I think you have a mistake with your algebra at the beginning.
I got which gives, eventually,
Also, regarding the first question - you can indeed show that the polynomial in your denominator has no zeroes in the unit circle using Rouche's theorem.
Hi everyone!
This is my first post in this forum. I've only just picked up math again, after more then 10 years off the bench. Keep that in mind while you're digging through my solving. I'm studying complex analysis and I've just solved, or tried to solve some integration problems. I'd really like someone to look at my solutions, however typing in LaTeX was somewhat gnarly, but I think I got the essence down.
This one I used Rouche's to find out roots in the denominator. The root inside unit circle can be used to calculate the Residues. However, I couldn't find any zeros within the unit circle, with leads to Res f(z) = 0.
With this one I changed the limits to Then I used and . This with Eulers formula, specially Cosines. Gave me:
I found two simple poles inside the unit circle, z = 0 and z = 2 - 3^0.5 Now I could calculate the residus with Hostpital's
Inserting figures.
Wish lead to, with a little help for wolframalpha
Consequentially
=
The numbers feels a bit strange, but this is what I got.
Please, feel free to comment!
OK... Thank you very much!
I'll redo the rewriting of the integral. Two questions.
1. - Is this a proper way of changing the limits?
2. About the Rouche stunt I was pulling. Is this a well mannered way of showing possible zeros of the polynom?
;
Now using - to show zeros within unit circle
This is not always true. Take, for example, f(z) = sinz and then this is clearly not true.
Here's what I did:
Where the first equality is correct since the integrand is an even function (verify it).
Now let and you get, using , that:
And now I see that I have an algebra mistake too I missed a .
Now you change variables and solve the integral over like you did before.
Not exactly. To use Rouche's theorem you have to have that for all such that2. About the Rouche stunt I was pulling. Is this a well mannered way of showing possible zeros of the polynom?
;
Now using - to show zeros within unit circle
To show this:
(for |z|=1)
And so
Thanks again!
I know that cosine is an even function... How would I go about verifying that more than just mentioning it when I rewrite the boundaries? I missed the angle = double angle, but I used it to simplify the integral. Well, I changed the boundaries but I missed out on rewriting the angles. Your rewriting really help me, the count feel down in the slot and made a clear cling.
Now I feel unsure of the Rouche thing... For instance, why do you show that why bigger or equally to 2 in particular?
Once again, thank you!
Just one thing. Your way of showing that cosine is even, cos -z = cos z, seems to be the same thing as saying, Cosine is an even function. If I really had to show that it's even maybe I could use the Taylor series for cosine to show that it's even, or Maclaurine since it's obviously centered at zero in this case. That gives me the terms of the series and they are clearly even.
I redid the integral and tried to break it down, so errors would be evident. Please point out any errors!
= = = = = even integer = = linearity =
Now I'm using and Eulers formulas, cosine. This gives: = = =
Finding the divisor's roots.
where is in the unit circle.
Residue for a simple pole
=
Now I use
=
Result