Extension of Half-Range Fourier Sine Function to Symmetric Range

Theorem
Let $\map f x$ be a real function defined on the interval $\openint 0 \lambda$.

Let $\map f x$ be represented by the half-range Fourier sine series $\map S x$:


 * $\map f x \sim \map S x = \ds \sum_{n \mathop = 1}^\infty b_n \sin \frac {n \pi x} \lambda$

where for all $n \in \Z_{> 0}$:
 * $b_n = \ds \frac 2 \lambda \int_0^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x$

Then $\map S x$ also represents the extension $g: \openint {-\lambda} \lambda \to \R$ of $f$, defined as:


 * $\forall x \in \openint {-\lambda} \lambda: \map g x = \begin {cases} \map f x & : x > 0 \\ -\map f {-x} & : x < 0 \\ 0 & : x = 0 \end {cases}$

Proof
It is apparent by inspection that:
 * $(1): \quad g$ is an extension of $f$
 * $(2): \quad g$ is an odd function.

Let $\map T x$ be the Fourier series representing $g$:


 * $\map g x \sim \map T x = \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$

where for all $n \in \Z_{> 0}$:


 * $a_n = \ds \frac 1 \lambda \int_{-\lambda}^\lambda \map g x \cos \frac {n \pi x} \lambda \rd x$


 * $b_n = \ds \frac 1 \lambda \int_{-\lambda}^\lambda \map g x \sin \frac {n \pi x} \lambda \rd x$

From Fourier Cosine Coefficients for Odd Function over Symmetric Range:
 * $\forall n \in \Z_{\ge 0}: a_n = 0$

and from Fourier Sine Coefficients for Odd Function over Symmetric Range:
 * $\forall n \in \Z_{>0}: b_n = \ds \frac 2 \lambda \int_0^\lambda \map g x \sin \frac {n \pi x} \lambda \rd x$

But on $\openint 0 \lambda$, by definition:
 * $\forall x \in \openint 0 \lambda: \map g x = \map f x$

Hence:
 * $\map T x = \ds \sum_{n \mathop = 1}^\infty \paren b_n \sin \frac {n \pi x} \lambda$

where:
 * $b_n = \ds \frac 2 \lambda \int_0^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x$

That is, $\map T x$ is the same as $\map S x$.

Hence the result.