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

2) Graph of natural number courtesy

3) Image of circle showing Pi as ratio of circumference to diameter by author Kjoonlee at

4) Graph of function of logarithm courtesy Krisnavedala at

5) * Natural number details can be read at

6) ** Pi details can be read at

7) *** Logarithm details can be read at

8. Title Image English: Sunrise over the bay, Little Gasparilla Island, Florida, Source courtesy  Author Mmacbeth at 


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.      






























































Author: Narinder

I am graduate in Electronics and Electrical Communication Engineering from Punjab Engineering College, Chandigarh. I worked as Scientist in Solid State Physics Lab DELHI then National Fertilizers Ltd , Bhatinda as Instrumentation Engineer, then Ministry of Labour, Employment, Training as Deputy Director. Thereafter, I joined Civil Services in the year 1986, worked in different capacities on administrative posts and retired on September 30, 2013 as Secretary to Government Punjab from Indian Administratve Services. My interests are Astronomy, Physics, History, Music, Law, Spirituality, Administration and writings. I believe in hard work, determination and consistency in efforts. I love to write on topics related to Astronomy Daily life experience and human sufferings. My favourite writers are Leo Tolstoy, Rabindranath Tagore, Mulk Raj Anand and Munshi Prem Chand.

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