What is the significance of [e^([pi]i)]+1=0 in mathematics?

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In summary, the equation [e^([pi]i)]+1=0 is a result of the connection between exponential functions and trigonometric functions, specifically through differential equations. This equation has significant implications in the field of complex numbers, as it shows that exponential, sine, and cosine functions are all equivalent from this perspective. The "spheres" mentioned in the conversation refer to the complex sphere, which is an enlarged plane with an added point at infinity. This equation showcases the unity and connection between seemingly unrelated numbers such as e, pi, and i.
  • #36
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?
 
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  • #37
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?
 
  • #38
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.
 
  • #39
mathwonk said:
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?
 
  • #40
tan(x) vs. e^(x) - pi^(-x)

Maybe this is getting off on a tangent (horrible pun intended :tongue2:), 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.
 
  • #41
mathwonk said:
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.

https://www.physicsforums.com/showthread.php?t=174527&highlight=euler's+identity
 
  • #42
polack said:
Maybe this is getting off on a tangent (horrible pun intended :tongue2:), 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.

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

https://www.physicsforums.com/showthread.php?t=174527&highlight=euler's+identity

How in the world did you find this thread? All previous posts were from three years ago!
 
  • #43
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.
 
  • #44
I alwyas thought that this equation was much more beautiful and mysterious:

[tex]i^2+j^2+k^2=i j k = -1[/tex]
 
  • #45
PRodQuanta said:
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]

?
 
  • #46
Healey01 said:
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
 
  • #47
It's just the defining equation of quaternions.

The algebra of it is a lot more interesting...
 
  • #48
How I found this three-year-old thread...

HallsofIvy said:
How in the world did you find this thread? All previous posts were from three years ago!

Why, with a Google http://www.google.com/search?hl=en&safe=active&q=+tan+e+pi+equation+theory"? :confused:) 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 "[URL [Broken] last theorem[/URL] from 1637--only 357 years later, but he didn't use the http://en.wikipedia.org/wiki/Internets_(colloquialism)" [Broken] :tongue2:.
 
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  • #49
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.
 
  • #50
It keeps going and going...

mathwonk said:
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 http://www.sciam.com/article.cfm?chanID=sa003&articleID=D55BA3C5-E7F2-99DF-3B0A5A55D41B63FB&ref=rss" 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.
 
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  • #51
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.
 
  • #52
Beautiful and useful too, phasors for AC circuit analysis are based on Euler's identity.
 
<H2>1. What is the significance of [e^([pi]i)]+1=0 in mathematics?</H2><p>The equation [e^([pi]i)]+1=0 is known as Euler's identity and it is considered one of the most beautiful and important equations in mathematics. It relates five fundamental mathematical constants: e (Euler's number), π (pi), i (the imaginary unit), 1 (the multiplicative identity), and 0 (the additive identity).</p><H2>2. How does [e^([pi]i)]+1=0 relate to complex numbers?</H2><p>The equation [e^([pi]i)]+1=0 is an example of a complex number, which is a number that contains both a real and an imaginary part. The imaginary part in this equation is [pi]i, which represents a rotation in the complex plane. The real part is 1, which represents a shift along the real axis. Together, these components create a complex number that has a magnitude of 1 and an angle of π radians, which is the same as 180 degrees.</p><H2>3. How is [e^([pi]i)]+1=0 derived?</H2><p>The derivation of [e^([pi]i)]+1=0 involves using Euler's formula, which states that e^(ix) = cos(x) + i*sin(x). By substituting x with π, we get e^(iπ) = cos(π) + i*sin(π). Since cos(π) = -1 and sin(π) = 0, we can rewrite the equation as e^(iπ) = -1. Then, by raising both sides to the power of i, we get [e^([pi]i)] = -1^i. Finally, using the definition of i as the square root of -1, we can rewrite -1^i as e^([pi]i) = cos([pi]*i) + i*sin([pi]*i). Simplifying further, we get e^([pi]i) = cos([pi]) + i*sin([pi]). Since cos([pi]) = -1 and sin([pi]) = 0, we end up with [e^([pi]i)]+1=0.</p><H2>4. How is [e^([pi]i)]+1=0 used in real life?</H2><p>Although [e^([pi]i)]+1=0 may seem like a purely theoretical equation, it has many practical applications in fields such as physics, engineering, and signal processing. For example, it is used in the study of electromagnetic waves and in the analysis of alternating current circuits. It also has applications in Fourier analysis, which is used to break down complex signals into simpler components. Additionally, Euler's identity has been used in cryptography and in the development of computer algorithms.</p><H2>5. Can [e^([pi]i)]+1=0 be generalized to other numbers?</H2><p>Yes, Euler's identity can be generalized to other numbers by using the same principles. For example, [e^(2iπ)]+1=0 is another form of the equation, which relates the number 2 to the other constants. This can be extended to any real number n, where [e^(niπ)]+1=0. This generalization is known as De Moivre's formula and it is used in trigonometry and complex analysis.</p>

1. What is the significance of [e^([pi]i)]+1=0 in mathematics?

The equation [e^([pi]i)]+1=0 is known as Euler's identity and it is considered one of the most beautiful and important equations in mathematics. It relates five fundamental mathematical constants: e (Euler's number), π (pi), i (the imaginary unit), 1 (the multiplicative identity), and 0 (the additive identity).

2. How does [e^([pi]i)]+1=0 relate to complex numbers?

The equation [e^([pi]i)]+1=0 is an example of a complex number, which is a number that contains both a real and an imaginary part. The imaginary part in this equation is [pi]i, which represents a rotation in the complex plane. The real part is 1, which represents a shift along the real axis. Together, these components create a complex number that has a magnitude of 1 and an angle of π radians, which is the same as 180 degrees.

3. How is [e^([pi]i)]+1=0 derived?

The derivation of [e^([pi]i)]+1=0 involves using Euler's formula, which states that e^(ix) = cos(x) + i*sin(x). By substituting x with π, we get e^(iπ) = cos(π) + i*sin(π). Since cos(π) = -1 and sin(π) = 0, we can rewrite the equation as e^(iπ) = -1. Then, by raising both sides to the power of i, we get [e^([pi]i)] = -1^i. Finally, using the definition of i as the square root of -1, we can rewrite -1^i as e^([pi]i) = cos([pi]*i) + i*sin([pi]*i). Simplifying further, we get e^([pi]i) = cos([pi]) + i*sin([pi]). Since cos([pi]) = -1 and sin([pi]) = 0, we end up with [e^([pi]i)]+1=0.

4. How is [e^([pi]i)]+1=0 used in real life?

Although [e^([pi]i)]+1=0 may seem like a purely theoretical equation, it has many practical applications in fields such as physics, engineering, and signal processing. For example, it is used in the study of electromagnetic waves and in the analysis of alternating current circuits. It also has applications in Fourier analysis, which is used to break down complex signals into simpler components. Additionally, Euler's identity has been used in cryptography and in the development of computer algorithms.

5. Can [e^([pi]i)]+1=0 be generalized to other numbers?

Yes, Euler's identity can be generalized to other numbers by using the same principles. For example, [e^(2iπ)]+1=0 is another form of the equation, which relates the number 2 to the other constants. This can be extended to any real number n, where [e^(niπ)]+1=0. This generalization is known as De Moivre's formula and it is used in trigonometry and complex analysis.

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