Create Fourier-Series of f(x) = x if 0 < x < Pi and 0 if Pi < x < 2*Pi
$begingroup$
I tried the following to create the Fourier-series of the function:
$$ f(x) = begin{cases} x & 0<x<pi \ 0 & pi < x < 2 pi end{cases}$$
This is what I tried:
$$a_0 = frac{1}{pi}intlimits_0^pi x dx = frac{1}{pi}cdotleft(frac{x^2}{2}|_0^piright) = frac{1}{pi}cdotfrac{pi^2}{2} = frac{pi^2}{2cdotpi} = frac{pi}{2}$$
Is that right?
$$a_n = frac{1}{pi}intlimits_0^pi x cdot cos(nx) dx = frac{1}{pi}cdotleft(frac{x cdot sin(nx)}{n} + frac{cos(nx)}{n^2}right)|_0^pi$$
Assuming that $sin(nx)$ is always $0$ and $cos(nx)$ is $(-1)^n$. Is that ok??
$$ = frac{1}{pi}cdotleft(frac{0}{n} + frac{(-1)^n}{n^2}right) = frac{1}{pi}cdotleft(frac{(-1)^n}{n^2}right) = frac{(-1)^n}{pi n^2}$$
That's already wrong I'm sure!
$$b_n = frac{1}{pi}intlimits_0^pi x sin(nx) dx = frac{1}{pi}cdotleft(frac{x cdot cos(nx)}{n} + frac{sin(nx)}{n^2}right)|_0^pi$$
Assuming that $sin(nx)$ is always $0$ and $cos(nx)$ is $(-1)^n$. Is that ok??
$$ = frac{1}{pi}cdotleft(frac{pi cdot (-1)^n}{n}right) = frac{picdot(-1)^n}{pi n} = frac{(-1)^n}{n}$$ voila, thats wrong again :-)
A solution would be great, but I also need to understand what was my mistake.
Thank you in advance!
fourier-series
edited Jun 17 '13 at 10:21
Yoni Rozenshein
5,3201634
asked Jun 17 '13 at 10:08
Kevin MuhlerKevin Muhler
1015
$endgroup$
add a comment |
$begingroup$
I tried the following to create the Fourier-series of the function:
$$ f(x) = begin{cases} x & 0<x<pi \ 0 & pi < x < 2 pi end{cases}$$
This is what I tried:
$$a_0 = frac{1}{pi}intlimits_0^pi x dx = frac{1}{pi}cdotleft(frac{x^2}{2}|_0^piright) = frac{1}{pi}cdotfrac{pi^2}{2} = frac{pi^2}{2cdotpi} = frac{pi}{2}$$
Is that right?
$$a_n = frac{1}{pi}intlimits_0^pi x cdot cos(nx) dx = frac{1}{pi}cdotleft(frac{x cdot sin(nx)}{n} + frac{cos(nx)}{n^2}right)|_0^pi$$
Assuming that $sin(nx)$ is always $0$ and $cos(nx)$ is $(-1)^n$. Is that ok??
$$ = frac{1}{pi}cdotleft(frac{0}{n} + frac{(-1)^n}{n^2}right) = frac{1}{pi}cdotleft(frac{(-1)^n}{n^2}right) = frac{(-1)^n}{pi n^2}$$
That's already wrong I'm sure!
$$b_n = frac{1}{pi}intlimits_0^pi x sin(nx) dx = frac{1}{pi}cdotleft(frac{x cdot cos(nx)}{n} + frac{sin(nx)}{n^2}right)|_0^pi$$
Assuming that $sin(nx)$ is always $0$ and $cos(nx)$ is $(-1)^n$. Is that ok??
$$ = frac{1}{pi}cdotleft(frac{pi cdot (-1)^n}{n}right) = frac{picdot(-1)^n}{pi n} = frac{(-1)^n}{n}$$ voila, thats wrong again :-)
A solution would be great, but I also need to understand what was my mistake.
Thank you in advance!
fourier-series
edited Jun 17 '13 at 10:21
Yoni Rozenshein
5,3201634
asked Jun 17 '13 at 10:08
Kevin MuhlerKevin Muhler
1015
$endgroup$
$begingroup$
This is an old post, but for anyone coming to this now: one issue I see is the integral for a0 at the beginning. It should be done from -pi to pi, not zero to pi.
$endgroup$
– Jake
Dec 5 '18 at 5:05
$begingroup$
@Jake This is obviously not an issue at all.
$endgroup$
– Did
Dec 5 '18 at 7:14
add a comment |
$begingroup$
I tried the following to create the Fourier-series of the function:
$$ f(x) = begin{cases} x & 0<x<pi \ 0 & pi < x < 2 pi end{cases}$$
This is what I tried:
$$a_0 = frac{1}{pi}intlimits_0^pi x dx = frac{1}{pi}cdotleft(frac{x^2}{2}|_0^piright) = frac{1}{pi}cdotfrac{pi^2}{2} = frac{pi^2}{2cdotpi} = frac{pi}{2}$$
Is that right?
$$a_n = frac{1}{pi}intlimits_0^pi x cdot cos(nx) dx = frac{1}{pi}cdotleft(frac{x cdot sin(nx)}{n} + frac{cos(nx)}{n^2}right)|_0^pi$$
Assuming that $sin(nx)$ is always $0$ and $cos(nx)$ is $(-1)^n$. Is that ok??
$$ = frac{1}{pi}cdotleft(frac{0}{n} + frac{(-1)^n}{n^2}right) = frac{1}{pi}cdotleft(frac{(-1)^n}{n^2}right) = frac{(-1)^n}{pi n^2}$$
That's already wrong I'm sure!
$$b_n = frac{1}{pi}intlimits_0^pi x sin(nx) dx = frac{1}{pi}cdotleft(frac{x cdot cos(nx)}{n} + frac{sin(nx)}{n^2}right)|_0^pi$$
Assuming that $sin(nx)$ is always $0$ and $cos(nx)$ is $(-1)^n$. Is that ok??
$$ = frac{1}{pi}cdotleft(frac{pi cdot (-1)^n}{n}right) = frac{picdot(-1)^n}{pi n} = frac{(-1)^n}{n}$$ voila, thats wrong again :-)
A solution would be great, but I also need to understand what was my mistake.
Thank you in advance!
fourier-series
edited Jun 17 '13 at 10:21
Yoni Rozenshein
5,3201634
asked Jun 17 '13 at 10:08
Kevin MuhlerKevin Muhler
1015
$endgroup$
I tried the following to create the Fourier-series of the function:
$$ f(x) = begin{cases} x & 0<x<pi \ 0 & pi < x < 2 pi end{cases}$$
This is what I tried:
$$a_0 = frac{1}{pi}intlimits_0^pi x dx = frac{1}{pi}cdotleft(frac{x^2}{2}|_0^piright) = frac{1}{pi}cdotfrac{pi^2}{2} = frac{pi^2}{2cdotpi} = frac{pi}{2}$$
Is that right?
$$a_n = frac{1}{pi}intlimits_0^pi x cdot cos(nx) dx = frac{1}{pi}cdotleft(frac{x cdot sin(nx)}{n} + frac{cos(nx)}{n^2}right)|_0^pi$$
Assuming that $sin(nx)$ is always $0$ and $cos(nx)$ is $(-1)^n$. Is that ok??
$$ = frac{1}{pi}cdotleft(frac{0}{n} + frac{(-1)^n}{n^2}right) = frac{1}{pi}cdotleft(frac{(-1)^n}{n^2}right) = frac{(-1)^n}{pi n^2}$$
That's already wrong I'm sure!
$$b_n = frac{1}{pi}intlimits_0^pi x sin(nx) dx = frac{1}{pi}cdotleft(frac{x cdot cos(nx)}{n} + frac{sin(nx)}{n^2}right)|_0^pi$$
Assuming that $sin(nx)$ is always $0$ and $cos(nx)$ is $(-1)^n$. Is that ok??
$$ = frac{1}{pi}cdotleft(frac{pi cdot (-1)^n}{n}right) = frac{picdot(-1)^n}{pi n} = frac{(-1)^n}{n}$$ voila, thats wrong again :-)
A solution would be great, but I also need to understand what was my mistake.
Thank you in advance!
fourier-series
fourier-series
edited Jun 17 '13 at 10:21
Yoni Rozenshein
5,3201634
asked Jun 17 '13 at 10:08
Kevin MuhlerKevin Muhler
1015
edited Jun 17 '13 at 10:21
Yoni Rozenshein
5,3201634
asked Jun 17 '13 at 10:08
Kevin MuhlerKevin Muhler
1015
edited Jun 17 '13 at 10:21
Yoni Rozenshein
5,3201634
edited Jun 17 '13 at 10:21
Yoni Rozenshein
5,3201634
edited Jun 17 '13 at 10:21
Yoni Rozenshein
5,3201634
5,3201634
asked Jun 17 '13 at 10:08
Kevin MuhlerKevin Muhler
1015
asked Jun 17 '13 at 10:08
Kevin MuhlerKevin Muhler
1015
asked Jun 17 '13 at 10:08
Kevin MuhlerKevin Muhler
1015
1015
$begingroup$
This is an old post, but for anyone coming to this now: one issue I see is the integral for a0 at the beginning. It should be done from -pi to pi, not zero to pi.
$endgroup$
– Jake
Dec 5 '18 at 5:05
$begingroup$
@Jake This is obviously not an issue at all.
$endgroup$
– Did
Dec 5 '18 at 7:14
add a comment |
$begingroup$
This is an old post, but for anyone coming to this now: one issue I see is the integral for a0 at the beginning. It should be done from -pi to pi, not zero to pi.
$endgroup$
– Jake
Dec 5 '18 at 5:05
$begingroup$
@Jake This is obviously not an issue at all.
$endgroup$
– Did
Dec 5 '18 at 7:14
$begingroup$
This is an old post, but for anyone coming to this now: one issue I see is the integral for a0 at the beginning. It should be done from -pi to pi, not zero to pi.
$endgroup$
– Jake
Dec 5 '18 at 5:05
$begingroup$
This is an old post, but for anyone coming to this now: one issue I see is the integral for a0 at the beginning. It should be done from -pi to pi, not zero to pi.
$endgroup$
– Jake
Dec 5 '18 at 5:05
$begingroup$
@Jake This is obviously not an issue at all.
$endgroup$
– Did
Dec 5 '18 at 7:14
$begingroup$
@Jake This is obviously not an issue at all.
$endgroup$
– Did
Dec 5 '18 at 7:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Plugging in $x=0$ and $x=pi$ in $sin (nx)$ indeed gives you $0$.
Plugging in $x=pi$ in $cos (nx)$ indeed gives you $(-1)^n$.
But don't forget that plugging in $x=0$ in $cos (nx)$ gives you $1$.
Also, in the integral for $b_n$, there seems to be a minus sign missing after the integration by parts. Recheck your calculation.
answered Jun 17 '13 at 10:24
Yoni RozensheinYoni Rozenshein
5,3201634
$endgroup$
$begingroup$
But don't forget that plugging in x=0 in cos(nx) gives you 1. --> Yes, that was an Issue in the an calculation. Should I correct this finding in the initial question or what would be the proper way on Stackexchange?
$endgroup$
– Kevin Muhler
Jun 17 '13 at 12:27
$begingroup$
But I don't get where the minus in bn might be missing?
$endgroup$
– Kevin Muhler
Jun 17 '13 at 12:28
$begingroup$
What is the anti-derivative of $sin x$?
$endgroup$
– Yoni Rozenshein
Jun 17 '13 at 13:13
$begingroup$
Since you asked about a mistake in your calculation, it is better to leave the mistake as-is in the question, so future readers will know what the question and answer were about.
$endgroup$
– Yoni Rozenshein
Jun 17 '13 at 13:14
$begingroup$
Uuups, right, the anti-derivative is: -(x * cos(nx) / n)) + (sin(nx) / n²)
$endgroup$
– Kevin Muhler
Jun 17 '13 at 13:38
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
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active
oldest
votes
$begingroup$
Plugging in $x=0$ and $x=pi$ in $sin (nx)$ indeed gives you $0$.
Plugging in $x=pi$ in $cos (nx)$ indeed gives you $(-1)^n$.
But don't forget that plugging in $x=0$ in $cos (nx)$ gives you $1$.
Also, in the integral for $b_n$, there seems to be a minus sign missing after the integration by parts. Recheck your calculation.
answered Jun 17 '13 at 10:24
Yoni RozensheinYoni Rozenshein
5,3201634
$endgroup$
$begingroup$
But don't forget that plugging in x=0 in cos(nx) gives you 1. --> Yes, that was an Issue in the an calculation. Should I correct this finding in the initial question or what would be the proper way on Stackexchange?
$endgroup$
– Kevin Muhler
Jun 17 '13 at 12:27
$begingroup$
But I don't get where the minus in bn might be missing?
$endgroup$
– Kevin Muhler
Jun 17 '13 at 12:28
$begingroup$
What is the anti-derivative of $sin x$?
$endgroup$
– Yoni Rozenshein
Jun 17 '13 at 13:13
$begingroup$
Since you asked about a mistake in your calculation, it is better to leave the mistake as-is in the question, so future readers will know what the question and answer were about.
$endgroup$
– Yoni Rozenshein
Jun 17 '13 at 13:14
$begingroup$
Uuups, right, the anti-derivative is: -(x * cos(nx) / n)) + (sin(nx) / n²)
$endgroup$
– Kevin Muhler
Jun 17 '13 at 13:38
add a comment |
$begingroup$
Plugging in $x=0$ and $x=pi$ in $sin (nx)$ indeed gives you $0$.
Plugging in $x=pi$ in $cos (nx)$ indeed gives you $(-1)^n$.
But don't forget that plugging in $x=0$ in $cos (nx)$ gives you $1$.
Also, in the integral for $b_n$, there seems to be a minus sign missing after the integration by parts. Recheck your calculation.
answered Jun 17 '13 at 10:24
Yoni RozensheinYoni Rozenshein
5,3201634
$endgroup$
$begingroup$
But don't forget that plugging in x=0 in cos(nx) gives you 1. --> Yes, that was an Issue in the an calculation. Should I correct this finding in the initial question or what would be the proper way on Stackexchange?
$endgroup$
– Kevin Muhler
Jun 17 '13 at 12:27
$begingroup$
But I don't get where the minus in bn might be missing?
$endgroup$
– Kevin Muhler
Jun 17 '13 at 12:28
$begingroup$
What is the anti-derivative of $sin x$?
$endgroup$
– Yoni Rozenshein
Jun 17 '13 at 13:13
$begingroup$
Since you asked about a mistake in your calculation, it is better to leave the mistake as-is in the question, so future readers will know what the question and answer were about.
$endgroup$
– Yoni Rozenshein
Jun 17 '13 at 13:14
$begingroup$
Uuups, right, the anti-derivative is: -(x * cos(nx) / n)) + (sin(nx) / n²)
$endgroup$
– Kevin Muhler
Jun 17 '13 at 13:38
add a comment |
$begingroup$
Plugging in $x=0$ and $x=pi$ in $sin (nx)$ indeed gives you $0$.
Plugging in $x=pi$ in $cos (nx)$ indeed gives you $(-1)^n$.
But don't forget that plugging in $x=0$ in $cos (nx)$ gives you $1$.
Also, in the integral for $b_n$, there seems to be a minus sign missing after the integration by parts. Recheck your calculation.
answered Jun 17 '13 at 10:24
Yoni RozensheinYoni Rozenshein
5,3201634
$endgroup$
Plugging in $x=0$ and $x=pi$ in $sin (nx)$ indeed gives you $0$.
Plugging in $x=pi$ in $cos (nx)$ indeed gives you $(-1)^n$.
But don't forget that plugging in $x=0$ in $cos (nx)$ gives you $1$.
Also, in the integral for $b_n$, there seems to be a minus sign missing after the integration by parts. Recheck your calculation.
answered Jun 17 '13 at 10:24
Yoni RozensheinYoni Rozenshein
5,3201634
answered Jun 17 '13 at 10:24
Yoni RozensheinYoni Rozenshein
5,3201634
answered Jun 17 '13 at 10:24
Yoni RozensheinYoni Rozenshein
5,3201634
answered Jun 17 '13 at 10:24
Yoni RozensheinYoni Rozenshein
5,3201634
5,3201634
$begingroup$
But don't forget that plugging in x=0 in cos(nx) gives you 1. --> Yes, that was an Issue in the an calculation. Should I correct this finding in the initial question or what would be the proper way on Stackexchange?
$endgroup$
– Kevin Muhler
Jun 17 '13 at 12:27
$begingroup$
But I don't get where the minus in bn might be missing?
$endgroup$
– Kevin Muhler
Jun 17 '13 at 12:28
$begingroup$
What is the anti-derivative of $sin x$?
$endgroup$
– Yoni Rozenshein
Jun 17 '13 at 13:13
$begingroup$
Since you asked about a mistake in your calculation, it is better to leave the mistake as-is in the question, so future readers will know what the question and answer were about.
$endgroup$
– Yoni Rozenshein
Jun 17 '13 at 13:14
$begingroup$
Uuups, right, the anti-derivative is: -(x * cos(nx) / n)) + (sin(nx) / n²)
$endgroup$
– Kevin Muhler
Jun 17 '13 at 13:38
add a comment |
$begingroup$
But don't forget that plugging in x=0 in cos(nx) gives you 1. --> Yes, that was an Issue in the an calculation. Should I correct this finding in the initial question or what would be the proper way on Stackexchange?
$endgroup$
– Kevin Muhler
Jun 17 '13 at 12:27
$begingroup$
But I don't get where the minus in bn might be missing?
$endgroup$
– Kevin Muhler
Jun 17 '13 at 12:28
$begingroup$
What is the anti-derivative of $sin x$?
$endgroup$
– Yoni Rozenshein
Jun 17 '13 at 13:13
$begingroup$
Since you asked about a mistake in your calculation, it is better to leave the mistake as-is in the question, so future readers will know what the question and answer were about.
$endgroup$
– Yoni Rozenshein
Jun 17 '13 at 13:14
$begingroup$
Uuups, right, the anti-derivative is: -(x * cos(nx) / n)) + (sin(nx) / n²)
$endgroup$
– Kevin Muhler
Jun 17 '13 at 13:38
$begingroup$
But don't forget that plugging in x=0 in cos(nx) gives you 1. --> Yes, that was an Issue in the an calculation. Should I correct this finding in the initial question or what would be the proper way on Stackexchange?
$endgroup$
– Kevin Muhler
Jun 17 '13 at 12:27
$begingroup$
But don't forget that plugging in x=0 in cos(nx) gives you 1. --> Yes, that was an Issue in the an calculation. Should I correct this finding in the initial question or what would be the proper way on Stackexchange?
$endgroup$
– Kevin Muhler
Jun 17 '13 at 12:27
$begingroup$
But I don't get where the minus in bn might be missing?
$endgroup$
– Kevin Muhler
Jun 17 '13 at 12:28
$begingroup$
But I don't get where the minus in bn might be missing?
$endgroup$
– Kevin Muhler
Jun 17 '13 at 12:28
$begingroup$
What is the anti-derivative of $sin x$?
$endgroup$
– Yoni Rozenshein
Jun 17 '13 at 13:13
$begingroup$
What is the anti-derivative of $sin x$?
$endgroup$
– Yoni Rozenshein
Jun 17 '13 at 13:13
$begingroup$
Since you asked about a mistake in your calculation, it is better to leave the mistake as-is in the question, so future readers will know what the question and answer were about.
$endgroup$
– Yoni Rozenshein
Jun 17 '13 at 13:14
$begingroup$
Since you asked about a mistake in your calculation, it is better to leave the mistake as-is in the question, so future readers will know what the question and answer were about.
$endgroup$
– Yoni Rozenshein
Jun 17 '13 at 13:14
$begingroup$
Uuups, right, the anti-derivative is: -(x * cos(nx) / n)) + (sin(nx) / n²)
$endgroup$
– Kevin Muhler
Jun 17 '13 at 13:38
$begingroup$
Uuups, right, the anti-derivative is: -(x * cos(nx) / n)) + (sin(nx) / n²)
$endgroup$
– Kevin Muhler
Jun 17 '13 at 13:38
add a comment |
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$begingroup$
This is an old post, but for anyone coming to this now: one issue I see is the integral for a0 at the beginning. It should be done from -pi to pi, not zero to pi.
$endgroup$
– Jake
Dec 5 '18 at 5:05
$begingroup$
@Jake This is obviously not an issue at all.
$endgroup$
– Did
Dec 5 '18 at 7:14