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The equivalence between exponential form of Fourier series and trigonometric form of fourier series. I do not know How they are equivalent, could anyone explain this for me please?

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    $\begingroup$ How much do you know about $L^2[0,2\pi]$? $\endgroup$ – user228113 May 19 '17 at 10:04
  • $\begingroup$ I know to the extent that the partial sum of Fourier series is convergent in L^2. $\endgroup$ – Emptymind May 19 '17 at 10:24
  • $\begingroup$ I'm not sure I understand. The relation $e^{in \theta} = \cos( n \theta) + i sin(n \theta)$ allows us to convert a exponential Fourier serie into a trigonometric serie and vice-versa. Was that your question ? $\endgroup$ – user171326 May 19 '17 at 10:43
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My attempt to answer your question

Given that $\sqrt{-1} = j$, the exponential fourier series can describe periodic functions, with period $T_c$, that take on complex values. It can be denoted as follows

$$ f(t) = \sum_{n = - \infty}^{\infty}X_ne^{jn\omega_c t} $$

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$$ Xn = \frac{\left< f(t), e^{jn\omega_c t} \right>_{PP}}{{\| e^{jn\omega_c t} \|}^2} = \frac{1}{T}\int_{a -T/2}^{a + T/2} f(t)e^{-jn\omega_c t}dt $$

$$ X_n = |X_n|\angle \theta_n $$

Here, $f(t)$ can take on complex values because both $X_n$ and $e^{jn\omega_c t}$ take on complex values. If you want to describe a complex function with other functions, those other functions have to be complex too. (Notice how the inner product takes the complex conjugate of the second function in $X_n$)

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The trigonometric fourier series can only describe periodic functions, with period $T_c$, that take on real values (The major difference). It can be denoted as follows

$$ f(t) = a_0 + \sum_{n = 1}^{\infty}a_n cos(n\omega_c t) + \sum_{n = 1}^{\infty}b_n cos(n\omega_c t)$$

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$$ a_v = \frac{a_0}{2} = \frac{\left< f(t), 1 \right>_{PP}}{{\| 1 \|}^2} = \frac{1}{T}\int_{a-T/2}^{a+T/2} f(t)dt $$

$$ a_n = \frac{\left< f(t), cos( n\omega_c t) \right>_{PP}}{{\| cos( n\omega_c t) \|}^2} = \frac{2}{T}\int_{a-T/2}^{a+T/2} f(t)cos( n\omega_c t)dt $$

$$ b_n = \frac{\left< f(t), sin( n\omega_c t) \right>_{PP}}{{\| sin( n\omega_c t) \|}^2} = \frac{2}{T}\int_{a-T/2}^{a+T/2} f(t)sin( n\omega_c t)dt $$

Here the trigonometric functions, and the constant $1$, are restricted to take on only real values, and thus $f(t)$ can only take on real values.

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My attempt to show the connection between the two fourier series

For real values of $f(t)$ the exponential fourier series can be simplified to

$$ f(t) = X_0 + 2\sum_{n = 1}^{\infty}|X_n|cos(n\omega_c t + \theta_n) $$

This should remind you of the compact trigonometric fourier series which can be denoted as follows

$$ f(t) = \frac{C_0 cos(\theta_0)}{2} + \sum_{n = 1}^{\infty}C_ncos(n\omega_c t + \theta_n) $$

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$$ C_n = \sqrt{(a_n)^2 + (b_n)^2} $$

$$ \theta_n = \begin{cases} -arctan(\frac{b_n}{a_n}) &; \space a_n \geq 0 \\ \pi -arctan(\frac{b_n}{a_n}) &; \space a_n \lt 0 \end{cases} $$

We can confirm the first term, knowing $b_0 = 0$, by looking at the following evaluation

$$ \theta_0 = \begin{cases} 0 &; \space a_n \geq 0 \\ \pi &; \space a_n \lt 0 \end{cases} $$

$$ \theta_0 = sign(a_n)$$

$$ C_0 = \sqrt{ (a_n)^2} = |a_n|$$

$$ C_0 cos(\theta_0) = |a_n| sign(a_0) = a_0 $$

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We can thus make the following observation regarding the different coefficients

$$ C_n = 2|X_n| \space , \space \space n \geq 0 $$

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Conclusion

From the above definitions, we can derive the following combining definition

$$ X_n = \frac{a_n - jb_n}{2} $$

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