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

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

RomeroBest ResponseYou've already chosen the best response.0
that means I can split cos(x) into two different functions right?
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
what do you mean split up cosx ? that looks a bit like the hperbolic coshx...
 one year ago

RomeroBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

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

shivam_bhallaBest ResponseYou've already chosen the best response.0
Write e^ix =cos(x)+i sin(x)
 one year ago

estudierBest ResponseYou've already chosen the best response.1
e^(ix) = cos x + i sin x (Euler's Formula) e^(ix) = cos(x) + i sin (x) = cos x i sin x Add them.
 one year ago
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