test whether two polynomials are shift equivalent - Maple Help

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PolynomialTools[ShiftEquivalent] - test whether two polynomials are shift equivalent

 Calling Sequence ShiftEquivalent(f,g,x) ShiftEquivalent(f,g,x,T)

Parameters

 f, g - polynomials in x x - indeterminate T - (optional) type

Description

 • The ShiftEquivalent command determines whether the two polynomials $f,g$ are shift equivalent w.r.t. the variable x, that is, whether there is an $h$ independent of x satisfying $\mathrm{lc}\left(g\right)f\left(x+h\right)=\mathrm{lc}\left(f\right)g\left(x\right)$, where $\mathrm{lc}$ denotes the leading coefficient with respect to x. It returns $h$, if it exists, and otherwise FAIL.
 • If the optional argument T is specified, then ShiftEquivalent returns FAIL even if $h$ exists but is not of type T. This is more efficient than first calling ShiftEquivalent without the optional argument and then checking whether the return value is of type T.
 • It is assumed that both input polynomials are collected w.r.t. the variable x.
 • If $f,g$ are nonconstant w.r.t. x, then $h$ is uniquely determined. If both are nonzero and constant w.r.t. $x$, or if both are zero, then the return value is 0.

Examples

 > $\mathrm{with}\left(\mathrm{PolynomialTools}\right):$
 > $\mathrm{ShiftEquivalent}\left({x}^{2}+x+1,{x}^{2}-x+1,x\right)$
 ${-}{1}$ (1)
 > $\mathrm{Translate}\left({x}^{2}+x+1,x,\right)$
 ${{x}}^{{2}}{-}{x}{+}{1}$ (2)
 > $\mathrm{ShiftEquivalent}\left({x}^{2}+1,{x}^{2}-x+1,x\right)$
 ${\mathrm{FAIL}}$ (3)

 > $\mathrm{ShiftEquivalent}\left(2x-1,x+\frac{1}{2},x\right)$
 ${1}$ (4)
 > $\mathrm{Translate}\left(2x-1,x,\right)$
 ${1}{+}{2}{}{x}$ (5)
 > $\mathrm{ShiftEquivalent}\left(2x-1,x,x\right)$
 $\frac{{1}}{{2}}$ (6)
 > $\mathrm{ShiftEquivalent}\left(2x-1,x,x,'\mathrm{integer}'\right)$
 ${\mathrm{FAIL}}$ (7)
 > $\mathrm{ShiftEquivalent}\left(x,x+n,x\right)$
 ${n}$ (8)