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I want to ask Question about the Fourier series in continuous time domain while reading a book signals and systems Alan Oppenheim. I have confusion in understanding the statement on page 189 of its 2nd edition.
To derive alternative form of Fourier series, we first rearrange the summation in Eq.3.25 as $$x(t) = \sum^{+\infty}_{k\ =\ 1} [a_k e^{jk\omega_0t}+a_{-k} e^{-jk\omega_0t}]$$

where Eq. 3.25$$x(t) = \sum^{+\infty}_{k\ =\ -\infty} a_k e^{jk\omega_0t}$$ I want to know two things
1) How summation from -infinity to +infinity changes to 1 to +infinity

2) How we get this term $$a_{-k} e^{-jk\omega_0t}$$ in the equation.

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You just need to split the sum into the positive and the negative indices (without forgetting the index $k=0$):

$$x(t)=a_0+\sum_{k=1}^{\infty}a_ke^{jk\omega_0 t}+\sum_{k=-\infty}^{-1}a_ke^{jk\omega_0 t}\tag{1}$$

By simply changing the sign of the index $k$ in the second sum in $(1)$ you get

$$\begin{align}x(t)&=a_0+\sum_{k=1}^{\infty}a_ke^{jk\omega_0 t}+\sum_{k=1}^{\infty}a_{-k}e^{-jk\omega_0 t}\\&= a_0+\sum_{k=1}^{\infty}\left[a_ke^{jk\omega_0 t}+a_{-k}e^{-jk\omega_0 t}\right]\end{align}\tag{2}$$

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  • $\begingroup$ we can change the sign of index K by multiplying -1 to summation? $\endgroup$ – Aadnan Farooq A Oct 5 '15 at 12:46
  • $\begingroup$ @AadnanFarooqA: No, you just change the sign both in the sum and in elements over which you sum: e.g., $\sum_{k=-2}^{-1}f_k=\sum_{k=1}^2f_{-k}=f_{-2}+f_{-1}$ $\endgroup$ – Matt L. Oct 5 '15 at 13:06
  • $\begingroup$ Can you please answer my this question $\endgroup$ – Aadnan Farooq A Oct 5 '15 at 13:55

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