If one expands a function $phi(x,v(theta),t)$ to complex Fourier: $sum_{k=-infty}^{infty} phi_k(x,t)...
$begingroup$
If one expands a function $phi(x,v(theta),t)$ to complex Fourier series marked like: $$phi(x,v(theta),t)=sum_{k=-infty}^{infty} phi_k(x,t) e^{iktheta}$$
then why/how does argument $v$ "disappear"?
Intuitively,
$$phi_k(x,t)=frac{1}{2pi} int_{-pi}^{pi} e^{-int} phi(x,v(theta),t) dsomething$$
Would the notation used perhaps suggest that $dsomething=dv$?
And that by "going backwards" one "recovers" $v$ to $phi(x,v,t)$, when one computes the complex Fourier series?
fourier-series
asked Dec 23 '18 at 17:52
mavaviljmavavilj
2,81911137
$endgroup$
add a comment |
$begingroup$
If one expands a function $phi(x,v(theta),t)$ to complex Fourier series marked like: $$phi(x,v(theta),t)=sum_{k=-infty}^{infty} phi_k(x,t) e^{iktheta}$$
then why/how does argument $v$ "disappear"?
Intuitively,
$$phi_k(x,t)=frac{1}{2pi} int_{-pi}^{pi} e^{-int} phi(x,v(theta),t) dsomething$$
Would the notation used perhaps suggest that $dsomething=dv$?
And that by "going backwards" one "recovers" $v$ to $phi(x,v,t)$, when one computes the complex Fourier series?
fourier-series
asked Dec 23 '18 at 17:52
mavaviljmavavilj
2,81911137
$endgroup$
add a comment |
$begingroup$
If one expands a function $phi(x,v(theta),t)$ to complex Fourier series marked like: $$phi(x,v(theta),t)=sum_{k=-infty}^{infty} phi_k(x,t) e^{iktheta}$$
then why/how does argument $v$ "disappear"?
Intuitively,
$$phi_k(x,t)=frac{1}{2pi} int_{-pi}^{pi} e^{-int} phi(x,v(theta),t) dsomething$$
Would the notation used perhaps suggest that $dsomething=dv$?
And that by "going backwards" one "recovers" $v$ to $phi(x,v,t)$, when one computes the complex Fourier series?
fourier-series
asked Dec 23 '18 at 17:52
mavaviljmavavilj
2,81911137
$endgroup$
If one expands a function $phi(x,v(theta),t)$ to complex Fourier series marked like: $$phi(x,v(theta),t)=sum_{k=-infty}^{infty} phi_k(x,t) e^{iktheta}$$
then why/how does argument $v$ "disappear"?
Intuitively,
$$phi_k(x,t)=frac{1}{2pi} int_{-pi}^{pi} e^{-int} phi(x,v(theta),t) dsomething$$
Would the notation used perhaps suggest that $dsomething=dv$?
And that by "going backwards" one "recovers" $v$ to $phi(x,v,t)$, when one computes the complex Fourier series?
fourier-series
fourier-series
asked Dec 23 '18 at 17:52
mavaviljmavavilj
2,81911137
asked Dec 23 '18 at 17:52
mavaviljmavavilj
2,81911137
edited Dec 23 '18 at 19:04
mavavilj
asked Dec 23 '18 at 17:52
mavaviljmavavilj
2,81911137
asked Dec 23 '18 at 17:52
mavaviljmavavilj
2,81911137
asked Dec 23 '18 at 17:52
mavaviljmavavilj
2,81911137
2,81911137
add a comment |
add a comment |
1 Answer
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As given there are added variables and missing variables.
If given a function of three variables one may expand a function about one of the variables, as in:
$$phi(x,t,theta) = sum_{n=0}^{infty} phi_{n}(x,t) , e^{i n theta}.$$
In an integral case a variable can be removed, or evaluated, as in:
$$int_{a}^{b} phi(x,t,theta) , dtheta = sigma(x,t),$$
where $sigma$ is the function of two variables after integration.
answered Dec 23 '18 at 18:44
LeucippusLeucippus
19.6k102871
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$begingroup$
As given there are added variables and missing variables.
If given a function of three variables one may expand a function about one of the variables, as in:
$$phi(x,t,theta) = sum_{n=0}^{infty} phi_{n}(x,t) , e^{i n theta}.$$
In an integral case a variable can be removed, or evaluated, as in:
$$int_{a}^{b} phi(x,t,theta) , dtheta = sigma(x,t),$$
where $sigma$ is the function of two variables after integration.
answered Dec 23 '18 at 18:44
LeucippusLeucippus
19.6k102871
$endgroup$
add a comment |
$begingroup$
As given there are added variables and missing variables.
If given a function of three variables one may expand a function about one of the variables, as in:
$$phi(x,t,theta) = sum_{n=0}^{infty} phi_{n}(x,t) , e^{i n theta}.$$
In an integral case a variable can be removed, or evaluated, as in:
$$int_{a}^{b} phi(x,t,theta) , dtheta = sigma(x,t),$$
where $sigma$ is the function of two variables after integration.
answered Dec 23 '18 at 18:44
LeucippusLeucippus
19.6k102871
$endgroup$
add a comment |
$begingroup$
As given there are added variables and missing variables.
If given a function of three variables one may expand a function about one of the variables, as in:
$$phi(x,t,theta) = sum_{n=0}^{infty} phi_{n}(x,t) , e^{i n theta}.$$
In an integral case a variable can be removed, or evaluated, as in:
$$int_{a}^{b} phi(x,t,theta) , dtheta = sigma(x,t),$$
where $sigma$ is the function of two variables after integration.
answered Dec 23 '18 at 18:44
LeucippusLeucippus
19.6k102871
$endgroup$
As given there are added variables and missing variables.
If given a function of three variables one may expand a function about one of the variables, as in:
$$phi(x,t,theta) = sum_{n=0}^{infty} phi_{n}(x,t) , e^{i n theta}.$$
In an integral case a variable can be removed, or evaluated, as in:
$$int_{a}^{b} phi(x,t,theta) , dtheta = sigma(x,t),$$
where $sigma$ is the function of two variables after integration.
answered Dec 23 '18 at 18:44
LeucippusLeucippus
19.6k102871
answered Dec 23 '18 at 18:44
LeucippusLeucippus
19.6k102871
answered Dec 23 '18 at 18:44
LeucippusLeucippus
19.6k102871
answered Dec 23 '18 at 18:44
LeucippusLeucippus
19.6k102871
19.6k102871
add a comment |
add a comment |
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