In mathematics, there’s this one term known as identity. Identity in mathematical context is defined as “an equation which is true regardless of what values are substituted for any variables, if there are any variables at all.” To help you get a better grasp of the idea, especially for the ones not familiar with this area of mathematics, these are a number of examples:
- 1+2=3
- (x+y)(x-y)=x²-y²
- sin(2x)=2sin(x)cos(x)
In this post, I’m going to prove Euler’s identity using Taylor series expansion as the tool. Euler’s identity says that
e^(iπ) + 1 = 0
- e: Euler’s number (approximately 2.71828)
- i: imaginary number (defined as the square root of -1)
- π: pi (approximately 3.14159)
What is a Taylor series? A Taylor series is a function’s expansion about a point (in graphical representative). Below is the Taylor series expansion formula:
f(x+a) = f(a) + x¹f’(a)/1! + x²f’’(a)/2! + x³f’’’(a)/3! + x⁴f’’’’(a)/4! + …
The apostrophe marks written next to almost every “f” letter denote derivatives. The number of them corresponds to the degree of derivation. We’ll get back to that in a minute. For now, let’s experiment with the eries a bit; let’s substitute a with the value 0. When a=0, the formula slightly changes. This new formula below is called Maclaurin series:
f(x) = f(0) + x¹f’(0)/1! + x²f’’(0)/2! + x³f’’’(0)/3! + x⁴f’’’’(0)/4! + …
Using the previously obtained Maclaurin series expansion, we can now proceed to proving Euler’s identity. First, let us apply Maclaurin expansion on these 3 functions:
- f(x) = sin(x)
- f(x) = cos(x)
- f(x) = e^(x)
As promised before, I’ll explain e bit about derivatives. In mathematics, derivative is defined as the rate of change of a function with respect to a variable. To help you understand a bit, a physical example of derivative is velocity. Velocity (with unit, let’s say, m/s) is a derivation of position/distance(m) against time (s). In human language, velocity is the rate of change of position with respect to time. Makes sense, right? Ok, now let’s carry on the task of proving Euler’s identity on the Maclaurin expansion section by first deriving all three previously mentioned functions.
Derivatives of sin(x) :
sin’(x) = cos (x)
sin’’(x) = cos’(x) = -sin(x)
sin’’’(x) = cos’’(x) = -sin’(x) = -cos(x)
sin’’’’(x) = cos’’’(x) = -sin’’(x) = -cos’(x) = sin(x)
Derivatives of cos(x) :
cos’(x) = -sin(x)
cos’’(x) = -sin’(x) = -cos(x)
cos’’’(x) = -sin’’(x) = -cos’(x) = sin(x)
cos’’’’(x) = -sin’’’(x) = -cos’’(x) = sin’(x) = cos(x)
Derivatives of e^x :
[e^(x)]’ = e^(x)
[e^(x)]’’ = [e^(x)]’ = e^(x)
[e^(x)]’’’ = [e^(x)]’’ = [e^(x)]’ = e^(x)
[e^(x)]’’’’ = [e^(x)]’’’ = [e^(x)]’’ = [e^(x)]’ = e^(x)
Now that we have obtained the required degrees of derivations, we can apply Maclaurin expansions on each of them respectively as follows.
Maclaurin expansion of sin(x) :
f(x) = sin(x)
sin(x) = sin(0) + x¹cos(0)/1! — x²sin(0)/2! — x³cos(0)/3! + x⁴sin(0)/4! + …
sin(x) = 0 + x¹*1/1 — x²*0/2 — x³*1/6 + x⁴*0/24 + …
sin(x) = 0 + x¹/1 — x³/6 + …
Maclaurin expansion of cos(x) :
f(x) = cos(x)
cos(x) = cos(0) — x¹sin(0)/1! — x²cos(0)/2! + x³sin(0)/3! + x⁴cos(0)/4! — …
cos(x) = 1 — x¹*0/1 — x²*1/2 + x³*0/6 + x⁴*1/24 — …
cos(x) = 1 — x²/2 +x⁴/24 — …
Maclaurin expansion of e^(x) :
f(x) = e^(x)
e^(x) = e⁰ + x¹e⁰/1! + x²e⁰/2! + x³e⁰/3! + x⁴e⁰/4! + …
e^(x) = 1 + x¹*1/1 + x²*1/2 + x³*1/6 + x⁴*1/24 + …
e^(x) = 1 + x/1 + x²/2 + x³/6 + x⁴/24 + …
Now let’s tidy them results up a bit, in terms of visual display of course. These are the results after that “trimming” process:
e^(x) = 1 + x¹/1 + x²/2 + x³/6 + x⁴/24 + …
sin(x) = 0 + x¹/1 — x³/6 + …
cos(x) = 1 — x²/2 +x⁴/24 — …
And now we can start to put i in the mix. To refresh your memory, remember that i is defined as the square root of -1, or rewritten
i² = -1 , therefore
i³ = i²*i = -1*i = -i
i⁴ = i³*i = -i*i = -(-1) = 1
Next, insert i into the previously established expansions of functions :
e^(ix) = 1 + i¹x¹/1 + i²x²/2 + i³x³/6 + i⁴x⁴/24 + …
e^(ix) = 1 + ix/1 — x²/2 — ix³/6 + x⁴/24 + …
i*sin(x) = 0 + i¹x¹/1 — i³x³/6 + …
i*sin(x) = 0 + ix/1 — ix³/6 + …
cos(x) = 1 — x²/2 + x⁴/24 — …
We obtain :
cos(x) + i*sin(x) = (1+0) + ix/1 — x²/2 — ix³/6 + x⁴/24 + …
cos(x) + i*sin(x) = 1 + ix/1 — x²/2 — ix³/6 + x⁴/24 + …
and since : e^(ix) = 1 + ix/1 — x²/2 — ix³/6 + x⁴/24 + …
Therefore,
e^(ix) = cos(x) + i*sin(x)
This is actually the general form of Euler’s identity or Euler’s equation. This relation provides a great help in understanding and playing with complex numbers. But don’t worry, we’re not going there, not today. Back to our task, let’s take a look again at the established equation. We are going to make things, ehm, mathematically aesthetic (holy lord Pablo Escobar, so cheesy and yet so glorious, that “mathematically aesthetic” phrase). It’s a common knowledge among people who have already studied even the tiniest bit of trigonometry that
- cos(π) = 1, and
- sin(π) = 0
So, substituting x with π:
e^(ix) = cos(x) + i*sin(x)
x = π
e^(iπ) = cos (π) + i*sin(π)
e^(iπ) = -1 + 0
e^(iπ) = -1
e^(iπ) + 1 = 0
There, we (and again, by “we” I mean “I” with all of you my fellow readers doing nothing but watching the whole damn thing unfold) managed to prove Euler’s identity with the assistance of Taylor (actually Maclaurin) series expansion. Congratulations for your ability to read this post down ‘till the end despite in the end you do understand everything I’ve written above or understand nothing of them at all. Oh and by the way, the final form of Euler’s identity in this post is considered beautiful by mathematicians because it contains Euler’s number (e) and pi (π) which are irrational (in mathematics, means numbers that can’t be written in simple fractions), an imaginary number (i) which is…uh…imaginary (i define it as a number that no physical condition can represent the concept), and one (1) from the real number family. Adding to the cocktail’s flavor for you to taste, here’s the identity written in human language: there’s this irrational thing (e) to the power of another thing which is both unreal and irrational (iπ) that if you add it with one (1) is going to give you nothing (0). Man, how I love semiotics and semantics when translating from mathematical language to human language (in this case, that human language happens to be English). Maybe, just maybe, next time I’ll do something in the opposite direction. Just for my personal pleasure, as usual, of course. Ha! And dear some internet mathematicians, please stop throwing cheesy jokes in your videos. Thank you.
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