Results 1 to 5 of 5

Math Help - hyperbolics

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    7

    hyperbolics

    what is cosix? how can an angle be imaginary?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by swaha View Post
    what is cosix? how can an angle be imaginary?
    In the complex plane, we define \cos z=\frac{e^{iz}+e^{-iz}}{2}.


    Letting z=ix, we have \cos ix=\frac{e^{i(ix)}+e^{-i(ix)}}{2}=\frac{e^{i^2x}+e^{-i^2x}}{2}=\frac{e^{-x}+e^{-(-x)}}{2}=\frac{e^x+e^{-x}}{2}=\cosh x

    Thus, when evaluating \cos ix at an angle x, its the same as evaluating \cosh x.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,412
    Thanks
    1328
    Quote Originally Posted by swaha View Post
    what is cosix? how can an angle be imaginary?
    An angle can't be imaginary! But the "trig functions", sine, cosine, etc. are not always related to angles. Sine and cosine, in particular are the "ideal" periodic functions and are often used to model periodic phenomena that have nothing to do with angles (although engineers will still insist upon talking about the "phase angle" of an electrical current where there is no actual angle involved).

    The easiest thing to do with complex numbers is to multiply and add them so for more complicated functions it is best to work with power series expansions:
    e^x= 1+ x+ \frac{1}{2}x^2+ \cdot\cdot\cdot+ \frac{1}{n!}x^n+ \cdot\cdot\cdot
    cos(x)= 1- \frac{1}{2!}x^2+ \cdot\cdot\cdot+ \frac{(-1)^n}{(2n)!}x^{2n}+ \cdot\cdot\cdot
    sin(x)= x- \frac{1}{31}x^3+ \cdot\cdot\cdot+ \frac{(-1)^n}{(2n+1)!}x^{2n+1}

    If you replace x by "ix" in the first, you get
    e^{ix}= 1+ ix+ \frac{1}{2}(ix)^2+ \cdot\cdot\cdot+ \frac{1}{n!}(ix)^n+ \cdot\cdot\cdot
    But i^2= -1, i^3= (i^2)(i)= -i, i^4= (i^2)(i^2)= (-1)(-1)= 1 and then i^5= (i^4)i= i, etc.

    e^{ix}= 1+ ix- \frac{1}{2}x^2- i\frac{1}{3!}x^3+ \frac{1}{4!}x^4+ \cdot\cdot\cdot
    so all of the "odd" terms have "i" while the "even" terms do not. Separating "real" and "imaginary" parts,
    e^{ix}= (1- \frac{1}{2}x^2+ \frac{1}{4!}x^4+ \cdot\cdot\cdot)+ i(x- \frac{1}{3!}x^3+ \frac{1}{5!}x^5+ \cdot\cdot\cdot)
    or e^{ix}= cos(x)+ i sin(x). Replacing x by -x, e^{-ix}= cos(-x)+ i sin(-x)= cos(x)+ i sin(x), since cos(-x)= cos(x) and sin(-x)= -sin(x).
    Adding those two equations, e^{ix}+ e^{-ix}= cos(x)+ i sin(x)+ cos(x)- i sin(x)= 2 cos(x) so cos(x)= \frac{e^{ix}+ e^{-ix}}{2}. In particular, if we now replace x by ix, we have cos(ix)= \frac{e^{-x}+ e^{x}}{2}= cosh(x).

    (Which is why you titled this "hyperbolics", right?)
    Last edited by mr fantastic; September 6th 2009 at 04:20 AM. Reason: Added a latex tag.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2009
    Posts
    7
    what does the term e^ix mean? what is the significance of raising soth to imaginary power?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,412
    Thanks
    1328
    Quote Originally Posted by swaha View Post
    what does the term e^ix mean? what is the significance of raising soth to imaginary power?
    It means
    e^{ix}= 1+ ix+ \frac{1}{2}(ix)^2+ \cdot\cdot\cdot+ \frac{1}{n!}(ix)^n+ \cdot\cdot\cdot
    just as I said above.

    From those same formulas you can get e^{ix}= cos(x)+ i sin(x), where, since x is now a real number, cos(x) and sin(x) have their usual significance.
    Last edited by mr fantastic; September 6th 2009 at 04:20 AM. Reason: Fixed a latex tag
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: March 19th 2010, 03:48 PM
  2. hyperbolics!
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 6th 2010, 02:19 AM
  3. Hyperbolics
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 11th 2009, 02:34 PM
  4. Hyperbolics
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 19th 2008, 01:57 PM
  5. can't integrate powers of hyperbolics
    Posted in the Calculus Forum
    Replies: 5
    Last Post: August 9th 2007, 05:38 AM

Search Tags


/mathhelpforum @mathhelpforum