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In several texts, I see that the Fourier series of a function $f=f(t)$ with period $2L$ is defined by

$$Sf(t):=a_0+\sum_{n=1}^{\infty}\left[a_n\cos\frac{n\pi t}{L}+b_n\sin\frac{n\pi t}{L}\right]$$

I wonder why this is not written this way instead:

$$Sf(t):=\sum_{n=0}^{\infty}\left[a_n\cos\frac{n\pi t}{L}+b_n\sin\frac{n\pi t}{L}\right]$$

Thanks in advance for the clarification.

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  • $\begingroup$ I guess it doesn't hurt, it's just that $b_0$ doesn't mean anything because $\sin 0 = 0$. $\endgroup$ – axblount Sep 14 '12 at 20:25
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    $\begingroup$ The integral used to calculate $a_0$ is "off" from that of the other $a_n$ by a factor of $2$. Personally I find that separating it makes this easier to remember. $\endgroup$ – Robert Mastragostino Sep 14 '12 at 20:27
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    $\begingroup$ If you want the coefficients $a_n$ to be $\displaystyle \frac{1}{L}\int_{-L}^L f(t)\cos(\frac{n\pi t}{L})$ for every $n\geq 0$ then the good formula has to be $Sf(t)=\frac{a_0}{2}+\sum_{n\geq 1}a_n\cos(\frac{n\pi t}{L})+b_n\sin(\frac{n\pi t}{L})$. $\endgroup$ – Bebop Sep 14 '12 at 20:28
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    $\begingroup$ Forget about sin and cosine and get to know the exponential function then everything becomes symmetric :-) $\endgroup$ – Fabian Sep 14 '12 at 21:55
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People usually choose to handle the $a_0$ in these two ways (for real Fourier series):

\begin{gather} f(t)=\color{blue}{a_0}+\sum_{n=1}^{\infty}\left[a_n\cos\frac{n\pi t}{L}+b_n\sin\frac{n\pi t}{L}\right], \quad -L<t<L,\\ a_0=\color{blue}{{1\over 2L}}\int_{-L}^L f(t)\,dt, \quad a_n=\color{red}{{1\over L}}\int_{-L}^L f(t)\cos(n\pi t/L)\,dt,\quad b_n=\color{red}{{1\over L}}\int_{-L}^L f(t)\sin(n\pi t/L)\,dt \end{gather}

(which can certainly be written as $\sum_{n=0}^{\infty}\left[a_n\cos\frac{n\pi t}{L}+b_n\sin\frac{n\pi t}{L}\right]$, although in my experience this is not common)

vs.

\begin{gather} f(t)=\color{blue}{{1\over 2}a_0}+\sum_{n=1}^{\infty}\left[a_n\cos\frac{n\pi t}{L}+b_n\sin\frac{n\pi t}{L}\right], \quad -L<t<L,\\ a_0=\color{blue}{{1\over L}}\int_{-L}^L f(t)\,dt, \quad a_n=\color{red}{{1\over L}}\int_{-L}^L f(t)\cos(n\pi t/L)\,dt,\quad b_n=\color{red}{{1\over L}}\int_{-L}^L f(t)\sin(n\pi t/L)\,dt \end{gather}

They are of course equivalent.

The benefit of the latter formulation is that the factors on the front of the integrals is all the same, but this also requires you to remember the ${1\over 2}$ in front of the $a_0$ in the series.

Hope that helps.

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