Today morning, while pondering over first and second derivative of a curve, I wandered to the subject of Euler’s formula. Such is the beauty of this formula that Richard P. Feynman considered it as the most remarkable formula in mathematics. Even readers of the Mathematical Intelligencer voted it “the Most Beautiful Mathematical Formula Ever”.

*Euler’s Formula
*

In simple words, it i*s e^ix = cos x + i sin x. If x is pi then e^i.pi = -1 or e^i.pi +1 = 0 where e is natural number and i square root of -1. *

This formula owed to great man Leonhard Euler who was a Swiss mathematician, physicist, astronomer, logician and engineer. His contribution to mathematics is so broad based, if we were to minus his contributions, mathematics would come to its nascent stage.

Euler’s formula relates to complex numbers which have real and imaginary parts. Real part is realisable, measurable and practical whereas imaginary part as the name says, is unreal, unmeasurable and impractical. Imaginary parts pertains to square root of minus one. This imaginary quantity, appearing in complex analysis, is denoted by alphabet, ‘i’. Further if a quantity has only imaginary part, it is called pure imaginary, if it has only real part it is called real and if both, it is called complex quantity.Two multiplied by two or 2^2 is solvable as 4 but 2^i that is real quantity raised to imaginary power, is not solvable but Euler made it possible.

*Expression Of A Quantity In Hyperbolic Functions And Trigonometric Functions
*

A thought comes to ones mind, if “e” raised to power imaginary quantity say i. x, is solvable to real and imaginary parts as cos x, i sin x, respectively then “e” raised to power some real quantity may also be solvable to some real parts like cos x, sin x or some other functions. Also sin x and cos x are quantities which can have maximum value 1, therefore, solving e^x may result in too large a quantity to be coped up by sin and cosine of an angle. Hyperbolic functions sinh x, cosh x constituted of e^x and e^-x, it seems, can solve the purpose in such cases. We will examine this aspect in due course.

*Independent Approach
*

I submit, I am attempting to solve Euler’s formula by independent approach which is altogether different from settled methods, I have not perused the method, the great mathematician adopted and I am also not using Taylor’s or Maclaurin’s series for proving the formula. Expansions of functions in power series are based on my own method and I also subscribe to the notion that all functions are expandable in power series without using settled methods except a few. Those few I am not discussing here, being off the topic.

*Defining “e” And “e^x”
*

With this background and information, I proceed to define ‘e’. ‘e’ is a natural number which equals the quantity (1 + 1/n)^n when n tends to infinity. By applying Binomial theorem,

the quantity (1 + 1/n)^n can be expanded as

limit n—–infinity, [ 1 +n.1/.n +1/n^2 (n)(n-1)/2! + 1/n^3 (n)(n-1)(n-2)/3! +1/n^4(n)(n-1)(n-2)(n-3)/4! +…upto infinity].

Limit n—- infinity or 1/n – 0, [ 1+ 1+ (1-1/n)/2! + (1-1/n)(1-2/n)/3! …..up to infinity,

= [1+ 1+ ½! 1/3! +……….upto infinity].

Therefore, *e = 1+ 1+ ½! 1/3! +……….upto infinity.*

Similarly, e^x can be expressed as

limit n–> infinity, (1 + 1/n)^n.x.

By applying Binomial theorem,

limit n—–>infinity, [ 1 +n.x.1/.n +1/n^2 (nx)(nx -1)/2! + 1/n^3 (n.x)(nx- 1)(nx -2)/3! +1/n^4(nx )(nx -1)(nx -2)(nx-3)/4! +…upto infinity].

Limit n—-> infinity or 1/n –> 0, [ 1+ x+ x(x-1/n)/2! +x. (x -1/n)(x-2/n)/3! …..up to infinity

= [1+ x +x^2. ½! + x^3.1/3! +……….upto infinity]. Therefore,

*e^x = 1+ x + ½!.x^2 +1/3! .x^3.+……….upto infinity…,…………………(1).
*

* Hyperbolic Functions
*

This e^x along with 1/e^x are the constituents of hyperbolic functions where

*Sinh x = (e^x – e^-x)/2,…………………………………….(2)
*

*Cosh x = (e^x + e^-x)/2,…………………………………….(3)
*

*Tanh x = (e^x / e^-x)/(e^x + e^-x).
*

Other functions I am not mentioning as these will not be used in our discussion and analysis. But for the knowledge of learned readers, these are more or less analogous to trigonometric functions.

If we add equation (3) to (2) then

*Cosh x + Sinh x = (e^x + e^-x)/2 + (e^x – e^-x)/2 = e^x………..(4)
*

Therefore, e^x can be written as Cosh x + Sinh x. Coming to the discussion where I was probing the *feasibility of solving natural number *raised to power real quantity in terms of hyperbolic functions, it is clear now that* it is solvable in hyperbolic functions of cosh x and sinh x as hyperbolic functions can have values more than 1. *

*Expressing Any Quantity In Hyperbolic Function *

I am giving here examples to illustrate how e raised to power any quantity real or imaginary is expressible in cosh and Sinh functions. Let n be the quantity and say y = n.

On taking log from both sides, we get

Log y = log n,

or y = e^log n = cosh (log n) + sinh (log n) using equation (4). This is also equal to

cos (i log n) – i sin (i log n) using equations (12) and (13) as we will see later on.

If quantity is y = n^i then log y = i log n. Taking antilog,

y = e^(i. log n) Repeating the same procedure of earlier case,

y = cosh (i. log n) + sinh (i.log n)

= cos (log n) +i sin (log n) using equations (10) and (11). This will prove in due course.

*Expressing e^ix In Hyperbolic Functions
*

If we put x as i.x, then from equation 4,

e^i.x = cosh i.x + sinh i.x………………………………….(4/1)

On differentiating both sides, we get

i.e^i.x = i.sinh x + i cosh ix

At x = 0,

i = 0 + i. cosh 0.

That means *cosh 0 must be real as coefficient of i on left side is 1 which is real. That further means cosh ix must be real.
*

If we put x as i.x in equation (1), we get

e^ix = 1+ ix – ½!.x^2 – ix^3.1/3! .x^3.+ ix^4/4!…………..upto infinity and on separating real and imaginary parts,

e^i.x =* (1- ½!.x^2 + ¼!.x^4 – 1/6!.x^6 +…………upto infinity)
*

*+ i ( x -1/3!.x^3 + 1/5!.x^5- 1/7!.x^7 +…….up to infinity)…………(5)
*

*Expansion Of Cos x
*

I submit that any function can be expressed as power series. Let

*Cos x = a0. + a1.x + a2.x^2 + a3.x^3 + a4.x^4+………. up to infinity…………………………………(6). This is general form of power series for all functions but here it is being considered for cos x.
*

When x =0, 1 = a0, on putting x = 0 in Equation (6).

On differentiating equation (6) and putting x = 0, we get

0 = a1.

On differentiating second time and putting x = 0, we get

-1 = 2.a2 or a2 = – ½!.

On differentiating third time and putting x = 0, we get

a3 = 0.

On differentiating fourth time and putting x = 0, we get

a4 = 1/4! …………….so on

Putting these values of a0, a1, a2, a3, a4……… in equation (6), we get

*Cos x = 1- x^2/2! + x^4/4! + ………………..upto infinity …..(7),
*

On differentiating equation (7), we get

-Sin x = -x + x^3/3!- x^5/5!+ ………………up to infinity.

Or *sin x = = x – x^3/3! + x^5/5!+ ……………up to infinity…(8).
*

On putting these values of cos x and sin x in equation (5), we get

*e^i x = cos x + i sin x………………………………………….(9).
*

*That proves Euler’s formula by independent method.*

Also from equation 9,

(cos x + i sin x)^2 = (e^ix)^2 = cos 2x + i.sin 2x.

Or (cos x + i sin x)^2 = cos 2x + i sin 2x. By mathematical induction,

(*cos x + i sin x)^n = cos n.x+ i sin n.x.
*

*This proves De Moivre’s theorem.
*

*Hyperbolic Identities
*

We have already proved in equation (4/1) that

e^i.x = cosh i.x + sinh i.x and we have also proved cosh ix is a real part.

On comparing real and imaginary parts of equation (9) with equation (4/1), we get

*cosh ix = cos x …………………………………………….(10)
*

*and sinh ix = i sin x…………………………………….(11)
*

If we put ix = z, in equations (10) and (11), we get

*cosh z = cos z/i = cos (-iz) = cos iz……………….(12)
*

*sinh z = i sin z/i =i. sin (-iz) = – i.sin iz……………..(13)
*

To avoid the article from being lengthy, I am not proving other identities which otherwise can easily be proved using the above Identities.

*Conclusions*

1.Euler’s formula that e^ix = cos x + i sin x, has been proved in equation 9 independently without resorting to the method adopted by great mathematician Euler.

2. e^x, cos x and sin x has been expanded without resorting to Taylor’s or Maclaurin’s series expansion.

3. Hyperbolic identities

cosh ix = cos x,

sinh ix = i.sin x,

cosh x = cos ix and

sinh x = -i sin ix,

have been proved based on independent derivation of Euler’s formula.

4. It is stated that Euler’s formula that e^ix = cos x + i sin x, is applicable only when power of natural number e is imaginary but if the power is real, e^x can not be expanded in cosines and sine of angle. Whereas formula e^x = cosh x + sinh x as derived in equation (4), is applicable whether x is real or imaginary. If x is real, e^x = cosh x + sinh x = cos ix – i sin ix. And if x is imaginary, then e^ix = cosh ix + sinh ix = cos x + i sin x.

5. Any quantity real or imaginary can be expressed in hyperbolic functions of cosh and sinh and also in trigonometric ratios of cosine and sine of angle.

Before parting with these derivations, I submit for information of my learned readers that *all functions can be expressed in power series the way I did for e^x, cos x and sin x. Once expanded, desired operations like additions, subtractions, multiplications, divisions, differentiation, integration can be performed. For example, from the power series expansion of e^x, cos x, sin x as solved by me, one can determine, summation of (e^x + sin x), subtraction of (e^x – sin x), multiplication of (e^x.sin x), division of (e^x/sin x), differentiation ie d/dx(e^x.cos x), Integration of (e^x.sin x) with respect to x. It is easy, one can do it provided one starts doing it on ones note book.
*

Reference:

Three-dimensional figure of Euler’s formula image curtesy https://commons.m.wikimedia.org/wiki/File:Euler%27s_Formula_c.png#mw-jump-to-license

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*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, Physics, Mathematics, Yoga, Humanity etc.
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