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DiscreteShift ​

The forward-shift operator: DiscreteShift(f, n) is f with every free n replaced by n + 1.

✓ checked against that system · ! checked, with disagreements (hover a chip)Wolfram LanguageDiscreteShift!
DiscreteShift(f, n)f with every free occurrence of the symbol n replaced by n + 1.
DiscreteShift(f, {n, h})f with every free occurrence of n replaced by n + h -- an h-step shift, given as the {n, h} pair.
Typed signatures
TypePackageApplies
(expression, expression) -> expressionanalyticalways

Domain: Elementary functions

Details
  • A substitution, not an arithmetic evaluation -- f need not itself be a declared function; DiscreteShift(a(n), n) is a(n + 1) even though a has no definition.
  • The h-step form takes n and h together as the second argument, {n, h} -- not a third argument.

Examples

Basic #

One-step forward shift

live ↗
InDiscreteShift(a(n),n)\mathrm{DiscreteShift}(a(n), n)Outa(n+1)a(n+1)
live ↗
InDiscreteShift(sin⁡((−27i)π),[i,10])\mathrm{DiscreteShift}(\sin((-\frac{2}{7}i)π), \bigl\lbrack i, 10\bigr\rbrack)OutDiscreteShift(isinh⁡(−2π7),[i,10])\mathrm{DiscreteShift}(i\sinh(-\frac{2π}{7}), \bigl\lbrack i, 10\bigr\rbrack)
live ↗
InDiscreteShift(sin⁡(17×2πk3),[k,10])\mathrm{DiscreteShift}(\sin(\frac{1}{7}×2πk^3), \bigl\lbrack k, 10\bigr\rbrack)Outsin⁡(17×2π(k+10)3)\sin(\frac{1}{7}×2π(k+10)^3)
live ↗
InDiscreteShift(Fibonacci(k),[k,5])\mathrm{DiscreteShift}(\mathrm{Fibonacci}(k), \bigl\lbrack k, 5\bigr\rbrack)OutFibonacci(k+5)\mathrm{Fibonacci}(k+5)
live ↗
InDiscreteShift(k!,[k,5])\mathrm{DiscreteShift}(k!, \bigl\lbrack k, 5\bigr\rbrack)Out(k+5)!(k+5)!
live ↗
InDiscreteShift(f′(x),k)\mathrm{DiscreteShift}(f^{\prime}(x), k)Outf′(x)f^{\prime}(x)
Scope #

The h-step form -- {n, h}, not a third argument

live ↗
InDiscreteShift(a(n),[n,2])\mathrm{DiscreteShift}(a(n), \bigl\lbrack n, 2\bigr\rbrack)Outa(n+2)a(n+2)
live ↗
InDiscreteShift(⌊k⌋,k)\mathrm{DiscreteShift}(\lfloor k\rfloor, k)Out⌊k+1⌋\lfloor k+1\rfloor
live ↗
InDiscreteShift(round(k),[k,3])\mathrm{DiscreteShift}(\mathrm{round}(k), \bigl\lbrack k, 3\bigr\rbrack)Outround(k+3)\mathrm{round}(k+3)
live ↗
InDiscreteShift(2k,k)\mathrm{DiscreteShift}(2^{k}, k)Out2k+12^{k+1}
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InDiscreteShift(k4+11k3+20k2,[k,3])\mathrm{DiscreteShift}(k^4+11k^3+20k^2, \bigl\lbrack k, 3\bigr\rbrack)Out(k+3)4+11(k+3)3+20(k+3)2(k+3)^4+11(k+3)^3+20(k+3)^2
live ↗
InDiscreteShift(k2+5k−7,k)\mathrm{DiscreteShift}(\frac{k^2+5}{k-7}, k)Out(k+1)2+5k−6\frac{(k+1)^2+5}{k-6}
live ↗
InDiscreteShift(FractionalPart(k),k)\mathrm{DiscreteShift}(\mathrm{FractionalPart}(k), k)OutFractionalPart(k+1)\mathrm{FractionalPart}(k+1)
live ↗
InDiscreteShift(Γ(k),[k,3])\mathrm{DiscreteShift}(Γ(k), \bigl\lbrack k, 3\bigr\rbrack)OutΓ(k+3)Γ(k+3)
live ↗
InDiscreteShift(CatalanNumber(k),[k,3])\mathrm{DiscreteShift}(\mathrm{CatalanNumber}(k), \bigl\lbrack k, 3\bigr\rbrack)OutCatalanNumber(k+3)\mathrm{CatalanNumber}(k+3)
About
Examples15
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