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







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












$begingroup$


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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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






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












$begingroup$

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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036024%2ftwo-different-results-with-residue-calculus-for-the-integral-int-02-pi-fr%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036024%2ftwo-different-results-with-residue-calculus-for-the-integral-int-02-pi-fr%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
Read more

File:Tiny Toon Adventures Wacky Sports JP Title.png

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! File:Tiny Toon Adventures Wacky Sports JP Title.png From The Cutting Room Floor Jump to: navigation, search File File history File usage No higher resolution available. Tiny_Toon_Adventures_Wacky_Sports_JP_Title.png ‎ (160 × 144 pixels, file size: 3 KB, MIME type: image/png ) This image is a screenshot of a(n) Game Boy title screen. File history Click on a date/time to view the file as it appeared at that time. Date/Time Thumbnail Dimensions User Comment current 21:31, 20 August 2016 160 × 144 (3 KB) TheBluegeek (Talk | contribs) 17:25, 29 August 2013 160 × 144 (3 KB) Luigi-San (Talk | contribs) {{titlescreen|Game Boy}} You cannot overwrite this file. File usage The following page links to this file: Tiny Toon Advent
Read more