root is accidentally determined. Therefore, we take that case, say, where the value of equation is not zero but is positive or negative. To proceed further to find a root, a theorem is put forth here under.

Theorem: There is a value nearby its root that makes an equation on threshold positive or negative and increment or decrement by one in that value, reverses the sign of the equation then its root lies between that value and the value increased or decreased by one.

Let f(x) = a0x^n +a1x^n-1 +………………= 0, is an equation,

then there will be a value α, that will make the function

a0 α^n + a1 α^n-1 +……… at threshold positive or negative

and on putting the value as α -1 or α + 1 in equation, either the value of a0 (α+1)^n + a1 (α+1)^n-1 +……… will be opposite in sign


the value a0 (α-1)^n + a1 (α-1)^n-1 +………will be opposite in sign to that of a0 α^n + a1 α^n-1 +………

By applying this theorem and putting a random value of x and then noting the sign of equation and then decreasing or increasing its value by one so that a stage reaches when its sign is reversed, approximate value of its root lying between α and α- 1 or α and α+ 1 can be found out. This will bring us closer to the value of the root of equation.

Transformation Of Cubic Equation

Coming to equation (6), let the value of its root approximated is m. Then there will be a quantity y which will always be less than 1, such that

y+ m will be its exact root. On putting this value in equation (6), it takes the form

(y+ m)^3 + p.(y+ m)^2 + q.(y+ m) + r = 0.

On simplification, a cubic equation in y is formed as

y^3 +y^2(3m +p) + y.( 3m^2 + 2pm + q) + m^3+ p. m^2 + q.m + r = 0.  

This transformed equation can be rewritten as

y^3 + a.y^2 + b.y+ c =0…………………………………………………(7),

where a = 3m +p,

           b = 3m^2 + 2pm + q,

           c = m^3+ p. m^2 + q.m + r.

Only discernible difference of this equation from equation (6), is that the cubic equation in y has a root which is less than 1 and root of this equation is related to equation (6) by

x = y + m.

Determination Of Root By Combination Of Two Fractions

To ascertain value of y, we can write the equation (7) as

y^3 + a.y^2 = -(b.y+ c),

or y = -(b.y + c)/ y.(1+ a.y).

Expanding In Partial Fraction

Above equation has numerator of degree one and denominator of degree two and, therefore, can be decomposed to partial fractions as

y =-(b.y + c)/ y.(1+ a.y) = -[ A/y + B/ (1+ a. y)]

Or by + c = A(1+ a.y) + B.y.

On comparing coefficient of y and constant term,

b = A.a + B and c = A.

On simplification,

A = c and B = b – a.c.

Therefore, y = – [ c/y + (b- ac)/(1+ a.y)

                        = (a.c- b)/(1+ a.y) – c/y…………………………….(8).

Equation (8) has two fractions (a.c- b)/(1+ a.y) and – c/y which have y in in the denominators indicating that these will continue infinitely. On successively putting the value of y as given by equation (8) in equation (8) itself, it will form never ending fractions as

y = (a.c- b)/[1+ a. {(a.c- b)/(1+ a.y) – c/y}] – c/{ (a.c- b)/(1+ a.y) – c/y}.

To show that this fraction is continuing infinitely, I put sign ……. In place of y.

y = (a.c- b)/[1+ a. {(a.c- b)/(1+ a…..) – c/…….}] – c/{ (a.c- b)/(1+ a……)} – c/……..}……………………………..,..(9).

y, therefore, can be calculated from convergent pn/qn as we discussed earlier where n will be number of times we want to consider partial quotients of fractions depending upon accuracy of result desired.

Alternatively, the equation (7) can also be rearranged so as to make it

y(y- s).(y- t) = – c + u.y  

by factorising LHS of equation y^3 + a.y^2 = -(b.y+ c).

y therefore equals (- c + u.y)/(y- s).(y- t).

This can be decomposed to partial fraction as

y = A/(s- y) + B/(s- y)

On successively putting the value of y as A/(s- y) + B/(s- t) in itself, it will form never ending fractions as

 = A/[s- {A/(s- y) + B/(s- y)}]

  + B/[t- {A/(s- y) + B/(s- y)}].

Since y is being successively substituted, dots are marked in place of y showing its infinite continuation. Then

y = = A/[s- {A/(s- …..)+ B/(s- …..)}]

  + B/[t- {A/(s- …….) + B/(s- ……..)}]…………………………………(9/1)

This method will be preferred as it can indicate approximate value of y by putting y as zero in right hand side and will be used in all calculations.

Determination Of A Root Without Approximation

Notwithstanding, the method where a root was approximated, a cubic equation can also be solved straight forward.

Let the equation be

x^3 + px^2 + qx + r =0.

It can also be written as x^3 + ax^2+ bx^2 + abx + q’x + r = 0

where p = a+ b and q = ab + q’.

This equation can also be factorised as

x(x+ a).(x+ b) = – (q’x + r).

This can then be decomposed to partial fraction as

x = A/(a+ x) + B/(b+ x),

where A and B can be determined by comparing coefficients of x and constant term. This is also an infinite continued fractions as x figures in both left hand side and right hand side. On successive substitution of value of x as A/(a+ x) + B/(b+ x) in its own equation, we get continued fraction as

= A/[a+ {A/(a+ x) + B/(b+ x)}]

+ B/[b+ A/(a+ x) + B/(b+ x)}]

= A/[a+ {A/(a+ …..) + B/(b+ …..)}]

+ B/[b+ {A/(a+ …..) + B/(b+ ….)}]…………………………..(9/2)

However when we compare fraction (9/2) with (9/1) and (9), we find that y (x in case of fraction 9/2), is always less than 1 but it may be more than 1 in case of equation 9/2. Y, in case of fractions 9 and 9/1 can be ignored but in fraction 9/2, it may result in crude approximation particularly in earlier convergents like p0/q0 and p1/q1. With this fact in mind, method of approximation of root is preferred and recommended.

Theory And Concept

If in a mathematical equation, its right hand side contains a part or whole of the quantity in its left hand side then the equation regenerates left hand side in right hand side and the regenerated quantity again contains left hand side, and regeneration again occurs and the cycle continues infinitely. The equation thus forms a close loop that continues infinitely.

For example, equation x = 1/(1+ x) has x in left hand side and 1/(1+ x) in right hand side. The output 1/(1+x) contains the input x. That means x in denominator of 1/(1+ x) can be replaced by 1/(1+ x) and the equation takes the form

x = 1/{1+ 1/(1+ x)}.  

Again output or right hand side RHS has x and this can again be substituted by 1/(1+ x) and this will again contain x in RHS. Such a quantity is infinitely continued.

If x = 1+ x^2,

then x in RHS can be substituted by 1+ x^2 and

x = 1+ (1+ x)^2 = 1 +1 + 2x + x^2.

On successive substitution, a power series is generated.

However, when x is of the form A/(B + x), it is then a fraction with numerator A and denominator 1/(B+ x). Since it also contains x in RHS, it is also an infinitely continued fraction similar to the one as mentioned at (1) with the only difference that the present fraction continues infinitely and has

a0 = a1 = ……. b0= b1 = …… = 1.

But an equation of the form,

x= A/(B + x)

can be simplified as

x^2 + Bx – A = 0,

and this forms a quadratic equation in x. This deduces us to make a statement that infinite continued fraction is always expressible in quadratic equation. This quadratic equation has roots

= ½. [ – B + (B^2 + 4 A)^1/2] or

1/2.[- B – (B^2 + 4 A)^1/2].  

If the quantity (B^2 + 4 A) is a perfect square, the roots will not have any square root term. Otherwise these will always have a square root term.

Such square root quantities are irrational as these can never be calculated exactly and their decimal portion continues infinitely. Such quantities give birth to fractions that continue endlessly. Being born form quadratic equations, these are called quadratic irrationalities or algebraic irrationalities.

With these concepts in mind, Great Mathematician Lagrange gave the theorem,

“Every periodic continued fraction represents a quadratic irrational number and every quadratic irrational number is represented by a periodic continued fraction.” 

Period of a function has already been defined earlier by me. Every infinite continued fraction is periodic, that means quadratic equation is essential requirement for infinite continued fractions. But in this paper, cubic equations which are a degree higher than quadratic are being dealt. Does it not then contradict Lagrange theorem?

This is the moot question which will be convincingly settled here. A cubic equation in general is expressed as

y^3 + ly^2 +m y + n =0,

and its approximate root can always be estimated as it has already been explained in detail. Let the root approximated is y’ then on substituting

y = x + y’, a cubic equation can be found out. Let this equation be

x^3 + px^2 + qx + r =0.

It can also be written on rearrangement as

  x = -r/(x^2+ px+ q)………………………………………………………(T)

This equation as it is and without any operation and in its present form, is not expandable to continued fraction, is admittedly true. Obviously, it contains a quadratic equation in denominator and that will result in a cubic equation whereas Lagrange Theorem requires formation of quadratic equation. A quadratic equation will only form when denominator of equation (T) contains x in first degree. Therefore, it is essential to transform it so that it may contain x in first degree in its denominator.

That is possible if denominator -r/(x^2+ px+ q) is decomposed into partial fractions as that will give it the form A/(a+ x) + B/(b+ x) where a+ x and b+ x correspond to first degree in x.

By equating -r/(x^2+ px+ q) = A/(a+ x) + B/ (b+ x) and comparing coefficient of x and constant term, A and B can be determined.

Then x can be written as

x = A/(a+ x) + B/(b+ x).

In calculus, while integrating a quantity of the form -r/(x^2+ px+ q) where x is variable, it is split up in partial fractions as A/(a+ x) + B/(b+ x) and each part is integrated separately without affecting the result. Same procedure is applied here.

It is explicit that the value of x is related to both partial fractions A/(a+ x) and B/(b+ x). It is also clear, x when approximated in the form p0/q0 will result as sum of A/a and B/b and this will also be true when x is small. As each A/(a+ x) and B/(b+ x) has a term x, these are also interconnected. Approximation of x as p/q, improves as number of quotients considered are increased, since both interconnected fractions are continued infinitely and also solutions given in examples corroborate the convergence. Variation in x on account of improvement in approximation will cause changes in A/(a+ x) and B/(b+ x). As improvement in approximation is affected by both A/(a+ x) and B/(b+ x), changes in A/(a+ x) will cause changes in B/(b+ x) and vice versa. Conclusion is that both A/(a+ x) and B/(b+ x) although separate are integrated with each other so that their resultant effect is same as that of -r/(x^2+ px+ q).

In other words, this can also be stated that A/(a+ x) and B/(b+ x) are interrelated two continued fractions and approximation of x on account of A/(a+ x) also affects B/(b+ x) and vice versa. This is due to the fact that expansion of A/(a+ x) into infinite continued fraction contains term B/(b+ x) and likewise expansion of B/(b+ x) into continued fraction contains term A/(a+ x).

Apart from this, except for their interrelation, A/(a+ x) is an infinite continued fraction with periodicity one and similarly B/(b+ x) is also an infinite continued fraction with periodicity one. These two fractions behaves like normal periodic fractions. However, it is their interrelation, that brings the effect of cubic equation.

Had there been a single periodic continued fraction formed from cubic equation, it would have violated Lagrange Theorem. On the other hand, formation of two periodic fractions has not only solved cubic equation but has kept principle employed in the theorem intact. It is a special case where a cubic equation is solvable and also meets the requirement of the theorem.

It is thus essential, for solving cubic equation by continued fractions, there must be two interrelated periodic continued fractions.  

While solving cubic equation, two periodic interrelated continued convergent fractions were generated from A/(a+ x) and B/(b+ x) and how these two Infinite continued fractions are convergent have been dealt while solving cubic equations in example section.

This concept is also applicable to Quartic ( degree four) equations but for solving such equations, three interrelated continued fractions will be needed. These three partial fractions A/(a+x) + B/(b+x) + C/(c+ x) can always be expanded and x can be solved from equation

x = A/(a+x) + B/(b+x) + C/(c+ x)

as it was solved in case of cubic equation.

How terms A/(a+ x) and B/(b+ x) progress and what their algorithm is shown in figure.

{Figure 1 showing progression of terms of A/(a+ x) and B/(b+ x) in algorithm of x = A/(a+ x) and B/(b+ x }

Red line corresponds to term A/(a+ x) and black to B/(b+ x). Point A that corresponds to p0/q0 is first approximation point where x approximation is A/a + B/b. At this point, x equals A/(a+ x) + B/(b + x) and approximation is made by neglecting x, otherwise also x is small due to the fact transformation has been done after estimating the root. As the terms of fraction increase , points B of both branches (p1/q1) are reached, these contain four terms and at points C of four branches ( approximation p2/q2), these contains 8 terms. In this way, number of terms increases in geometric progression. However the figure is drawn up to point E, thereafter, dotted lines indicate that it is continuing up to infinity. The figure illustrates how the terms A/(a+ x), and B/(b+ x) are generated in geometrical progression.

Examples And Applications

Keeping in mind, the theory described above, cubic equation f(x) = x^3- 6x^2+ 10x -9 = 0 is solved here.

Applying theorem as stated above, at x = 4,

f(4) = 64 – 96 + 40 – 9 = – 1

and at x = 4 + 1 or 5,

f(5) = 125 – 150 + 50 – 9 = 41.

Obviously, when x increases from 4 to 5, equation changes sign from negative to positive. That means one of the root of the equation lies between 4 and 5. Let the root be 4 + y where y will be less than 1. 4+ y then must satisfy the equation,

(y + 4)^3 – 6. (y + 4)^2 + 10 (y + 4) -9 = 0.

On simplification,

y^3 + 6y^2 + 10 y -1 =0……………………………..(10).

Or y = 1/( y^2 + 6y + 10).

Since one of the root lies between 4 and 5, that means value of y is always less than 1. If we assume y to be very small, then rough estimate of y is,  

y = 1/(0+ 0 + 10) =1/10 or .1 and x will be equal to 4.1.

But for finding exact value of y, continued fractions will be resorted to. Equation

y^3+ 6y^2 + 10y -1 = 0, 

can be written as

y(y^2+ 6y + 8)+ 2y -1 = 0 on splitting 10 y into 8y + 2y so as to facilitate factors of y^2+ 6y + 8 as (y + 4).(y+ 2).

The equation therefore can be written as

y(y + 4).(y+ 2) = 1- 2y,

y = (1-2y)/(y+ 4).(y+ 2).

This can be decomposed in to partial fraction as

y =(1-2y)/(y+ 4).(y+ 2) = A/(y+ 2) + B/(y+ 4)

= (5/2)/(2+ y) – (9/2)/(4+ y)

Both these fractions contain y in the denominator and therefore, can be expanded as infinite continued fractions on successive substituting the value of y in the above equation.

y = (5/2)/[2+ {(5/2)/(2+ y)} – (9/2)/(4+ y)}] – (9/2)/[(4+ {(5/2)/(2+ y) – (9/2)/(4+ y)}]………………………………………………(11)

y = convergent p0/qo = .125

  = convergent p1/q1 = .0855

  = convergent p2/q2 = .094.

The pattern of above values indicates that the difference

p0/q0 – p1/q1 = .0395,

p1/q1- p2/q2 = -.0085, 

is alternating from negative to positive as p/q convergent increases. Swing for first correction is .0395 and for second, correction is -.0085, it is analogous to swing of span of physical balance which also experiences shortening amplitude swing from its central position when an object of same mass as that the standard weight is kept on the span.  

Convergence of y is also illustrated in figure 2 which is not drawn according to scale. Proving of convergence from this example establishes that the combined effect of two interrelated infinite continued fractions completely and correctly theorise the solution of cubic equations. It is illustrated in figure 2 that as approximation p/q of x pertains to large convergent, x attains the exact value from 0.125 to .0855 to .094 to……………….exact value.

Also second part Of figure 2, shows how swing reduces to zero from .0395 to -0.0085 to …………….0.

                      Figure 2 shows improvement in approximation and swing with increase in number of convergents

It is also worthwhile to check deviation of calculated value of root from its actual value. At y = .094, function y^3+ 6y^2 + 10y -1 has value of .00131 and at .0945, it has value of .0006 and assuming, x as .0945 satisfying the equation, value of x will calculate as 4 + y = 4.095. But y from continued fractions calculates as 4 + .094. Therefore at p2/q2, calculations by continued fraction has an error of .122 percent. As the number of terms for calculations will increase, value of root improves. That further proves that method of determining roots of a cubic equation as theorised, does not suffer from any infirmities.

Calculating Cube Root Of A Quantity by Continued Fractions

Applying the theory as described above for finding a root of cubic equation, let us calculate cube root of 9. It is explicit that 9^1/3 equals approximately 2. Three will not be good approximation as 3^3 makes 27 which differs by 18 from 9 whereas 2^3 makes 8 and that differs by 1 from 9. Let x which will be less than 1, be the addendum that makes

x + 2 = 9^1/3.

On cubing both sides, we get

x^3 + 6x^2 + 12 x + 8 = 9.

Or x^3 + 6x^2 + 12 x = 1.

Or x = 1/( x^2 + 6x + 12).

The fact that at x small and less than 1, terms containing x can be ignored. That gives approximate value of x as 1/12 but we are interested in determining the actual value of x, therefore on rearranging, the equation can be written as

x^3 + 6x^2 + 8 x = 1- 4x.

Or x(x+ 4).(x+ 2) = 1- 4x.

Or x = (1- 4x)/(x+ 4).(x+ 2).

On expanding it to partial fraction,

x= (1- 4x)/(x+ 4).(x+ 2) = A/(x+ 4) + B/(x+ 2).

Or x = (9/2)/(2+x) – (17/2)/(4 +x).

In the right hand side RHS, both the partial fractions have x, therefore, on successive substitution of

x in RHS as (9/2)/(2+x) – (17/2)/(4 +x),

we get

x = (9/2)/[{2+{(9/2)/(2+………..)}– {(17/2)/(4 +……….. )}]

– (17/2)/[4 +{ (9/2)/(2+………….) – (17/2)/(4 +……)}]……………………(12),

Where sign dots indicates that fraction is continuing infinitely.

Since x is small and less than 1 as explained earlier, first approximation

x = p0/q0 = 1/8 = 0.125,  

second approximation,

x = p1/q1 = 0.0570410,

third approximation,

x = p2/q2 = 0.07357277.

It is explicit from above, as p/q convergent increases employing more terms, difference p0/q0 – p1/q1, p1/q1- p2/q2 alternates as convergent improves. That corroborates, continued infinite fraction is convergent and is correctly defined by infinite continued fraction (12).

It is also worthwhile to check deviation of calculated value of cube root 9 from its actual value. At approximated value 2 + .07357277 = 2.07357277, its cube that is (2.07357277)^3 equals 8.9157492 and therefore, has an error of .95 percent but as convergent p/q will increase, it reaches its exact and correct value. Here also pattern of values of p/q reveals that convergence is analogous to swings of span of a physical balance. After a few swings, the span settles at its mid position as illustrated in figure 2.

Examples how to determine a root of quartic equation has not been given, however quartic equations can also be solved following the method described in section theory and concept.


1) NON-PERIODIC CONTINUED FRACTIONS FOR QUADRATIC IRRATIONALITIESMICHAEL O. OYENGO at http://www.math.illinois.edu/~mchlyng2/Non-Periodic%20continued%20fractions.pdf

2) The Topsy-Turvy World of Continued Fractions by MICHAEL O. OYENGO

3) General Method for Extracting Roots using (Folded) Continued Fractions by Manny Sardina

4) Graph of a cubic function with 3 real roots (where the curve crosses the horizontal axis—where y = 0). Here the function is f(x) = (x3 + 3×2 − 6x − 8)/4. Courtesy N Mori at https://en.m.wikipedia.org/wiki/Cubic_function

About Author:

Author is an Electronics and Electrical Communication Engineering graduate and was earlier Scientist, then Instrument Maintenance Engineer, then Civil Servant in Indian Administrative Service (IAS). After retirement, he writes on subjects, Astronomy, Mathematics, Yoga, Humanity etc. 







Natural Number Is Not Much Natural


All functions are expressible by addition or multiplication of its small constituents. In fact, it is the pattern or way or nature of these small constituents or building blocks that determines its characteristics. This concept is also applicable to living beings. How these constituents can be extracted from the mathematical functions is a part of the subject I will be discussing. Notwithstanding these small building blocks, I will also discuss connected subjects ie roots of functions, infinite series, their applicability to the derivation of infinite product of functions particularly exponential function, definition and derivation of value of natural number using complex analysis and calculus. Last I will discuss why natural number are more unnatural than being natural. For information of learned readers, I will attempt to analyse independently and untraditionally.

Background Of Natural Number Reveals Money Lenders And Bankers Are Also Mathematicians

All exponential functions are based on natural number or Euler number denoted by ‘e’ and its use was so common that it went unnoticed from the eyes of John Napier discoverer of Logarithm. He while calculating logarithm of quantities, used natural number as their base but its value was later on accidentally determined while calculating compound interest. Money is always dearer and attracts most attention, probably this is the reason why a semi educated can calculate his merchandise better and more accurately than a moderately educated person. Credit to recognise the importance of natural number also goes to money lenders and bankers who were always striving to find ways to multiply principal by repeatedly compounding interest. It is interesting to note their ingenuity in such exponential compounding that we are today discussing natural number.

Compounding Of Interest And Natural Number

Let us briefly analyse the concept of compounding of interest. If P is principal amount, it after t years will become C if it is lent on interest at rate t per annum compounded yearly by formula

C = P.(1 + r/n)^n.t where n is number of times it is compounded in a year. If it compounded quarterly, the formula gets modified as

C = P.(1 + r/4)^4.t. As we go on increasing the number of compounding n in a year, C will go on on increasing. If the interest is compounded pet hour, or per minute or per second, n goes on increasing and with that C will also continue escalating till it reaches the maximum value 2.718 of principal amount. This maximum multiplier of C/P achieved when the interest was compounded INFINITE times in a year, is the value of natural number 2.718281828459045. What is the concept of natural number involved here will be discussed at appropriate time.

Roots Of A Function

With this background note, concept of roots and their applicability to determining the infinite factors of a function are taken up. Most functions have roots which are those values of the variables at which the functions will equal to zero. This quadratic equation x^2 – 5.x – 6 = 0, has roots (also called its zero) at x = 3 and x = 2. This equation can also be written as (x- 3).(x- 2) = 0 where (x- 3) and (x- 2) are the factors of the above quadratic equations. Number of roots depends upon the degree of equation or power of x.

Equation an.x^n + an-1.x^n-1 + an-2.x^n-2 + ……………a1.x1 + a0 =0 has n roots but if the equation is of form

a0 + a1.x^1 + a2.x^2 + a3.x^3+…………upto infinity, it will have infinite roots.

Applying above formula, (1-x^2)^1/7 = 0, will have two roots 1 and -1 but this can also be expanded in infinite terms by applying Binomial theorem and then this must have infinite roots. When it is expanded in infinite terms, it will have infinite roots but roots 1 and -1 will also be its roots. Since infinite expansion is always approximate, its infinite roots will also be approximate. It is submitted that such infinite summations are seldom used in extracting roots. But mathematician Euler used roots to find the value of pi. I give another example of the equation 1/(1- x), it has only one root at x equal to infinity but this equation can also be expanded and roots approximated.

Functions Can Be Derived From Their Roots

Like wise if we know the roots of an equation, we can determine the equation. If a, b, c, ….. are n roots of an equation, the equation will be given by (x- a).(x- b).(x- c)………….upto n terms = 0.

Trigonometric Functions And Their Roots

Let us take the case of sin x to determine its infinite product. To form its equation, we must know its roots or value of x that makes sin x equal to zero.

Sin x = 0, when x is 0, pi, 2pi, 3pi so on or we can write this x = k.pi where x is any integer from 0 onwards. That means x = k.pi are roots of sin x. Similarly,

sin x = 0, when x is 0 or – pi, -2pi, -3pi so on or we can write this x = -k.pi where x is any negative integer. That means x = -k.pi are also roots of sin x.

That further means x ( pi – x).(pi- x).(2pi+ x).(2pi+ x).(3pi+ x).(3pi- x)……………..up to Infinity = 0 is an equation of sin x.

Or sin x = x (1 – x^2/pi^2). (1 – x^2/4pi^2) (1 – x^2/9pi^2)……………so on upto infinity.

Or in short, sin x = Infinite Product x.(1- x^2/(k.pi)^2 where k is an integer from 0 to Infinity.

Taking the case of cos x, I submit cos x equals zero when x is of the form (2k- 1) pi/2 or – (2k- 1).pi/2 where k is any integer between 0 to Infinity. Therefore,

cos x = Infinite product (1 – 4.x^2/[(2k- 1).pi]^2,

where k is an integer from 0 to Infinity.

Roots Of Exponential Function e^x

Let us attempt to solve the infinite products of “e” or “e^x”. For that we need to find the roots of “e^x” as we have found in the cases of sin x and cos x.

I submit, e^x equals to 0 when x tends to negative Infinity. That makes e^x as e^-∞ or 1/e^∞ which is equal to 1/∞ or 0 when x tends to -∞. Next is periodicity of e^x, periodicity is the interval or angle after which the function completes its full cycle and restart from its corresponding point. Sin x, cos x, tan x etc are all periodic as these on completing one full cycle of 2pi angle, restart from their corresponding position. Sin (x + 2pi) equals sin x, cos (x+ 2pi) = cos x, tan (x+ 2pi) = tan x likewise.

                                     Image curve of exponential function

In case of exponential functions, if we change x to x + 2.pi.i where i is square root of -1 then e^(x+2.pi.i) equals e^x(cos 2.pi + i sin 2pi) = e^x. That means, such functions have periodicity of 2.pi.i angle. As we have seen root is -∞, it will not make any change to -∞, if we add or subtract 2.k.pi to infinity. That means it has only one root at x = -∞ and that repeats infinitely. Let n is a quantity that tends to ∞ and -∞ is recurring root, we can say -n is the root of e^x.


 e^x = [1- x/(-n)][1- x/(-n)] [1- x/(-n)] ……….up to Infinity.

On simplification,

e^x = (1+ x/n) (1+ x/n) (1+ x/n) (1+ x/n)………….up to Infinity.

Or e^x = (1+ x/n)^n………………………………….(1)

where n tends to Infinity.

Let p = n/x or n = p.x, then as x tends to ∞, p also tends to Infinity. Putting n = p.x in equation (1), 

we get 

e^x = (1+ 1/p)^p.x…………………….. (2) 

 = [(1+ 1/p)^p]^x where p tends to ∞. That means

e = (1+ 1/p)^p………………………………………….(3)

 where p tends to ∞. This is the definition of e, we have been taught in class rooms. Here the basis why e equals (1+ 1/p)^p where p tends to Infinity is derived from the roots of e^x.

If we expand equation (1) by Binomial Theorem and simplify, we get,

e^x = 1 + x + x^2/2! + x^3/3! + x^4/4!+……………….(4).

On putting x = 1,

e = 1 + 1 + 1/2! + 1/3! + 1/4!+……………….(5).

Coming to the compounding of interest, equation

 C = P.(1 + r/n)^t is analysed. It is submitted n is enlarged by increasing the frequency of compounding. In t years, total number of compoundIng will be t.n and this equation transforms to

C = P.(1 + r/n)^n.t.

 Let r/n =1/q when n ie number of compounding in a year is very large, q will also be large. And the equation will further transform to

C = P.(1 + 1/q)^t.q.r =P[(1 + 1/q)^q]^t.r = P.e^t.r when q is very large, since (1+ 1/q)^q = e, when q is very very large. But when t.r equals 1, C will be e times P. That proves the interlink of compounding of interest with natural number “e”.

What are the terms of infinite product of e^x?

Examination of equation (1), reveals that it is (1+ x/n) where x tends to Infinity, and it repeats infinitely. Can we consider, (1+ x/n) as 1, since when n reaches infinity, x/n tends to zero. Its individual limiting factor can not be considered particularly due to the fact that it is repeating n times where n tends to infinity. Even a quantity as small as tending to zero on adding to 1 after infinite multiplication as (1+ x/n)^n will culminate in a number as large as e.

If n in equation (1) is changed to -n, (1+ x/n)^-n will equal to 1/e^x.

If n is changed to n/2, (1+ x/n)^n/2 will equal to e^x/2.

If the infinite product is of the form (1- x/n)^n, then it will equal to e^-x,

If it is of form (1-1/x.n)^n, then it will be equal to e^-1/x. How these are done, is left to the readers to calculate by method of substitution. One can also do more manipulations and determine the values of this infinite product.

Determining “e” From Complex Analysis

Before more methods are taken up to find infinite products of e^x, natural number e (which is no other than e^x where x is one) is calculated independently from complex analysis where all derivations have been done ab initio. Almost similar exercise was also carried out in my earlier submission titled “The Jewel In The Crown Of Mathematics” at https://narinderkw.wordpress.com/2017/06/26/the-jewel-in-the-crown-of-mathematics/

I am therefore reluctant to repeat independent derivations of series expansion of log (1+ x), tan inverse x and also Binomial expansion. Those want to go through, may click the link.

Log (1+ i tan x can be expanded as

= i tan x – (-1). Tan^2 x/2 – i tan^3 x/3– tan^4 x/4 + i tan^5 x/5 +…………..

= i ( tan x – tan^3 x/3 + tan^5 x/5 ………..) + ( tan^2 x/2 – tan^4 x/4 + tan^6 x/6 ……) on separating imaginary and real parts.

Also log (1+ i tan x) = log (cos x + i sin x)/cos x

= log (cos x + i sin x) – log cos x. Therefore,

log (cos x + i sin x) – log cos x = i ( tan x – tan^3 x/3 + tan^5 x/5………..) + ( tan^2 x/2– tan^4 x/4 + tan^6 x/6 ……).

For all real x, cos x is real, therefore, log cos x is also always real.

Comparing imaginary parts,

Log ( cos x + i sin x) = i(tan x – tan^3 x/3+ tan^5 x/5 ……)

But tan inverse x = x – x^3/3 + x^5/5 -x^7/7+……….. up to Infinity.

Or x + 2k.pi = tan x – (tan x)^3/3 + (tan x)^5/5 – (tan x)^7/7….up to Infinity.

Taking its principle value as i.x,

log (cos x +i sin x) = i.x.

Taking antilog of both sides,

cos x + i sin x = e^i.x…………………………………………(6),

On putting x equal to pi/2 in equation (6),

cos pi/2 + i.sin pi/2 = i = e^i.pi/2.

On raising power i of both sides,

i^i = e^-pi/2 or e = i^(-2.i/pi).

On squaring i^I, i^2i = (-1)^i = e^-pi or e = (-1)^(-i/pi).

Also putting i.x as x in equation (6),

e^x = cosh x – sinh x. That is the definitions of cosh x and sinh x.

Also on putting x as 1,

e = Cosh 1 – sinh 1

Determining “e” From Calculus

Another approach to find e is based on the function y = x^x. Explicitly, it appears from function, with increase in x, x^x also increases exponentially and as x tends to Infinity x^x will also approach Infinity. But crucial points to examine when when x =0 and other when x^x will have minimum value. This function can also be written as

log y = x.log x.

 When x—- 0, log y = log x/(1/x) and log x/(1/x) takes the form Infinity/infinity, thus L Hospital rule is applicable. On differentiating numerator and denominator with respect to x,

log y = (1/x)/(-1/x^2) = -x = 0 as x ——0.

Or y = e^0 = 1.

For finding its minima, x^x is differentiated with respect to x, that is

1/y.dy/dx = x.1/x + log x. 1

dy/dx = x^x(1+ log x).

It is submitted that whenever there is maxima or minima, slope or dy/dx of the function becomes zero then it changes sign which are ascertained from second derivative. When slope will be zero then

x^x(1+ log x) = 0.

That means either x^x = 0, or (1+ log x) = 0. But when x^x is zero, it does not give definite value of x, therefore, only option is

(1+ log x) = 0.

Or log x = – 1 or x = e^-1 = 1/e.

Thus minimum value of function is (1/e)^1/e.

From this, it is deduced that reciprocal of natural number (1/e) is that value of x at which the function x^x has the minimum value (1/e)^1/e).

Another Method From Calculus

Still there is another method to find the value of e from the function of x^1/x where x —-infinity.

Let y = x^1/x.

On taking log from both sides,

log y = (1/x).log x which is of form Infinity/Infinity. Applying L Hospital rule by differentiating numerator and denominator,

log y = (1/x)/1 = 1/x = 0 as x tends to – Infinity.

Or y = x^1/x = e^0 = 1.

Similarly it can be proved, when x —– 0, x^1/x will tend to 0 and when x — Infinity, it has value 1. Let us find its maximum value by equating dy/dx = 0.

1/y.dy/dx = (1/x).(1/x) + log x ( -1/x^2).

Or dy/dx = (x^1/x)/x^2(1- log x) = 0.

                   Global maximum of x^1/x occurs at x = e.

Therefore log x = 1 or x = e. On plotting graph or double differentiating, it can be proved that function x^1/x has maximum value e^1/e at x = e.

From this, it is deduced that natural number is that value of x at which function x^1/x has maximum value (e^1/e).

Still Another Method

Another function y = (1/n.x)^sin n.x when x -0, is also examined for determining “e”.

Taking log and differentiating by applying L Hospital rule,

 log y = n.cos n.x/n which equals 1 at x —0.

Or function (1/n.x)^sin n.x equals e, when x — 0.

In this manner, e can be calculated by taking limits of similar exponential functions.

Determining e From Summation Of Series

e can also be determined from equation (5) on summing up its few terms depending upon accuracy of value e.

Infinite Product Of “e^x”

In the opening para of this article, it was mentioned that all functions can be expressed as summation or product of infinite terms. e^x (or any other function) can also be expressed as

= a0(1+ a1.x). (1+ a2.x^2). (1+ a3.x^3). (1+ a4.x^4). (1+ a5.x^5)…….upto infinity……………………………(7),

where a1, a2, a3, a4 …….. coefficients of x, x^2, x^3, x^4 ………

On putting x. = 0, in equation (7), we get,

1 = a0.

For determining other coefficients, four or five terms of right hand side may be multiplied and differentiated successively after putting x =0 at each differentiation. Alternatively, taking logarithm of both sides and thereafter successive differentiation after putting x as zero at each differentiation, these coefficients can be ascertained. These methods are similar to Taylor’s series expansion.

The coefficients can also be found by successive division of expanded infinite series of e^x as 1 + x + x^2/2! + x^3/3! + x^4/4!……….by 1+x then by next term containing x^2 after deciding a2, a3, … in such a way that it eliminates the numerator term after term. The coefficients in the present case are,

 a1 = 1, a2 = ½, a3 = -1/3, a4 = 3/8……..

On putting these in equation (7), it transforms to

e^x = (1+ x). (1+ 1/2.x^2). (1-1/3.x^3). (1+ 3/8.x^4)………….upto infinity……………………………(8).

This is an independent way of determining infinite product of e^x. It can be observed from this infinite product that as power of x increased, value of higher coefficients will decrease. To calculate “e”, x is replaced by 1 and first few factors are multiplied.

e = (2).(3/2).(2/3).(11/8) or 2.75.

When the number of multiplying factors are considered appreciably large, value of e will improve.

In the end, I submit that there would hardly been any field where natural number does not find its use. It may be living science, physical science, statistics, finance, demography, shape of galaxy, or growth of net worth of companies, exponential functions (where e is the base) will always come into play. It reminds me of the legendary man who was not only wise but was mathematically genius and was fully aware of the enormity of exponential growth that he humbly asked for grains of rice according to squares of chess board, “Oh emperor, my wishes are simple. I only wish for this. Give me one grain of rice for the first square of the chessboard, two grains for the next square, four for the next, eight for the next and so on for all 64 squares, with each square having double the number of grains as the square before.”

It was natural to fulfil the wish by the emperor or any person in his place as very few people know how large are 2^63 + 2^62+ 2^61+………….2^2+ 2^1 +2^0. Here 2^63 seems achievable to any person but this is so large a quantity that the rice in the entire universe may fall short if one were to gift this. The emperor naturally accepted the wish, little realising unnatural hugeness of rice and failed to fulfil it. Guessing exponential term based on natural number is not so natural. It is the field where intuitive calculations fail and unnatural happens.


1. https://en.m.wikipedia.org/wiki/E_(mathematical_constant)#/media/File%3AXth_root_of_x.svg 2. Image curve of exponential function courtesy author Peter John Acklam at https://commons.m.wikimedia.org/wiki/File:Exp.svg#mw-jump-to-license                              3. http://www.mathsisfun.com/numbers/e-eulers-number.html
                                                                 4. Reference Rice and chess game board at http://www.dr-mikes-math-games-for-kids.com/rice-and-chessboard.html

About Author

Author is an Electronics and Electrical Communication Engineering graduate and was earlier Scientist, then Instrument Maintenance Engineer, then Civil Servant in Indian Administrative Service (IAS). After retirement, he writes on subjects, Astronomy, Mathematics, Yoga, Humanity etc.