The Most Beautiful Formula in all of Mathematics: Euler’s Identity: e^(πi) +1 = 0
mathematics·@daut44·
0.000 HBDThe Most Beautiful Formula in all of Mathematics: Euler’s Identity: e^(πi) +1 = 0
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](https://steemit.com/mathematics/@daut44/an-introduction-to-mathematical-proofs-part-1-basic-logic-and-truth-tables) [Part 2](https://steemit.com/mathematics/@daut44/an-introduction-to-mathematical-proofs-part-2-conditionals-deduction-biconditionals-proofs-of-equivalence) [Part 3](https://steemit.com/mathematics/@daut44/an-introduction-to-mathematical-proofs-part-3-universal-existential-and-unique-existential-quantifiers) ## 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: 1. log a - log b = log (a/b) 2. 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). https://s9.postimg.io/ffd1c4tin/Screen_Shot_2016_09_01_at_1_14_40_PM.png *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. https://s17.postimg.io/wdpywdz9r/Screen_Shot_2016_09_01_at_10_53_45_AM.png 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). https://upload.wikimedia.org/wikipedia/commons/thumb/7/71/Sine_cosine_one_period.svg/2000px-Sine_cosine_one_period.svg.png *source: [Wikipedia](https://upload.wikimedia.org/wikipedia/commons/thumb/7/71/Sine_cosine_one_period.svg/2000px-Sine_cosine_one_period.svg.png)* ### 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.** https://s12.postimg.io/ngwps1tct/Screen_Shot_2016_09_01_at_11_48_29_AM.png *source: [Wikipedia](https://upload.wikimedia.org/wikipedia/commons/thumb/7/71/Euler's_formula.svg/2000px-Euler's_formula.svg.png)* 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. 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