Cancel Inverses - Maple Help

SolveTools

 CancelInverses
 normalize expression by canceling functions inverse to each other

 Calling Sequence CancelInverses(expr) CancelInverses(expr,'safe')

Parameters

 expr - expression 'safe' - 'safe'

Description

 • The CancelInverses command performs cancellation of the functions that are inverse to each other.
 • The following simplifications are made:

$\mathrm{trig}\left(\mathrm{arctrig}\left(\mathrm{subexpr}\right)\right)\to \mathrm{subexpr}$

$\mathrm{arctrig}\left(\mathrm{trig}\left(\mathrm{subexpr}\right)\right)\to \mathrm{subexpr}$

$\mathrm{ln}\left({ⅇ}^{\mathrm{subexpr}}\right)\to \mathrm{subexpr}$

${ⅇ}^{\mathrm{ln}\left(\mathrm{subexpr}\right)}\to \mathrm{subexpr}$

$\mathrm{ln}\left(\mathrm{LambertW}\left({ⅇ}^{\mathrm{subexpr}}\right)\right)\to \mathrm{subexpr}-\mathrm{LambertW}\left({ⅇ}^{\mathrm{subexpr}}\right)$

${ⅇ}^{\mathrm{integer}\mathrm{LambertW}\left(\mathrm{subexpr}\right)}\to \frac{{\mathrm{subexpr}}^{\mathrm{integer}}}{\mathrm{LambertW}{\left(\mathrm{subexpr}\right)}^{\mathrm{integer}}}$

$1-\mathrm{sin}{\left(\mathrm{subexpr}\right)}^{2}\to \mathrm{cos}{\left(\mathrm{subexpr}\right)}^{2}$

$1-\mathrm{cos}{\left(\mathrm{subexpr}\right)}^{2}\to \mathrm{sin}{\left(\mathrm{subexpr}\right)}^{2}$

${\mathrm{subexpr}}^{\frac{1}{2}}\to \sqrt{\mathrm{subexpr}}$

${\left({\mathrm{subexpr}}^{\mathrm{rational}}\right)}^{\mathrm{integer}}\to {\mathrm{subexpr}}^{\mathrm{rational}\mathrm{integer}}$

 Any combinations of the previous simplifications are made.  Also, arctan of two arguments is simplified if it contains sin(subexpr).
 Note: Not all simplifications are valid everywhere. You should be aware of this when calling CancelInverses, if option 'safe' is given, then only valid simplifications are applied.

Examples

 > $\mathrm{with}\left(\mathrm{SolveTools}\right):$
 > $\mathrm{CancelInverses}\left(1+{ⅇ}^{-{x}^{2}}\right)$
 ${1}{+}\frac{{1}}{{{ⅇ}}^{{{x}}^{{2}}}}$ (1)
 > $\mathrm{CancelInverses}\left(\mathrm{arccos}\left(\mathrm{cos}\left(x-y\cdot 2\right)\right)\right)$
 ${x}{-}{2}{}{y}$ (2)
 > $\mathrm{CancelInverses}\left(\mathrm{tan}\left(\mathrm{arctan}\left(x-y\cdot 2\right)\right)\right)$
 ${x}{-}{2}{}{y}$ (3)
 > $\mathrm{CancelInverses}\left(\mathrm{ln}\left({ⅇ}^{x-y\cdot 2}\right)\right)$
 ${x}{-}{2}{}{y}$ (4)
 > $\mathrm{CancelInverses}\left({ⅇ}^{\mathrm{ln}\left(x-y\cdot 2\right)}\right)$
 ${x}{-}{2}{}{y}$ (5)
 > $\mathrm{CancelInverses}\left(\mathrm{ln}\left(\mathrm{LambertW}\left({ⅇ}^{a+{b}^{2}+{c}^{3}}\right)\right)\right)$
 ${{c}}^{{3}}{+}{{b}}^{{2}}{+}{a}{-}{\mathrm{LambertW}}{}\left({{ⅇ}}^{{{c}}^{{3}}{+}{{b}}^{{2}}{+}{a}}\right)$ (6)
 > $\mathrm{CancelInverses}\left({ⅇ}^{\mathrm{LambertW}\left(a+{b}^{2}+{c}^{3}\right)}\right)$
 $\frac{{{c}}^{{3}}{+}{{b}}^{{2}}{+}{a}}{{\mathrm{LambertW}}{}\left({{c}}^{{3}}{+}{{b}}^{{2}}{+}{a}\right)}$ (7)
 > $\mathrm{CancelInverses}\left(g\left(1-{\mathrm{cos}\left(x+y+z\right)}^{2}\right)\right)$
 ${g}{}\left({{\mathrm{sin}}{}\left({x}{+}{y}{+}{z}\right)}^{{2}}\right)$ (8)
 > $\mathrm{CancelInverses}\left({\left(-5\right)}^{\frac{1}{2}}\right)$
 ${I}{}\sqrt{{5}}$ (9)
 > $\mathrm{CancelInverses}\left({\left({\left(x\mathrm{sin}\left(x\right)\right)}^{\frac{1}{2}}\right)}^{3}\right)$
 ${\left({x}{}{\mathrm{sin}}{}\left({x}\right)\right)}^{{3}}{{2}}}$ (10)
 > $\mathrm{CancelInverses}\left({ⅇ}^{\mathrm{ln}\left(x\right)+\mathrm{ln}\left(y+z\right)+t}\right)$
 ${{ⅇ}}^{{t}}{}\left({y}{+}{z}\right){}{x}$ (11)
 > $\mathrm{CancelInverses}\left({ⅇ}^{2\mathrm{ln}\left(x\right)+3\mathrm{ln}\left(y+z\right)-4t}\right)$
 ${{ⅇ}}^{{-}{4}{}{t}}{}{\left({y}{+}{z}\right)}^{{3}}{}{{x}}^{{2}}$ (12)
 > $\mathrm{CancelInverses}\left({ⅇ}^{2\mathrm{ln}\left(\mathrm{sin}\left(x\right)\right)}\right)$
 ${{\mathrm{sin}}{}\left({x}\right)}^{{2}}$ (13)
 > $\mathrm{CancelInverses}\left(\mathrm{ln}\left({\left({ⅇ}^{ck}\right)}^{z}{ⅇ}^{bt}\right)\right)$
 ${z}{}{c}{}{k}{+}{b}{}{t}$ (14)
 > $\mathrm{CancelInverses}\left(\mathrm{ln}\left({z}^{\frac{3}{2}}x\right)\right)$
 ${\mathrm{ln}}{}\left({x}\right){+}\frac{{3}{}{\mathrm{ln}}{}\left({z}\right)}{{2}}$ (15)
 > $\mathrm{CancelInverses}\left(\mathrm{ln}\left({\left({z}^{\frac{3}{2}}\right)}^{5}x\right)\right)$
 ${\mathrm{ln}}{}\left({x}\right){+}\frac{{15}{}{\mathrm{ln}}{}\left({z}\right)}{{2}}$ (16)
 > $\mathrm{CancelInverses}\left(\mathrm{ln}\left({\left({ⅇ}^{z}\right)}^{\mathrm{cos}\left(x\right)}\right)\right)$
 ${\mathrm{cos}}{}\left({x}\right){}{z}$ (17)
 > $\mathrm{CancelInverses}\left(\mathrm{ln}\left({\left({s}^{f\left(x\right)}\right)}^{g\left(z\right)}\right)\right)$
 ${g}{}\left({z}\right){}{f}{}\left({x}\right){}{\mathrm{ln}}{}\left({s}\right)$ (18)
 > $\mathrm{CancelInverses}\left({x}^{\frac{x+1}{\mathrm{ln}\left(x\right)}+\frac{x-1}{\mathrm{ln}\left(x\right)}}\right)$
 ${{ⅇ}}^{{2}{}{x}}$ (19)
 > $\mathrm{CancelInverses}\left(\mathrm{arctan}\left(-\mathrm{sin}\left(b\right),\mathrm{cos}\left(b\right)\right)\right)$
 ${-}{b}$ (20)
 > $\mathrm{CancelInverses}\left(\mathrm{arctan}\left(\mathrm{cos}\left(b\right),\mathrm{sin}\left(b\right)\right)\right)$
 $\frac{{\mathrm{\pi }}}{{2}}{-}{b}$ (21)
 > $\mathrm{CancelInverses}\left(\mathrm{ln}\left({ⅇ}^{x}\right)+\mathrm{cos}\left(\mathrm{arccos}\left(x\right)\right)\right)$
 ${2}{}{x}$ (22)
 > $\mathrm{CancelInverses}\left(\mathrm{ln}\left({ⅇ}^{x}\right)+\mathrm{cos}\left(\mathrm{arccos}\left(x\right)\right),'\mathrm{safe}'\right)$
 ${\mathrm{ln}}{}\left({{ⅇ}}^{{x}}\right){+}{x}$ (23)