MHB Fourier series damped driven oscillator ODE

Dustinsfl
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$$
-\sum_{n = 0}^{\infty}n^2\omega^2C_ne^{in\omega t} + 2\beta\sum_{n = 0}^{\infty}in\omega C_ne^{in\omega t} + \omega_0^2\sum_{n = 0}^{\infty}C_ne^{in\omega t} = \sum_{n = 0}^{\infty}f_ne^{in\omega t}
$$
How can I justify removing the summations and solving for $C_n$?
$$
-n^2\omega^2C_ne^{in\omega t} + 2\beta in\omega C_ne^{in\omega t} + \omega_0^2C_ne^{in\omega t} = f_ne^{in\omega t}
$$
 
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dwsmith said:
$$
-\sum_{n = 0}^{\infty}n^2\omega^2C_ne^{in\omega t} + 2\beta\sum_{n = 0}^{\infty}in\omega C_ne^{in\omega t} + \omega_0^2\sum_{n = 0}^{\infty}C_ne^{in\omega t} = \sum_{n = 0}^{\infty}f_ne^{in\omega t}
$$
How can I justify removing the summations and solving for $C_n$?
$$
-n^2\omega^2C_ne^{in\omega t} + 2\beta in\omega C_ne^{in\omega t} + \omega_0^2C_ne^{in\omega t} = f_ne^{in\omega t}
$$
Combining the summations, you can write this as $$\sum_{n = 0}^{\infty}\bigl(-n^2\omega^2C_n + 2\beta in\omega C_n + \omega_0^2 C_n \bigr) e^{in\omega t} = \sum_{n = 0}^{\infty}f_ne^{in\omega t}.$$
Now use the fact that a (reasonably well-behaved) function has a unique Fourier expansion to conclude that the coefficients on each side must be the same, to conclude that $-n^2\omega^2C_n + 2\beta in\omega C_n + \omega_0^2 C_n = f_n$ for each $n.$
 
Opalg said:
Combining the summations, you can write this as $$\sum_{n = 0}^{\infty}\bigl(-n^2\omega^2C_n + 2\beta in\omega C_n + \omega_0^2 C_n \bigr) e^{in\omega t} = \sum_{n = 0}^{\infty}f_ne^{in\omega t}.$$
Now use the fact that a (reasonably well-behaved) function has a unique Fourier expansion to conclude that the coefficients on each side must be the same, to conclude that $-n^2\omega^2C_n + 2\beta in\omega C_n + \omega_0^2 C_n = f_n$ for each $n.$

Could I also just multiple through by $\frac{1}{2\pi}\overline{e^{in\omega t}}$, and by Sturm-Liouville, the summation integrates to 1?
 
Last edited:
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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