Compute $frac{1}{2pi}int_0^{2pi}f(t)g(t)dt$ where $g(t)= sumlimits_{n=1}^infty frac{(-1)^n}{n}sin(nt)$
$begingroup$
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
edited Dec 18 '18 at 21:34
Did
248k23224462
asked Dec 18 '18 at 13:00
L200123L200123
855
$endgroup$
|
show 1 more comment
$begingroup$
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
edited Dec 18 '18 at 21:34
Did
248k23224462
asked Dec 18 '18 at 13:00
L200123L200123
855
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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
edited Dec 18 '18 at 21:34
Did
248k23224462
asked Dec 18 '18 at 13:00
L200123L200123
855
$endgroup$
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
edited Dec 18 '18 at 21:34
Did
248k23224462
asked Dec 18 '18 at 13:00
L200123L200123
855
edited Dec 18 '18 at 21:34
Did
248k23224462
asked Dec 18 '18 at 13:00
L200123L200123
855
edited Dec 18 '18 at 21:34
Did
248k23224462
edited Dec 18 '18 at 21:34
Did
248k23224462
edited Dec 18 '18 at 21:34
Did
248k23224462
248k23224462
asked Dec 18 '18 at 13:00
L200123L200123
855
asked Dec 18 '18 at 13:00
L200123L200123
855
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
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,
$$
frac{1}{2pi}int_{0}^{2pi}f(t)g(t)dt = frac{7}{4}.
$$
answered Dec 18 '18 at 17:10
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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,
$$
frac{1}{2pi}int_{0}^{2pi}f(t)g(t)dt = frac{7}{4}.
$$
answered Dec 18 '18 at 17:10
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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,
$$
frac{1}{2pi}int_{0}^{2pi}f(t)g(t)dt = frac{7}{4}.
$$
answered Dec 18 '18 at 17:10
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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,
$$
frac{1}{2pi}int_{0}^{2pi}f(t)g(t)dt = frac{7}{4}.
$$
answered Dec 18 '18 at 17:10
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
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,
$$
frac{1}{2pi}int_{0}^{2pi}f(t)g(t)dt = frac{7}{4}.
$$
answered Dec 18 '18 at 17:10
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Dec 18 '18 at 17:10
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Dec 18 '18 at 17:10
DisintegratingByPartsDisintegratingByParts
59.4k42580
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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