The shift operator (often denoted as $E$, $S$, or $\mathrm{\tau }$) acts on functions by adding +1 to the independent variable (often denoted as $n$ or $x$). If for example the shift operator is denoted with $E$, and the independent variable by $x$, then $E$ is the operator that sends an expression $u\left(x\right)$ to $u\left(x+1\right)$.

One can choose the name of the shift operator by assigning it to _Env_LRE_tau, and the name of the independent variable by assigning it to _Env_LRE_x. If these environment variables are assigned then they will be used to denote the shift operator and independent variable.

An operator $L$ in $C\left(x\right)$[$E$] can be written as $L$ = $a_n\left(x\right)E^n$ + ... + $a_0\left(x\right)$ for rational functions $a_i\left(x\right)$ in $C\left(x\right)$. If the dependent variable dvar is for example $u\left(x\right)$, then the equation $L\left(u\left(x\right)\right)=0$ is the recurrence relation $a_n\left(x\right)u\left(x+n\right)$ + ... + $a_0\left(x\right)u\left(x\right)$ = 0. So a difference operator represents a linear homogeneous recurrence relation. Converting representations can be done with the RecurrenceToOperator and OperatorToRecurrence commands.

The product $L$ := $\mathrm{MultiplyOperators}\left(\mathrm{L1}\,\mathrm{L2}\right)$ corresponds to composition of linear difference operators. For example, if $E$ is the shift operator and $x$ is the independent variable, then $E$ will send any expression $u\left(x\right)$ to $u\left(x+1\right)$, while the operator $x$ sends $u\left(x\right)$ to $xu\left(x\right)$. The product $xE$ sends $u\left(x\right)$ to $x\left(E\left(u\left(x\right)\right)\right)$ = $x\left(u\left(x+1\right)\right)$ = $xu\left(x+1\right)$ while the product $E$ $x$ sends $u\left(x\right)$ to $E\left(x\left(u\left(x\right)\right)\right)$ = $E\left(xu\left(x\right)\right)$ = $\left(x+1\right)u\left(x+1\right)$. So $E$ $x$ acts the same as $\left(x+1\right)E$, which means that the operator $E$ $x$ equals the operator $\left(x+1\right)E$.

If $L$ := $\mathrm{MultiplyOperators}\left(\mathrm{L1}\,\mathrm{L2}\right)$ and if $u\left(x\right)$ is a solution of $\mathrm{L2}$, in other words $\mathrm{L2}\left(u\left(x\right)\right)=0$, then $L\left(u\left(x\right)\right)$ = $\mathrm{L1}\left(\mathrm{L2}\left(u\left(x\right)\right)\right)$ = $\mathrm{L1}\left(0\right)$ = $0$. So right-factors of $L$ are important for solving $L$ because solutions of right-factors are also a solutions of $L$ (this is not true for left-factors, which is why GCLD/LCRM/LeftDivision are omitted here).

The assignment $L$ := $\mathrm{LCLM}\left(\mathrm{L1}\,\mathrm{L2}\right)$ computes the Least Common Left Multiple of operators $\mathrm{L1}$ and $\mathrm{L2}$, which means that $\mathrm{L1}$ and $\mathrm{L2}$ are right-factors of $L$, and $L$ is minimal with this property. Then the solution space of $L$ is the sum of the solution spaces of $\mathrm{L1}$ and $\mathrm{L2}$. The same functionality is provided by gfun[`rec+rec`]. Difference operators are also a special case of Ore operators.

The assignment $L$ := $\mathrm{GCRD}\left(\mathrm{L1}\,\mathrm{L2}\right)$ computes the Greatest Common Right Divisor of $\mathrm{L1}$ and $\mathrm{L2}$, which means that $L$ is a right-factor of both $\mathrm{L1}$ and $\mathrm{L2}$, and is maximal with this property. Then the solution space of $L$ is the intersection of the solution spaces of $\mathrm{L1}$ and $\mathrm{L2}$.

One may specify more than two operators in MultiplyOperators, GCRD, or LCLM. For instance, $L$ := $\mathrm{LCLM}\left(\mathrm{L1}\,\mathrm{L2}\,\mathrm{L3}\right)$ is the Least Common Left Multiple of $\mathrm{L1}$, $\mathrm{L2}$, $\mathrm{L3}$, so solutions of $L$ are sums of solutions of $\mathrm{L1}$, $\mathrm{L2}$, and $\mathrm{L3}$.

The assignment $Q,R$ := $\mathrm{RightDivision}\left(\mathrm{L1}\,\mathrm{L2}\right)$ right-divides $\mathrm{L1}$ by $\mathrm{L2}$. This means that $\mathrm{L1}=Q\mathrm{L2}+R$ where the order of $R$ is less than that of $\mathrm{L2}$. $R$ will be $0$ if and only if $\mathrm{L2}$ is a right-factor of $\mathrm{L1}$.

$\mathrm{with}\left(\mathrm{LREtools}\right)\:$

$\mathrm{\_Env\_LRE\_tau}≔E\;$$\mathrm{\_Env\_LRE\_x}≔x$

$\mathrm{\_Env\_LRE\_tau}≔E$

$\mathrm{\_Env\_LRE\_x}≔x$

$\mathrm{MultiplyOperators}\left(E\,x\right)$

$\left(x+1\right)E$

$\mathrm{L1}≔E^2-x$

$\mathrm{L1}≔E^2-x$

$\mathrm{L2}≔E^2-E-x$

$\mathrm{L2}≔E^2-E-x$

$L≔\mathrm{LCLM}\left(\mathrm{L1}\,\mathrm{L2}\right)$

$L≔E^4-E^3+\left(-2x-3\right)E^2+\left(x+1\right)E+x^2+x$

$\mathrm{L3}≔\mathrm{MultiplyOperators}\left(E-x\,\mathrm{L1}\right)$

$\mathrm{L3}≔E^3-x{E}^2+\left(-x-1\right)E+x^2$

$\mathrm{GCRD}\left(L\,\mathrm{L3}\right)$

$E^2-x$

$Q,R≔\mathrm{RightDivision}\left(L\,\mathrm{L3}\right)$

$Q,R≔E+x,\left(x^2-x-1\right)E^2-x^3+x^2+x$

$\mathrm{Lnew}≔\mathrm{MultiplyOperators}\left(Q\,\mathrm{L3}\right)+R$

$\mathrm{Lnew}≔E^4-E^3+\left(-x^2-x-2\right)E^2+\left(x+1\right)E+\left(x^2-x-1\right)E^2+x^2+x$

$\mathrm{expand}\left(L-\mathrm{Lnew}\right)$

$0$

$\mathrm{Op}≔\mathrm{RecurrenceToOperator}\left(u\left(n+3\right)+2u\left(n+1\right)+5nu\left(n\right)=0\,u\left(n\right)\right)$

$\mathrm{Op}≔E^3+2E+5x$

$\mathrm{rec}≔\mathrm{OperatorToRecurrence}\left(\mathrm{Op}\,u\left(n\right)\right)$

$\mathrm{rec}≔u\left(n+3\right)+2u\left(n+1\right)+5nu\left(n\right)=0$

The LREtools[MultiplyOperators], LREtools[GCRD], LREtools[LCLM], LREtools[RightDivision], LREtools[RecurrenceToOperator] and LREtools[OperatorToRecurrence] commands were introduced in Maple 2021.

For more information on Maple 2021 changes, see Updates in Maple 2021.