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If a function $f$ is defined on $[-L, L]$, then its Fourier series is given by $$ \begin{aligned} a _ { 0 } & = \frac { 1 } { L } \int _ { - L } ^ { L } f \left( x \right) d x \\ a _ { n } & = \frac { 1 } { L } \int _ { - L } ^ { L } f \left( x \right) \cos \left( \frac { n \pi x } { L } \right) d x \\ b _ { n } & = \frac { 1 } { L } \int _ { - L } ^ { L } f \left( x \right) \sin \left( \frac { n \pi x } { L } \right) d x \end{aligned} $$

This means that the even though the function is defined on $[L, L]$, the fourier series representation is over $(-\infty, \infty)$ and with period $2L$. However, if a function is defined on $[0, 2L]$, then the formulas changes slightly, $$ \begin{aligned} a _ { 0 } & = \frac { 1 } { L } \int _ { 0 } ^ { 2 L } f \left( x \right) d x \\ a _ { n } & = \frac { 1 } { L } \int _ { 0 } ^ { 2 L } f \left( x \right) \cos \left( \frac { n \pi x } { L } \right) d \\ b _ { n } & = \frac { 1 } { L } \int _ { 0 } ^ { 2 L } f \left( x \right) \sin \left( \frac { n \pi x} { L } \right) d \end{aligned} $$

See the difference in the limits of integration. I see how the equations are derived but I am missing the fundamental connection of how the "period" comes into play here. In particular, I am not understanding the statement

For f(x) periodic with period 2L, any interval $(x_0,x_0+2L)$ can be used, with the choice being one of convenience or personal preference

Tyler Hilton
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1 Answers1

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Suppose $T$ is a period of $f$, then $\underline {\text{for each }A \in \mathbb R }$ there is some $n \in \mathbb Z$ s.t. $nT \leqslant A < (n+1)T$, then $A = nT +s$ for some $s \in [0, T)$, so $$ \boxed {\int_A^{T+A} f (t)\mathrm dt}= \int_{nT+s}^{(n+1)T+s} f(t )\mathrm dt =\int_{s-T}^s f(u+(n+1)T)\mathrm du [t = u + (n+1)T] = \int_{s-T}^s f(u)\mathrm du \;[\text{periodicity}] = \left(\int_{s-T}^0 + \int_0^s\right) f = \int_{s-T}^0 f(t+T)\mathrm dt +\int_0^s f \;[\text{periodicity}] = \left(\int_s^T + \int_0^s\right) f = \boxed {\int_0^T f}\;. $$

xbh
  • 8,835
  • Thanks, I get this but I don't get how the limits of integration has changed to $2L$ while the coefficient is still $1/L$ and even inside the trig function, the division is by $L$. – Tyler Hilton Dec 10 '18 at 04:44
  • Then you could just extend $f$ on to $\overline f\colon \mathbb R \to \mathbb R$ by periodicity $2L$ , then apply the formula for $\overline f$ on $[-L, L]$, then change the limits of integration. – xbh Dec 10 '18 at 04:50