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deltro300111 03-14-2006 11:43 AM

Anyone know why this works?
 
pi = ln(-e)-1/sqrt(-1)

The original equation is

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

I just isolated pi.

Tasselfoot 03-14-2006 12:30 PM

Re: Anyone know why this works?
 
Yes. I know why.

aperson 03-14-2006 01:24 PM

Re: Anyone know why this works?
 
You're just rearranging Euler's identity with some added terms:


talisman 03-14-2006 01:27 PM

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 03-14-2006 01:35 PM

Re: Anyone know why this works?
 
damn smart ****s

aperson 03-14-2006 01:49 PM

Re: Anyone know why this works?
 
Quote:

Originally Posted by talisman
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 03-14-2006 01:57 PM

Re: Anyone know why this works?
 
Oh hey this topic came up almost a year ago.

Quote:

Originally Posted by Kefit, a year or so 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 03-14-2006 04:31 PM

Re: Anyone know why this works?
 
Shame on me for not being familiar with that too.

deltro300111 03-14-2006 09:40 PM

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 03-14-2006 11:12 PM

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 03-15-2006 12:10 AM

Re: Anyone know why this works?
 
Quote:

Originally Posted by GuidoHunter
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 03-15-2006 12:31 AM

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 03-15-2006 01:01 AM

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 03-15-2006 01:52 AM

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 03-15-2006 01:57 AM

Re: Anyone know why this works?
 
Quote:

Originally Posted by stealthkidd
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.

iggymatrixcounter 03-16-2006 12:38 PM

Re: Anyone know why this works?
 
ROFL owned.

But I don't remember negative logs in my algebra class. (Maybe I slept that day or something.)

GuidoHunter 03-16-2006 05:17 PM

Re: Anyone know why this works?
 
No, I'm pretty sure studying complex algebra in algebra II isn't a normal thing.

All we ever did was know a little about complex roots of polynomials. We never got into anything in the real-complex plane, branch cutting, or anything like that.

--Guido

http://andy.mikee385.com

deltro300111 03-17-2006 05:01 AM

Re: Anyone know why this works?
 
Quote:

Originally Posted by stealthkidd
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

YOU ARE COOL BECAUSE YOU ARE STUPID.

Now fµck off. Now that it's out there no one cares.

flamingspinach 03-20-2006 12:37 AM

Re: Anyone know why this works?
 
I'm probably not noticing something, but aren't you not allowed to rearrange alternating series?

-fs

GuidoHunter 03-20-2006 01:39 AM

Re: Anyone know why this works?
 
What exactly do you mean by rearranging? The terms are all just added, if that's what you're meaning...

--Guido

http://andy.mikee385.com


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