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Trigonometric Fourier series is given as

$$x(t)=\frac {a_0} 2 + \sum \limits _{m=1} ^\infty (a_m \cos \frac {2 \pi m t} T + b_m \sin \frac {2 \pi m t} T) \quad(1)$$

Polar FS is given as

$$x(t)=\sum \limits _{m=0} ^\infty c_m \cos({w_m t + \theta_m}) \quad (2)$$

Exponential FS is given as

$$x(t) = \sum\limits_{m=-\infty}^{+\infty} c_m e^ \frac{j2 \pi m t}{T_0} \quad (3)$$

If anyone notices, the time period range of exponential Fourier series is from $-\infty$ to + $\infty$. While other two types of Fourier series ( trigonometric and polar ) have time range from 0 to $+\infty$ . Why this is so?

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Cosine is an even function; sine is odd. Using negative indices for these functions adds no information because, for example, $$5\sin 2x+ 4\sin(-2x) = \sin 2x$$ On the other hand, $e^{-2ix}$ is not just a constant multiple of $e^{2ix}$. And we need both of these to express $\sin 2x$, as in $$\sin 2x = \frac1{2i}(e^{2ix} - e^{-2ix})$$


A more precise answer is that each of the families $$\{\cos nx : n=0,1,2,\dots\}\cup \{\sin nx : n=1,2,\dots\} \tag1 $$ and $$\{\exp(i nx) : n\in\mathbb{Z}\}$$ is a basis for the space of trigonometric polynomials. Each of them is linearly independent; also, any element of (1) can be expressed as a linear combination of the elements of (2) and the other way around. We would not get a basis by allowing negative $n$ in (1), or by disallowing them in (2).

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  • $\begingroup$ I took T to be $2\pi$ for simplicity, if this is your concern. $\endgroup$ – user147263 Jun 8 '16 at 4:22

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