
#1
Jan2505, 08:48 PM

P: 30

e^(2*pi*i)




#2
Jan2505, 10:13 PM

P: 418

[tex]e^{\left( i\theta\right) } =\cos\theta +i \sin\theta[/tex] [tex]e^{\left(i2\pi\right) } =\cos2\pi +i \sin2\pi[/tex] [tex]e^{\left(i2\pi\right) } =1 + 0[/tex] [tex]e^{\left(i2\pi\right) } =1[/tex] 



#3
Jan3105, 04:45 AM

P: 25

Well the way you get this is to look at the infinite polynomial expansion of
e^x, cos(x) and sin(x) e^x= 1+ x/1! + x^2/2! + x^3/3! + x^4/4! .... + x^k/k! + ..........etc... cos(x) = 1  x^2/2! + x^4/4!  x^6/6! + x^8/8!  x^10/10! +............ etc.... sin(x) = x  x^3/3! + x^5/5!  x^7/7! + x^9/9!  x^11/11! ...............etc.... This is how your calculator probably calculates sin(x) and cos(x) ... These series can be derived quite easily, and if your curious I can explain how you can get these infinite series expressions for the functions e^x, sin(x) and cos(x). Now if you substitute i*x for x into e^x, where i = square root of 1. Note that : i = i i^2 = 1 i^3 = i i^4 = 1 i^5 = i i^6 = 1 i^7 = i i^8 = 1 ect in this repeating fasion then e^(i*x) = 1 + i*x/1! + (i*x)^2/2! + (i*x)^3/3! + (i*x)^4/4! +(i*x)^5/5!..... =1 +ix/1!  x^2/2! ix^3/3! + x^4/4! + ix^5/5!  x^6/6!... This series can be spit into two infinite series e^(i*x) = 1  x^2/2! + x^4/4!  x^6/6! + x^8/8!......... = cos(x) + i*( x  x^3/3! + x^5/5!  x^7/7! + x^9/9! .... = i*sin(x) therefor e^ (i*x) = cos(x) + i*sin(x) Then if you substitute Pi for x you have e^(i*Pi) = cos(Pi) + i*sin(Pi) = 1 + 0 = 1 This is a very beautiful and elegant formula discovered by Euler I think. The formula also has a very nice geometric representation when you view it in the complex plane, and this is generally an easier way to understand the formula if you dont like all the algebra. 



#4
Jan3105, 08:44 AM

P: 84

How does that evaluate 1??
taylor series
something that is also interesting i^i= e^(ipi)^(i^1/2)=(i^2)^(i^(1/2)= e^(pi/2) 



#5
Feb305, 11:52 AM

Sci Advisor
HW Helper
P: 9,421

one also knows the unique solution of f''+f = 0, f(0)= a, and f'(0) = b, is
acos(x) + b sin(x). But e^(ix) solves it with a = 1, b = i. so e^(ix) = cos(x)+isin(x). 



#6
Mar505, 10:29 AM

P: 77

I havent seen this before...
but though i follow the concluding above, I ran in some trouble a moment. ago... My first thought was that it is quite close to the Eulers formula e^(pi*i) +1 =0 , so I stared to fiddle with it in hope of getting something funny out of it.. but i found this... e^(pi*i) + 1 = 0 e^(2*pi*i) = 1 e^(pi*i) = 1 (e^(pi*i))^2 = 1  (e^(pi*i)) = (e^(pi*i))^2 SQRT( (e^(pi*i)) ) = e^(pi*i) The problem now is that on the left side there is a SQRT of a negative number (about 23,14) which means it is a complex number... while on the other side we have a real postive number (about 23,14)...... Have I made some mathematcial error, or how can this be...? 



#7
Mar505, 10:53 AM

P: 509





#8
Mar505, 12:17 PM

P: 77

ok.... got it now...
guessed already in beforehand that Euler probably knows more than me :) 


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