Fourier coefficients of $sin{(2 pi f_0 t)}$












0












$begingroup$


I have to calculate the Fourier coefficients of $ sin (2 pi f_0 t) $.



I try to apply the definition and afterwards I start to rewrite $sin$ with Euler formulas:



$$ frac{1}{f_0}int_{frac{-pi}{2}}^{frac{pi}{2}} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt .$$
I obtained $ frac{1}{2f_0i} [pi + frac{sin(4pi f_0) frac{pi}{2}}{2pi f_0}]$. But the result should be $frac{1}{2}e^{-ifrac{pi}{2}} $ if k = 1
$frac{1}{2}e^{ifrac{pi}{2}} $ if $k = -1$ and $0$ if $k $ is different from $1$.



Can you help me to understand how to solve this exercise ?






fourier-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean to write $sin 2 pi f_0 t$?
    $endgroup$
    – Kevin
    Dec 19 '18 at 15:39










  • $begingroup$
    Yes , sorry i’m Italian and not very good with LaTex
    $endgroup$
    – Elena Martini
    Dec 19 '18 at 15:52
















0












$begingroup$


I have to calculate the Fourier coefficients of $ sin (2 pi f_0 t) $.



I try to apply the definition and afterwards I start to rewrite $sin$ with Euler formulas:



$$ frac{1}{f_0}int_{frac{-pi}{2}}^{frac{pi}{2}} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt .$$
I obtained $ frac{1}{2f_0i} [pi + frac{sin(4pi f_0) frac{pi}{2}}{2pi f_0}]$. But the result should be $frac{1}{2}e^{-ifrac{pi}{2}} $ if k = 1
$frac{1}{2}e^{ifrac{pi}{2}} $ if $k = -1$ and $0$ if $k $ is different from $1$.



Can you help me to understand how to solve this exercise ?






fourier-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean to write $sin 2 pi f_0 t$?
    $endgroup$
    – Kevin
    Dec 19 '18 at 15:39










  • $begingroup$
    Yes , sorry i’m Italian and not very good with LaTex
    $endgroup$
    – Elena Martini
    Dec 19 '18 at 15:52














0












0








0





$begingroup$


I have to calculate the Fourier coefficients of $ sin (2 pi f_0 t) $.



I try to apply the definition and afterwards I start to rewrite $sin$ with Euler formulas:



$$ frac{1}{f_0}int_{frac{-pi}{2}}^{frac{pi}{2}} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt .$$
I obtained $ frac{1}{2f_0i} [pi + frac{sin(4pi f_0) frac{pi}{2}}{2pi f_0}]$. But the result should be $frac{1}{2}e^{-ifrac{pi}{2}} $ if k = 1
$frac{1}{2}e^{ifrac{pi}{2}} $ if $k = -1$ and $0$ if $k $ is different from $1$.



Can you help me to understand how to solve this exercise ?






fourier-series







share|cite|improve this question











$endgroup$




I have to calculate the Fourier coefficients of $ sin (2 pi f_0 t) $.



I try to apply the definition and afterwards I start to rewrite $sin$ with Euler formulas:



$$ frac{1}{f_0}int_{frac{-pi}{2}}^{frac{pi}{2}} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt .$$
I obtained $ frac{1}{2f_0i} [pi + frac{sin(4pi f_0) frac{pi}{2}}{2pi f_0}]$. But the result should be $frac{1}{2}e^{-ifrac{pi}{2}} $ if k = 1
$frac{1}{2}e^{ifrac{pi}{2}} $ if $k = -1$ and $0$ if $k $ is different from $1$.



Can you help me to understand how to solve this exercise ?






fourier-series




fourier-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 15:40









quid♦

37.2k95193




37.2k95193










asked Dec 19 '18 at 15:33









Elena MartiniElena Martini

113




113












  • $begingroup$
    Do you mean to write $sin 2 pi f_0 t$?
    $endgroup$
    – Kevin
    Dec 19 '18 at 15:39










  • $begingroup$
    Yes , sorry i’m Italian and not very good with LaTex
    $endgroup$
    – Elena Martini
    Dec 19 '18 at 15:52


















  • $begingroup$
    Do you mean to write $sin 2 pi f_0 t$?
    $endgroup$
    – Kevin
    Dec 19 '18 at 15:39










  • $begingroup$
    Yes , sorry i’m Italian and not very good with LaTex
    $endgroup$
    – Elena Martini
    Dec 19 '18 at 15:52
















$begingroup$
Do you mean to write $sin 2 pi f_0 t$?
$endgroup$
– Kevin
Dec 19 '18 at 15:39




$begingroup$
Do you mean to write $sin 2 pi f_0 t$?
$endgroup$
– Kevin
Dec 19 '18 at 15:39












$begingroup$
Yes , sorry i’m Italian and not very good with LaTex
$endgroup$
– Elena Martini
Dec 19 '18 at 15:52




$begingroup$
Yes , sorry i’m Italian and not very good with LaTex
$endgroup$
– Elena Martini
Dec 19 '18 at 15:52










1 Answer
1






active

oldest

votes


















1












$begingroup$

The intuition of Fourier series is that a periodic function can be expressed as the sum of sinusoidal functions with frequencies equal to $kf_0$, or multiples of the fundamental frequency (which is $f_0$). Since the given function is a(one) sinusoid, $sin(2pi{f_0}t)$, we should immediately get that the non-zero Fourier coefficients will be only at $k=pm 1$, so



$c_k = 0$ at $kneqpm1 $.



Now, for $k=1$, we proceed to the definition (formula is copied from your question, but corrected):
$$c_k = frac{1}{T_0}int_{0}^{T_0} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt.$$
Where $T_0 = 1/f_0$. Solving that, we have:



$$begin{align}
c_k &= frac{1}{2i T_0}int_{0}^{T_0} {(e^{2 i pi f_0 t}e^{-2 i pi f_0 t} - e^{-2 i pi f_0 t}e^{-2 i pi f_0 t})}dt\ &= frac{1}{2i T_0}int_{0}^{T_0} 1 - e^{-4 i pi f_0 t}dt\
...&= frac{1}{2i T_0}(T_0 - frac{1-e^{-4ipi}}{4ipi f_0})
end{align}$$



but $e^{-4pi i} = cos(-4pi) + i sin(-4pi) = 1 + 0i = 1$, so we are left with $$c_k = frac{1}{2i T_0} (T_0) = frac{1}{2i}.$$



The given answer ($c_k=frac{1}{2} e^{-i frac{pi}{2}}$ for $k=1$), will just be equal to $frac{1}{2i}$, because $e^{-i frac{pi}{2}} = cos(-frac{pi}{2}) + i sin(-frac{pi}{2}) = 0 + (-1)i= -i = frac{1}{i}.$



The case for $k=-1$ is similar to the computation above.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you very much ! your answer was very precise and thanks above all for writing all the steps, in this way I managed to understand much better. finally I understood why already at first sight you could already find the coefficients + and - 1 and the Fourier series explained in a simple way. the only thing that still is not clear to me are the extremes of the integral? why I start from 0 and I do not go to $- pi / 2 $ ? outside of $ pm sin $, the sin function repeats itself
    $endgroup$
    – Elena Martini
    Dec 19 '18 at 17:53












  • $begingroup$
    The only important thing about the extremes of the integrals is that they should cover one period of the input function. So the difference between the extremes should be $1/f_0$ which we defined as $T_0$. You can actually start from $-pi/2$, but the upper extreme should be $(-pi/2)+{T_0}$, and that will not make our calculations easier.
    $endgroup$
    – Poypoyan
    Dec 19 '18 at 18:08












  • $begingroup$
    Again thank you so so much!! now I will do other exercises in order to practice and I hope to improve :)
    $endgroup$
    – Elena Martini
    Dec 19 '18 at 18:15












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1 Answer
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1 Answer
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1












$begingroup$

The intuition of Fourier series is that a periodic function can be expressed as the sum of sinusoidal functions with frequencies equal to $kf_0$, or multiples of the fundamental frequency (which is $f_0$). Since the given function is a(one) sinusoid, $sin(2pi{f_0}t)$, we should immediately get that the non-zero Fourier coefficients will be only at $k=pm 1$, so



$c_k = 0$ at $kneqpm1 $.



Now, for $k=1$, we proceed to the definition (formula is copied from your question, but corrected):
$$c_k = frac{1}{T_0}int_{0}^{T_0} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt.$$
Where $T_0 = 1/f_0$. Solving that, we have:



$$begin{align}
c_k &= frac{1}{2i T_0}int_{0}^{T_0} {(e^{2 i pi f_0 t}e^{-2 i pi f_0 t} - e^{-2 i pi f_0 t}e^{-2 i pi f_0 t})}dt\ &= frac{1}{2i T_0}int_{0}^{T_0} 1 - e^{-4 i pi f_0 t}dt\
...&= frac{1}{2i T_0}(T_0 - frac{1-e^{-4ipi}}{4ipi f_0})
end{align}$$



but $e^{-4pi i} = cos(-4pi) + i sin(-4pi) = 1 + 0i = 1$, so we are left with $$c_k = frac{1}{2i T_0} (T_0) = frac{1}{2i}.$$



The given answer ($c_k=frac{1}{2} e^{-i frac{pi}{2}}$ for $k=1$), will just be equal to $frac{1}{2i}$, because $e^{-i frac{pi}{2}} = cos(-frac{pi}{2}) + i sin(-frac{pi}{2}) = 0 + (-1)i= -i = frac{1}{i}.$



The case for $k=-1$ is similar to the computation above.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you very much ! your answer was very precise and thanks above all for writing all the steps, in this way I managed to understand much better. finally I understood why already at first sight you could already find the coefficients + and - 1 and the Fourier series explained in a simple way. the only thing that still is not clear to me are the extremes of the integral? why I start from 0 and I do not go to $- pi / 2 $ ? outside of $ pm sin $, the sin function repeats itself
    $endgroup$
    – Elena Martini
    Dec 19 '18 at 17:53












  • $begingroup$
    The only important thing about the extremes of the integrals is that they should cover one period of the input function. So the difference between the extremes should be $1/f_0$ which we defined as $T_0$. You can actually start from $-pi/2$, but the upper extreme should be $(-pi/2)+{T_0}$, and that will not make our calculations easier.
    $endgroup$
    – Poypoyan
    Dec 19 '18 at 18:08












  • $begingroup$
    Again thank you so so much!! now I will do other exercises in order to practice and I hope to improve :)
    $endgroup$
    – Elena Martini
    Dec 19 '18 at 18:15
















1












$begingroup$

The intuition of Fourier series is that a periodic function can be expressed as the sum of sinusoidal functions with frequencies equal to $kf_0$, or multiples of the fundamental frequency (which is $f_0$). Since the given function is a(one) sinusoid, $sin(2pi{f_0}t)$, we should immediately get that the non-zero Fourier coefficients will be only at $k=pm 1$, so



$c_k = 0$ at $kneqpm1 $.



Now, for $k=1$, we proceed to the definition (formula is copied from your question, but corrected):
$$c_k = frac{1}{T_0}int_{0}^{T_0} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt.$$
Where $T_0 = 1/f_0$. Solving that, we have:



$$begin{align}
c_k &= frac{1}{2i T_0}int_{0}^{T_0} {(e^{2 i pi f_0 t}e^{-2 i pi f_0 t} - e^{-2 i pi f_0 t}e^{-2 i pi f_0 t})}dt\ &= frac{1}{2i T_0}int_{0}^{T_0} 1 - e^{-4 i pi f_0 t}dt\
...&= frac{1}{2i T_0}(T_0 - frac{1-e^{-4ipi}}{4ipi f_0})
end{align}$$



but $e^{-4pi i} = cos(-4pi) + i sin(-4pi) = 1 + 0i = 1$, so we are left with $$c_k = frac{1}{2i T_0} (T_0) = frac{1}{2i}.$$



The given answer ($c_k=frac{1}{2} e^{-i frac{pi}{2}}$ for $k=1$), will just be equal to $frac{1}{2i}$, because $e^{-i frac{pi}{2}} = cos(-frac{pi}{2}) + i sin(-frac{pi}{2}) = 0 + (-1)i= -i = frac{1}{i}.$



The case for $k=-1$ is similar to the computation above.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you very much ! your answer was very precise and thanks above all for writing all the steps, in this way I managed to understand much better. finally I understood why already at first sight you could already find the coefficients + and - 1 and the Fourier series explained in a simple way. the only thing that still is not clear to me are the extremes of the integral? why I start from 0 and I do not go to $- pi / 2 $ ? outside of $ pm sin $, the sin function repeats itself
    $endgroup$
    – Elena Martini
    Dec 19 '18 at 17:53












  • $begingroup$
    The only important thing about the extremes of the integrals is that they should cover one period of the input function. So the difference between the extremes should be $1/f_0$ which we defined as $T_0$. You can actually start from $-pi/2$, but the upper extreme should be $(-pi/2)+{T_0}$, and that will not make our calculations easier.
    $endgroup$
    – Poypoyan
    Dec 19 '18 at 18:08












  • $begingroup$
    Again thank you so so much!! now I will do other exercises in order to practice and I hope to improve :)
    $endgroup$
    – Elena Martini
    Dec 19 '18 at 18:15














1












1








1





$begingroup$

The intuition of Fourier series is that a periodic function can be expressed as the sum of sinusoidal functions with frequencies equal to $kf_0$, or multiples of the fundamental frequency (which is $f_0$). Since the given function is a(one) sinusoid, $sin(2pi{f_0}t)$, we should immediately get that the non-zero Fourier coefficients will be only at $k=pm 1$, so



$c_k = 0$ at $kneqpm1 $.



Now, for $k=1$, we proceed to the definition (formula is copied from your question, but corrected):
$$c_k = frac{1}{T_0}int_{0}^{T_0} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt.$$
Where $T_0 = 1/f_0$. Solving that, we have:



$$begin{align}
c_k &= frac{1}{2i T_0}int_{0}^{T_0} {(e^{2 i pi f_0 t}e^{-2 i pi f_0 t} - e^{-2 i pi f_0 t}e^{-2 i pi f_0 t})}dt\ &= frac{1}{2i T_0}int_{0}^{T_0} 1 - e^{-4 i pi f_0 t}dt\
...&= frac{1}{2i T_0}(T_0 - frac{1-e^{-4ipi}}{4ipi f_0})
end{align}$$



but $e^{-4pi i} = cos(-4pi) + i sin(-4pi) = 1 + 0i = 1$, so we are left with $$c_k = frac{1}{2i T_0} (T_0) = frac{1}{2i}.$$



The given answer ($c_k=frac{1}{2} e^{-i frac{pi}{2}}$ for $k=1$), will just be equal to $frac{1}{2i}$, because $e^{-i frac{pi}{2}} = cos(-frac{pi}{2}) + i sin(-frac{pi}{2}) = 0 + (-1)i= -i = frac{1}{i}.$



The case for $k=-1$ is similar to the computation above.






share|cite|improve this answer









$endgroup$



The intuition of Fourier series is that a periodic function can be expressed as the sum of sinusoidal functions with frequencies equal to $kf_0$, or multiples of the fundamental frequency (which is $f_0$). Since the given function is a(one) sinusoid, $sin(2pi{f_0}t)$, we should immediately get that the non-zero Fourier coefficients will be only at $k=pm 1$, so



$c_k = 0$ at $kneqpm1 $.



Now, for $k=1$, we proceed to the definition (formula is copied from your question, but corrected):
$$c_k = frac{1}{T_0}int_{0}^{T_0} frac{e^{2 i pi f_0 t} - e^{-2 i pi f_0 t}}{2i}e^{-2 i pi f_0 t}dt.$$
Where $T_0 = 1/f_0$. Solving that, we have:



$$begin{align}
c_k &= frac{1}{2i T_0}int_{0}^{T_0} {(e^{2 i pi f_0 t}e^{-2 i pi f_0 t} - e^{-2 i pi f_0 t}e^{-2 i pi f_0 t})}dt\ &= frac{1}{2i T_0}int_{0}^{T_0} 1 - e^{-4 i pi f_0 t}dt\
...&= frac{1}{2i T_0}(T_0 - frac{1-e^{-4ipi}}{4ipi f_0})
end{align}$$



but $e^{-4pi i} = cos(-4pi) + i sin(-4pi) = 1 + 0i = 1$, so we are left with $$c_k = frac{1}{2i T_0} (T_0) = frac{1}{2i}.$$



The given answer ($c_k=frac{1}{2} e^{-i frac{pi}{2}}$ for $k=1$), will just be equal to $frac{1}{2i}$, because $e^{-i frac{pi}{2}} = cos(-frac{pi}{2}) + i sin(-frac{pi}{2}) = 0 + (-1)i= -i = frac{1}{i}.$



The case for $k=-1$ is similar to the computation above.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 17:13









PoypoyanPoypoyan

473411




473411












  • $begingroup$
    thank you very much ! your answer was very precise and thanks above all for writing all the steps, in this way I managed to understand much better. finally I understood why already at first sight you could already find the coefficients + and - 1 and the Fourier series explained in a simple way. the only thing that still is not clear to me are the extremes of the integral? why I start from 0 and I do not go to $- pi / 2 $ ? outside of $ pm sin $, the sin function repeats itself
    $endgroup$
    – Elena Martini
    Dec 19 '18 at 17:53












  • $begingroup$
    The only important thing about the extremes of the integrals is that they should cover one period of the input function. So the difference between the extremes should be $1/f_0$ which we defined as $T_0$. You can actually start from $-pi/2$, but the upper extreme should be $(-pi/2)+{T_0}$, and that will not make our calculations easier.
    $endgroup$
    – Poypoyan
    Dec 19 '18 at 18:08












  • $begingroup$
    Again thank you so so much!! now I will do other exercises in order to practice and I hope to improve :)
    $endgroup$
    – Elena Martini
    Dec 19 '18 at 18:15


















  • $begingroup$
    thank you very much ! your answer was very precise and thanks above all for writing all the steps, in this way I managed to understand much better. finally I understood why already at first sight you could already find the coefficients + and - 1 and the Fourier series explained in a simple way. the only thing that still is not clear to me are the extremes of the integral? why I start from 0 and I do not go to $- pi / 2 $ ? outside of $ pm sin $, the sin function repeats itself
    $endgroup$
    – Elena Martini
    Dec 19 '18 at 17:53












  • $begingroup$
    The only important thing about the extremes of the integrals is that they should cover one period of the input function. So the difference between the extremes should be $1/f_0$ which we defined as $T_0$. You can actually start from $-pi/2$, but the upper extreme should be $(-pi/2)+{T_0}$, and that will not make our calculations easier.
    $endgroup$
    – Poypoyan
    Dec 19 '18 at 18:08












  • $begingroup$
    Again thank you so so much!! now I will do other exercises in order to practice and I hope to improve :)
    $endgroup$
    – Elena Martini
    Dec 19 '18 at 18:15
















$begingroup$
thank you very much ! your answer was very precise and thanks above all for writing all the steps, in this way I managed to understand much better. finally I understood why already at first sight you could already find the coefficients + and - 1 and the Fourier series explained in a simple way. the only thing that still is not clear to me are the extremes of the integral? why I start from 0 and I do not go to $- pi / 2 $ ? outside of $ pm sin $, the sin function repeats itself
$endgroup$
– Elena Martini
Dec 19 '18 at 17:53






$begingroup$
thank you very much ! your answer was very precise and thanks above all for writing all the steps, in this way I managed to understand much better. finally I understood why already at first sight you could already find the coefficients + and - 1 and the Fourier series explained in a simple way. the only thing that still is not clear to me are the extremes of the integral? why I start from 0 and I do not go to $- pi / 2 $ ? outside of $ pm sin $, the sin function repeats itself
$endgroup$
– Elena Martini
Dec 19 '18 at 17:53














$begingroup$
The only important thing about the extremes of the integrals is that they should cover one period of the input function. So the difference between the extremes should be $1/f_0$ which we defined as $T_0$. You can actually start from $-pi/2$, but the upper extreme should be $(-pi/2)+{T_0}$, and that will not make our calculations easier.
$endgroup$
– Poypoyan
Dec 19 '18 at 18:08






$begingroup$
The only important thing about the extremes of the integrals is that they should cover one period of the input function. So the difference between the extremes should be $1/f_0$ which we defined as $T_0$. You can actually start from $-pi/2$, but the upper extreme should be $(-pi/2)+{T_0}$, and that will not make our calculations easier.
$endgroup$
– Poypoyan
Dec 19 '18 at 18:08














$begingroup$
Again thank you so so much!! now I will do other exercises in order to practice and I hope to improve :)
$endgroup$
– Elena Martini
Dec 19 '18 at 18:15




$begingroup$
Again thank you so so much!! now I will do other exercises in order to practice and I hope to improve :)
$endgroup$
– Elena Martini
Dec 19 '18 at 18:15


















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Plaza Victoria

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Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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