Complex integration is giving the wrong answer by a factor of two

[LCSphysicist]

Homework Statement

.

Relevant Equations

.

$$\int_{0}^{2\pi } (1+2cost)^{n}cos(nt) dt$$

$$e^{it} = z, izdt = dz$$

$$\oint (1+e^{it}+e^{-it})^{n}\frac{e^{nit}+e^{-nit}}{2} \frac{dz}{iz} = \oint (1+z+z^{-1})^{n}\frac{z^{n}+z^{-n}}{2} \frac{dz}{iz}$$

$$\oint (z+z^{2}+1)^{n}\frac{z^{2n}+1}{z^{2n+1}} \frac{dz}{2i} = \pi Res = \pi \frac{d^{2n}}{(2n)! dz^{2n}}((z+z^{2}+1)^{n}(z^{2n}+1))$$

$$= \pi \sum \begin{pmatrix}
2n\\ \beta
\end{pmatrix} \frac{d^{\beta} {(z^{2n}+1)}}{dz^{\beta}(2n)!} \frac{d^{2n-\beta}}{dz^{2n-\beta}}(z+z^{2}+1)^{n}$$

Since the pole is at z=0, all terms above vanish apart from $\beta = 2n$, in this case:

$$= \pi \frac{d^{2n} {(z^{2n}+1)}}{(2n)!dz^{2n}} \frac{d^{0}}{dz^{0}}(z+z^{2}+1)^{n} $$
$$= \pi \frac{(2n)!}{(2n)!} (1) = \pi $$

The answer provided by the book is $2 \pi$

I can't really find the error i supposedly did :/

 

[Charles Link]

Take a look at the integral and integrand to figure out the residue term without the cumbersome derivatives: You are just interested in $z^{2n}$, (in the numerator), and you have $z^{2n}+1$, multiplied by $(z^2+z+1)^n$. For the $(z^2+z+1)^n$, you need the $z^{2n}$ and the $1$, (the end terms on either side in the expansion). None of the other terms in the expansion will give a $z^{2n}$ term when multiplied by $z^{2n}+1$. When multiplied through and summed, it is clear that you get $2 z^{2n}$ as the term in the numerator that will give you the residue term, i.e. $\frac{b}{z}$, where $b=2$.

 

[LCSphysicist]

Take a look at the integral and integrand to figure out the residue term without the cumbersome derivatives: You are just interested in $z^{2n}$, (in the numerator), and you have $z^{2n}+1$, multiplied by $(z^2+z+1)^n$. For the $(z^2+z+1)^n$, you need the $z^{2n}$ and the $1$, (the end terms on either side in the expansion). None of the other terms in the expansion will give a $z^{2n}$ term when multiplied by $z^{2n}+1$. When multiplied through and summed, it is clear that you get $2 z^{2n}$ as the term in the numerator that will give you the residue term, i.e. $\frac{b}{z}$, where $b=2$.

I can see your approach. But i really would like to know what error did i comitted :c i still did't find it

 

[pasmith]

You have omitted
$$\frac{1}{(2n)!}\binom{2n}{0} \frac{d^{0}}{dz^0}(z^{2n}+1)\frac{d^{2n}}{dz^{2n}} ((z^2 + z + 1)^n) = \frac{1}{(2n)!}(z^{2n} + 1)(2n)!.$$

 

[Charles Link]

I can see your approach. But i really would like to know what error did i comitted :c i still did't find it

I think you also get a term from $\beta=0$ in the derivative summation. (The product rule with multiple derivatives is clumsy, but I think that's the term you missed).

Edit: and I see @pasmith agrees with me=he posted just as I was posting.

 

[hutchphd]

Homework Statement:: .
Relevant Equations:: .

$$\int_{0}^{2\pi } (1+2cost)^{n}cos(nt) dt$$

$$e^{it} = z, izdt = dz$$

$$\oint (1+e^{it}+e^{-it})^{n}\frac{e^{nit}+e^{-nit}}{2} \frac{dz}{iz} = \oint (1+z+z^{-1})^{n}\frac{z^{n}+z^{-n}}{2} \frac{dz}{iz}$$

$$\oint (z+z^{2}+1)^{n}\frac{z^{2n}+1}{z^{2n+1}} \frac{dz}{2i} = \pi Res = \pi \frac{d^{2n}}{(2n)! dz^{2n}}((z+z^{2}+1)^{n}(z^{2n}+1))$$

$$= \pi \sum \begin{pmatrix}
2n\\ \beta
\end{pmatrix} \frac{d^{\beta} {(z^{2n}+1)}}{dz^{\beta}(2n)!} \frac{d^{2n-\beta}}{dz^{2n-\beta}}(z+z^{2}+1)^{n}$$

Since the pole is at z=0, all terms above vanish apart from $\beta = 2n$, in this case:

$$= \pi \frac{d^{2n} {(z^{2n}+1)}}{(2n)!dz^{2n}} \frac{d^{0}}{dz^{0}}(z+z^{2}+1)^{n} $$
$$= \pi \frac{(2n)!}{(2n)!} (1) = \pi $$

The answer provided by the book is $2 \pi$

I can't really find the error i supposedly did :/

Why is it not $2\pi i Res$ in
$$\oint (z+z^{2}+1)^{n}\frac{z^{2n}+1}{z^{2n+1}} \frac{dz}{2i} = \pi Res = \pi \frac{d^{2n}}{(2n)! dz^{2n}}((z+z^{2}+1)^{n}(z^{2n}+1))$$

The residue at z=0 is $$\frac 1 i$$

 

[vela]

The OP simplified using the factor of $2i$ in the denominator of the integrand. It made me look twice too.

 

[hutchphd]

I still don't see it..
$$ \oint (1+z+z^{-1})^{n}\frac{z^{n}+z^{-n}}{2} \frac{dz}{iz}=2\pi i~Res \left[(1+z+z^{-1})^{n}\frac{z^{n}+z^{-n}}{2} (-i)\right]_{z=0} $$ $$=2\pi$$

 

[vela]

The OP is finding the residue of the integrand without the factor of $2i$.

 

[hutchphd]

Apologies I don't know what you mean. The value of the integral is $2\pi$ and this is how you find it. He gets the wrong value...

 

[Charles Link]

Apologies I don't know what you mean. The value of the integral is $2\pi$ and this is how you find it. He gets the wrong value...

The OP was somewhat sloppy in changing the function of interest by dropping a factor of $1/2 i$, ( using it to cancel the $2i$ of $2 \pi i$). In any case, he uses a derivative formula to isolate the term of interest in the series, (the coefficient $b$ of the $\frac{b}{z }$ term), and when he did, he miscalculated the derivative.

 

[hutchphd]

Wow that is a strange manipulation. Got it. Thanks.
I know it was mentioned by @Charles Link but the OP needs to understand his error was occasioned by doing the problem in a most ill-advised manner. That needs to be the take-away IMHO. The Residue Theorem is troo voodoo.

 

[Charles Link]

In response to the previous post by @hutchphd , I do see a way to work this problem without using the residue theorem, where you can use $\int\limits_{0}^{2 \pi} e^{imt} \, dt=0$ for all integers $m$, except $m=0$.

 

[hutchphd]

Don't get me wrong I love Cauchy, but I always feel that I am summoning the occult. Sticking pins in the complex plane as it were. It works so well. I suppose when I really feel at home with Euler's Equation the feeling will abate. But anyway no derivatives are necessary to get to the residue term in this case.