Half-Range Fourier Sine Series over Negative Range

From ProofWiki
Jump to navigation Jump to search

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:

$\map S x = -\map f {-x}$



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