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Can someone please show me why e[itex]^{-ikx}[/itex] +

  1. Sep 25, 2011 #1
    Can someone please show me why e[itex]^{-ikx}[/itex] + e[itex]^{ikx}[/itex] simplifies to 2e[itex]^{ikx}[/itex] instead of 1+e[itex]^{ikx}[/itex]??
     
    Last edited: Sep 25, 2011
  2. jcsd
  3. Sep 25, 2011 #2

    mathman

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    Re: Simplifying

    It doesn't at all. It is equal to 2cos(kx).
     
  4. Sep 25, 2011 #3
    Re: Simplifying

    Ok, yes, that's true and I can get there if I change e[itex]^{ikx}[/itex] to cos(kx) + i*sin(kx). But I'm having exponent issues if I don't change it into sine and cosine (i.e. [itex]\frac{1}{e^{ikx}}[/itex] + [itex]\frac{e^{ikx}}{1}[/itex] ). Wouldn't that simplify to 1 + e[itex]^{ikx}[/itex]?
     
  5. Sep 25, 2011 #4

    Char. Limit

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    Re: Simplifying

    Well, neither of your answers are really correct. Your answer is already in about as simple a form as it can get... maybe this could be simpler?

    [tex]e^{-i k x} \left( 1 + e^{2 i k x} \right)[/tex]
     
  6. Sep 25, 2011 #5

    symbolipoint

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    Re: Simplifying

    I worked through the initial expression and also came to [itex]1+e^{ikx} [/itex]
     
  7. Sep 25, 2011 #6

    Char. Limit

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    Re: Simplifying

    Show us your work. You're making a mistake here somewhere.
     
  8. Sep 25, 2011 #7

    symbolipoint

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    Re: Simplifying

    I just noticed: You edited the original post for a sign change. I worked with what you wrote originally, e^(-ikx) - e^(ikx), [itex]e^{-ikx}-e^{ikx} [/itex]
     
  9. Sep 25, 2011 #8

    symbolipoint

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    Re: Simplifying

    As the addition instead of the subtraction, this process:
    Using t=kx,

    cos(-t)+i*sin(-t) + cos(t)+ i*sin(t)
    = cos(t) - i*sin(t) + cos(t) +i*sin(t)
    = cos(t) + cos(t) + i*sin(t) - i*sin(t)
    = 2cos(t)

    Please check my understanding of these steps, since I cannot think of a way to transform that back to exponential form.
     
  10. Sep 25, 2011 #9

    Char. Limit

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    Re: Simplifying

    Yup, you got that part right. The thing is, you started with the exponential form for 2cos(t). You started with

    [tex]e^{- i t} + e^{i t}[/tex]

    and ended with 2 cos(t). What you started with IS the exponential form.
     
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