How does that evaluate 1??


by xenogizmo
Tags: evaluate
xenogizmo
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#1
Jan25-05, 08:48 PM
P: 30
e^(2*pi*i)
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NeutronStar
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#2
Jan25-05, 10:13 PM
P: 418
Quote Quote by xenogizmo
e^(2*pi*i)
It comes from Euler's formula

[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]
damoclark
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#3
Jan31-05, 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.

tongos
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#4
Jan31-05, 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)
mathwonk
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#5
Feb3-05, 11:52 AM
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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).
strid
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#6
Mar5-05, 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...?
hypermorphism
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#7
Mar5-05, 10:53 AM
P: 509
Quote 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.
strid
strid is offline
#8
Mar5-05, 12:17 PM
P: 77
ok.... got it now...

guessed already in before-hand that Euler probably knows more than me :)


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