The discussion revolves around the general solution of the function \(\psi(x) = Ae^{-ikx}\) in the context of calculus and ordinary differential equations (ODEs). Participants are exploring the transformation of this expression into a form involving sine and cosine functions.
Some participants have provided insights into the relationship between the exponential function and trigonometric functions, referencing Euler's theorem. There is an ongoing exploration of the mathematical reasoning behind the transformation, with some participants expressing clarity on the topic.
There is a mention of a second-order differential equation related to the general solution, but the initial inquiry does not explicitly involve a differential equation. Participants are navigating assumptions about the nature of the problem and the definitions involved.
Char. Limit said:This is mainly done by rewriting e^(i u) as cos(u) + i sin(u), I think.
HallsofIvy said:There was NO differential equation in anything you wrote!
It is true that "Asin(kx)+ Bcos(kx)" is the general solution of the second order differential equation y''+ k^2y= 0. Other than that, I can't see what you are asking.
Chaste said:alright, let me rephrase my question more clearly this time.
How does Ae-ikx transform into Asin(kx)+ Bcos(kx)?
Char. Limit said:Let e-ikx = cos(kx) - i sin(kx). Then Ae-ikx = Acos(kx) - i Asin(kx). Now let B=-iA. Then Ae-ikx = Acos(kx)+Bsin(kx).
And the first statement is Euler's Theorem, or something like that. It's easily proven using the Taylor series for the two functions.