# How does [e^([pi]i)]+1=0?

by Muon12
Tags: epii
 P: n/a Perhaps we could compare with $$\pi^{ie}$$... Is i ever used in the power of other numbers than e? Any use to to doing this?
 Sci Advisor HW Helper P: 9,407 Look. pi^(ie) is just e^(ie ln(pi)), so NO number is ever used as an exponent for bases other than e. That is to say, e is the universal base for all exponents. i.e. [(haha) perhaps I should say in russian: "tau yest" instead of "id est"]\\anyway: for any a, we have a^b = e^(bln(a)). Let's start with e^i. Note that since e^(it) = cos(t) + isin(t), that then e^i = cos(1)+i sin(1), not at all an interesting or elementary number at least not to me. Thus in the same spirit, pi^(ie) = e^(ie ln(pi)) = cos(e ln(pi)) + i sin(e ln(pi)), apparently a similarly ugly number. If anyone can give a nice interpretation of this number, my hat is off, and I would enjoy seeing it.
HW Helper
PF Gold
P: 11,983
 Quote by mathwonk Thus in the same spirit, pi^(ie) = e^(ie ln(pi)) = cos(e ln(pi)) + i sin(e ln(pi)), apparently a similarly ugly number. If anyone can give a nice interpretation of this number, my hat is off, and I would enjoy seeing it.
Would you still consider it an UGLY number?
P: 3
Maybe this is getting off on a tangent (horrible pun intended ), but I'm exploring how similar the graphs of tan(x) and $$e^{x}-\pi^{-x}$$ are... this may weave back to the earlier observations about path integrals from mathwonk:

 To see this, just notice that e^z is the inverse of ln(z), which is the path integral of 1/z which means the value varies according to how the path winds around 0 and infinity. On the other hand tan(z) is the inverse of arctan(z) = the path integral of 1/(1+z^2), which is determined by how many times the path winds around i and -i. I.e. 1/((1+z^2) is actually continuous at infinity and single valued there, so the two functions (if I got this right) seem to differ only by a mobius transformation which interchanges the pair 0 and infinity, for i and -i.
...so I'm curious if this goes anywhere new.
P: 1,636

I actually attempted to prove Euler's identity using a different way. This was discussed recently.

http://www.physicsforums.com/showthr...r%27s+identity
Math
Emeritus
Thanks
PF Gold
P: 38,711
 Quote by polack Maybe this is getting off on a tangent (horrible pun intended ), but I'm exploring how similar the graphs of tan(x) and $$e^{x}-\pi^{-x}$$ are... this may weave back to the earlier observations about path integrals from mathwonk: ...so I'm curious if this goes anywhere new.
 Quote by waht I actually attempted to prove Euler's identity using a different way. This was discussed recently. http://www.physicsforums.com/showthr...r%27s+identity
How in the world did you find this thread? All previous posts were from three years ago!
 P: 1,060 Hello all. Just a bit of trivia but in someway I feel descriptive of the power, beauty and simplicity of the formula. I once saw it referred to as A Mathematical Poem. Matheinste.
 P: 640 I alwyas thought that this equation was much more beautiful and mysterious: $$i^2+j^2+k^2=i j k = -1$$
P: 2,265
 Quote by PRodQuanta Just so you can see it in latex: $$e^{(pi)i}+1=0$$ Paden Roder
wouldn't it be

$$e^{i \pi} + 1 = 0$$

?
P: 1,060
 Quote by Healey01 I alwyas thought that this equation was much more beautiful and mysterious: $$i^2+j^2+k^2=i j k = -1$$

Your equation is certainly mysterious. When I find out what it means it might also be beautiful.

Mateinste
 P: 372 It's just the defining equation of quaternions. The algebra of it is a lot more interesting...
P: 3
 Quote by HallsofIvy How in the world did you find this thread? All previous posts were from three years ago!
Why, with a Google search of course! (or is that Googol? ) I was curious if someone had already invented the wheel I was working on... so I searched for it.

This sort of rediscovery isn't too unusual... there's Wile's 1994 rediscovery of Fermat's last theorem from 1637--only 357 years later, but he didn't use the internets [sic] .
 Sci Advisor HW Helper P: 9,407 exponentiation is a homomorphism from a ddition to multiplication, and surjects onto the positive reals for sure, and neither the value nor the derivative is ever zero. hence i claim it is a topological covering map from the complexes to the non zero complkexes, hence must wrap around the origin, and be periodic for some value. i.e. it must take the values 1 and -1 infinitely often. it remains only to find the value x such that e^x = -1. that follows from trig (eulers style via power series expansions of sin, cos). i think this is a new answer to an old question.
P: 3
 Quote by mathwonk exponentiation is a homomorphism from a ddition to multiplication, and surjects onto the positive reals for sure, and neither the value nor the derivative is ever zero. hence i claim it is a topological covering map from the complexes to the non zero complkexes, hence must wrap around the origin, and be periodic for some value. i.e. it must take the values 1 and -1 infinitely often. it remains only to find the value x such that e^x = -1. that follows from trig (eulers style via power series expansions of sin, cos). i think this is a new answer to an old question.
So it wraps like a mobius strip then? No edges or ends... periodic, as you say. I feel the flow; circular. So $$\pi$$ is to circle/sphere as e is to exponentiation? I'm still working on the graph of tan(x) compared to $$e^{x} - \pi^{-x}$$... the "-" implies an i in there somewhere.
 Sci Advisor HW Helper P: 9,407 im just saying there is essentially no other way to map the plane onto the punctured plane with derivative non zero, except to go around and around, so it has to hit the same point more than once, i.e. it hits 1 in finitely often, and also -1. i see it like a spiral staircase.
 P: 1 Beautiful and useful too, phasors for AC circuit analysis are based on Euler's identity.