Trignometric Fourier series representation of a continous time signal
alright, let's review a little bit of Euler before we get to the Fourier.

$$ e^{j \theta} \ = \ \cos(\theta) \ + \ j \sin(\theta) $$

from that you can get

$$ \cos(\theta) = \frac{e^{j \theta} + e^{-j \theta}}{2} \quad\quad\quad \sin(\theta) = \frac{e^{j \theta} - e^{-j \theta}}{2j} $$

so now let's look at Eq (1)

$$ \begin{align}
x(t) \ &amp;= \ A_0 \ + \ 2 \sum_{k=1}^{\infty} A_k \cos(k \omega_0 t) - B_k \sin(k \omega_0 t) \\
 &amp;= \ A_0 \ + \ 2 \sum_{k=1}^{\infty} A_k \frac{e^{j k \omega_0 t} + e^{-j k \omega_0 t}}{2} \ - \ B_k \frac{e^{j k \omega_0 t} - e^{-j k \omega_0 t}}{2j} \\
 &amp;= \ A_0 \ + \ \sum_{k=1}^{\infty} (A_k + jB_k) e^{j k \omega_0 t} \ + \ (A_k - jB_k) e^{-j k \omega_0 t} \\
 &amp;= \ \sum_{k=-\infty}^{\infty} c_k \ e^{j k \omega_0 t} \\
\end{align} $$

where

$$ c_k =
\begin{cases}
A_{-k} - jB_{-k}, &amp; \quad \text{for } k &lt; 0 \\
A_0, &amp; \quad\quad k = 0 \\
A_k + jB_k, &amp; \quad\quad k &gt; 0 
\end{cases} $$

and, going in the other direction,

$$ \begin{array}{lcl}
A_0 &amp; = &amp; c_0 \\
A_k &amp; = &amp; \Re\{c_k \} \quad\quad \text{for } k&gt;0\\
B_k &amp; = &amp; \Im\{c_k \} \quad\quad\quad k&gt;0
\end{array} $$

note that for real $A_k$ and real $B_k$, then

$$ c_{-k} = c_k^* = \text{"complex conjugate of } c_k \text{ "}$$

so, for Eq (1), the simplicity of having the real and imaginary parts of the $c_k$ coefficients be simply $A_k$ and $B_k$ (for $k \ge 0$) is the motivation for the convention of the leading "2" before the summation and for the minus sign.  it's just a convention.  this convention is useful because it's much easier to derive the coefficients $c_k$ than it is to derive the coefficients $A_k$ and $B_k$.

$$ c_k \ = \ \frac{\omega_0}{2 \pi}\int_{t_0 - \pi/\omega_0}^{t_0 + \pi/\omega_0} x(t) \ e^{-j k \omega_0 t} \ dt $$

where $ -\infty &lt; t_0 &lt; +\infty $ can be any convenient real value.

the convention for Eq. (2) comes about from looking at Fourier series first as a real analysis problem without the use of complex variables.  then this is a simple first statement:

$$x(t) \ = \ a_0 \ + \ \sum_{k=1}^{\infty} a_k \cos(k \omega_0 t) + b_k \sin(k \omega_0 t)$$

as you can see $a_k$ and $b_k$ are related to $A_k$ and $B_k$ in a very simple and straight forward manner.  you can derive that simple relationship.  but the formulae for getting $a_k$ and $b_k$ from $x(t)$ (and $\omega_0$) is less straight forward to derive and to express.  if fact, it's two equations, not one simpler equation.