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Power Series

  1. Oct 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Given:

    [tex] sin(x) = \frac{e^{ix}-e^{-ix}}{2} [/tex]

    Show that [tex] sin(x) [/tex] can be written as:

    [tex] sin(x) = \sum_{n=0}^n \frac{x^{(2n+1)}}{(2n+1)!} [/tex]



    2. Relevant equations


    [tex] e^x = \sum_{n=0}^n \frac{x^{n}}{(n)!} [/tex]

    3. The attempt at a solution

    I'm unsure how to treat the imaginary number in order to get a power expansion for [tex] e^{ix} [/tex]

    I know that it's not the same as [tex] e^{ax} [/tex] where a is just a constant. Any help on the complex part would be greatly appreciated!
     
    Last edited: Oct 27, 2013
  2. jcsd
  3. Oct 27, 2013 #2

    Ray Vickson

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    How is it different from ##e^{ax}## with ##a = i##? More to the point: who told you it is different?
     
  4. Oct 27, 2013 #3
    I'm just going by the fundamental law for imaginary multiplication. i.e.

    [tex] i = \sqrt{-1} [/tex]
    [tex] i^2 = -1 [/tex]
    [tex] i^3 = -i [/tex]
    [tex] i^4 = 1 [/tex]

    Therefore, if a = i, the expansion has to follow the above pattern?
     
  5. Oct 27, 2013 #4
    Wait, I guess I never got the formula for

    [tex] e^{ax} [/tex] in the first place now that I think about it. Good catch. Sorry about that!

    Is [tex] e^{ax} = [\sum_{n=0}^n \frac{(ax)^{n}}{(n)!}] [/tex] correct?
     
    Last edited: Oct 27, 2013
  6. Oct 27, 2013 #5
    I think it's

    [tex] e^{ix} = \sum_{n=0}^n \frac{(ix)^{n}}{(n)!} [/tex]

    therefore,

    [tex] \frac{e^{ix}}{2} = \frac{1}{2}\sum_{n=0}^n \frac{(ix)^{n}}{(n)!} [/tex]

    [tex] \frac{e^{-ix}}{2} = \frac{1}{2}\sum_{n=0}^n \frac{(-ix)^{n}}{(n)!} [/tex]

    Then I show that:

    [tex] \frac{1}{2}\sum_{n=0}^n \frac{(ix)^{n}}{(n)!} - \frac{1}{2}\sum_{n=0}^n \frac{(-ix)^{n}}{(n)!} = \sum_{n=0}^n \frac{x^{(2n+1)}}{(2n+1)!} [/tex]

    Right?
     
  7. Oct 27, 2013 #6

    Dick

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    You would do that except you've got some bad information in the problem statement. sin(x)=(e^(ix)-e^(-ix))/(2i) not over 2. And there should be an alternating sign in the series. Or are you really trying to expand the hyperbolic sine, sinh(x)?
     
  8. Oct 28, 2013 #7
    Sorry, it's late and I'm missing the details.

    He writes that

    [tex] sin(x) = \frac{e^{ix} - e^{-ix}}{2} [/tex]

    can be represented in power expanded form as:

    [tex] sin(x) = \sum_{n=0}^n (-1)^n \frac{x^{(2n+1)}}{(2n+1)!} [/tex]

    so you were sight, I was missing the alternating part in the final formula. I looked it over twice and there is no 2i, just 2? Maybe he made a mistake?
     
  9. Oct 28, 2013 #8

    Dick

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    Yes, he made a mistake. e^(ix)=cos(x)+i*sin(x) and e^(-ix)=exp(i*(-x))=cos(-x)+i*sin(-x)=cos(x)-i*sin(x). So e^(ix)-e^(-ix)=(2i)*sin(x).
     
  10. Oct 28, 2013 #9
    According to his notes: [tex] cosh(x) = \frac{e^{x} - e ^{-x}}{2} [/tex]

    He didn't give us sinh(x), or cos(x). Just assigned the sin(x) as I stated above with the alternating addition. It has to be 2i to cancel out in order to get the final formula.

    I've been trying to cancel/combine the expansion terms for 45 mins now and couldn't get his formula.

    I will bring it to his attention. Thank you!!
     
    Last edited: Oct 28, 2013
  11. Oct 28, 2013 #10

    Dick

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    The cosh formula is wrong too. cosh(x)=(e^x+e^(-x))/2. Either whoever is writing the notes or transcibing the notes has some attention problems.
     
  12. Oct 28, 2013 #11
    Wow, I'm looking at his pdf and he has cosh(x) written once in your format, and then once again with the -. This is a 3 hour lecture and he was talking for 2 hrs and then just rushed through the last part in 45 mins.

    One more thing, is his formula for cos(x) = (e^(ix) + e(-ix))/2 right? Or should this also have 2i in the denominator? I will bring all of this to his attention tomorrow.
     
  13. Oct 28, 2013 #12

    Dick

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    Ah, at last a correct formula. Yes, that one works.
     
  14. Oct 28, 2013 #13
    Thank you,

    So, just to summarize:

    [tex] sin(x) = \frac{e^{ix}-e^{-ix}}{2i} [/tex]

    [tex] sinh(x) = \frac{e^{x}-e^{-x}}{2} [/tex]

    [tex] cos(x) = \frac{e^{ix}+e^{-ix}}{2} [/tex]

    [tex] cosh(x) = \frac{e^{x}+e^{-x}}{2} [/tex]



    Thanks!
     
  15. Oct 28, 2013 #14

    Dick

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    Sure, and [tex] sinh(x) = \frac{e^{x}-e^{-x}}{2} [/tex] if you want to complete the list. The sin and cos formulas follow from deMoivre's expression ##e^{ix}=cos(x)+isin(x)##, you don't have memorize them all you can derive them.
     
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