When I was 14 I was entranced by an equation: a mathematical identity so beautiful yet simple that it managed to combine 4 of the most important numbers in math along with an imaginary number and nothing else into a single formula.
How can numbers even be taken to an imaginary power? I sought after proofs, but in the late 1990s the internet was not replete with the information it is today, and with a knowledge of mathematics only through Trigonometry, I did not understand them, but my obsession with the equation remained.
The result was not fun for my math teachers: I would routinely answer questions in terms of e^(πi). For instance, if a question on a test had a correct answer of 40, I would write -40*e^(πi). My teachers soon grew tired of my bad jokes, but my incredulousness over the existence of this identity would last another 2 years until 11th grade, when I was able to prove it using my newfound knowledge of calculus.
You may have noticed my series An Introduction to Mathematical proofs, where so far I taught the background necessary to start learning proof techniques in 3 parts.
I’ve decided to do this blog before actually teaching proof techniques. This blog is a more advanced proof that can use some of those techniques, however I decided to do it first, and not just because I love this equation.
This proof requires some heavy calculus knowledge to do rigorously, something I wouldn’t expect everyone to have beforehand, so in that series I’ll concentrate on proofs that people can understand much easier without domain knowledge in advanced areas of mathematics. To fully break down every detail of this proof would require I teach 2 courses on Calculus, something I'll leave to @team-leibniz, but I’ll step through the important parts to show why this equation is true.
You can read my other series here:
Part 1
Part 2
Part 3
Before we prove that e^(πi) = -1, we must learn and prove some background material first.
What is e exactly?
e is not only an irrational number, meaning it cannot be represented as a ratio of integers (fraction), but e is also a transcendental number: it is not the root of a polynomial equation with integer coefficients. Another example of a transcendental number is π.
e is defined as the limit as n goes to ∞ of (1+1/n)^n.
when n = 1, (1+1/n)^n = 2
when n = 2, (1+1/n)^n = 2.25
when n = 3, (1+1/n)^n = 2.37037...
when n = 4, (1+1/n)^n = 2.44140625
when n = 5, (1+1/n)^n = 2.48832
when n = 10, (1+1/n)^n = 2.5937424601
when n = 100, (1+1/n)^n = 2.7048138294
when n = 1000, (1+1/n)^n = 2.7169239322
As n goes to ∞, this series converges, and is roughly equal to 2.71828
Also recall the following properties of logarithms:
- log a - log b = log (a/b)
- n * log b = log (b)ⁿ
The first step of our proof involves finding the derivative, or the slope of the line, of ln(x).
LaTex is an amazing text editor, but the font is a little small.
Given the above lemma that says the derivative of ln(x) = 1/x, we can now prove that the derivative of e^x is e^x.
Let’s look at the derivative of ln ( e^x ) in two different ways. The first way uses a property of logarithms and the second way uses the chain rule.
This is an astonishing result: at every point of the graph of e^x, the slope is also e^x. But so is the second derivative, and the third, ad infinitum. If f(n) = e^x, then f’(x) = e^x, f’’(x) = e^x, and so on forever.
This is amazing, but there is a polynomial which satisfies this property too.
Let f(x) = 1 + x + x²/2 + x³/3! + x⁴/4! + x⁵/5! +…
Then f’(x) = 0 + 1 + x + x²/2 + x³/3! + x⁴/4! + x⁵/5! +…
And since f(x) = f’(x), we can continue on down to f’’(x), f’’’(x) ad infinitum as well.
There is only one unique function which is equal to its derivative: e^x. Multiples of it have the same property (0, e^x, 2e^x,etc) but there are no other functions that satisfy this property. Thus, the function we just described is also e^x.
Substituting (ix) in for (x), we can now calculate what e^(ix) is in general form based on e^x = 1 + x + x²/2 + x³/3! + x⁴/4! + x⁵/5! +…
e^(ix) = 1 + ix + i²x²/2 + i³x³/3! + i⁴x⁴/4! + i⁵x⁵/5! +…
Remember the powers of i are cyclical:
i⁰ = 1
i = i
i² = -1
i³ = -i
i⁴ = 1
i⁵ = i
Substituting in those values for the powers of i, we can rewrite e^(ix):
e^(ix) = 1 + ix - x²/2 - ix³/3! + x⁴/4! + ix⁵/5! +…
We have seemingly hit an impasse. But progress is made when comparing the Maclaurin series expansions for cos(x) and sin(x) with the expansion of e^(ix).
source: Wikipedia
The basic setup of a Maclaurin series
Maclaurin series is a representation of a function as an infinite set of terms about the point 0:
f(x) = f(0) + f’(0)x + f’’(0) x²/2 + f’’’(0) x³/3! + f’’’’(0) x⁴/4! + f’’’’’(0) x⁵/5! +...
Let’s first consider cos(x) and it’s derivatives and their values at 0 to calculate it’s Maclaurin series.
f(x) = cos x f(0) = 1
f’(x) = -sin x f’(0) = 0
f’’(x) = -cos x f’’(0) = -1
f’’’(x) = sin x f’’’(0) = 0
f’’’’(x) = cos x f’’’’(0) = 1
So because some of the terms cancel out due to the odd numbered derivatives being 0 at the point 0, the Maclaurin series for cos x = 1 - x²/2 + x⁴/4! - x⁶/6! +….
Now let’s look at the Maclaurin series for sin x.
f(x) = sin x f(0) = 0
f’(x) = cos x f’(0) = 1
f’’(x) = -sin x f’’(0) = 0
f’’’(x) = -cos x f’’’(0) = -1
f’’’’(x) = sin x f’’’’(0) = 0
Again some terms cancel off, so the Maclaurin series for sin x = x - x³/3! + x⁵/5! - ...
Let’s take another look at e^(ix)
e^(ix) = 1 + ix - x²/2 - ix³/3! + x⁴/4! + ix⁵/5! +…
Let’s instead group the terms by even powers and odd powers
e^(ix) = (1 - x²/2 + x⁴/4! - x⁶/6! +….) + (ix - ix³/3! + ix⁵/5! - …)
We can also pull an i out of the second half of this expansion:
e^(ix) = (1 - x²/2 + x⁴/4! - x⁶/6! +….) + i * (x - x³/3! + x⁵/5! - …)
The first half is equal to the expansion for cos x, and the second half is equal to the expansion of sin x!
Thus, e^(ix) = cos x + i * sin x. This is known as Euler’s Formula.
source: Wikipedia
Physicist Richard Feynman on Euler’s Formula:
The most remarkable formula in mathematics.
But our version is not yet complete. Substituting in the value of π for x finishes off the proof of Euler’s Identity:
e^(πi) = cos π + i * sin π = -1 + i * 0 = -1!
Therefore, e^(πi) + 1 = 0.
This identity combines two transcendental numbers (π and e) along with an imaginary number, 1, and 0 into a beautifully simple formula. I hope it brings you as much amazement as it brought 14 year old Ryan.
My name is Ryan Daut and I'd love to have you as a follower. Click here to go to my page, then click 
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Loved the intro to this piece...then the math started and...showers. haha
If you enjoyed this, here are a few other fun math blogs I enjoyed:
What Is Chaos Theory by @complexring
An introduction to Group Theory by @complexring
Group Theory part 2 by @complexring
Pascal's Triangle by @team-leibniz
The Birthday Paradox by @team-leibniz
Gabriel's Horn by @team-leibniz
Post any others you guys have enjoyed in the comments below!
So, true story. I had to get talked out of getting this equation as a tattoo at one point in my life. I kind of still want it even now, to tell you the truth!
I remember falsely learning too about euler and how this formula made him believe in "higher powers" in the universe. Turns out that is a tall tale. But, if you had to have a hall of fame of numbers, everything in this identity would be WELL represented! Great piece!
I don't have a tattoo because there's nothing I'm sure I'd want on my body forever, but this is a pretty great idea for one. I'm jumping on team tattoo here and think you should do it!
Oh gosh, this is essentially a rundown of calculus 2. I remember learning about Maclaurin series way back when I took this in undergraduate. I always found calculus to be challenging. In fact this material resulted in one of my few B's. Nice job with this post. Keep it up man!
I always think of it that way: Rotate by 180° on the complex plane and you arrive at the real numbered one. Now substract one and you're done :-)
If e is a one dimensional smooth growth proccess, then the rotation of e^i*phi is the two dimensional equivalent giving us a perfect circle :-)
Interesting. Upvoted and followed!
Hi Ryan,
You can possibly play with the size of the equation in latex to make it fit nicer to the eyes. Or maybe suggest to the developers to enable tex for writing-up posts and comments!
Cheers,
B>
Love it.
Nice :-)
Very nice article. I still have problems to this day to fathom the imaginary numbers. They look so surreal, it's easy to work with them in formulas, but to understand what they are, they are beyond my imagination.