The applyop command manipulates the selected parts of an expression. The first argument, f, is applied to the operands of e specified by i.

If i is an integer, applyop( f, i, e) applies f to the ith operand of e. This is equivalent to subsop( i = f(op( i, e)), e). For example, if the value of e is the sum $x+y+z$, applyop( f, 2, e) computes $x+f\left(y\right)+z$.

If i is a list of integers, the call applyop( f, i, e) is equivalent to subsop( i = f(op( i, e)), e). This allows you to manipulate any suboperand of an expression.

If i is a set, f is applied simultaneously to all operands of e specified in the set. Note: applyop( f, {}, e) returns e.

Any additional arguments xk are passed as additional arguments to f in the order given.

$p≔y^2-2y-3$

$p≔y^2-2y-3$

$\mathrm{applyop}\left(f\,2\,p\right)$

$y^2+f\left(-2y\right)-3$

$\mathrm{applyop}\left(f\,2\,p\,\mathrm{x1}\,\mathrm{x2}\right)$

$y^2+f\left(-2y\,\mathrm{x1}\,\mathrm{x2}\right)-3$

$\mathrm{applyop}\left(f\,\left[2\,2\right]\,p\right)$

$y^2-2f\left(y\right)-3$

$\mathrm{applyop}\left(f\,\left\{2\,3\right\}\,p\right)$

$y^2+f\left(-2y\right)+f\left(\mathrm{-3}\right)$

$\mathrm{applyop}\left(\mathrm{abs}\,\left\{3\,\left[2\,1\right]\right\}\,p\right)$

$y^2+2y+3$

$e≔\left(z+1\right)\ln \left(z\left(z^2-2\right)\right)$

$e≔\left(z+1\right)\ln \left(z\left(z^2-2\right)\right)$

$\mathrm{expand}\left(e\right)$

$\ln \left(z\left(z^2-2\right)\right)z+\ln \left(z\left(z^2-2\right)\right)$

$\mathrm{applyop}\left(\mathrm{expand}\,\left[2\,1\right]\,e\right)$

$\left(z+1\right)\ln \left(z^3-2z\right)$

$\mathrm{applyop}\left(\mathrm{factor}\,\left[2\,1\right]\,e\,\mathrm{real}\right)$

$\left(z+1\right)\ln \left(z\left(z+1.414213562\right)\left(z-1.414213562\right)\right)$