Half-Range Fourier Sine Series over Negative Range

Theorem
Let $f \left({x}\right)$ be a real function defined on the interval $\left[{0 \,.\,.\, l}\right]$.

Let $f$ be expressed using the half-range Fourier sine series over $\left[{0 \,.\,.\, l}\right]$:


 * $\displaystyle S \left({x}\right) \sim \sum_{n \mathop = 1}^\infty b_n \sin \frac {n \pi x} l$

where:
 * $b_n = \displaystyle \frac 2 l \int_0^l f \left({x}\right) \sin \frac {n \pi x} l \, \mathrm d x$

for all $n \in \Z_{\ge 0}$.

Then over the interval $\left[{-l \,.\,.\, 0}\right]$, $S \left({x}\right)$ takes the values:
 * $S \left({x}\right) = -f \left({-x}\right)$

That is, the real function expressed by the half-range Fourier sine series over $\left[{0 \,.\,.\, l}\right]$ is an odd function over $\left[{-l \,.\,.\, l}\right]$.

Proof
From Fourier Series for Odd Function over Symmetric Range, $S \left({x}\right)$ is the Fourier series of an odd real function over the interval $\left[{1 \,.\,.\, l}\right]$.

We have that $S \left({x}\right) \sim f \left({x}\right)$ over $\left[{0 \,.\,.\, l}\right]$.

Thus over $\left[{-l \,.\,.\, 0}\right]$ it follows that $S \left({x}\right) = -f \left({-x}\right)$.