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.
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),
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.
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
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.