Half-Range Fourier Sine Series over Negative Range
Theorem
Let $\map f x$ be a real function defined on the open real interval $\openint 0 \lambda$.
Let $f$ be expressed using the half-range Fourier sine series over $\openint 0 \lambda$:
- $\ds \map S x \sim \sum_{n \mathop = 1}^\infty b_n \sin \frac {n \pi x} \lambda$
where:
- $b_n = \ds \frac 2 \lambda \int_0^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x$
for all $n \in \Z_{\ge 0}$.
Then over the interval $\openint {-\lambda} 0$, $\map S x$ takes the values:
- $\map S x = -\map f {-x}$
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That is, the real function expressed by the half-range Fourier sine series over $\openint 0 \lambda$ is an odd function over $\openint {-\lambda} \lambda$.
Proof
From Fourier Series for Odd Function over Symmetric Range, $\map S x$ is the Fourier series of an odd real function over the interval $\openint 0 \lambda$.
We have that $\map S x \sim \map f x$ over $\openint 0 \lambda$.
Thus over $\openint {-\lambda} 0$ it follows that:
- $\map S x = -\map f {-x}$
$\blacksquare$
Sources
- 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Chapter One: $\S 5$. Half-Range Sine Series