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

Muon12

Oh, okay. That helps make some more sense of it. Thanks.

mathwonk

Are some posters unaware of the previous posts? The same comments and proofs are occurring three or four times, as if they had not already been presented. indeed in the very first answer to this question i both gave the equation e^ix = cos x + i sin x, and proved it, using uniqueness of solutions of differential equations. the second answer or so gave the taylor series explanation. and yet it is all cycling over again like e^z. As i predicted, people like answering this question, apparently much more than reading previous answers.

If something new is forthcoming, besides the taylor series or diff eq answer, I would be interested. perhaps a path integral. since e^z is inverse to the path integral of 1/z, i guess we could ask why the path integral if 1/z from 1 to -1, equals i <pi>. but that integral has an exact real part, and an imaginary part equivalent to dtheta, so one does get arg(-1) = i<pi> + 2n<pi>.

i admit that one is not so original either. any more?

Gonzolo

Perhaps we could compare with [tex]\pi^{ie}[/tex]...

Is i ever used in the power of other numbers than e? Any use to to doing this?

mathwonk

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.

arildno

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?

polack

Maybe this is getting off on a tangent (horrible pun intended

), but I'm exploring how similar the graphs of tan(x) and [tex]e^{x}-\pi^{-x}[/tex] 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.

waht

Quote by mathwonk

Are some posters unaware of the previous posts? The same comments and proofs are occurring three or four times, as if they had not already been presented. indeed in the very first answer to this question i both gave the equation e^ix = cos x + i sin x, and proved it, using uniqueness of solutions of differential equations. the second answer or so gave the taylor series explanation. and yet it is all cycling over again like e^z. As i predicted, people like answering this question, apparently much more than reading previous answers.

If something new is forthcoming, besides the taylor series or diff eq answer, I would be interested. perhaps a path integral. since e^z is inverse to the path integral of 1/z, i guess we could ask why the path integral if 1/z from 1 to -1, equals i <pi>. but that integral has an exact real part, and an imaginary part equivalent to dtheta, so one does get arg(-1) = i<pi> + 2n<pi>.

i admit that one is not so original either. any more?

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

http://www.physicsforums.com/showthr...r%27s+identity

HallsofIvy

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 [tex]e^{x}-\pi^{-x}[/tex] 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!

matheinste

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.

K.J.Healey

I alwyas thought that this equation was much more beautiful and mysterious:

[tex]i^2+j^2+k^2=i j k = -1[/tex]

rbj

Quote by PRodQuanta

Just so you can see it in latex:

[tex]e^{(pi)i}+1=0[/tex]

Paden Roder

wouldn't it be

[tex] e^{i \pi} + 1 = 0 [/tex]

?

matheinste

Quote by Healey01

I alwyas thought that this equation was much more beautiful and mysterious:

[tex]i^2+j^2+k^2=i j k = -1[/tex]

Thanks for your reply Healey01.

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

Mateinste

ZioX

It's just the defining equation of quaternions.

The algebra of it is a lot more interesting...

polack

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]

.

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.

polack

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 [tex]\pi[/tex] is to circle/sphere as e is to exponentiation? I'm still working on the graph of tan(x) compared to [tex]e^{x} - \pi^{-x}[/tex]... the "-" implies an i in there somewhere.

mathwonk

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.

Muon12	How does [e^([pi]i)]+1=0? Oh, okay. That helps make some more sense of it. Thanks.

Gonzolo	Perhaps we could compare with [tex]\pi^{ie}[/tex]... Is i ever used in the power of other numbers than e? Any use to to doing this?

matheinste	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.

K.J.Healey	I alwyas thought that this equation was much more beautiful and mysterious: [tex]i^2+j^2+k^2=i j k = -1[/tex]

ZioX	It's just the defining equation of quaternions. The algebra of it is a lot more interesting...

mathwonk Recognitions: Homework Help Science Advisor	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.