Facebook Page
Twitter
RSS
+ Reply to Thread
Results 1 to 10 of 10
  1. MHB Seeker
    MHB Global Moderator
    MHB Math Scholar
    MHB Coder
    I like Serena's Avatar
    Status
    Online
    Join Date
    Mar 2012
    Location
    Netherlands
    Posts
    6,418
    Thanks
    4,354 times
    Thanked
    12,118 times
    Thank/Post
    1.888
    Awards
    MHB Model Helper Award (2016)  

MHB Best Ideas (2016)  

MHB LaTeX Award (2016)  

MHB Calculus Award (2014)  

MHB Discrete Mathematics Award (Jul-Dec 2013)
    #1
    I was looking at the header of MHB and I suddenly thought... wait!
    What does it mean?!

    I mean, it says:
    $e^{i \pi} + 1 = 0$

    Now, I understand that if you calculate $e^{i \pi} + 1$ that you get 0.
    But you can also rewrite it as:
    $-1 = e^{i \pi}$

    But how is that correct?
    Isn't $-1$ multivalued, as in $e^{i\pi(1+2k)}$, while $e^{i \pi}$ is single valued?!

    Or am I misunderstanding what the equality sign actually means?

  2. MHB Journeyman
    Chris L T521's Avatar
    Status
    Offline
    Join Date
    Jan 2012
    Location
    Chicago, IL
    Posts
    976
    Thanks
    2,263 times
    Thanked
    2,631 times
    Thank/Post
    2.696
    #2
    Quote Originally Posted by ILikeSerena View Post
    I was looking at the header of MHB and I suddenly thought... wait!
    What does it mean?!

    I mean, it says:
    $e^{i \pi} + 1 = 0$

    Now, I understand that if you calculate $e^{i \pi} + 1$ that you get 0.
    But you can also rewrite it as:
    $-1 = e^{i \pi}$

    But how is that correct?
    Isn't $-1$ multivalued, as in $e^{i\pi(1+2k)}$, while $e^{i \pi}$ is single valued?!

    Or am I misunderstanding what the equality sign actually means?
    Well, yes, technically $e^{i(2k-1)\pi}=-1$ due to the fact that the complex exponential is $2\pi$-periodic; but when people think of Euler's formula, we just consider the base case when $k=1$ (or if you consider Euler's formula $e^{i\theta}=\cos\theta+i\sin\theta$, letting $\theta=\pi$ leads to the famous formula $e^{i\pi}+1=0$).

  3. MHB Seeker
    MHB Global Moderator
    MHB Math Scholar
    MHB Coder
    I like Serena's Avatar
    Status
    Online
    Join Date
    Mar 2012
    Location
    Netherlands
    Posts
    6,418
    Thanks
    4,354 times
    Thanked
    12,118 times
    Thank/Post
    1.888
    Awards
    MHB Model Helper Award (2016)  

MHB Best Ideas (2016)  

MHB LaTeX Award (2016)  

MHB Calculus Award (2014)  

MHB Discrete Mathematics Award (Jul-Dec 2013)
    #3 Thread Author
    Quote Originally Posted by Chris L T521 View Post
    Well, yes, technically $e^{i(2k-1)\pi}=-1$ due to the fact that the complex exponential is $2\pi$-periodic; but when people think of Euler's formula, we just consider the base case when $k=1$ (or if you consider Euler's formula $e^{i\theta}=\cos\theta+i\sin\theta$, letting $\theta=\pi$ leads to the famous formula $e^{i\pi}+1=0$).
    Hmm, as long as I don't fill in $\theta$, the left and right hand sides of Euler's formula are single valued.
    The multi valued aspect is "hidden" in $\theta$.

    When you fill in $\theta=\pi$ and evaluate the right hand side the result would be that both sides remain single valued.

    It seems there are two versions of $-1$.
    A single valued version (as in the famous formula) and a multi valued version.
    It seems to me they should be distinguished somehow.

  4. MHB Master
    MHB Site Helper
    MHB Math Scholar
    Deveno's Avatar
    Status
    Offline
    Join Date
    Feb 2012
    Location
    just south of canada
    Posts
    1,964
    Thanks
    419 times
    Thanked
    4,365 times
    Thank/Post
    2.223
    Awards
    MHB Advanced Algebra Award (2015)  

MHB Advanced Algebra Award (Jul-Dec 2013)
    #4
    i don't see how $-1$ can be "multi-valued". i'm willing to hear your thoughts, though.

  5. MHB Seeker
    MHB Global Moderator
    MHB Math Scholar
    MHB Coder
    I like Serena's Avatar
    Status
    Online
    Join Date
    Mar 2012
    Location
    Netherlands
    Posts
    6,418
    Thanks
    4,354 times
    Thanked
    12,118 times
    Thank/Post
    1.888
    Awards
    MHB Model Helper Award (2016)  

MHB Best Ideas (2016)  

MHB LaTeX Award (2016)  

MHB Calculus Award (2014)  

MHB Discrete Mathematics Award (Jul-Dec 2013)
    #5 Thread Author
    Quote Originally Posted by Deveno View Post
    i don't see how $-1$ can be "multi-valued". i'm willing to hear your thoughts, though.
    Well, I ran recently into a problem that said:
    How can I write $(-e)^{i\pi}$ in $a+bi$ form?

    This is the sort of problem where you get a break-down of the use of powers with complex numbers.
    It brings up the question whether you want the principal value or the multi valued one. And even if you only want to have the principal value, you still have to consider the effect of the other possible values so as to avoid making mistakes.

    In particular I would split off a -1 and consider its multi valued form.
    Last edited by I like Serena; January 28th, 2013 at 20:07.

  6. MHB Master
    MHB Site Helper
    MHB Math Scholar
    Deveno's Avatar
    Status
    Offline
    Join Date
    Feb 2012
    Location
    just south of canada
    Posts
    1,964
    Thanks
    419 times
    Thanked
    4,365 times
    Thank/Post
    2.223
    Awards
    MHB Advanced Algebra Award (2015)  

MHB Advanced Algebra Award (Jul-Dec 2013)
    #6
    Alright, I buy that...but the problem isn't with "$-1$" per se, it's with how complex exponentiation is defined.

    If we try to define:

    $z^w = e^{\log(z)w}$

    the trouble we run into is with the "log" function, NOT the "exp" function. The complex exponential function is defined and analytic EVERYWHERE, but the complex logarithm is NOT so well-behaved.

    On the real numbers, the exponential function is injective, and so its inverse is single-valued wherever defined. On the complex numbers, however, the complex exponential is no longer injective, but "periodic in $i$". So this means that logarithm is "multi-valued" (much like "square root" is for real numbers, to get a function, we need to "pick a branch", we normally pick the "top half of the parabola").

    The expression:

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

    is unambiguous. The expression:

    $\log(-1) = i\pi$

    is not. In complex numbers, the square root function is also "multi-valued" with a vengence (we no longer have the option of picking "the positive one") leading to many amusing proofs that $-1 = 1$.

  7. MHB Master
    MHB Math Helper

    Status
    Offline
    Join Date
    Jan 2012
    Posts
    1,299
    Thanks
    130 times
    Thanked
    2,064 times
    Thank/Post
    1.589
    #7
    Quote Originally Posted by ILikeSerena View Post
    I was looking at the header of MHB and I suddenly thought... wait!
    What does it mean?!

    I mean, it says:
    $e^{i \pi} + 1 = 0$

    Now, I understand that if you calculate $e^{i \pi} + 1$ that you get 0.
    But you can also rewrite it as:
    $-1 = e^{i \pi}$

    But how is that correct?
    Isn't $-1$ multivalued, as in $e^{i\pi(1+2k)}$, while $e^{i \pi}$ is single valued?!

    Or am I misunderstanding what the equality sign actually means?
    You are overthinking this. Basically, $ \displaystyle \begin{align*} e^{i\pi \left( 1 + 2k \right)} \end{align*} $ will be -1 for all integer values of k. $ \displaystyle \begin{align*} e^{i\pi} \end{align*} $ is simply one of those values, i.e. where k = 0.

  8. MHB Seeker
    MHB Global Moderator
    MHB Math Scholar
    MHB Coder
    I like Serena's Avatar
    Status
    Online
    Join Date
    Mar 2012
    Location
    Netherlands
    Posts
    6,418
    Thanks
    4,354 times
    Thanked
    12,118 times
    Thank/Post
    1.888
    Awards
    MHB Model Helper Award (2016)  

MHB Best Ideas (2016)  

MHB LaTeX Award (2016)  

MHB Calculus Award (2014)  

MHB Discrete Mathematics Award (Jul-Dec 2013)
    #8 Thread Author
    I guess I'm wrestling with merging my knowledge of complex analysis with algebra theory.

    If we consider the field $\mathbb C$ with just addition and multiplication, it seems fine to consider each element single valued.

    However, when we introduce the logarithm or non-integer powers, this is no longer sufficient.
    As I see it know, each element of $\mathbb C$ must be considered multi valued in that context. Or more specifically, each element is an equivalence class. Each equivalence class maps to a subset of $\mathbb R^2$.
    One might consider using the congruence symbol in that context, just like it's used in number theory.

    So $e^{i \pi}$ would represent the equivalence class $\{e^{i \pi (1 + 2k)} : k \in \mathbb Z\}$.
    And as such it would be the same as the equivalence class represented by $-1$.

    In particular we can interpret the famous formula $e^{i\pi}+1=0$ both as single valued as well as multi valued.

    Furthermore $(-1)^i$ represents $\{e^{- \pi (1 + 2k)} : k \in \mathbb Z\}$, which is a set of discrete points on the real line.

    The complex number $1^\pi$ represents the unit circle, showing that an equivalence class can have an uncountable size.
    Last edited by I like Serena; January 29th, 2013 at 17:20.

  9. MHB Master
    MHB Site Helper
    MHB Math Scholar
    Deveno's Avatar
    Status
    Offline
    Join Date
    Feb 2012
    Location
    just south of canada
    Posts
    1,964
    Thanks
    419 times
    Thanked
    4,365 times
    Thank/Post
    2.223
    Awards
    MHB Advanced Algebra Award (2015)  

MHB Advanced Algebra Award (Jul-Dec 2013)
    #9
    Yes, when we consider complex exponentiation (which involves the log function "hidden" in it), we find that non-integer powers of complex numbers do not behave well once we get off the positive real line.

    This poses an interesting dilemma: we could define the complex logarithm so it's "cut" was the positive real line, but then we do not have a continuous "extension" of the real log function. Or we could make the "cut" at the negative reals (since the real log isn't defined there anyway) but then we lose the ability to extend $(-1)^n$ in a natural way.

    So what is done is to "make the complex numbers multi-valued" by creating a Riemann surface for log (and thus a complex number to a complex power) to be defined on. This Riemann surface looks much like a parking garage, and every complex number occurs on "different levels" (except for 0, where log(0) remains "undefinable"). I believe it is this Riemann surface that you are thinking about.

    To be more explicit, the surface $R$ comes with a natural projection $P:R \to \Bbb C^{\ast}$, and it is the pre-images of this projection that form your equivalence classes.
    Last edited by Deveno; January 29th, 2013 at 18:23.

  10. MHB Master
    MHB Site Helper
    MHB Math Scholar
    Deveno's Avatar
    Status
    Offline
    Join Date
    Feb 2012
    Location
    just south of canada
    Posts
    1,964
    Thanks
    419 times
    Thanked
    4,365 times
    Thank/Post
    2.223
    Awards
    MHB Advanced Algebra Award (2015)  

MHB Advanced Algebra Award (Jul-Dec 2013)
    #10
    This sounds rather complicated, but it's really NOT.

    Consider something that travels in a circular orbit. We might model its position at time $t$ by:

    $p(t) = (\cos(t),\sin(t))$

    This is fine as long as we just want to know: "where is it now?". But what if we want to know "how many times has it gone around?". Then we need a new coordinate to keep track of this as well:

    $f(t) = (\cos(t),\sin(t),t)$

    and our circle becomes a helix. If we consider all the circles of varying size, we get a "sheet of helices":

    $f(r,t) = (r\cos(t),r\sin(t),t),\ r > 0$

    This is the surface $R$. The projection $P$ is just $P(f(r,t)) = re^{it}$. So we are implicitly using this surface $R$ every time we use "polar coordinates" (we forget about the "other angles" outside of our normal $2\pi$ range, we project our helix onto a single orbit, a circle).

    This is an example of what topologists call "a covering space", and the maps that "translate" levels are called "deck transformations". In this case, such a translation is just a vertical shift of $2\pi k$ for an integer $k$, so the deck transformations form a group isomorphic to the integers (composition of deck transformations corresponds to adding different $k$'s). We can recover the "true" helix from a "shadow circle" by adding another piece of information: the winding number (this tells us which level of the parking garage we're really on, which orbit we're traversing THIS time).

    If we pick a range of $2\pi$ as our "home range", we can think of $R$ as "stacked like pancakes", with each copy of the "base plane" $\Bbb C^{\ast}$ lying in its own "integer address". The true beauty of $R$, though, is that it is a *smooth* surface, and there isn't any discontinuous "jump" from one level to the next, they flow into each other. And $\log$ can be defined "single-valued" as a function on $R$, and thus so can $g(z) = z^w$, which lets us obtain SOME value for $g(z)$ by projecting it down, and also reminds us why we have to be careful when calculating $z^w$: we might get a different answer than some other expression which our intuition with the reals would lead us to believe is the "same" (this is most easily seen by taking $w = \frac{1}{2}$).

Similar Threads

  1. Confusion with variables when solving a recurrence equation
    By marobin in forum Discrete Mathematics, Set Theory, and Logic
    Replies: 1
    Last Post: August 9th, 2012, 11:01

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards