e^(pi*i)+1=0

[jordan]

Hey, does anyone have any idea how this works? The one big thing to overcome is taking it to an imaginary power. Can anyone explain how that works? And how does an irrational number taken to a power ever make itself rational again? That seems rather counterintuitive.

[AL]

And how does an irrational number taken to a power ever make itself rational again?

Well, most obviously the square root of 2 squared is an irrational number taken to a power to make itself rational, so that example should do away with anything counterintuitive.

 

Hey, does anyone have any idea how this works? The one big thing to overcome is taking it to an imaginary power. Can anyone explain how that works?

As for e^(pi*i) + 1 = 0, Euler's Equation says

e^ix = cos(x) + i*sin(x). So naturally, e^i*pi = -1 (just plug pi in for x).

 

As for the proof of the equation, you use Taylor series expansions for e^x, cos(x) and sin(x), which are:

e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! +...

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

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

 

If you expand the formula for e^xi, you get:

e^xi = 1 + xi -x^2/2! - ix^3/3! + x^4/4!+...

= cos(x) + isin(x)

[Treadstone]

also, i is not irrational, its imaginary

[123rock]

...e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! +...

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

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

 

Is there any proof for that?

[bloodhound]

http://www.scienceforums.net/forums/showthread.php?t=4263

 

there is some problem with the Latex rendering. if you follow each step one by one u will get it. otherwise it seems that there is no spaces betweeen the implications

 

[edit]sorted the latex stuff.[/edit]

[bloodhound]

with the eulers formula you can get even more amazing results like

[math]i^{i}=e^{\frac{-\pi}{2}}[/math]

 

ill leave it as a simple exercise.

 

[edit] OOPS sorry..... didnt read the qeustion properly. thats not what was being asked.. [/edit]

[jordan]

Thanks for the link bloodhound.

 

As for e^(pi*i) + 1 = 0' date=' Euler's Equation says

e^ix = cos(x) + i*sin(x). So naturally, e^i*pi = -1 (just plug pi in for x).[/quote']

I'll ask a question about this part, though.

 

The i has absolutely no impact on the equation then, right? If, according to the theorem, the i is distributed before [math]sin{(\pi)}[/math] and the [math]sin{(\pi)}=0[/math] then it doesn't matter if it's i or 6 or 1 billion.

 

And does this proof only account for pi in the sense of radians? The [math]cos{(\pi)}[/math] is only -1 when we're talking in radians.

[Dave]

I'm afraid the i does have some impact. Try working out [math]e^{\pi}[/math] - I can assure you it's definately not negative

[jordan]

Well what's going on then?

 

According to his theorem, that would expand to [math]cos{(\pi)}+1*sin{(\pi)}[/math].

 

That would be -1+0 or -1.

 

What am I missing?

[AL]

Thanks for the link bloodhound.

 

 

I'll ask a question about this part' date=' though.

 

The i has absolutely no impact on the equation then, right? If, according to the theorem, the i is distributed before [math']sin{(\pi)}[/math] and the [math]sin{(\pi)}=0[/math] then it doesn't matter if it's i or 6 or 1 billion.

 

And does this proof only account for pi in the sense of radians? The [math]cos{(\pi)}[/math] is only -1 when we're talking in radians.

 

In the particular case of [math]x=\pi[/math], the i gets multiplied by zero, so the imaginary term disappears and you end up with a real number: -1. However, for other values of x, the i stays in there. Try out other values.

 

And yes, the formula works for degrees, provided you make the proper adjustment from radians first. Radians, though, are more useful for theoretical math.

[jordan]

Whenever [math]x=\pi[/math] then you get -1 no matter what is in place of the i?

 

dave has already provided a counter-example where that's not true. In his, it was [math]1*\pi[/math], and [math]e^{\pi}[/math] isn't -1.

[123rock]

The very definition of cosine is in terms of radians and thus the cosine of 1 radian or cos(1) would be cos(1/pi) since there are 2pi radians in a circle.