PX153 - J5 - symmetric and antisymmetric functions

if $f (x) = f (- x) : f (x)$ is symmetric. eg: $\cos n x$
if $f (x) = - f (- x) : f (x)$ is antisymmetric. eg: $\sin n x$
for symmetric functions: $b_{n} = 0 \forall n$
for anti-symmetric functions: $a_{n} = 0 \forall n$
symmetric functions have only cosine terms, and anti-symmetric functions have only sine terms in their fourier series

proof of the anti-symmetric case

\begin{aligned} a_{n} & = \frac{1}{π} \int_{- π}^{π} f (x) \cos (n x) d x \\ = \frac{1}{π} [\int_{- π}^{0} f (x) \cos (n x) d x + \int_{0}^{π} f (x) \cos (n x) d x] \\ l e t x^{'} = - x ⟹ d x^{'} = - d x \\ = \frac{1}{π} [\int_{π}^{0} - f (x^{'}) \cos (n x^{'}) (- d x^{'}) + \int_{0}^{π} f (x) \cos (n x) d x] \\ = \frac{1}{π} [- \int_{0}^{π} f (x) \cos (n x) d x + \int_{0}^{π} f (x) \cos (n x) d x] \\ = 0 \end{aligned}

\begin{aligned} a_{n} & = \frac{1}{π} \int_{- π}^{π} f (x) \sin (n x) d x \\ = \frac{1}{π} [\int_{- π}^{0} f (x) \sin (n x) d x + \int_{0}^{π} f (x) \sin (n x) d x] \\ l e t x^{'} = - x ⟹ d x^{'} = - d x \\ = \frac{1}{π} [\int_{π}^{0} f (x^{'}) (- \sin (n x^{'})) (- d x^{'}) + \int_{0}^{π} f (x) \sin (n x) d x] \\ = \frac{1}{π} [- \int_{0}^{π} f (x) \sin (n x) d x + \int_{0}^{π} f (x) \sin (n x) d x] \\ = 0 \end{aligned}