No, \cos \pi/6=\frac{1}{2}\sqrt{3}. Secondly I am not asking you to draw the position where cos is -1/2 in the unit circle exactly. The use in the unit circle is that you can see in which quadrant the solutions lay. When you draw a line radially outwards from the origin you can form a triangle...
Mark gave you the values where the cosine is -1/2 in the third quadrant and 1/2 in the fourth quadrant. You're interested in the values where cosine is -1/2, not 1/2.
The first thing you should do is draw a unit circle. Then sketch the places where the cosine is equal to -1/2 and find the...
I can see where the confusion is coming from. You're right that if we set u=x^2 then f(u)=f(x^2) \neq f(x). To avoid this confusion it's better to not do another substitution with the same letter . In the integral however u functions as a dummy variable. Perhaps your confusion goes away if you...
They are the same function, f(u), f(x),f(a),f(bla) etc are all the same 'functions'. It doesn't matter what name you give to the variable. Besides from your substitution u=x it immediately follows that f(u)=f(x).
No sorry about that. It is not zero and your result in post #24 is correct if you replace x^2/3 with a_0. Don't reply when you just got out of bed!
To clear it up:
b_n=-\frac{1}{2n}
This integral doesn't have a primitive in terms of elementary functions. To evaluate this integral you can use the Raleigh-Jeans approximation by looking it up in your book or by using the Taylor series of the exponent up to first order.
It is almost correct.
The part between brackets is zero, this should not be the case.
The x^2/3 in front of the sum isn't correct either, it should be a_0 instead.
I don't think so. That is not an example of a unitary matrix that is Hermitian. You just wrote the definition of a unitary matrix in another form.
Definition of a unitary matrix: UU^\dagger=I. Then we multiply both sides with the inverse of U, which gives us...