eli5 Euler’s Theorem?

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I’ve heard this is “the most beautiful and useful equation in all of math. I’ve watched videos, but still don’t understand what it does.

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4 Answers

Anonymous 0 Comments

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Anonymous 0 Comments

People like it for it’s aesthetic value. By most measures, the five most important constants in math are:

– The additive identity, 0
– The multiplicative identity, 1
– The imaginary unit, i
– The base of the exponential function, which is ubiquitous
– pi, which is *weirdly* ubiquitous

The identity e^πi + 1 = 0 brings all those constants together in a shocking way. Why is it shocking? Well…

All these constants are defined independently of one another. π and e are (very) irrational numbers that at first glance are entirely unrelated, and i isn’t even *real*. What does a complex-valued exponent even *mean*?

You’d be forgiven for thinking, then, that it’s absurd for e^πi to equal an integer. Like, what? why? how?

Nowadays mathematicians know that the exponential function e^z is defined in a very pleasing and natural way on imaginary (and therefore complex) inputs, and that π is *intimately* related to that function, which is why π shows up in a lot of places where e^x shows up. Still, you have to imagine how ridiculous this must have seemed to the mathematicians that first made the connection.

By the way, the full Euler’s formula is much more general. It says

– e^ix = cos(x) + i(sin(x))

This is genuinely useful, to the point where nowadays the functions sin and cos are actually *defined* by this relation, rather than their geometric interpretation.

Anonymous 0 Comments

Step 1:

Here’s how you derive the form of the Taylor power series expansion for a function:
https://math.stackexchange.com/questions/481661/simplest-proof-of-taylors-theorem.

The Taylor series is particularly useful because it allows you to write any non-polynomial function as an infinite series of powers of a linear term (ie (x-a)) with some constant in front of each term in the series (you can also Taylor-expand polynomials, but you’ll just get back the same polynomial with the terms grouped differently as a finite series).

Of course, functions like e^(x), with a being a constant, and sin(x) and cos (x) are non-polynomials and that’s why Taylor series are applied onto them.

Step 2:

https://proofwiki.org/wiki/Power_Series_Expansion_for_Sine_Function

Compute the Taylor series for e^(z), sin(x), and cos(x). Use the cases that at some values for x, sin or cos are 0 or 1 or -1 (ie cos(0) = 1, sin(0) = 1) and that the derivative of cos or sin leads you to the other or the negative of the other.

Use the case that derivative of e^(z) = e^(z), being itself.

Step 3:

Note that if you set z = i •x, the Taylor power series of e^(z) gives you a mix of real and imaginary terms that you can match perfectly with the Taylor series of cos(x), sin(x) in the form e^(I • x) = cos(x) + i • sin(x)

Step 4:

Set x = pi. cos(pi) = -1. sin(pi) = 0, which zeroes the term i • sin(pi). Since the right side of the equation is -1 and the right side equals the left side, e^(i • pi) = -1.

Anonymous 0 Comments

Rather than think of points on the XY plane as being comprised of separate x and y values, think of those points as being comprised of some length r and some degree of rotation. If you consider a point as referencing some degree of rotation, you’ll be able to relate the trigonometric functions (sin, cos) to that point and the XY plane overall.

Then it simply becomes letting the Y-axis represent the imaginary component of some complex number and the Y-value representing the magnitude of that imaginary component.