Recent content by VertexOperator

Then how do I find the discriminant of the quartic equation? Do I even need the discriminant?

Because when it has multiplicity 2 the discriminant is 0 :(

When I assume we have one real root of multiplicity 2 I get a^2=-4(b+2), is that right?

I believe that the rules of the homework help forum are too strict so I think it would be great if people can give more obvious directions so that problems are solved faster.

I ignored some because Euler's formula isn't covered by the syllabus. Maybe we should use cis2x+(cis2x)^2+...+(cis2x)^n as a suitable geometric series. I was able to do the integral without using the series but that isn't acceptable for this question. Let I_n=\int_0^{\pi/2}\frac{\sin...

Isn't the real part the cos2nx series?

But why is \int_{0}^{\frac{\pi }{2}}\frac{e^{i(2n+1)x}-e^{-i(2n+1)x}}{e^{ix}-e^{-ix}}dx equal to the cos2x series? Isn't this the imaginary part of e^2ix+e^4ix+...+e^2nix and the cos2x series the real part?

The denominator is equal to 2isinx and the numerator =2isin(2n+1)x? How can I integrate the integral I have in post 20 though, it looks very hard :(

I know, you are not allowed to solve the question form me but if the student is stuck shouldn't you give some hints? :(

I know this is dumb but this is what I did :( \int_{0}^{\frac{\pi }{2}}\frac{e^{i(2n+1)x}-e^{-i(2n+1)x}}{e^{ix}-e^{-ix}}dx

How can I make use of these identities now?

Ok, so I get \frac{e^{ix}+e^{-ix}}{2} =cosx \frac{e^{ix}-e^{-ix}}{2} =sinx

So can you guys tell me what to do next :)

oh yes it should be x^2+1/x^2.

I get e^2ix=cos2x+ison2x e^4ix=cos4x+ison4x . . . e^nix=cosnx+isonnx I still don't see how this can help with the integral :(