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

by Muon12
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Gonzolo
#37
Oct1-04, 12:37 PM
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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
#38
Oct1-04, 10:09 PM
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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
#39
Oct2-04, 05:12 AM
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Quote 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
#40
Jul17-07, 07:24 PM
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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
#41
Jul17-07, 10:51 PM
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Quote Quote by mathwonk View Post
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
#42
Jul18-07, 05:38 AM
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Quote Quote by polack View Post
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 Quote by waht View Post
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
#43
Jul18-07, 10:10 AM
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.
K.J.Healey
#44
Jul18-07, 10:26 AM
P: 641
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
#45
Jul18-07, 10:40 AM
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Quote Quote by PRodQuanta View Post
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
#46
Jul18-07, 12:16 PM
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Quote Quote by Healey01 View Post
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
#47
Jul18-07, 12:58 PM
P: 371
It's just the defining equation of quaternions.

The algebra of it is a lot more interesting...
polack
#48
Jul18-07, 01:32 PM
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Quote Quote by HallsofIvy View Post
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
#49
Jul18-07, 02:38 PM
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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
#50
Jul18-07, 02:57 PM
P: 3
Quote Quote by mathwonk View Post
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
#51
Jul19-07, 12:57 PM
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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.
widewombat
#52
Nov10-09, 06:23 AM
P: 1
Beautiful and useful too, phasors for AC circuit analysis are based on Euler's identity.


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