These procedures test the parity of the expression f, that is, whether it is even or odd, with respect to x.

More precisely, $\mathrm{type}\left(f\left(x\right)\,\mathrm{evenfunc}\left(x\right)\right)$ returns true if Maple can determine that $f\left(x\right)-f\left(-x\right)$ is zero and false otherwise.  Maple performs this test by calling Testzero on this difference. If false is returned, that is not a guarantee that f is not even: it may be zero, but if so, Testzero is not strong enough to recognize that the result is zero. Type $\mathrm{oddfunc}\left(x\right)$ is determined in a similar way but using the expression $f\left(x\right)+f\left(-x\right)$.

$\mathrm{type}\left(x^2\,\mathrm{evenfunc}\left(x\right)\right)$

$\mathrm{true}$

$\mathrm{type}\left(x^2{y}^3+2\,\mathrm{evenfunc}\left(x\right)\right)$

$\mathrm{true}$

$\mathrm{type}\left(\sin \left(x\right)\,\mathrm{oddfunc}\left(x\right)\right)$

$\mathrm{true}$

$\mathrm{type}\left(\frac{x^2}{x-1}\,\mathrm{oddfunc}\left(x\right)\right)$

$\mathrm{false}$

$f≔\mathrm{piecewise}\left(x=0\,1\,\frac{1}{x^2}\right)$

$f≔\left\{\begin{array}{cc}1& x=0\\ \frac{1}{x^2}& \mathrm{otherwise}\end{array}\right.$

$\mathrm{type}\left(f\,\mathrm{evenfunc}\left(x\right)\right)$

$\mathrm{false}$

$\mathrm{Normalizer}≔\mathrm{simplify}$

$\mathrm{Normalizer}≔\mathrm{simplify}$

$\mathrm{type}\left(f\,\mathrm{evenfunc}\left(x\right)\right)$

$\mathrm{true}$