e^(i x pi) = -1

[w/e]

Wuuuuuut?!

 

[Ghan]

Yes. That is indeed true, though I have never seen the proof for it. Which would be interesting.

 

[ElderKingpin]

what?? I is -1 square rooted, and that isnt a real number.

 

[w/e]

what?? I is -1 square rooted, and that isnt a real number.

EXACTLY. An irrational number raised to another irrational number multiplied by an imaginary number equals a negative integer. Preposterous. But correct.

 

[uberfoop]

It's actually quite simple if you understand taylor series. This is, I think, the most comprehensible proof of Euler's Formula e^(ix)=cos(x)+isin(x), which is why the thread title statement is true.

I'll summarize the proof on this page as well to perhaps keep page toggling down if anyone is interested, note though that the important part is at the bottom of the post:

The taylor series for e^x is:
e^x=1 + x + (x^2)/2! + (x^3)/3! + (x^4)/4! + ... (x^n)/n!

Now plug in ix to x to get the taylor series for e^(ix):
e^(ix)=1 + ix + ((ix)^2)/2! + ((ix)^3)/3! + ((ix)^4)/4! + ....

Apply the power to the i to bring it out:
e^(ix)=1 + ix - (x^2)/2! - i(x^3)/3! + (x^4)/4! + i(x^5)/5! - (x^6)/6! + ...

Note that if you group the terms with i together and those without i together and them factor out the i from the group with the i's in it...
e^(ix)=(1 - (x^2)/x! + (x^4)/4! - (x^6)/6! + ...) + i(x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...)

...You wind up with the red being the taylor series for cosine and blue being the taylor series for sine. Thus, you can justify replacing the series with the functions themselves, giving Euler's Formula:

e^(ix)=cos(x)+isin(x)


//=========================

So we now have the formula e^(ix)=cos(x)+isin(x).
Now, we're looking for the value of e^(ix) when x is pi, thus e^(i*pi). Well, plug in pi for x to get:
e^(i*pi)=cos(pi)+i*sin(pi)
e^(i*pi)=-1 + i*0
e^(i*pi)=-1

 

[Darthfett]

Wow, I just learned the Tayler Series stuff about 2 weeks ago.

http://xkcd.com/179/ - Pretty much sums it up.

 

[ReVolver]

Somebody saw the Simpsons...

 

[w/e]

It was in a simpsons ep? My math teacher told me about it...

 

[Lyerae]

It's actually quite simple if you understand taylor series. This is, I think, the most comprehensible proof of Euler's Formula e^(ix)=cos(x)+isin(x), which is why the thread title statement is true.

I'll summarize the proof on this page as well to perhaps keep page toggling down if anyone is interested, note though that the important part is at the bottom of the post:

The taylor series for e^x is:
e^x=1 + x + (x^2)/2! + (x^3)/3! + (x^4)/4! + ... (x^n)/n!

Now plug in ix to x to get the taylor series for e^(ix):
e^(ix)=1 + ix + ((ix)^2)/2! + ((ix)^3)/3! + ((ix)^4)/4! + ....

Apply the power to the i to bring it out:
e^(ix)=1 + ix - (x^2)/2! - i(x^3)/3! + (x^4)/4! + i(x^5)/5! - (x^6)/6! + ...

Note that if you group the terms with i together and those without i together and them factor out the i from the group with the i's in it...
e^(ix)=(1 - (x^2)/x! + (x^4)/4! - (x^6)/6! + ...) + i(x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...)

...You wind up with the red being the taylor series for cosine and blue being the taylor series for sine. Thus, you can justify replacing the series with the functions themselves, giving Euler's Formula:

e^(ix)=cos(x)+isin(x)


//=========================

So we now have the formula e^(ix)=cos(x)+isin(x).
Now, we're looking for the value of e^(ix) when x is pi, thus e^(i*pi). Well, plug in pi for x to get:
e^(i*pi)=cos(pi)+i*sin(pi)
e^(i*pi)=-1 + i*0
e^(i*pi)=-1

MY BRAIN HURTS AFTER READING THAT. :banghead:
Seriously though. I can't even begin to understand that.

 

[Ghan]

Yes, well.

When you are in higher math for a while, you begin to understand. I managed to understand what the deal was without even reading it fully. That doesn't mean I could necessarily replicate the proof, but I know what it is doing.

 

[Tru_Power22]

I think I just had a stroke reading this thread.

Cool though.

 

[Ninva]

Someone told me about this while we were caroling. I just felt like sharing.

 

[ElderKingpin]

isnt math amazing?

 

[Monovertex]

I have no idea what are the Taylor's series, but, considering

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

true from the beginning (because I have no knowledge of Taylor's series ), it's pretty easy to understand the demonstration.

And honest to be, it's not that amazing to me, I mean i is from the beginning a weird number, if you think about it. Heck, it doesn't even exists, it's imaginary. Raising numbers to an imaginary power should result in some pretty weird shit xD.