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Romero

  • 2 years ago

Explain this identity.

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  1. dpaInc
    • 2 years ago
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    hmm... is that e^(ix) + e^(ix) so = 2e^(ix) on the right side?

  2. Romero
    • 2 years ago
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    \[\cos(x)=\frac{e^{ix}+e^{-ix}}{2} \]

  3. Romero
    • 2 years ago
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    that means I can split cos(x) into two different functions right?

  4. dpaInc
    • 2 years ago
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    what do you mean split up cosx ? that looks a bit like the hperbolic coshx...

  5. Romero
    • 2 years ago
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    Well there is a problem where I try to show if cos(x) is an eigenfunction of an operator. I did the operation and found out it wasn't but then my book says if it's not a the eigenfunction that it must be a superposition eigenfunction so it splits it into two functions. I just dont see how it splits or how this superposition will be the eigenfunction.

  6. Romero
    • 2 years ago
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    Can you at least show me how \[\cos(x)=\frac{e^{ix}+e^{-ix}}{2} \]

  7. dpaInc
    • 2 years ago
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    this is guess, but try doing the series expansion for e^(ix) + e^(-ix). i have a feeling the imaginary terms will cancel.

  8. shivam_bhalla
    • 2 years ago
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    Write e^ix =cos(x)+i sin(x)

  9. estudier
    • 2 years ago
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    e^(ix) = cos x + i sin x (Euler's Formula) e^(-ix) = cos(-x) + i sin (-x) = cos x- i sin x Add them.

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