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Electrical Engineering student. Life is pretty good, but boring.

Alex Lamb @Al6200

Age 34, Male

Studying Engineering

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Proof of Euler's formula

Posted by Al6200 - August 20th, 2008


Well, I have nothing else to do, so I'm going to prove Euler's formula for all of you Newgrounder's benefit.

What's really interesting about Euler's formula is that it's so counter intuitive, that it is not uncommon for people to present it as proof of God's existence. It says that the natural constant, e, multiplied by itself x times the square root of negative times is equal to the square root of negative one times the ratio of a right triangle's opposite side to its hypotenuse if the triangle's angle is x, plus the ratio of a right triangle's adjacent side to its hypotenuse if the triangle's angle is x.

Or in math: e^(i * t) = i * sin(t) + cos(t)

Anyway, unto the proof:

i = sqrt( -1 ) and t be some variable:

y(t) = e^(i * t)

y'(t) = i * e^(i * t)

y''(t) = -1 * e^(i * t)

y(t) + y''(t) = 0

Taking the laplace transform of both sides:

L[ y(t) ] + L[ y''(t) ] = L[ 0 ]

y(s) + s * y'(s) - y'(0) = 0
y(s) + s^2 * y(s) - s * y(0) - y'(0) = 0

Alright, now going back to our equations

y(0) = 1
y'(0) = i
y''(0) = -1

Plugging those in we get

y(s) + s^2 * y(s) - s - i = 0

y(s) * (1 + s^2) - s - i = 0

y(s) = s / (1 + s^2) + i / (1 + s^2)

Taking the inverse Laplace transform of both sides (which is usually done through tables, since the transform itself is really ugly), we find that:

y(t) = cos(t) + i * sin(t)

y(t) = e^(i * t)

e^(i * t) = cos(t) + i * sin(t)

So there you have it, no magic, just maths.


Comments

to... many... numbers... *starts twitching*

I wish got all the kewl math Programmers and Engineers get :'(

Hmmm... Yes. This is important for electrical engineering.

The power that comes out of ordinary wall sockets is called "AC current". In other words, the voltage that it provides is a cosine curve: cos(Bx + C), where B and C are constants (which I think depend on what country you're in).

By Euler's Identity, you can represent the voltage curve as the real part of e^(ix). So:

Real[ e^( (a +bi) x) ] = cos(Bx + C)

So you end up representing a voltage source as a complex number: a + bi, which corresponds to a certain cosine curve.

This is INCREDIBLY useful, because you can use a concept called "Impedence", which follows Ohm's law. For some impedance Z, a current I, and a voltage V:

V = I * Z

A resistor has a real positive impedance (meaning that it flattens the cosine curve, but does not shift it). A capacitor has a negative imaginary impedence, and an inductor has a positive imaginary impedence.

So if you have a voltage source V = 3 + 5i, and an inductor of 5 Henries, then the current that flows through the circuit will be I = (3 + 5i)/(5wi) where w is a constant. This is really quite awesome.

looks complex, how could someone think this is proof of god though?

Well, if you write

e^(Pi * i), you get -1.

So basically if you put 3 of the weirdest and most fascinating numbers that exist into one equation, you get out...

negative one, the most dull and boring number out there...

So, wait, because of the length of a triangle side, god is real?

No, but it is a beautiful relationship that one would not expect to exist.

e is a weird number.

Or is it i that is the weird number?

That's...beautiful.

Indeed.

Hey, I'm doing differential equations right now! Woot! I'll get back to you once I understand the laplace thing. PS, your icon is great.

Yes, I'm in Calc 4 now.

Four dimensional volumes. Not fun.