Writing Sine and Cosine functions as Fourier Series The Next CEO of Stack OverflowFourier...
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Writing Sine and Cosine functions as Fourier Series
The Next CEO of Stack OverflowFourier cosine series and sum helpIs it possible for cosine functions to have Fourier sine series expressions or sine functions to have Fourier cosine series expressions?Fourier Sine Series and Cosine SeriesFinding Fourier cosine series of sine functionIs Fourier transform still writing a function as a series of sines and cosines?Fourier sine and cosine series of these functions?Why is the Fourier Series of an even signal the Fourier cosine series?Even and odd functions with Fourier seriesWhy are the Fourier Series an orthogonal basis?Confusion about Fourier sine/cosine series
$begingroup$
What are the fourier series for: $sin(pi*x)+cos(3pi x)$ and $sin(3x)$
I was taught that the purpose of the Fourier series was to describe periodic functions in the form of an infinite sum of cosines and sines. But if the function is already in the form of sin and cos such as above, then wouldn't it be redundant to express it in the form it is already in? I can see that the functions above are not in the form of a series, but how can you put sin and cosine into a series of itself? Is the question a trick question and the answer just the function provided?
fourier-series
edited Nov 27 '15 at 9:07
Nizar
2,40921123
asked Nov 27 '15 at 8:52
VanGoVanGo
10516
$endgroup$
add a comment |
$begingroup$
What are the fourier series for: $sin(pi*x)+cos(3pi x)$ and $sin(3x)$
I was taught that the purpose of the Fourier series was to describe periodic functions in the form of an infinite sum of cosines and sines. But if the function is already in the form of sin and cos such as above, then wouldn't it be redundant to express it in the form it is already in? I can see that the functions above are not in the form of a series, but how can you put sin and cosine into a series of itself? Is the question a trick question and the answer just the function provided?
fourier-series
edited Nov 27 '15 at 9:07
Nizar
2,40921123
asked Nov 27 '15 at 8:52
VanGoVanGo
10516
$endgroup$
$begingroup$
If you denote $mathcal{F}(f)(x)$ the Fourier series for $f$, then simply $mathcal{F}(sin)(x) = sin(x)$, $mathcal{F}(cos)(x) = cos(x)$, the series are themselves. That's it.
$endgroup$
– Klaramun
Nov 27 '15 at 8:55
$begingroup$
@Klaramun what does it mean when you say that the Fourier series is a series with 'integer weights'?
$endgroup$
– VanGo
Nov 27 '15 at 8:59
$begingroup$
Ok now I have seen the edit. The second function $sin(3x)$ is the fourier series itself, but not the first one. Recall the Fourier series is a serie of the form $a_n sinBig(frac{2 pi n x}{T} Big) + b_n cosBig(frac{2 pi n x}{T} Big)$ where $T$ is the period of the function. So the first you need to fins is the period of such a function, and then express this as a sum of these sines and cosines.
$endgroup$
– Klaramun
Nov 27 '15 at 9:03
$begingroup$
I have now seen the 2nd edit you made. I have posted an answer if it helps, but it was more oriented on the last function you had in your question. Still, hope it helps.
$endgroup$
– Klaramun
Nov 27 '15 at 9:17
add a comment |
$begingroup$
What are the fourier series for: $sin(pi*x)+cos(3pi x)$ and $sin(3x)$
I was taught that the purpose of the Fourier series was to describe periodic functions in the form of an infinite sum of cosines and sines. But if the function is already in the form of sin and cos such as above, then wouldn't it be redundant to express it in the form it is already in? I can see that the functions above are not in the form of a series, but how can you put sin and cosine into a series of itself? Is the question a trick question and the answer just the function provided?
fourier-series
edited Nov 27 '15 at 9:07
Nizar
2,40921123
asked Nov 27 '15 at 8:52
VanGoVanGo
10516
$endgroup$
What are the fourier series for: $sin(pi*x)+cos(3pi x)$ and $sin(3x)$
I was taught that the purpose of the Fourier series was to describe periodic functions in the form of an infinite sum of cosines and sines. But if the function is already in the form of sin and cos such as above, then wouldn't it be redundant to express it in the form it is already in? I can see that the functions above are not in the form of a series, but how can you put sin and cosine into a series of itself? Is the question a trick question and the answer just the function provided?
fourier-series
fourier-series
edited Nov 27 '15 at 9:07
Nizar
2,40921123
asked Nov 27 '15 at 8:52
VanGoVanGo
10516
edited Nov 27 '15 at 9:07
Nizar
2,40921123
asked Nov 27 '15 at 8:52
VanGoVanGo
10516
edited Nov 27 '15 at 9:07
Nizar
2,40921123
edited Nov 27 '15 at 9:07
Nizar
2,40921123
edited Nov 27 '15 at 9:07
Nizar
2,40921123
2,40921123
asked Nov 27 '15 at 8:52
VanGoVanGo
10516
asked Nov 27 '15 at 8:52
VanGoVanGo
10516
asked Nov 27 '15 at 8:52
VanGoVanGo
10516
10516
$begingroup$
If you denote $mathcal{F}(f)(x)$ the Fourier series for $f$, then simply $mathcal{F}(sin)(x) = sin(x)$, $mathcal{F}(cos)(x) = cos(x)$, the series are themselves. That's it.
$endgroup$
– Klaramun
Nov 27 '15 at 8:55
$begingroup$
@Klaramun what does it mean when you say that the Fourier series is a series with 'integer weights'?
$endgroup$
– VanGo
Nov 27 '15 at 8:59
$begingroup$
Ok now I have seen the edit. The second function $sin(3x)$ is the fourier series itself, but not the first one. Recall the Fourier series is a serie of the form $a_n sinBig(frac{2 pi n x}{T} Big) + b_n cosBig(frac{2 pi n x}{T} Big)$ where $T$ is the period of the function. So the first you need to fins is the period of such a function, and then express this as a sum of these sines and cosines.
$endgroup$
– Klaramun
Nov 27 '15 at 9:03
$begingroup$
I have now seen the 2nd edit you made. I have posted an answer if it helps, but it was more oriented on the last function you had in your question. Still, hope it helps.
$endgroup$
– Klaramun
Nov 27 '15 at 9:17
add a comment |
$begingroup$
If you denote $mathcal{F}(f)(x)$ the Fourier series for $f$, then simply $mathcal{F}(sin)(x) = sin(x)$, $mathcal{F}(cos)(x) = cos(x)$, the series are themselves. That's it.
$endgroup$
– Klaramun
Nov 27 '15 at 8:55
$begingroup$
@Klaramun what does it mean when you say that the Fourier series is a series with 'integer weights'?
$endgroup$
– VanGo
Nov 27 '15 at 8:59
$begingroup$
Ok now I have seen the edit. The second function $sin(3x)$ is the fourier series itself, but not the first one. Recall the Fourier series is a serie of the form $a_n sinBig(frac{2 pi n x}{T} Big) + b_n cosBig(frac{2 pi n x}{T} Big)$ where $T$ is the period of the function. So the first you need to fins is the period of such a function, and then express this as a sum of these sines and cosines.
$endgroup$
– Klaramun
Nov 27 '15 at 9:03
$begingroup$
I have now seen the 2nd edit you made. I have posted an answer if it helps, but it was more oriented on the last function you had in your question. Still, hope it helps.
$endgroup$
– Klaramun
Nov 27 '15 at 9:17
$begingroup$
If you denote $mathcal{F}(f)(x)$ the Fourier series for $f$, then simply $mathcal{F}(sin)(x) = sin(x)$, $mathcal{F}(cos)(x) = cos(x)$, the series are themselves. That's it.
$endgroup$
– Klaramun
Nov 27 '15 at 8:55
$begingroup$
If you denote $mathcal{F}(f)(x)$ the Fourier series for $f$, then simply $mathcal{F}(sin)(x) = sin(x)$, $mathcal{F}(cos)(x) = cos(x)$, the series are themselves. That's it.
$endgroup$
– Klaramun
Nov 27 '15 at 8:55
$begingroup$
@Klaramun what does it mean when you say that the Fourier series is a series with 'integer weights'?
$endgroup$
– VanGo
Nov 27 '15 at 8:59
$begingroup$
@Klaramun what does it mean when you say that the Fourier series is a series with 'integer weights'?
$endgroup$
– VanGo
Nov 27 '15 at 8:59
$begingroup$
Ok now I have seen the edit. The second function $sin(3x)$ is the fourier series itself, but not the first one. Recall the Fourier series is a serie of the form $a_n sinBig(frac{2 pi n x}{T} Big) + b_n cosBig(frac{2 pi n x}{T} Big)$ where $T$ is the period of the function. So the first you need to fins is the period of such a function, and then express this as a sum of these sines and cosines.
$endgroup$
– Klaramun
Nov 27 '15 at 9:03
$begingroup$
Ok now I have seen the edit. The second function $sin(3x)$ is the fourier series itself, but not the first one. Recall the Fourier series is a serie of the form $a_n sinBig(frac{2 pi n x}{T} Big) + b_n cosBig(frac{2 pi n x}{T} Big)$ where $T$ is the period of the function. So the first you need to fins is the period of such a function, and then express this as a sum of these sines and cosines.
$endgroup$
– Klaramun
Nov 27 '15 at 9:03
$begingroup$
I have now seen the 2nd edit you made. I have posted an answer if it helps, but it was more oriented on the last function you had in your question. Still, hope it helps.
$endgroup$
– Klaramun
Nov 27 '15 at 9:17
$begingroup$
I have now seen the 2nd edit you made. I have posted an answer if it helps, but it was more oriented on the last function you had in your question. Still, hope it helps.
$endgroup$
– Klaramun
Nov 27 '15 at 9:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The Fourier series express a periodic function in terms of sines and cosines, but with a precise relation between their periods. Let me explain.
If $f$ is a periodic function with period $T$ (i.e. $f(x+T) = f(x)$), then roughly speaking (I'll not worry about convergence issues) its Fourier series will be
begin{equation*}
begin{split}
mathcal{F}(f)(x) & = sum_{n=1}^{+infty} a_n sinBig( frac{2pi n}{T}xBig) + b_n cosBig( frac{2pi n}{T}xBig) = \
& = underbrace{Big( a_1 sinBig( frac{2pi}{T}xBig) + b_1 cosBig( frac{2pi}{T}xBig) Big)}_{text{First term, period } T_0} + underbrace{Big( a_2 sinBig( frac{2pi·2}{T}xBig) + b_2 cosBig( frac{2pi· 2}{T}xBig) Big)}_{text{Second term, period } 1/2·T_0} + cdots
end{split}
end{equation*}
So the relation between the periods of each term of the series is that they decrease in $frac{1}{n}$ at the $n$-th term with respect to the first one, the periods are related in a rational way.
In the case of the function $cos(pi x) + sin(3 x)$, the periods of the terms are not related in a rational way (in fact the first one is rational, but not the second one), so you must compute its Fourier series, which will not be the same as the function itself (first you need to determine the period $T$ of such a function).
answered Nov 27 '15 at 9:15
KlaramunKlaramun
78149
$endgroup$
$begingroup$
I end up getting an integral with 2 cosines multiplied together, each with a different period. Should I just integrate by parts from here?
$endgroup$
– VanGo
Nov 27 '15 at 9:59
$begingroup$
You re-edited your function, it ends up being $sin(pi x) + cos(3 pi x)$; it is its Fourier series itlsef because of what I just told you.
$endgroup$
– Klaramun
Nov 27 '15 at 10:04
add a comment |
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$begingroup$
The Fourier series express a periodic function in terms of sines and cosines, but with a precise relation between their periods. Let me explain.
If $f$ is a periodic function with period $T$ (i.e. $f(x+T) = f(x)$), then roughly speaking (I'll not worry about convergence issues) its Fourier series will be
begin{equation*}
begin{split}
mathcal{F}(f)(x) & = sum_{n=1}^{+infty} a_n sinBig( frac{2pi n}{T}xBig) + b_n cosBig( frac{2pi n}{T}xBig) = \
& = underbrace{Big( a_1 sinBig( frac{2pi}{T}xBig) + b_1 cosBig( frac{2pi}{T}xBig) Big)}_{text{First term, period } T_0} + underbrace{Big( a_2 sinBig( frac{2pi·2}{T}xBig) + b_2 cosBig( frac{2pi· 2}{T}xBig) Big)}_{text{Second term, period } 1/2·T_0} + cdots
end{split}
end{equation*}
So the relation between the periods of each term of the series is that they decrease in $frac{1}{n}$ at the $n$-th term with respect to the first one, the periods are related in a rational way.
In the case of the function $cos(pi x) + sin(3 x)$, the periods of the terms are not related in a rational way (in fact the first one is rational, but not the second one), so you must compute its Fourier series, which will not be the same as the function itself (first you need to determine the period $T$ of such a function).
answered Nov 27 '15 at 9:15
KlaramunKlaramun
78149
$endgroup$
$begingroup$
I end up getting an integral with 2 cosines multiplied together, each with a different period. Should I just integrate by parts from here?
$endgroup$
– VanGo
Nov 27 '15 at 9:59
$begingroup$
You re-edited your function, it ends up being $sin(pi x) + cos(3 pi x)$; it is its Fourier series itlsef because of what I just told you.
$endgroup$
– Klaramun
Nov 27 '15 at 10:04
add a comment |
$begingroup$
The Fourier series express a periodic function in terms of sines and cosines, but with a precise relation between their periods. Let me explain.
If $f$ is a periodic function with period $T$ (i.e. $f(x+T) = f(x)$), then roughly speaking (I'll not worry about convergence issues) its Fourier series will be
begin{equation*}
begin{split}
mathcal{F}(f)(x) & = sum_{n=1}^{+infty} a_n sinBig( frac{2pi n}{T}xBig) + b_n cosBig( frac{2pi n}{T}xBig) = \
& = underbrace{Big( a_1 sinBig( frac{2pi}{T}xBig) + b_1 cosBig( frac{2pi}{T}xBig) Big)}_{text{First term, period } T_0} + underbrace{Big( a_2 sinBig( frac{2pi·2}{T}xBig) + b_2 cosBig( frac{2pi· 2}{T}xBig) Big)}_{text{Second term, period } 1/2·T_0} + cdots
end{split}
end{equation*}
So the relation between the periods of each term of the series is that they decrease in $frac{1}{n}$ at the $n$-th term with respect to the first one, the periods are related in a rational way.
In the case of the function $cos(pi x) + sin(3 x)$, the periods of the terms are not related in a rational way (in fact the first one is rational, but not the second one), so you must compute its Fourier series, which will not be the same as the function itself (first you need to determine the period $T$ of such a function).
answered Nov 27 '15 at 9:15
KlaramunKlaramun
78149
$endgroup$
$begingroup$
I end up getting an integral with 2 cosines multiplied together, each with a different period. Should I just integrate by parts from here?
$endgroup$
– VanGo
Nov 27 '15 at 9:59
$begingroup$
You re-edited your function, it ends up being $sin(pi x) + cos(3 pi x)$; it is its Fourier series itlsef because of what I just told you.
$endgroup$
– Klaramun
Nov 27 '15 at 10:04
add a comment |
$begingroup$
The Fourier series express a periodic function in terms of sines and cosines, but with a precise relation between their periods. Let me explain.
If $f$ is a periodic function with period $T$ (i.e. $f(x+T) = f(x)$), then roughly speaking (I'll not worry about convergence issues) its Fourier series will be
begin{equation*}
begin{split}
mathcal{F}(f)(x) & = sum_{n=1}^{+infty} a_n sinBig( frac{2pi n}{T}xBig) + b_n cosBig( frac{2pi n}{T}xBig) = \
& = underbrace{Big( a_1 sinBig( frac{2pi}{T}xBig) + b_1 cosBig( frac{2pi}{T}xBig) Big)}_{text{First term, period } T_0} + underbrace{Big( a_2 sinBig( frac{2pi·2}{T}xBig) + b_2 cosBig( frac{2pi· 2}{T}xBig) Big)}_{text{Second term, period } 1/2·T_0} + cdots
end{split}
end{equation*}
So the relation between the periods of each term of the series is that they decrease in $frac{1}{n}$ at the $n$-th term with respect to the first one, the periods are related in a rational way.
In the case of the function $cos(pi x) + sin(3 x)$, the periods of the terms are not related in a rational way (in fact the first one is rational, but not the second one), so you must compute its Fourier series, which will not be the same as the function itself (first you need to determine the period $T$ of such a function).
answered Nov 27 '15 at 9:15
KlaramunKlaramun
78149
$endgroup$
The Fourier series express a periodic function in terms of sines and cosines, but with a precise relation between their periods. Let me explain.
If $f$ is a periodic function with period $T$ (i.e. $f(x+T) = f(x)$), then roughly speaking (I'll not worry about convergence issues) its Fourier series will be
begin{equation*}
begin{split}
mathcal{F}(f)(x) & = sum_{n=1}^{+infty} a_n sinBig( frac{2pi n}{T}xBig) + b_n cosBig( frac{2pi n}{T}xBig) = \
& = underbrace{Big( a_1 sinBig( frac{2pi}{T}xBig) + b_1 cosBig( frac{2pi}{T}xBig) Big)}_{text{First term, period } T_0} + underbrace{Big( a_2 sinBig( frac{2pi·2}{T}xBig) + b_2 cosBig( frac{2pi· 2}{T}xBig) Big)}_{text{Second term, period } 1/2·T_0} + cdots
end{split}
end{equation*}
So the relation between the periods of each term of the series is that they decrease in $frac{1}{n}$ at the $n$-th term with respect to the first one, the periods are related in a rational way.
In the case of the function $cos(pi x) + sin(3 x)$, the periods of the terms are not related in a rational way (in fact the first one is rational, but not the second one), so you must compute its Fourier series, which will not be the same as the function itself (first you need to determine the period $T$ of such a function).
answered Nov 27 '15 at 9:15
KlaramunKlaramun
78149
edited Nov 27 '15 at 10:02
answered Nov 27 '15 at 9:15
KlaramunKlaramun
78149
answered Nov 27 '15 at 9:15
KlaramunKlaramun
78149
answered Nov 27 '15 at 9:15
KlaramunKlaramun
78149
78149
$begingroup$
I end up getting an integral with 2 cosines multiplied together, each with a different period. Should I just integrate by parts from here?
$endgroup$
– VanGo
Nov 27 '15 at 9:59
$begingroup$
You re-edited your function, it ends up being $sin(pi x) + cos(3 pi x)$; it is its Fourier series itlsef because of what I just told you.
$endgroup$
– Klaramun
Nov 27 '15 at 10:04
add a comment |
$begingroup$
I end up getting an integral with 2 cosines multiplied together, each with a different period. Should I just integrate by parts from here?
$endgroup$
– VanGo
Nov 27 '15 at 9:59
$begingroup$
You re-edited your function, it ends up being $sin(pi x) + cos(3 pi x)$; it is its Fourier series itlsef because of what I just told you.
$endgroup$
– Klaramun
Nov 27 '15 at 10:04
$begingroup$
I end up getting an integral with 2 cosines multiplied together, each with a different period. Should I just integrate by parts from here?
$endgroup$
– VanGo
Nov 27 '15 at 9:59
$begingroup$
I end up getting an integral with 2 cosines multiplied together, each with a different period. Should I just integrate by parts from here?
$endgroup$
– VanGo
Nov 27 '15 at 9:59
$begingroup$
You re-edited your function, it ends up being $sin(pi x) + cos(3 pi x)$; it is its Fourier series itlsef because of what I just told you.
$endgroup$
– Klaramun
Nov 27 '15 at 10:04
$begingroup$
You re-edited your function, it ends up being $sin(pi x) + cos(3 pi x)$; it is its Fourier series itlsef because of what I just told you.
$endgroup$
– Klaramun
Nov 27 '15 at 10:04
add a comment |
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$begingroup$
If you denote $mathcal{F}(f)(x)$ the Fourier series for $f$, then simply $mathcal{F}(sin)(x) = sin(x)$, $mathcal{F}(cos)(x) = cos(x)$, the series are themselves. That's it.
$endgroup$
– Klaramun
Nov 27 '15 at 8:55
$begingroup$
@Klaramun what does it mean when you say that the Fourier series is a series with 'integer weights'?
$endgroup$
– VanGo
Nov 27 '15 at 8:59
$begingroup$
Ok now I have seen the edit. The second function $sin(3x)$ is the fourier series itself, but not the first one. Recall the Fourier series is a serie of the form $a_n sinBig(frac{2 pi n x}{T} Big) + b_n cosBig(frac{2 pi n x}{T} Big)$ where $T$ is the period of the function. So the first you need to fins is the period of such a function, and then express this as a sum of these sines and cosines.
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– Klaramun
Nov 27 '15 at 9:03
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I have now seen the 2nd edit you made. I have posted an answer if it helps, but it was more oriented on the last function you had in your question. Still, hope it helps.
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– Klaramun
Nov 27 '15 at 9:17