# How does that evaluate 1?

by xenogizmo
Tags: evaluate
 P: 30 e^(2*pi*i)
P: 418
 Quote by xenogizmo e^(2*pi*i)
It comes from Euler's formula

$$e^{\left( i\theta\right) } =\cos\theta +i \sin\theta$$

$$e^{\left(i2\pi\right) } =\cos2\pi +i \sin2\pi$$

$$e^{\left(i2\pi\right) } =1 + 0$$

$$e^{\left(i2\pi\right) } =1$$
 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.
 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)
 Sci Advisor HW Helper P: 9,498 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).
 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...?
P: 506
 Quote by strid ... 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...?
Two things. The designation -x does not mean that the quantity -x is a negative real number. Secondly, the f(x)=x^2 function is not globally invertible, so you cannot always assume that if x^2=a^2, then x=a. Thus, in your example, SQRT( -(e^(pi*i)) ) = SQRT(-(-1)) = SQRT(1) = 1. 1 is one square root of (e^(pi*i))^2 = 1. The other square root is -1.
 P: 77 ok.... got it now... guessed already in before-hand that Euler probably knows more than me :)

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