Two different results with residue calculus for the integral $int_{0}^{2pi}frac{cos(3x)}{5-4cos(x)}dx$. What...












3












$begingroup$


So I have to evaluate the following integral :$$int_{0}^{2pi}frac{cos(3x)}{5-4cos(x)}dx$$ So I solve as usual with the residue theorem and by using $cos(3x)= Re(e^{3ix})$, $cos(x)=frac{e^{ix}+e^{-ix}}{2}$ and $z=e^{ix}$, but then I have some troubles evaluating the residues. I get as poles 2 and $frac{1}{2}$ where only $frac{1}{2}$ is in our domain (circle of radius 1).



When I calculate the residue using $frac{g(z)}{f'(z)}$, I get $frac{frac{1}{2^3}}{5-2}=frac{1}{24}$ (I put the i outside, that's not the problem here) but when I use $lim_{z to frac{1}{2}} (z-frac{1}{2})frac{z^3}{-(z-2)(z-frac{1}{2})}$, I get $frac{1}{12}$ so that when I multiply by $2 pi $ (again the i is not the problem here), I get $frac{pi}{6}$. WolframAlpha gets $frac{pi}{12}=2pi *frac{1}{24}$ so my second way of finding the residue ($frac{1}{12}$-->$frac{pi}{6}$) seems to be wrong. My question is, what did I do wrong ?






integration complex-analysis residue-calculus







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  • 1




    $begingroup$
    Please write down the expression you got for $f$. What is $g$ in your answer? If you don't give details of your answer we cannot point out mistakes.
    $endgroup$
    – Kavi Rama Murthy
    Dec 11 '18 at 23:57












  • $begingroup$
    So $f$ is $5z-2z^2-2$ and $g$ is $z^3$ ($=e^{3ix}$ because we let $z=e^{ix}$)
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:01










  • $begingroup$
    @EeveeTrainer Maybe that would work, but it's much simpler if we make the substitutions $cos(x)=frac{e^{ix}+e^{-ix}}{2}$, $cos(3x)=Re(e^{3ix})$ and $z=e^{ix}$ and then find our poles so that we can use the residue theorem in an easier way.
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:09


















3












$begingroup$


So I have to evaluate the following integral :$$int_{0}^{2pi}frac{cos(3x)}{5-4cos(x)}dx$$ So I solve as usual with the residue theorem and by using $cos(3x)= Re(e^{3ix})$, $cos(x)=frac{e^{ix}+e^{-ix}}{2}$ and $z=e^{ix}$, but then I have some troubles evaluating the residues. I get as poles 2 and $frac{1}{2}$ where only $frac{1}{2}$ is in our domain (circle of radius 1).



When I calculate the residue using $frac{g(z)}{f'(z)}$, I get $frac{frac{1}{2^3}}{5-2}=frac{1}{24}$ (I put the i outside, that's not the problem here) but when I use $lim_{z to frac{1}{2}} (z-frac{1}{2})frac{z^3}{-(z-2)(z-frac{1}{2})}$, I get $frac{1}{12}$ so that when I multiply by $2 pi $ (again the i is not the problem here), I get $frac{pi}{6}$. WolframAlpha gets $frac{pi}{12}=2pi *frac{1}{24}$ so my second way of finding the residue ($frac{1}{12}$-->$frac{pi}{6}$) seems to be wrong. My question is, what did I do wrong ?






integration complex-analysis residue-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Please write down the expression you got for $f$. What is $g$ in your answer? If you don't give details of your answer we cannot point out mistakes.
    $endgroup$
    – Kavi Rama Murthy
    Dec 11 '18 at 23:57












  • $begingroup$
    So $f$ is $5z-2z^2-2$ and $g$ is $z^3$ ($=e^{3ix}$ because we let $z=e^{ix}$)
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:01










  • $begingroup$
    @EeveeTrainer Maybe that would work, but it's much simpler if we make the substitutions $cos(x)=frac{e^{ix}+e^{-ix}}{2}$, $cos(3x)=Re(e^{3ix})$ and $z=e^{ix}$ and then find our poles so that we can use the residue theorem in an easier way.
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:09
















3












3








3





$begingroup$


So I have to evaluate the following integral :$$int_{0}^{2pi}frac{cos(3x)}{5-4cos(x)}dx$$ So I solve as usual with the residue theorem and by using $cos(3x)= Re(e^{3ix})$, $cos(x)=frac{e^{ix}+e^{-ix}}{2}$ and $z=e^{ix}$, but then I have some troubles evaluating the residues. I get as poles 2 and $frac{1}{2}$ where only $frac{1}{2}$ is in our domain (circle of radius 1).



When I calculate the residue using $frac{g(z)}{f'(z)}$, I get $frac{frac{1}{2^3}}{5-2}=frac{1}{24}$ (I put the i outside, that's not the problem here) but when I use $lim_{z to frac{1}{2}} (z-frac{1}{2})frac{z^3}{-(z-2)(z-frac{1}{2})}$, I get $frac{1}{12}$ so that when I multiply by $2 pi $ (again the i is not the problem here), I get $frac{pi}{6}$. WolframAlpha gets $frac{pi}{12}=2pi *frac{1}{24}$ so my second way of finding the residue ($frac{1}{12}$-->$frac{pi}{6}$) seems to be wrong. My question is, what did I do wrong ?






integration complex-analysis residue-calculus







share|cite|improve this question











$endgroup$




So I have to evaluate the following integral :$$int_{0}^{2pi}frac{cos(3x)}{5-4cos(x)}dx$$ So I solve as usual with the residue theorem and by using $cos(3x)= Re(e^{3ix})$, $cos(x)=frac{e^{ix}+e^{-ix}}{2}$ and $z=e^{ix}$, but then I have some troubles evaluating the residues. I get as poles 2 and $frac{1}{2}$ where only $frac{1}{2}$ is in our domain (circle of radius 1).



When I calculate the residue using $frac{g(z)}{f'(z)}$, I get $frac{frac{1}{2^3}}{5-2}=frac{1}{24}$ (I put the i outside, that's not the problem here) but when I use $lim_{z to frac{1}{2}} (z-frac{1}{2})frac{z^3}{-(z-2)(z-frac{1}{2})}$, I get $frac{1}{12}$ so that when I multiply by $2 pi $ (again the i is not the problem here), I get $frac{pi}{6}$. WolframAlpha gets $frac{pi}{12}=2pi *frac{1}{24}$ so my second way of finding the residue ($frac{1}{12}$-->$frac{pi}{6}$) seems to be wrong. My question is, what did I do wrong ?






integration complex-analysis residue-calculus




integration complex-analysis residue-calculus






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share|cite|improve this question













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edited Jan 7 at 19:30







Poujh

















asked Dec 11 '18 at 23:50









PoujhPoujh

569516




569516








  • 1




    $begingroup$
    Please write down the expression you got for $f$. What is $g$ in your answer? If you don't give details of your answer we cannot point out mistakes.
    $endgroup$
    – Kavi Rama Murthy
    Dec 11 '18 at 23:57












  • $begingroup$
    So $f$ is $5z-2z^2-2$ and $g$ is $z^3$ ($=e^{3ix}$ because we let $z=e^{ix}$)
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:01










  • $begingroup$
    @EeveeTrainer Maybe that would work, but it's much simpler if we make the substitutions $cos(x)=frac{e^{ix}+e^{-ix}}{2}$, $cos(3x)=Re(e^{3ix})$ and $z=e^{ix}$ and then find our poles so that we can use the residue theorem in an easier way.
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:09
















  • 1




    $begingroup$
    Please write down the expression you got for $f$. What is $g$ in your answer? If you don't give details of your answer we cannot point out mistakes.
    $endgroup$
    – Kavi Rama Murthy
    Dec 11 '18 at 23:57












  • $begingroup$
    So $f$ is $5z-2z^2-2$ and $g$ is $z^3$ ($=e^{3ix}$ because we let $z=e^{ix}$)
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:01










  • $begingroup$
    @EeveeTrainer Maybe that would work, but it's much simpler if we make the substitutions $cos(x)=frac{e^{ix}+e^{-ix}}{2}$, $cos(3x)=Re(e^{3ix})$ and $z=e^{ix}$ and then find our poles so that we can use the residue theorem in an easier way.
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:09










1




1




$begingroup$
Please write down the expression you got for $f$. What is $g$ in your answer? If you don't give details of your answer we cannot point out mistakes.
$endgroup$
– Kavi Rama Murthy
Dec 11 '18 at 23:57






$begingroup$
Please write down the expression you got for $f$. What is $g$ in your answer? If you don't give details of your answer we cannot point out mistakes.
$endgroup$
– Kavi Rama Murthy
Dec 11 '18 at 23:57














$begingroup$
So $f$ is $5z-2z^2-2$ and $g$ is $z^3$ ($=e^{3ix}$ because we let $z=e^{ix}$)
$endgroup$
– Poujh
Dec 12 '18 at 0:01




$begingroup$
So $f$ is $5z-2z^2-2$ and $g$ is $z^3$ ($=e^{3ix}$ because we let $z=e^{ix}$)
$endgroup$
– Poujh
Dec 12 '18 at 0:01












$begingroup$
@EeveeTrainer Maybe that would work, but it's much simpler if we make the substitutions $cos(x)=frac{e^{ix}+e^{-ix}}{2}$, $cos(3x)=Re(e^{3ix})$ and $z=e^{ix}$ and then find our poles so that we can use the residue theorem in an easier way.
$endgroup$
– Poujh
Dec 12 '18 at 0:09






$begingroup$
@EeveeTrainer Maybe that would work, but it's much simpler if we make the substitutions $cos(x)=frac{e^{ix}+e^{-ix}}{2}$, $cos(3x)=Re(e^{3ix})$ and $z=e^{ix}$ and then find our poles so that we can use the residue theorem in an easier way.
$endgroup$
– Poujh
Dec 12 '18 at 0:09












2 Answers
2






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oldest

votes


















1












$begingroup$

Well, you arrived at: $$I=int_{|z|=1}frac{z^3}{5-4left(frac{z+ z^{-1}}{2}right)}frac{dz}{iz}$$
With some algebra the integral reduces to : $$int_{|z|=1}frac{z^3}{5z-2(z^2+1)}frac{dz}{i}=frac{1}{i}int_{|z|=1}frac{z^3}{-2z^2+5z-2}dz=$$
$$=frac{1}{i}int_{|z|=1}frac{z^3}{-(z-2)(2z-1)}dz=frac{1}{2i}int_{|z|=1}frac{z^3}{-(z-2)(z-tfrac12)}dz$$
Now here you can see that when you factored that denominator you forgot that additional $frac12$ and everything should be fine now.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for your help, but if I use the wollowing way, what am I doing wrong ? $frac{-5 pm sqrt{25-16}}{-4}$ = $frac{5 pm sqrt{25-16}}{4}$ = $frac{1}{2}$ and $2$. In this case, where did the extra factor of 2 go ?
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:17








  • 1




    $begingroup$
    If you're curious, I used the exact method (second one) here: math.stackexchange.com/q/2820433/515527
    $endgroup$
    – Zacky
    Dec 12 '18 at 0:17






  • 1




    $begingroup$
    I think I understand what you did altough I have not used that formula recently. You found the roots, then used this: For a polynomial $ax^2+bx+c$ if we have two roots it can be rewritten as: $a(x-x_1)(x-x_2)$, right? Well you forgot to multiply by $a$ then.
    $endgroup$
    – Zacky
    Dec 12 '18 at 0:22








  • 1




    $begingroup$
    In general I preffer to have the coefficient of $z^2$ to be $1$. Here is an alternative to find the factorization: $$-2z^2+5z-2=-frac{1}{2}((2z)^2-5(2z)+4)$$ Now I denote $(2z)=y$ $$-frac{1}{2}(y^2-5y+4)=-frac12 (y^2-y-4y+4)=-frac12 y(y-1)-4(y-1)=-frac12(y-4)(y-1)$$ Now returing to $z$ we have: $$-frac12(2z-4)(2z-1)=-2(2z-1)(z-frac12)$$
    $endgroup$
    – Zacky
    Dec 12 '18 at 0:40








  • 1




    $begingroup$
    Okay, thanks a lot ! I never thought it could be something as "basic" as that haha
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:45



















1












$begingroup$

Starting with a geometric series:
$$frac{1}{2-e^{ix}} = sum_{ngeq 0}frac{1}{2^{n+1}}e^{nix}$$
we may consider the product between this series and its conjugate:



$$begin{eqnarray*}frac{1}{5-4cos x}&=&sum_{m,ngeq 0}frac{1}{2^{m+n+2}}e^{(n-m)ix}\&=&frac{1}{3}+sum_{n>mgeq 0}frac{1}{2^{m+n+1}}cos((n-m)x)\&=&frac{1}{3}+sum_{dgeq 1}cos(dx)sum_{mgeq 0}frac{1}{2^{2m+d+1}}\&=&frac{1}{3}+sum_{dgeq 1}frac{cos(dx)}{2^d}sum_{mgeq 0}frac{1}{2^{2m+1}}\&=&frac{1}{3}+frac{2}{3}sum_{dgeq 1}frac{cos(dx)}{2^d}end{eqnarray*} $$
to get the whole Fourier cosine series of $frac{1}{5-4cos x}$. Then by the orthogonality relations
$$ int_{0}^{2pi}frac{cos(3x)}{5-4cos(x)},dx = frac{2pi}{3}cdotfrac{1}{2^3}=color{red}{frac{pi}{12}}.$$
This approach might look as an overkill, but it is pretty useful in understanding the asymptotic behaviour of the Fourier coefficients of $frac{1}{left(Rpmcosthetaright)^alpha}$ when $alphainfrac{1}{2}+mathbb{N}$.






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  • $begingroup$
    Thanks, I just have one question. What exactly did you do in the step $int_{0}^{2pi}frac{cos(3x)}{5-4cos(x)}$=$frac{2pi}{3}frac{1}{2^3}$ ?
    $endgroup$
    – Poujh
    Dec 12 '18 at 8:08












  • $begingroup$
    Ah okay, it's the dot product of $cos(3x)$ and $frac{1}{5-4cos(x)}$ which you wrote as a serie just before and then you use, as said, the orthogonality. And then you integrate and you get $2pi$ times $frac{1}{2}((cos(3x-3x)+cos(3x+3x))$ from the cos(a)cos(b) (where a=b) trig identity, and cos(0) gives 1 while cos(6x) integrated from $0$ to $2pi$ gives $0$. And also, $frac{1}{3}*cos(3x)$ from $0$ to $2pi$ is also $0$. Makes sense, thank you !
    $endgroup$
    – Poujh
    Dec 12 '18 at 8:21












  • $begingroup$
    It seems like that you hate contour integration a lot, regarding some of your recent answers... :-)
    $endgroup$
    – Szeto
    Dec 12 '18 at 10:30










  • $begingroup$
    @Szeto: it is not that I hate it, but many times Fourier series or the Laplace transform allow to avoid the "hunt" for the correct integration contour, since it is somewhat already encoded in their application, and I find this pretty practical.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 12:48











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2 Answers
2






active

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Well, you arrived at: $$I=int_{|z|=1}frac{z^3}{5-4left(frac{z+ z^{-1}}{2}right)}frac{dz}{iz}$$
With some algebra the integral reduces to : $$int_{|z|=1}frac{z^3}{5z-2(z^2+1)}frac{dz}{i}=frac{1}{i}int_{|z|=1}frac{z^3}{-2z^2+5z-2}dz=$$
$$=frac{1}{i}int_{|z|=1}frac{z^3}{-(z-2)(2z-1)}dz=frac{1}{2i}int_{|z|=1}frac{z^3}{-(z-2)(z-tfrac12)}dz$$
Now here you can see that when you factored that denominator you forgot that additional $frac12$ and everything should be fine now.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for your help, but if I use the wollowing way, what am I doing wrong ? $frac{-5 pm sqrt{25-16}}{-4}$ = $frac{5 pm sqrt{25-16}}{4}$ = $frac{1}{2}$ and $2$. In this case, where did the extra factor of 2 go ?
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:17








  • 1




    $begingroup$
    If you're curious, I used the exact method (second one) here: math.stackexchange.com/q/2820433/515527
    $endgroup$
    – Zacky
    Dec 12 '18 at 0:17






  • 1




    $begingroup$
    I think I understand what you did altough I have not used that formula recently. You found the roots, then used this: For a polynomial $ax^2+bx+c$ if we have two roots it can be rewritten as: $a(x-x_1)(x-x_2)$, right? Well you forgot to multiply by $a$ then.
    $endgroup$
    – Zacky
    Dec 12 '18 at 0:22








  • 1




    $begingroup$
    In general I preffer to have the coefficient of $z^2$ to be $1$. Here is an alternative to find the factorization: $$-2z^2+5z-2=-frac{1}{2}((2z)^2-5(2z)+4)$$ Now I denote $(2z)=y$ $$-frac{1}{2}(y^2-5y+4)=-frac12 (y^2-y-4y+4)=-frac12 y(y-1)-4(y-1)=-frac12(y-4)(y-1)$$ Now returing to $z$ we have: $$-frac12(2z-4)(2z-1)=-2(2z-1)(z-frac12)$$
    $endgroup$
    – Zacky
    Dec 12 '18 at 0:40








  • 1




    $begingroup$
    Okay, thanks a lot ! I never thought it could be something as "basic" as that haha
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:45
















1












$begingroup$

Well, you arrived at: $$I=int_{|z|=1}frac{z^3}{5-4left(frac{z+ z^{-1}}{2}right)}frac{dz}{iz}$$
With some algebra the integral reduces to : $$int_{|z|=1}frac{z^3}{5z-2(z^2+1)}frac{dz}{i}=frac{1}{i}int_{|z|=1}frac{z^3}{-2z^2+5z-2}dz=$$
$$=frac{1}{i}int_{|z|=1}frac{z^3}{-(z-2)(2z-1)}dz=frac{1}{2i}int_{|z|=1}frac{z^3}{-(z-2)(z-tfrac12)}dz$$
Now here you can see that when you factored that denominator you forgot that additional $frac12$ and everything should be fine now.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for your help, but if I use the wollowing way, what am I doing wrong ? $frac{-5 pm sqrt{25-16}}{-4}$ = $frac{5 pm sqrt{25-16}}{4}$ = $frac{1}{2}$ and $2$. In this case, where did the extra factor of 2 go ?
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:17








  • 1




    $begingroup$
    If you're curious, I used the exact method (second one) here: math.stackexchange.com/q/2820433/515527
    $endgroup$
    – Zacky
    Dec 12 '18 at 0:17






  • 1




    $begingroup$
    I think I understand what you did altough I have not used that formula recently. You found the roots, then used this: For a polynomial $ax^2+bx+c$ if we have two roots it can be rewritten as: $a(x-x_1)(x-x_2)$, right? Well you forgot to multiply by $a$ then.
    $endgroup$
    – Zacky
    Dec 12 '18 at 0:22








  • 1




    $begingroup$
    In general I preffer to have the coefficient of $z^2$ to be $1$. Here is an alternative to find the factorization: $$-2z^2+5z-2=-frac{1}{2}((2z)^2-5(2z)+4)$$ Now I denote $(2z)=y$ $$-frac{1}{2}(y^2-5y+4)=-frac12 (y^2-y-4y+4)=-frac12 y(y-1)-4(y-1)=-frac12(y-4)(y-1)$$ Now returing to $z$ we have: $$-frac12(2z-4)(2z-1)=-2(2z-1)(z-frac12)$$
    $endgroup$
    – Zacky
    Dec 12 '18 at 0:40








  • 1




    $begingroup$
    Okay, thanks a lot ! I never thought it could be something as "basic" as that haha
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:45














1












1








1





$begingroup$

Well, you arrived at: $$I=int_{|z|=1}frac{z^3}{5-4left(frac{z+ z^{-1}}{2}right)}frac{dz}{iz}$$
With some algebra the integral reduces to : $$int_{|z|=1}frac{z^3}{5z-2(z^2+1)}frac{dz}{i}=frac{1}{i}int_{|z|=1}frac{z^3}{-2z^2+5z-2}dz=$$
$$=frac{1}{i}int_{|z|=1}frac{z^3}{-(z-2)(2z-1)}dz=frac{1}{2i}int_{|z|=1}frac{z^3}{-(z-2)(z-tfrac12)}dz$$
Now here you can see that when you factored that denominator you forgot that additional $frac12$ and everything should be fine now.






share|cite|improve this answer











$endgroup$



Well, you arrived at: $$I=int_{|z|=1}frac{z^3}{5-4left(frac{z+ z^{-1}}{2}right)}frac{dz}{iz}$$
With some algebra the integral reduces to : $$int_{|z|=1}frac{z^3}{5z-2(z^2+1)}frac{dz}{i}=frac{1}{i}int_{|z|=1}frac{z^3}{-2z^2+5z-2}dz=$$
$$=frac{1}{i}int_{|z|=1}frac{z^3}{-(z-2)(2z-1)}dz=frac{1}{2i}int_{|z|=1}frac{z^3}{-(z-2)(z-tfrac12)}dz$$
Now here you can see that when you factored that denominator you forgot that additional $frac12$ and everything should be fine now.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 12 '18 at 0:12

























answered Dec 12 '18 at 0:07









ZackyZacky

5,3111754




5,3111754












  • $begingroup$
    Thanks for your help, but if I use the wollowing way, what am I doing wrong ? $frac{-5 pm sqrt{25-16}}{-4}$ = $frac{5 pm sqrt{25-16}}{4}$ = $frac{1}{2}$ and $2$. In this case, where did the extra factor of 2 go ?
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:17








  • 1




    $begingroup$
    If you're curious, I used the exact method (second one) here: math.stackexchange.com/q/2820433/515527
    $endgroup$
    – Zacky
    Dec 12 '18 at 0:17






  • 1




    $begingroup$
    I think I understand what you did altough I have not used that formula recently. You found the roots, then used this: For a polynomial $ax^2+bx+c$ if we have two roots it can be rewritten as: $a(x-x_1)(x-x_2)$, right? Well you forgot to multiply by $a$ then.
    $endgroup$
    – Zacky
    Dec 12 '18 at 0:22








  • 1




    $begingroup$
    In general I preffer to have the coefficient of $z^2$ to be $1$. Here is an alternative to find the factorization: $$-2z^2+5z-2=-frac{1}{2}((2z)^2-5(2z)+4)$$ Now I denote $(2z)=y$ $$-frac{1}{2}(y^2-5y+4)=-frac12 (y^2-y-4y+4)=-frac12 y(y-1)-4(y-1)=-frac12(y-4)(y-1)$$ Now returing to $z$ we have: $$-frac12(2z-4)(2z-1)=-2(2z-1)(z-frac12)$$
    $endgroup$
    – Zacky
    Dec 12 '18 at 0:40








  • 1




    $begingroup$
    Okay, thanks a lot ! I never thought it could be something as "basic" as that haha
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:45


















  • $begingroup$
    Thanks for your help, but if I use the wollowing way, what am I doing wrong ? $frac{-5 pm sqrt{25-16}}{-4}$ = $frac{5 pm sqrt{25-16}}{4}$ = $frac{1}{2}$ and $2$. In this case, where did the extra factor of 2 go ?
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:17








  • 1




    $begingroup$
    If you're curious, I used the exact method (second one) here: math.stackexchange.com/q/2820433/515527
    $endgroup$
    – Zacky
    Dec 12 '18 at 0:17






  • 1




    $begingroup$
    I think I understand what you did altough I have not used that formula recently. You found the roots, then used this: For a polynomial $ax^2+bx+c$ if we have two roots it can be rewritten as: $a(x-x_1)(x-x_2)$, right? Well you forgot to multiply by $a$ then.
    $endgroup$
    – Zacky
    Dec 12 '18 at 0:22








  • 1




    $begingroup$
    In general I preffer to have the coefficient of $z^2$ to be $1$. Here is an alternative to find the factorization: $$-2z^2+5z-2=-frac{1}{2}((2z)^2-5(2z)+4)$$ Now I denote $(2z)=y$ $$-frac{1}{2}(y^2-5y+4)=-frac12 (y^2-y-4y+4)=-frac12 y(y-1)-4(y-1)=-frac12(y-4)(y-1)$$ Now returing to $z$ we have: $$-frac12(2z-4)(2z-1)=-2(2z-1)(z-frac12)$$
    $endgroup$
    – Zacky
    Dec 12 '18 at 0:40








  • 1




    $begingroup$
    Okay, thanks a lot ! I never thought it could be something as "basic" as that haha
    $endgroup$
    – Poujh
    Dec 12 '18 at 0:45
















$begingroup$
Thanks for your help, but if I use the wollowing way, what am I doing wrong ? $frac{-5 pm sqrt{25-16}}{-4}$ = $frac{5 pm sqrt{25-16}}{4}$ = $frac{1}{2}$ and $2$. In this case, where did the extra factor of 2 go ?
$endgroup$
– Poujh
Dec 12 '18 at 0:17






$begingroup$
Thanks for your help, but if I use the wollowing way, what am I doing wrong ? $frac{-5 pm sqrt{25-16}}{-4}$ = $frac{5 pm sqrt{25-16}}{4}$ = $frac{1}{2}$ and $2$. In this case, where did the extra factor of 2 go ?
$endgroup$
– Poujh
Dec 12 '18 at 0:17






1




1




$begingroup$
If you're curious, I used the exact method (second one) here: math.stackexchange.com/q/2820433/515527
$endgroup$
– Zacky
Dec 12 '18 at 0:17




$begingroup$
If you're curious, I used the exact method (second one) here: math.stackexchange.com/q/2820433/515527
$endgroup$
– Zacky
Dec 12 '18 at 0:17




1




1




$begingroup$
I think I understand what you did altough I have not used that formula recently. You found the roots, then used this: For a polynomial $ax^2+bx+c$ if we have two roots it can be rewritten as: $a(x-x_1)(x-x_2)$, right? Well you forgot to multiply by $a$ then.
$endgroup$
– Zacky
Dec 12 '18 at 0:22






$begingroup$
I think I understand what you did altough I have not used that formula recently. You found the roots, then used this: For a polynomial $ax^2+bx+c$ if we have two roots it can be rewritten as: $a(x-x_1)(x-x_2)$, right? Well you forgot to multiply by $a$ then.
$endgroup$
– Zacky
Dec 12 '18 at 0:22






1




1




$begingroup$
In general I preffer to have the coefficient of $z^2$ to be $1$. Here is an alternative to find the factorization: $$-2z^2+5z-2=-frac{1}{2}((2z)^2-5(2z)+4)$$ Now I denote $(2z)=y$ $$-frac{1}{2}(y^2-5y+4)=-frac12 (y^2-y-4y+4)=-frac12 y(y-1)-4(y-1)=-frac12(y-4)(y-1)$$ Now returing to $z$ we have: $$-frac12(2z-4)(2z-1)=-2(2z-1)(z-frac12)$$
$endgroup$
– Zacky
Dec 12 '18 at 0:40






$begingroup$
In general I preffer to have the coefficient of $z^2$ to be $1$. Here is an alternative to find the factorization: $$-2z^2+5z-2=-frac{1}{2}((2z)^2-5(2z)+4)$$ Now I denote $(2z)=y$ $$-frac{1}{2}(y^2-5y+4)=-frac12 (y^2-y-4y+4)=-frac12 y(y-1)-4(y-1)=-frac12(y-4)(y-1)$$ Now returing to $z$ we have: $$-frac12(2z-4)(2z-1)=-2(2z-1)(z-frac12)$$
$endgroup$
– Zacky
Dec 12 '18 at 0:40






1




1




$begingroup$
Okay, thanks a lot ! I never thought it could be something as "basic" as that haha
$endgroup$
– Poujh
Dec 12 '18 at 0:45




$begingroup$
Okay, thanks a lot ! I never thought it could be something as "basic" as that haha
$endgroup$
– Poujh
Dec 12 '18 at 0:45











1












$begingroup$

Starting with a geometric series:
$$frac{1}{2-e^{ix}} = sum_{ngeq 0}frac{1}{2^{n+1}}e^{nix}$$
we may consider the product between this series and its conjugate:



$$begin{eqnarray*}frac{1}{5-4cos x}&=&sum_{m,ngeq 0}frac{1}{2^{m+n+2}}e^{(n-m)ix}\&=&frac{1}{3}+sum_{n>mgeq 0}frac{1}{2^{m+n+1}}cos((n-m)x)\&=&frac{1}{3}+sum_{dgeq 1}cos(dx)sum_{mgeq 0}frac{1}{2^{2m+d+1}}\&=&frac{1}{3}+sum_{dgeq 1}frac{cos(dx)}{2^d}sum_{mgeq 0}frac{1}{2^{2m+1}}\&=&frac{1}{3}+frac{2}{3}sum_{dgeq 1}frac{cos(dx)}{2^d}end{eqnarray*} $$
to get the whole Fourier cosine series of $frac{1}{5-4cos x}$. Then by the orthogonality relations
$$ int_{0}^{2pi}frac{cos(3x)}{5-4cos(x)},dx = frac{2pi}{3}cdotfrac{1}{2^3}=color{red}{frac{pi}{12}}.$$
This approach might look as an overkill, but it is pretty useful in understanding the asymptotic behaviour of the Fourier coefficients of $frac{1}{left(Rpmcosthetaright)^alpha}$ when $alphainfrac{1}{2}+mathbb{N}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, I just have one question. What exactly did you do in the step $int_{0}^{2pi}frac{cos(3x)}{5-4cos(x)}$=$frac{2pi}{3}frac{1}{2^3}$ ?
    $endgroup$
    – Poujh
    Dec 12 '18 at 8:08












  • $begingroup$
    Ah okay, it's the dot product of $cos(3x)$ and $frac{1}{5-4cos(x)}$ which you wrote as a serie just before and then you use, as said, the orthogonality. And then you integrate and you get $2pi$ times $frac{1}{2}((cos(3x-3x)+cos(3x+3x))$ from the cos(a)cos(b) (where a=b) trig identity, and cos(0) gives 1 while cos(6x) integrated from $0$ to $2pi$ gives $0$. And also, $frac{1}{3}*cos(3x)$ from $0$ to $2pi$ is also $0$. Makes sense, thank you !
    $endgroup$
    – Poujh
    Dec 12 '18 at 8:21












  • $begingroup$
    It seems like that you hate contour integration a lot, regarding some of your recent answers... :-)
    $endgroup$
    – Szeto
    Dec 12 '18 at 10:30










  • $begingroup$
    @Szeto: it is not that I hate it, but many times Fourier series or the Laplace transform allow to avoid the "hunt" for the correct integration contour, since it is somewhat already encoded in their application, and I find this pretty practical.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 12:48
















1












$begingroup$

Starting with a geometric series:
$$frac{1}{2-e^{ix}} = sum_{ngeq 0}frac{1}{2^{n+1}}e^{nix}$$
we may consider the product between this series and its conjugate:



$$begin{eqnarray*}frac{1}{5-4cos x}&=&sum_{m,ngeq 0}frac{1}{2^{m+n+2}}e^{(n-m)ix}\&=&frac{1}{3}+sum_{n>mgeq 0}frac{1}{2^{m+n+1}}cos((n-m)x)\&=&frac{1}{3}+sum_{dgeq 1}cos(dx)sum_{mgeq 0}frac{1}{2^{2m+d+1}}\&=&frac{1}{3}+sum_{dgeq 1}frac{cos(dx)}{2^d}sum_{mgeq 0}frac{1}{2^{2m+1}}\&=&frac{1}{3}+frac{2}{3}sum_{dgeq 1}frac{cos(dx)}{2^d}end{eqnarray*} $$
to get the whole Fourier cosine series of $frac{1}{5-4cos x}$. Then by the orthogonality relations
$$ int_{0}^{2pi}frac{cos(3x)}{5-4cos(x)},dx = frac{2pi}{3}cdotfrac{1}{2^3}=color{red}{frac{pi}{12}}.$$
This approach might look as an overkill, but it is pretty useful in understanding the asymptotic behaviour of the Fourier coefficients of $frac{1}{left(Rpmcosthetaright)^alpha}$ when $alphainfrac{1}{2}+mathbb{N}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, I just have one question. What exactly did you do in the step $int_{0}^{2pi}frac{cos(3x)}{5-4cos(x)}$=$frac{2pi}{3}frac{1}{2^3}$ ?
    $endgroup$
    – Poujh
    Dec 12 '18 at 8:08












  • $begingroup$
    Ah okay, it's the dot product of $cos(3x)$ and $frac{1}{5-4cos(x)}$ which you wrote as a serie just before and then you use, as said, the orthogonality. And then you integrate and you get $2pi$ times $frac{1}{2}((cos(3x-3x)+cos(3x+3x))$ from the cos(a)cos(b) (where a=b) trig identity, and cos(0) gives 1 while cos(6x) integrated from $0$ to $2pi$ gives $0$. And also, $frac{1}{3}*cos(3x)$ from $0$ to $2pi$ is also $0$. Makes sense, thank you !
    $endgroup$
    – Poujh
    Dec 12 '18 at 8:21












  • $begingroup$
    It seems like that you hate contour integration a lot, regarding some of your recent answers... :-)
    $endgroup$
    – Szeto
    Dec 12 '18 at 10:30










  • $begingroup$
    @Szeto: it is not that I hate it, but many times Fourier series or the Laplace transform allow to avoid the "hunt" for the correct integration contour, since it is somewhat already encoded in their application, and I find this pretty practical.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 12:48














1












1








1





$begingroup$

Starting with a geometric series:
$$frac{1}{2-e^{ix}} = sum_{ngeq 0}frac{1}{2^{n+1}}e^{nix}$$
we may consider the product between this series and its conjugate:



$$begin{eqnarray*}frac{1}{5-4cos x}&=&sum_{m,ngeq 0}frac{1}{2^{m+n+2}}e^{(n-m)ix}\&=&frac{1}{3}+sum_{n>mgeq 0}frac{1}{2^{m+n+1}}cos((n-m)x)\&=&frac{1}{3}+sum_{dgeq 1}cos(dx)sum_{mgeq 0}frac{1}{2^{2m+d+1}}\&=&frac{1}{3}+sum_{dgeq 1}frac{cos(dx)}{2^d}sum_{mgeq 0}frac{1}{2^{2m+1}}\&=&frac{1}{3}+frac{2}{3}sum_{dgeq 1}frac{cos(dx)}{2^d}end{eqnarray*} $$
to get the whole Fourier cosine series of $frac{1}{5-4cos x}$. Then by the orthogonality relations
$$ int_{0}^{2pi}frac{cos(3x)}{5-4cos(x)},dx = frac{2pi}{3}cdotfrac{1}{2^3}=color{red}{frac{pi}{12}}.$$
This approach might look as an overkill, but it is pretty useful in understanding the asymptotic behaviour of the Fourier coefficients of $frac{1}{left(Rpmcosthetaright)^alpha}$ when $alphainfrac{1}{2}+mathbb{N}$.






share|cite|improve this answer











$endgroup$



Starting with a geometric series:
$$frac{1}{2-e^{ix}} = sum_{ngeq 0}frac{1}{2^{n+1}}e^{nix}$$
we may consider the product between this series and its conjugate:



$$begin{eqnarray*}frac{1}{5-4cos x}&=&sum_{m,ngeq 0}frac{1}{2^{m+n+2}}e^{(n-m)ix}\&=&frac{1}{3}+sum_{n>mgeq 0}frac{1}{2^{m+n+1}}cos((n-m)x)\&=&frac{1}{3}+sum_{dgeq 1}cos(dx)sum_{mgeq 0}frac{1}{2^{2m+d+1}}\&=&frac{1}{3}+sum_{dgeq 1}frac{cos(dx)}{2^d}sum_{mgeq 0}frac{1}{2^{2m+1}}\&=&frac{1}{3}+frac{2}{3}sum_{dgeq 1}frac{cos(dx)}{2^d}end{eqnarray*} $$
to get the whole Fourier cosine series of $frac{1}{5-4cos x}$. Then by the orthogonality relations
$$ int_{0}^{2pi}frac{cos(3x)}{5-4cos(x)},dx = frac{2pi}{3}cdotfrac{1}{2^3}=color{red}{frac{pi}{12}}.$$
This approach might look as an overkill, but it is pretty useful in understanding the asymptotic behaviour of the Fourier coefficients of $frac{1}{left(Rpmcosthetaright)^alpha}$ when $alphainfrac{1}{2}+mathbb{N}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 12 '18 at 12:49

























answered Dec 12 '18 at 2:50









Jack D'AurizioJack D'Aurizio

288k33280660




288k33280660












  • $begingroup$
    Thanks, I just have one question. What exactly did you do in the step $int_{0}^{2pi}frac{cos(3x)}{5-4cos(x)}$=$frac{2pi}{3}frac{1}{2^3}$ ?
    $endgroup$
    – Poujh
    Dec 12 '18 at 8:08












  • $begingroup$
    Ah okay, it's the dot product of $cos(3x)$ and $frac{1}{5-4cos(x)}$ which you wrote as a serie just before and then you use, as said, the orthogonality. And then you integrate and you get $2pi$ times $frac{1}{2}((cos(3x-3x)+cos(3x+3x))$ from the cos(a)cos(b) (where a=b) trig identity, and cos(0) gives 1 while cos(6x) integrated from $0$ to $2pi$ gives $0$. And also, $frac{1}{3}*cos(3x)$ from $0$ to $2pi$ is also $0$. Makes sense, thank you !
    $endgroup$
    – Poujh
    Dec 12 '18 at 8:21












  • $begingroup$
    It seems like that you hate contour integration a lot, regarding some of your recent answers... :-)
    $endgroup$
    – Szeto
    Dec 12 '18 at 10:30










  • $begingroup$
    @Szeto: it is not that I hate it, but many times Fourier series or the Laplace transform allow to avoid the "hunt" for the correct integration contour, since it is somewhat already encoded in their application, and I find this pretty practical.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 12:48


















  • $begingroup$
    Thanks, I just have one question. What exactly did you do in the step $int_{0}^{2pi}frac{cos(3x)}{5-4cos(x)}$=$frac{2pi}{3}frac{1}{2^3}$ ?
    $endgroup$
    – Poujh
    Dec 12 '18 at 8:08












  • $begingroup$
    Ah okay, it's the dot product of $cos(3x)$ and $frac{1}{5-4cos(x)}$ which you wrote as a serie just before and then you use, as said, the orthogonality. And then you integrate and you get $2pi$ times $frac{1}{2}((cos(3x-3x)+cos(3x+3x))$ from the cos(a)cos(b) (where a=b) trig identity, and cos(0) gives 1 while cos(6x) integrated from $0$ to $2pi$ gives $0$. And also, $frac{1}{3}*cos(3x)$ from $0$ to $2pi$ is also $0$. Makes sense, thank you !
    $endgroup$
    – Poujh
    Dec 12 '18 at 8:21












  • $begingroup$
    It seems like that you hate contour integration a lot, regarding some of your recent answers... :-)
    $endgroup$
    – Szeto
    Dec 12 '18 at 10:30










  • $begingroup$
    @Szeto: it is not that I hate it, but many times Fourier series or the Laplace transform allow to avoid the "hunt" for the correct integration contour, since it is somewhat already encoded in their application, and I find this pretty practical.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 12:48
















$begingroup$
Thanks, I just have one question. What exactly did you do in the step $int_{0}^{2pi}frac{cos(3x)}{5-4cos(x)}$=$frac{2pi}{3}frac{1}{2^3}$ ?
$endgroup$
– Poujh
Dec 12 '18 at 8:08






$begingroup$
Thanks, I just have one question. What exactly did you do in the step $int_{0}^{2pi}frac{cos(3x)}{5-4cos(x)}$=$frac{2pi}{3}frac{1}{2^3}$ ?
$endgroup$
– Poujh
Dec 12 '18 at 8:08














$begingroup$
Ah okay, it's the dot product of $cos(3x)$ and $frac{1}{5-4cos(x)}$ which you wrote as a serie just before and then you use, as said, the orthogonality. And then you integrate and you get $2pi$ times $frac{1}{2}((cos(3x-3x)+cos(3x+3x))$ from the cos(a)cos(b) (where a=b) trig identity, and cos(0) gives 1 while cos(6x) integrated from $0$ to $2pi$ gives $0$. And also, $frac{1}{3}*cos(3x)$ from $0$ to $2pi$ is also $0$. Makes sense, thank you !
$endgroup$
– Poujh
Dec 12 '18 at 8:21






$begingroup$
Ah okay, it's the dot product of $cos(3x)$ and $frac{1}{5-4cos(x)}$ which you wrote as a serie just before and then you use, as said, the orthogonality. And then you integrate and you get $2pi$ times $frac{1}{2}((cos(3x-3x)+cos(3x+3x))$ from the cos(a)cos(b) (where a=b) trig identity, and cos(0) gives 1 while cos(6x) integrated from $0$ to $2pi$ gives $0$. And also, $frac{1}{3}*cos(3x)$ from $0$ to $2pi$ is also $0$. Makes sense, thank you !
$endgroup$
– Poujh
Dec 12 '18 at 8:21














$begingroup$
It seems like that you hate contour integration a lot, regarding some of your recent answers... :-)
$endgroup$
– Szeto
Dec 12 '18 at 10:30




$begingroup$
It seems like that you hate contour integration a lot, regarding some of your recent answers... :-)
$endgroup$
– Szeto
Dec 12 '18 at 10:30












$begingroup$
@Szeto: it is not that I hate it, but many times Fourier series or the Laplace transform allow to avoid the "hunt" for the correct integration contour, since it is somewhat already encoded in their application, and I find this pretty practical.
$endgroup$
– Jack D'Aurizio
Dec 12 '18 at 12:48




$begingroup$
@Szeto: it is not that I hate it, but many times Fourier series or the Laplace transform allow to avoid the "hunt" for the correct integration contour, since it is somewhat already encoded in their application, and I find this pretty practical.
$endgroup$
– Jack D'Aurizio
Dec 12 '18 at 12:48


















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