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dpaInc
 2 years ago
Best ResponseYou've already chosen the best response.1hmm... is that e^(ix) + e^(ix) so = 2e^(ix) on the right side?

Romero
 2 years ago
Best ResponseYou've already chosen the best response.0\[\cos(x)=\frac{e^{ix}+e^{ix}}{2} \]

Romero
 2 years ago
Best ResponseYou've already chosen the best response.0that means I can split cos(x) into two different functions right?

dpaInc
 2 years ago
Best ResponseYou've already chosen the best response.1what do you mean split up cosx ? that looks a bit like the hperbolic coshx...

Romero
 2 years ago
Best ResponseYou've already chosen the best response.0Well 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.

Romero
 2 years ago
Best ResponseYou've already chosen the best response.0Can you at least show me how \[\cos(x)=\frac{e^{ix}+e^{ix}}{2} \]

dpaInc
 2 years ago
Best ResponseYou've already chosen the best response.1this is guess, but try doing the series expansion for e^(ix) + e^(ix). i have a feeling the imaginary terms will cancel.

shivam_bhalla
 2 years ago
Best ResponseYou've already chosen the best response.0Write e^ix =cos(x)+i sin(x)

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1e^(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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