Fourier series of piecewise function, special case
$begingroup$
I'm wondering how to find the Fourier series piecewise functions where the interval on which each of the partial functions are defined are unequal. For example:
$f(x)=begin{cases} x & 0 le x le 1 \ x^2 & 1 le xle piend{cases} $ (and let's just consider the Fourier sine series of $f$ for the sake of brevity).
What coefficient should be used before the integral of $b_n$? My guess is $1$ for $int_0^1 dots$, but I'm stuck with the coefficient before $int_1^pi dots$. Or can I just use $frac1pi$ for both integrals?
I have one more questions regarding the arguments of $sin$ in both integrals. We need $sin(npi x/L)$ when working on $[-L,L]$. But since the integrals have other lengths, I'm wondering whether I can use $sin(nx)$?
Thanks for the help!
real-analysis
asked Jan 17 at 10:33
ZacharyZachary
1559
$endgroup$
add a comment |
$begingroup$
I'm wondering how to find the Fourier series piecewise functions where the interval on which each of the partial functions are defined are unequal. For example:
$f(x)=begin{cases} x & 0 le x le 1 \ x^2 & 1 le xle piend{cases} $ (and let's just consider the Fourier sine series of $f$ for the sake of brevity).
What coefficient should be used before the integral of $b_n$? My guess is $1$ for $int_0^1 dots$, but I'm stuck with the coefficient before $int_1^pi dots$. Or can I just use $frac1pi$ for both integrals?
I have one more questions regarding the arguments of $sin$ in both integrals. We need $sin(npi x/L)$ when working on $[-L,L]$. But since the integrals have other lengths, I'm wondering whether I can use $sin(nx)$?
Thanks for the help!
real-analysis
asked Jan 17 at 10:33
ZacharyZachary
1559
$endgroup$
$begingroup$
Write down the general formula for the coefficients first (expressed through $f(x)$), then substitute your particular function. The only special thing for piecewise functions is that you would need to separate the integral into two parts. The $1/ pi$ factor will stay the same
$endgroup$
– Yuriy S
Jan 17 at 10:42
$begingroup$
Can I use $1/pi$ when the interval is $[0,pi]$? Or should I first extend the function to make sure it is odd (since the question is about the sine series) on $[-pi,pi]$?
$endgroup$
– Zachary
Jan 17 at 10:58
$begingroup$
Wait a minute, how are you going to extend the function and use the sine series if your function is neither even nor odd? I think you should work on $[0, pi]$ and use the full series with both sine and cosine terms
$endgroup$
– Yuriy S
Jan 17 at 11:23
$begingroup$
This is what I've done: $F(x) = begin{cases} -x^2 & -pi le x le -1 \ -x & -1 le x le 0 \ x & 0 le x le 1 \ x^2 & 1 le x le pi end{cases}$. Then $b_n = frac1pi int_{-pi}^{-pi} F(x) sin(nx)dx = frac2pi int_{0}^{pi}f(x)sin(nx)dx$. And then I separated the integral into two parts. Result: $f(x) = frac{-2}{pi} sum_{n ge 1} ( frac{sin n}{n^2} + (-1)^n frac{pi^2}{n} + frac{2}{n^3} ((-1)^n - cos n))sin nx$. Which seems correct.
$endgroup$
– Zachary
Jan 17 at 12:37
add a comment |
$begingroup$
I'm wondering how to find the Fourier series piecewise functions where the interval on which each of the partial functions are defined are unequal. For example:
$f(x)=begin{cases} x & 0 le x le 1 \ x^2 & 1 le xle piend{cases} $ (and let's just consider the Fourier sine series of $f$ for the sake of brevity).
What coefficient should be used before the integral of $b_n$? My guess is $1$ for $int_0^1 dots$, but I'm stuck with the coefficient before $int_1^pi dots$. Or can I just use $frac1pi$ for both integrals?
I have one more questions regarding the arguments of $sin$ in both integrals. We need $sin(npi x/L)$ when working on $[-L,L]$. But since the integrals have other lengths, I'm wondering whether I can use $sin(nx)$?
Thanks for the help!
real-analysis
asked Jan 17 at 10:33
ZacharyZachary
1559
$endgroup$
I'm wondering how to find the Fourier series piecewise functions where the interval on which each of the partial functions are defined are unequal. For example:
$f(x)=begin{cases} x & 0 le x le 1 \ x^2 & 1 le xle piend{cases} $ (and let's just consider the Fourier sine series of $f$ for the sake of brevity).
What coefficient should be used before the integral of $b_n$? My guess is $1$ for $int_0^1 dots$, but I'm stuck with the coefficient before $int_1^pi dots$. Or can I just use $frac1pi$ for both integrals?
I have one more questions regarding the arguments of $sin$ in both integrals. We need $sin(npi x/L)$ when working on $[-L,L]$. But since the integrals have other lengths, I'm wondering whether I can use $sin(nx)$?
Thanks for the help!
real-analysis
real-analysis
asked Jan 17 at 10:33
ZacharyZachary
1559
asked Jan 17 at 10:33
ZacharyZachary
1559
asked Jan 17 at 10:33
ZacharyZachary
1559
asked Jan 17 at 10:33
ZacharyZachary
1559
asked Jan 17 at 10:33
ZacharyZachary
1559
1559
$begingroup$
Write down the general formula for the coefficients first (expressed through $f(x)$), then substitute your particular function. The only special thing for piecewise functions is that you would need to separate the integral into two parts. The $1/ pi$ factor will stay the same
$endgroup$
– Yuriy S
Jan 17 at 10:42
$begingroup$
Can I use $1/pi$ when the interval is $[0,pi]$? Or should I first extend the function to make sure it is odd (since the question is about the sine series) on $[-pi,pi]$?
$endgroup$
– Zachary
Jan 17 at 10:58
$begingroup$
Wait a minute, how are you going to extend the function and use the sine series if your function is neither even nor odd? I think you should work on $[0, pi]$ and use the full series with both sine and cosine terms
$endgroup$
– Yuriy S
Jan 17 at 11:23
$begingroup$
This is what I've done: $F(x) = begin{cases} -x^2 & -pi le x le -1 \ -x & -1 le x le 0 \ x & 0 le x le 1 \ x^2 & 1 le x le pi end{cases}$. Then $b_n = frac1pi int_{-pi}^{-pi} F(x) sin(nx)dx = frac2pi int_{0}^{pi}f(x)sin(nx)dx$. And then I separated the integral into two parts. Result: $f(x) = frac{-2}{pi} sum_{n ge 1} ( frac{sin n}{n^2} + (-1)^n frac{pi^2}{n} + frac{2}{n^3} ((-1)^n - cos n))sin nx$. Which seems correct.
$endgroup$
– Zachary
Jan 17 at 12:37
add a comment |
$begingroup$
Write down the general formula for the coefficients first (expressed through $f(x)$), then substitute your particular function. The only special thing for piecewise functions is that you would need to separate the integral into two parts. The $1/ pi$ factor will stay the same
$endgroup$
– Yuriy S
Jan 17 at 10:42
$begingroup$
Can I use $1/pi$ when the interval is $[0,pi]$? Or should I first extend the function to make sure it is odd (since the question is about the sine series) on $[-pi,pi]$?
$endgroup$
– Zachary
Jan 17 at 10:58
$begingroup$
Wait a minute, how are you going to extend the function and use the sine series if your function is neither even nor odd? I think you should work on $[0, pi]$ and use the full series with both sine and cosine terms
$endgroup$
– Yuriy S
Jan 17 at 11:23
$begingroup$
This is what I've done: $F(x) = begin{cases} -x^2 & -pi le x le -1 \ -x & -1 le x le 0 \ x & 0 le x le 1 \ x^2 & 1 le x le pi end{cases}$. Then $b_n = frac1pi int_{-pi}^{-pi} F(x) sin(nx)dx = frac2pi int_{0}^{pi}f(x)sin(nx)dx$. And then I separated the integral into two parts. Result: $f(x) = frac{-2}{pi} sum_{n ge 1} ( frac{sin n}{n^2} + (-1)^n frac{pi^2}{n} + frac{2}{n^3} ((-1)^n - cos n))sin nx$. Which seems correct.
$endgroup$
– Zachary
Jan 17 at 12:37
$begingroup$
Write down the general formula for the coefficients first (expressed through $f(x)$), then substitute your particular function. The only special thing for piecewise functions is that you would need to separate the integral into two parts. The $1/ pi$ factor will stay the same
$endgroup$
– Yuriy S
Jan 17 at 10:42
$begingroup$
Write down the general formula for the coefficients first (expressed through $f(x)$), then substitute your particular function. The only special thing for piecewise functions is that you would need to separate the integral into two parts. The $1/ pi$ factor will stay the same
$endgroup$
– Yuriy S
Jan 17 at 10:42
$begingroup$
Can I use $1/pi$ when the interval is $[0,pi]$? Or should I first extend the function to make sure it is odd (since the question is about the sine series) on $[-pi,pi]$?
$endgroup$
– Zachary
Jan 17 at 10:58
$begingroup$
Can I use $1/pi$ when the interval is $[0,pi]$? Or should I first extend the function to make sure it is odd (since the question is about the sine series) on $[-pi,pi]$?
$endgroup$
– Zachary
Jan 17 at 10:58
$begingroup$
Wait a minute, how are you going to extend the function and use the sine series if your function is neither even nor odd? I think you should work on $[0, pi]$ and use the full series with both sine and cosine terms
$endgroup$
– Yuriy S
Jan 17 at 11:23
$begingroup$
Wait a minute, how are you going to extend the function and use the sine series if your function is neither even nor odd? I think you should work on $[0, pi]$ and use the full series with both sine and cosine terms
$endgroup$
– Yuriy S
Jan 17 at 11:23
$begingroup$
This is what I've done: $F(x) = begin{cases} -x^2 & -pi le x le -1 \ -x & -1 le x le 0 \ x & 0 le x le 1 \ x^2 & 1 le x le pi end{cases}$. Then $b_n = frac1pi int_{-pi}^{-pi} F(x) sin(nx)dx = frac2pi int_{0}^{pi}f(x)sin(nx)dx$. And then I separated the integral into two parts. Result: $f(x) = frac{-2}{pi} sum_{n ge 1} ( frac{sin n}{n^2} + (-1)^n frac{pi^2}{n} + frac{2}{n^3} ((-1)^n - cos n))sin nx$. Which seems correct.
$endgroup$
– Zachary
Jan 17 at 12:37
$begingroup$
This is what I've done: $F(x) = begin{cases} -x^2 & -pi le x le -1 \ -x & -1 le x le 0 \ x & 0 le x le 1 \ x^2 & 1 le x le pi end{cases}$. Then $b_n = frac1pi int_{-pi}^{-pi} F(x) sin(nx)dx = frac2pi int_{0}^{pi}f(x)sin(nx)dx$. And then I separated the integral into two parts. Result: $f(x) = frac{-2}{pi} sum_{n ge 1} ( frac{sin n}{n^2} + (-1)^n frac{pi^2}{n} + frac{2}{n^3} ((-1)^n - cos n))sin nx$. Which seems correct.
$endgroup$
– Zachary
Jan 17 at 12:37
add a comment |
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$begingroup$
Write down the general formula for the coefficients first (expressed through $f(x)$), then substitute your particular function. The only special thing for piecewise functions is that you would need to separate the integral into two parts. The $1/ pi$ factor will stay the same
$endgroup$
– Yuriy S
Jan 17 at 10:42
$begingroup$
Can I use $1/pi$ when the interval is $[0,pi]$? Or should I first extend the function to make sure it is odd (since the question is about the sine series) on $[-pi,pi]$?
$endgroup$
– Zachary
Jan 17 at 10:58
$begingroup$
Wait a minute, how are you going to extend the function and use the sine series if your function is neither even nor odd? I think you should work on $[0, pi]$ and use the full series with both sine and cosine terms
$endgroup$
– Yuriy S
Jan 17 at 11:23
$begingroup$
This is what I've done: $F(x) = begin{cases} -x^2 & -pi le x le -1 \ -x & -1 le x le 0 \ x & 0 le x le 1 \ x^2 & 1 le x le pi end{cases}$. Then $b_n = frac1pi int_{-pi}^{-pi} F(x) sin(nx)dx = frac2pi int_{0}^{pi}f(x)sin(nx)dx$. And then I separated the integral into two parts. Result: $f(x) = frac{-2}{pi} sum_{n ge 1} ( frac{sin n}{n^2} + (-1)^n frac{pi^2}{n} + frac{2}{n^3} ((-1)^n - cos n))sin nx$. Which seems correct.
$endgroup$
– Zachary
Jan 17 at 12:37