Anyone know why this works?

[deltro300111]

pi = ln(-e)-1/sqrt(-1)

The original equation is

ln(-e) = 1 + (pi)i

I just isolated pi.

[Tasselfoot]

Re: Anyone know why this works?

Yes. I know why.

[aperson]

Re: Anyone know why this works?

You're just rearranging Euler's identity with some added terms:

[talisman]

Re: Anyone know why this works?

I just typed the same thing as ap but vb swallowed it...

and that identity comes from some Calc II nonsense I believe, with series and whatnot.

[XXXsmittyXXX]

Re: Anyone know why this works?

damn smart ****s

[aperson]

Re: Anyone know why this works?

I just typed the same thing as ap but vb swallowed it...

and that identity comes from some Calc II nonsense I believe, with series and whatnot.

It's generally known as Euler's formula. I think you can prove it with Calc II series, but the first application I've seen of it was solving 2nd order linear ODEs that have complex conjugates as answers.

[Kefit]

Re: Anyone know why this works?

Oh hey this topic came up almost a year ago.

This equation is incredibly important to physics and differential equations. Lemme see if I can remember the proof:

First off, this proof uses infinite series. Put basically, any differentiable function can be represented by the sum of an infinite series. No, I don't expect you to know what that means. Just accept that:

cos(x) = 1 - (x^2)/(2!) + (x^4)/(4!) - (x^6)/(6!) + . . .

sin(x) = x - (x^3)/(3!) + (x^5)/(5!) - (x^7)/(7!) + . . .

e^x = 1 + x + (x^2)/(2!) + (x^3)/(3!) + . . .

It follows that:

e^(i*x) = 1 + (i*x) - (x^2)/(2!) - (i*x^3)/(3!) + (x^4)/(4!) + (i*x^5)/(5!) - (x^6)/(6!) + . . .

i*sin(x) = (i*x) - (i*x^3)/(3!) + (i*x^5)/(5!) - (i*x^7)/(7!) + . . .

Now something interesting happens when you add cos(x) and i*sin(x) together. Lemme see if I can illustrate this clearly:

.....cos(x) = 1..........- (x^2)/(2!) ......................+ (x^4)/(4!) .................- (x^6)/(6!) + . . .
+ i*sin(x) = ..(i*x)..................... - (i*x^3)/(3!) .................+ (i*x^5)/(5!) - . . .
---------------------------------------------------------------------------------------------------------------
....e^(i*x) = 1 + (i*x) - (x^2)/(2!) - (i*x^3)/(3!) + (x^4)/(4!) + (i*x^5)/(5!) - (x^6)/(6!) + . . .

Don't be discouraged if the idea of infinite series is unfamiliar to you - just take the series I gave for sin(x), cos(x), and e^x for granted, and everything else I show above follows from simple arithmetic on the series.

Anyway, this creates the general equation:

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

if x = PI, then:

e^(i*PI) = cos(PI) + i*sin(PI) = -1 + i*0 = -1

Shame on you ap for not being familiar with this.

[Kilgamayan]

Re: Anyone know why this works?

Shame on me for not being familiar with that too.

[deltro300111]

Re: Anyone know why this works?

this all came out when I was taking an Algebra II test, and trying to find one of the ever prevalent patterns. I plugged in ln(-e), and realized that the calculator spat out pi(i)+1, and I was like, "what the hell", and I tried to isolate e, but lacked the mad skills to do so. I know from where e is derived but never knew why.

[GuidoHunter]

Re: Anyone know why this works?

They didn't teach you what the log function looked like in Algebra II? What the hell were you trying to do taking the natural log of a negative number?

--Guido

http://andy.mikee385.com

[aperson]

Re: Anyone know why this works?

They didn't teach you what the log function looked like in Algebra II? What the hell were you trying to do taking the natural log of a negative number?

--Guido

http://andy.mikee385.com

I take it you've never heard of the complex coordinate plane have you.

[GuidoHunter]

Re: Anyone know why this works?

Funny, as I'm taking Functions of Complex Variables right now. Very cute.

Note how he's in algebra II. Also note that I said natural log. I am well aware that there are an infinite number of solutions to log(-x) for any x, and how log and Log mean very different things.

--Guido

http://andy.mikee385.com

[aperson]

Re: Anyone know why this works?

I definitely did complex logarithms in my algebra II class.

Although I didn't see them again until differential equations

[stealthkidd]

Re: Anyone know why this works?

HOLY CRAP!!! aperson u should get a new section for aperson's insanly hard questions becasue everyone under 18 or so has no ****ing clue wtf ur saying but ppl like it. this is because urs are mostly harder

[aperson]

Re: Anyone know why this works?

HOLY CRAP!!! aperson u should get a new section for aperson's insanly hard questions becasue everyone under 18 or so has no ****ing clue wtf ur saying but ppl like it. this is because urs are mostly harder

I'm 17.