Compute $frac{1}{2pi}int_0^{2pi}f(t)g(t)dt$ where $g(t)= sumlimits_{n=1}^infty frac{(-1)^n}{n}sin(nt)$












3












$begingroup$



Fourier analysis, compute integral $dfrac{1}{2pi}int_0^{2pi}f(t)g(t)dt$ where $g(t)= sum_{n=1}^infty dfrac{(-1)^n}{n}sin(nt)$ and $f(t)=cos(t)+7sin(2t)$.




I'm having slight problems solving this integral, can we use alternating series test to say that the sum converges in some way? It feels like the sum might converge to $0$ and that we in that case will get an integal that is just $0$, but that feels counterintuitive so I'm unsure. Is there some theorem that might help that I'm missing?






real-analysis integration fourier-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $f(t)$? And can't you use the closed form of $g(t)$?
    $endgroup$
    – metamorphy
    Dec 18 '18 at 13:06










  • $begingroup$
    Thanks! forgot to add $f(t)$ have now added it.
    $endgroup$
    – L200123
    Dec 18 '18 at 13:11






  • 1




    $begingroup$
    I don't know if you are supposed to recognize this, but I think $g(t)$ is the same as $-t/2$ (which makes this just an integration by parts problem).
    $endgroup$
    – Rellek
    Dec 18 '18 at 13:32










  • $begingroup$
    Remember that for $|x|<pi$, $$x^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos nx$$ So then you take $d/dx$ on both sides to get $$-2x=4sum_{ngeq1}frac{(-1)^n}{n}sin nx$$ so we have that $$g(x)=-frac{x}2$$
    $endgroup$
    – clathratus
    Dec 18 '18 at 18:34








  • 1




    $begingroup$
    @Rellek here $g(t)=-t/2$ when $tin[-pi,pi]$, but he is integrating in $[0,2pi]$, in this set $$g(t)=begin{cases}-t/2,&tin[0,pi)\pi-t/2,&tin(pi,2pi]\0,&t=piend{cases}$$
    $endgroup$
    – Masacroso
    Dec 18 '18 at 18:41
















3












$begingroup$



Fourier analysis, compute integral $dfrac{1}{2pi}int_0^{2pi}f(t)g(t)dt$ where $g(t)= sum_{n=1}^infty dfrac{(-1)^n}{n}sin(nt)$ and $f(t)=cos(t)+7sin(2t)$.




I'm having slight problems solving this integral, can we use alternating series test to say that the sum converges in some way? It feels like the sum might converge to $0$ and that we in that case will get an integal that is just $0$, but that feels counterintuitive so I'm unsure. Is there some theorem that might help that I'm missing?






real-analysis integration fourier-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $f(t)$? And can't you use the closed form of $g(t)$?
    $endgroup$
    – metamorphy
    Dec 18 '18 at 13:06










  • $begingroup$
    Thanks! forgot to add $f(t)$ have now added it.
    $endgroup$
    – L200123
    Dec 18 '18 at 13:11






  • 1




    $begingroup$
    I don't know if you are supposed to recognize this, but I think $g(t)$ is the same as $-t/2$ (which makes this just an integration by parts problem).
    $endgroup$
    – Rellek
    Dec 18 '18 at 13:32










  • $begingroup$
    Remember that for $|x|<pi$, $$x^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos nx$$ So then you take $d/dx$ on both sides to get $$-2x=4sum_{ngeq1}frac{(-1)^n}{n}sin nx$$ so we have that $$g(x)=-frac{x}2$$
    $endgroup$
    – clathratus
    Dec 18 '18 at 18:34








  • 1




    $begingroup$
    @Rellek here $g(t)=-t/2$ when $tin[-pi,pi]$, but he is integrating in $[0,2pi]$, in this set $$g(t)=begin{cases}-t/2,&tin[0,pi)\pi-t/2,&tin(pi,2pi]\0,&t=piend{cases}$$
    $endgroup$
    – Masacroso
    Dec 18 '18 at 18:41














3












3








3





$begingroup$



Fourier analysis, compute integral $dfrac{1}{2pi}int_0^{2pi}f(t)g(t)dt$ where $g(t)= sum_{n=1}^infty dfrac{(-1)^n}{n}sin(nt)$ and $f(t)=cos(t)+7sin(2t)$.




I'm having slight problems solving this integral, can we use alternating series test to say that the sum converges in some way? It feels like the sum might converge to $0$ and that we in that case will get an integal that is just $0$, but that feels counterintuitive so I'm unsure. Is there some theorem that might help that I'm missing?






real-analysis integration fourier-analysis







share|cite|improve this question











$endgroup$





Fourier analysis, compute integral $dfrac{1}{2pi}int_0^{2pi}f(t)g(t)dt$ where $g(t)= sum_{n=1}^infty dfrac{(-1)^n}{n}sin(nt)$ and $f(t)=cos(t)+7sin(2t)$.




I'm having slight problems solving this integral, can we use alternating series test to say that the sum converges in some way? It feels like the sum might converge to $0$ and that we in that case will get an integal that is just $0$, but that feels counterintuitive so I'm unsure. Is there some theorem that might help that I'm missing?






real-analysis integration fourier-analysis




real-analysis integration fourier-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 21:34









Did

248k23224462




248k23224462










asked Dec 18 '18 at 13:00









L200123L200123

855




855












  • $begingroup$
    What is $f(t)$? And can't you use the closed form of $g(t)$?
    $endgroup$
    – metamorphy
    Dec 18 '18 at 13:06










  • $begingroup$
    Thanks! forgot to add $f(t)$ have now added it.
    $endgroup$
    – L200123
    Dec 18 '18 at 13:11






  • 1




    $begingroup$
    I don't know if you are supposed to recognize this, but I think $g(t)$ is the same as $-t/2$ (which makes this just an integration by parts problem).
    $endgroup$
    – Rellek
    Dec 18 '18 at 13:32










  • $begingroup$
    Remember that for $|x|<pi$, $$x^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos nx$$ So then you take $d/dx$ on both sides to get $$-2x=4sum_{ngeq1}frac{(-1)^n}{n}sin nx$$ so we have that $$g(x)=-frac{x}2$$
    $endgroup$
    – clathratus
    Dec 18 '18 at 18:34








  • 1




    $begingroup$
    @Rellek here $g(t)=-t/2$ when $tin[-pi,pi]$, but he is integrating in $[0,2pi]$, in this set $$g(t)=begin{cases}-t/2,&tin[0,pi)\pi-t/2,&tin(pi,2pi]\0,&t=piend{cases}$$
    $endgroup$
    – Masacroso
    Dec 18 '18 at 18:41


















  • $begingroup$
    What is $f(t)$? And can't you use the closed form of $g(t)$?
    $endgroup$
    – metamorphy
    Dec 18 '18 at 13:06










  • $begingroup$
    Thanks! forgot to add $f(t)$ have now added it.
    $endgroup$
    – L200123
    Dec 18 '18 at 13:11






  • 1




    $begingroup$
    I don't know if you are supposed to recognize this, but I think $g(t)$ is the same as $-t/2$ (which makes this just an integration by parts problem).
    $endgroup$
    – Rellek
    Dec 18 '18 at 13:32










  • $begingroup$
    Remember that for $|x|<pi$, $$x^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos nx$$ So then you take $d/dx$ on both sides to get $$-2x=4sum_{ngeq1}frac{(-1)^n}{n}sin nx$$ so we have that $$g(x)=-frac{x}2$$
    $endgroup$
    – clathratus
    Dec 18 '18 at 18:34








  • 1




    $begingroup$
    @Rellek here $g(t)=-t/2$ when $tin[-pi,pi]$, but he is integrating in $[0,2pi]$, in this set $$g(t)=begin{cases}-t/2,&tin[0,pi)\pi-t/2,&tin(pi,2pi]\0,&t=piend{cases}$$
    $endgroup$
    – Masacroso
    Dec 18 '18 at 18:41
















$begingroup$
What is $f(t)$? And can't you use the closed form of $g(t)$?
$endgroup$
– metamorphy
Dec 18 '18 at 13:06




$begingroup$
What is $f(t)$? And can't you use the closed form of $g(t)$?
$endgroup$
– metamorphy
Dec 18 '18 at 13:06












$begingroup$
Thanks! forgot to add $f(t)$ have now added it.
$endgroup$
– L200123
Dec 18 '18 at 13:11




$begingroup$
Thanks! forgot to add $f(t)$ have now added it.
$endgroup$
– L200123
Dec 18 '18 at 13:11




1




1




$begingroup$
I don't know if you are supposed to recognize this, but I think $g(t)$ is the same as $-t/2$ (which makes this just an integration by parts problem).
$endgroup$
– Rellek
Dec 18 '18 at 13:32




$begingroup$
I don't know if you are supposed to recognize this, but I think $g(t)$ is the same as $-t/2$ (which makes this just an integration by parts problem).
$endgroup$
– Rellek
Dec 18 '18 at 13:32












$begingroup$
Remember that for $|x|<pi$, $$x^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos nx$$ So then you take $d/dx$ on both sides to get $$-2x=4sum_{ngeq1}frac{(-1)^n}{n}sin nx$$ so we have that $$g(x)=-frac{x}2$$
$endgroup$
– clathratus
Dec 18 '18 at 18:34






$begingroup$
Remember that for $|x|<pi$, $$x^2=frac{pi^2}3+4sum_{ngeq1}frac{(-1)^n}{n^2}cos nx$$ So then you take $d/dx$ on both sides to get $$-2x=4sum_{ngeq1}frac{(-1)^n}{n}sin nx$$ so we have that $$g(x)=-frac{x}2$$
$endgroup$
– clathratus
Dec 18 '18 at 18:34






1




1




$begingroup$
@Rellek here $g(t)=-t/2$ when $tin[-pi,pi]$, but he is integrating in $[0,2pi]$, in this set $$g(t)=begin{cases}-t/2,&tin[0,pi)\pi-t/2,&tin(pi,2pi]\0,&t=piend{cases}$$
$endgroup$
– Masacroso
Dec 18 '18 at 18:41




$begingroup$
@Rellek here $g(t)=-t/2$ when $tin[-pi,pi]$, but he is integrating in $[0,2pi]$, in this set $$g(t)=begin{cases}-t/2,&tin[0,pi)\pi-t/2,&tin(pi,2pi]\0,&t=piend{cases}$$
$endgroup$
– Masacroso
Dec 18 '18 at 18:41










1 Answer
1






active

oldest

votes


















1












$begingroup$

The sum for $g$ converges in $L^2$ because $sum_{n=1}^{infty}frac{1}{n^2} < infty$ and because ${ s_n(x)=frac{sin(nt)}{sqrt{pi}} }_{n=1}^{infty}$ is an orthonormal subset of $L^2[0,2pi]$. That is,
$$
g_N(t) = sum_{n=1}^{N}frac{(-1)^n}{n}sin(nt)
$$
converges in $L^2$ to some $gin L^2$. Because of this,
$$
int_{0}^{2pi}g(t)f(t)dt=lim_{N}int_{0}^{2pi}g_N(t)f(t)dt \
= sum_{n=1}^{infty}frac{(-1)^n}{n}int_{0}^{2pi}sin(nt)f(t)dt \
= sum_{n=1}^{infty}frac{(-1)^n}{n}int_{0}^{2pi}sin(nt)(cos(t)+7sin(2t))dt.
$$



The functions ${ 1,cos(t),sin(t),cos(2t),sin(2t),cdots }$ are mutually orthogonal. So the above reduces to



$$
frac{7}{2}int_0^{2pi}sin^2(2t)dt=frac{7pi}{2}.
$$
Therefore,
$$
frac{1}{2pi}int_{0}^{2pi}f(t)g(t)dt = frac{7}{4}.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    it is not enough clear that, because $g_n$ converges in $L^2$, we can exchange sum and integral. I will say that we can do this because $fin L^2$ also and the inner product in $L^2(Bbb R)$ is bounded and hence continuous, so $$lim_{ntoinfty}(g_n|f)=(g| f)$$
    $endgroup$
    – Masacroso
    Dec 20 '18 at 15:14













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

The sum for $g$ converges in $L^2$ because $sum_{n=1}^{infty}frac{1}{n^2} < infty$ and because ${ s_n(x)=frac{sin(nt)}{sqrt{pi}} }_{n=1}^{infty}$ is an orthonormal subset of $L^2[0,2pi]$. That is,
$$
g_N(t) = sum_{n=1}^{N}frac{(-1)^n}{n}sin(nt)
$$
converges in $L^2$ to some $gin L^2$. Because of this,
$$
int_{0}^{2pi}g(t)f(t)dt=lim_{N}int_{0}^{2pi}g_N(t)f(t)dt \
= sum_{n=1}^{infty}frac{(-1)^n}{n}int_{0}^{2pi}sin(nt)f(t)dt \
= sum_{n=1}^{infty}frac{(-1)^n}{n}int_{0}^{2pi}sin(nt)(cos(t)+7sin(2t))dt.
$$



The functions ${ 1,cos(t),sin(t),cos(2t),sin(2t),cdots }$ are mutually orthogonal. So the above reduces to



$$
frac{7}{2}int_0^{2pi}sin^2(2t)dt=frac{7pi}{2}.
$$
Therefore,
$$
frac{1}{2pi}int_{0}^{2pi}f(t)g(t)dt = frac{7}{4}.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    it is not enough clear that, because $g_n$ converges in $L^2$, we can exchange sum and integral. I will say that we can do this because $fin L^2$ also and the inner product in $L^2(Bbb R)$ is bounded and hence continuous, so $$lim_{ntoinfty}(g_n|f)=(g| f)$$
    $endgroup$
    – Masacroso
    Dec 20 '18 at 15:14


















1












$begingroup$

The sum for $g$ converges in $L^2$ because $sum_{n=1}^{infty}frac{1}{n^2} < infty$ and because ${ s_n(x)=frac{sin(nt)}{sqrt{pi}} }_{n=1}^{infty}$ is an orthonormal subset of $L^2[0,2pi]$. That is,
$$
g_N(t) = sum_{n=1}^{N}frac{(-1)^n}{n}sin(nt)
$$
converges in $L^2$ to some $gin L^2$. Because of this,
$$
int_{0}^{2pi}g(t)f(t)dt=lim_{N}int_{0}^{2pi}g_N(t)f(t)dt \
= sum_{n=1}^{infty}frac{(-1)^n}{n}int_{0}^{2pi}sin(nt)f(t)dt \
= sum_{n=1}^{infty}frac{(-1)^n}{n}int_{0}^{2pi}sin(nt)(cos(t)+7sin(2t))dt.
$$



The functions ${ 1,cos(t),sin(t),cos(2t),sin(2t),cdots }$ are mutually orthogonal. So the above reduces to



$$
frac{7}{2}int_0^{2pi}sin^2(2t)dt=frac{7pi}{2}.
$$
Therefore,
$$
frac{1}{2pi}int_{0}^{2pi}f(t)g(t)dt = frac{7}{4}.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    it is not enough clear that, because $g_n$ converges in $L^2$, we can exchange sum and integral. I will say that we can do this because $fin L^2$ also and the inner product in $L^2(Bbb R)$ is bounded and hence continuous, so $$lim_{ntoinfty}(g_n|f)=(g| f)$$
    $endgroup$
    – Masacroso
    Dec 20 '18 at 15:14
















1












1








1





$begingroup$

The sum for $g$ converges in $L^2$ because $sum_{n=1}^{infty}frac{1}{n^2} < infty$ and because ${ s_n(x)=frac{sin(nt)}{sqrt{pi}} }_{n=1}^{infty}$ is an orthonormal subset of $L^2[0,2pi]$. That is,
$$
g_N(t) = sum_{n=1}^{N}frac{(-1)^n}{n}sin(nt)
$$
converges in $L^2$ to some $gin L^2$. Because of this,
$$
int_{0}^{2pi}g(t)f(t)dt=lim_{N}int_{0}^{2pi}g_N(t)f(t)dt \
= sum_{n=1}^{infty}frac{(-1)^n}{n}int_{0}^{2pi}sin(nt)f(t)dt \
= sum_{n=1}^{infty}frac{(-1)^n}{n}int_{0}^{2pi}sin(nt)(cos(t)+7sin(2t))dt.
$$



The functions ${ 1,cos(t),sin(t),cos(2t),sin(2t),cdots }$ are mutually orthogonal. So the above reduces to



$$
frac{7}{2}int_0^{2pi}sin^2(2t)dt=frac{7pi}{2}.
$$
Therefore,
$$
frac{1}{2pi}int_{0}^{2pi}f(t)g(t)dt = frac{7}{4}.
$$






share|cite|improve this answer









$endgroup$



The sum for $g$ converges in $L^2$ because $sum_{n=1}^{infty}frac{1}{n^2} < infty$ and because ${ s_n(x)=frac{sin(nt)}{sqrt{pi}} }_{n=1}^{infty}$ is an orthonormal subset of $L^2[0,2pi]$. That is,
$$
g_N(t) = sum_{n=1}^{N}frac{(-1)^n}{n}sin(nt)
$$
converges in $L^2$ to some $gin L^2$. Because of this,
$$
int_{0}^{2pi}g(t)f(t)dt=lim_{N}int_{0}^{2pi}g_N(t)f(t)dt \
= sum_{n=1}^{infty}frac{(-1)^n}{n}int_{0}^{2pi}sin(nt)f(t)dt \
= sum_{n=1}^{infty}frac{(-1)^n}{n}int_{0}^{2pi}sin(nt)(cos(t)+7sin(2t))dt.
$$



The functions ${ 1,cos(t),sin(t),cos(2t),sin(2t),cdots }$ are mutually orthogonal. So the above reduces to



$$
frac{7}{2}int_0^{2pi}sin^2(2t)dt=frac{7pi}{2}.
$$
Therefore,
$$
frac{1}{2pi}int_{0}^{2pi}f(t)g(t)dt = frac{7}{4}.
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 18 '18 at 17:10









DisintegratingByPartsDisintegratingByParts

59.4k42580




59.4k42580












  • $begingroup$
    it is not enough clear that, because $g_n$ converges in $L^2$, we can exchange sum and integral. I will say that we can do this because $fin L^2$ also and the inner product in $L^2(Bbb R)$ is bounded and hence continuous, so $$lim_{ntoinfty}(g_n|f)=(g| f)$$
    $endgroup$
    – Masacroso
    Dec 20 '18 at 15:14




















  • $begingroup$
    it is not enough clear that, because $g_n$ converges in $L^2$, we can exchange sum and integral. I will say that we can do this because $fin L^2$ also and the inner product in $L^2(Bbb R)$ is bounded and hence continuous, so $$lim_{ntoinfty}(g_n|f)=(g| f)$$
    $endgroup$
    – Masacroso
    Dec 20 '18 at 15:14


















$begingroup$
it is not enough clear that, because $g_n$ converges in $L^2$, we can exchange sum and integral. I will say that we can do this because $fin L^2$ also and the inner product in $L^2(Bbb R)$ is bounded and hence continuous, so $$lim_{ntoinfty}(g_n|f)=(g| f)$$
$endgroup$
– Masacroso
Dec 20 '18 at 15:14






$begingroup$
it is not enough clear that, because $g_n$ converges in $L^2$, we can exchange sum and integral. I will say that we can do this because $fin L^2$ also and the inner product in $L^2(Bbb R)$ is bounded and hence continuous, so $$lim_{ntoinfty}(g_n|f)=(g| f)$$
$endgroup$
– Masacroso
Dec 20 '18 at 15:14




















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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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