In the last post, we talked about imaginary numbers and how they were defined. We said that the imaginary number i was defined as the square root of -1 and using i we could express all imaginary numbers. However, there is a more complex and intricate way to define imaginary numbers, with Euler’s Identity. In its general form and a form you may have seen before Euler’s formula looks like this-
e^(iπ)+1=0
This relationship is often touted as the most intricate as it includes all of the most important mathematical entities that are used throughout math. It has the natural log base e (Also referred to as Euler’s number), the imaginary number i, the ratio of the circumference of a circle to its diameter, π, and perhaps the two most important whole numbers 1 and 0.
Euler’s identity is just one very specific instance of Euler’s formula.
How to derive Euler’s formula?
Now that we can appreciate why Euler’s identity is so important, it is important to understand exactly how it was derived and what mathematical gymnastics does one have to do to get this identity that uses some of the most important constants in mathematics? The answer to that question is by using Taylor series.
To get to the above result, we will first look at the Taylor series for e^x.
e^x=1+x+(x^2/2!) +(x^3/3!)+(x^4/4!)+(x^5/5!)………..
In this formula, lets substitute i*x for x, so we get.
e^(i*x)=1+(i*x)+((i*x)^2/2!) +((i*x)^3/3!)+((i*x)^4/4!)+((i*x)^5/5!)………..
From the last lesson we know i^2=-1, i^3=-i, i^4, using this information
e^(i*x)=1+ix+(-x^2/2!) +(-ix^3/3!)+(x^4/4!)+(ix^5/5!) ………
And Now grouping we get
e^(i*x)=(1-(x^2/2!)+(x^4/4!) …….) + (ix -(ix^3/3!) +(ix^5/5!)……..)
Factor out i from the second term and we get
e^(i*x)=(1-(x^2/2!)+(x^4/4!) …….) + i(x -(x^3/3!) +(x^5/5!)……..)
Now, the two terms in this Taylor series should be familiar if you have studied Taylor series as they are simply the expansions of cos x (1-(x^2/2!)+(x^4/4!) …….) and sin x (x -(x^3/3!) +(x^5/5!)……..) so we can substitute cos x and sin x into the Taylor series and get-
e^(i*x)=cos x+ i*sin x
If we plug in π for x we get
e^(i*π)= cos π +i*sin π
e^(i*π)=1+i*0
e^(i*π)-1=0
This is how Euler’s identity was derived.
At the end of the day, Euler’s identity gives us a relationship between the most important constants in mathematics and while its derivation might be complicated, it is worth understanding.