Somebody want a fun math problem?

Prove this:
e^πi=-1
And yes, it's actually true.
No cheating and looking it up.

Comments

[DaemoN67]

Da buq does that even MEAN?!

[ScriptScrolls]

592948 Read "E to the quantity of pi times I is equal to negative one"

[StormHoof32]

you sir, make my brain hurt.

[DaemoN67]

592960
Mmkay, I'ma gonna think about this REAL hard...
Urgh, now my head hurts. How the hay am I gonna prove THAT?!
I'll wait for a mathematically inclined person to show up.

[Ducky911]

That my friend is difficult. But I can't see how it's true unless it is (e^pi)i = -1

[Ducky911]

So I can only assume that e^pi is equivalent to 1 so that when you multiply by the imaginary square root of -1 which is basically just a replacement for a negative in the first place, then you get -1. Am I right?

[ScriptScrolls]

593040>>593050
You need a basic knowledge of Taylor polynomials to make this happen; that's the only clue you get.

[ScriptScrolls]

593094 It doesn't work that way considering you're not squaring i, and no, e^pi is equal to about 23.

[DaemoN67]

593095
Taylor polynomials? Am I allowed to Wikipedia that?

[Ducky911]

Hmm. Then I'm stumped. I have no basic knowledge of Taylor polynomials... I'm a history major man...

[Ducky911]

Not sure why the hell I thought e^pi was equal to 1 though. That was stupid...

[ImChangingmynameforreaso]

I hate imaginary numbers!

[ScriptScrolls]

593112 images.wikia.com/happywheels/images/2/23/Okay-meme.jpg
593131 First thing I thought when I saw the problem too.

[ScriptScrolls]

593142 Welcome to my world; I HATE vector spaces which makes up about 75% of my final tomorrow.

[Dreamer of Dragons]

Just multiply both sides of the equation by zero.

[ScriptScrolls]

593666 Multiplying by 0 could mean 2=-13. No, you can't do that in any proof.

[awesomekik]

It won't let me post the entire thing, but I did a 6th grade math project on it (yes, calculus in 6th grade) and it's basically if you add the Taylor series (function expressed as a polynomial) of cos(x) to the Taylor series of sin(x) multiplied by i you get the Taylor series of e^ix. This means e^ix = cos(x) + isin(x). Plug in pi and you're done.

[awesomekik]

I'm so smart. I win.

[ScriptScrolls]

596987 Bingo, but I just started with the Taylor series for e^x and plugged in ipi and seperated the ones with an i and ones without an i

[awesomekik]

597002 Yeah, I'm a bit fuzzy on it, I was probably wrong up to the e^ix = cos(x) + isin(x). Comments from the mobile site aren't working btw.