PiecewiseExpand

Rewrite Abs, Sign, Min, Max, UnitStep, Clip, UnitBox, UnitTriangle and, over a bounded variable, Floor, Ceil, Round, IntegerPart, FractionalPart, Mod, Quotient and the waves inside $expr$ into Piecewise, recursively. Provided by @enumeratio/analytic.

✓ checked against that system · ! checked, with disagreements (hover a chip)Wolfram LanguagePiecewiseExpand!↗

PiecewiseExpand(expr)rewrite using whatever is already assumed (e.g. inside an enclosing Assuming).

PiecewiseExpand(expr, assumptions)assume $assumptions$ (scoped, as in Assuming) for the rewrite, then forget it.

Typed signatures

Type	Package	Applies
(expression, expression?) -> expression	analytic	always

Domain: Analytic

Details

• Abs, Sign and Argument expand only once their argument is known real — on a genuinely complex value they are not piecewise-comparable, so (matching Wolfram's own PiecewiseExpand) an argument whose realness isn't established is left untouched rather than guessed at. RealNumbers as the assumption makes every variable real.
• Min, Max, UnitStep, Clip, UnitBox and UnitTriangle need no assumption, as in Wolfram: they expand over any symbols. The branches, conditions and default follow Wolfram's: a condition between two unknowns is written x - y >= 0, one against a constant x <= 2, and a zero-valued branch is the default. What the assumptions decide of a condition is not written (UnitBox(x) on 0 < x < 2 is x <= 1/2), a branch they rule out goes, and when the branches cover what is left the last one is the default.
• Floor, Ceil, Round, IntegerPart, FractionalPart, Mod and Quotient, and SquareWave, TriangleWave and SawtoothWave, expand over an argument bounded on both sides by an assumption, one branch per stretch. The bound is read as an interval: 0 < x < 3, but also x^2 < 3, 0 < 2 x < 5, a constant such as Pi, or x < 3 y with 0 < y < 1. The argument is a symbol, linear in one (Floor(2 x)), or over an interval where it is monotone a square or square root of one (Floor(x^2) is bounded by Sqrt(2), Sqrt(3)). A modulus, or the multiple in a two-argument Floor, is a positive number, a constant, or a symbol: Mod(k, m) with 0 <= k <= 3 m reads the bound as 0 <= k/m <= 3 and writes its conditions on k/m. The values are written as Wolfram's Together writes them, in its order. A span of 100 stretches or more is left alone, as Wolfram leaves it.
• Rewrites recurse into subexpressions, so PiecewiseExpand(Abs(x) + 1, …) rewrites the Abs inside the sum. A sum, product or power of Piecewise values, or a Max, Min, Clip or step over them, is one Piecewise: a branch for each combination, values that agree sharing their conditions, a zero value (else the last in Wolfram's order) the default. Conditions that are bounds on one unknown merge into intervals (x <= 0 with x >= 0 is x == 0, and two intervals that do not touch stay apart as x > 0 || x < 0). A relation is not entered (Wolfram turns a Piecewise inside one into conditions).

Examples

Basic #

$|x| \to$ a two-branch Piecewise, once $x$ is known real

live ↗

In$\mathrm{PiecewiseExpand}(\vert x\vert, x∈ℝ)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack-x, x\lt0\bigr\rbrack\bigr\rbrack, x)$

$\operatorname{sign}(x)$ expanded into its three branches

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{sgn}(x), x∈ℝ)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack-1, x\lt0\bigr\rbrack, \bigl\lbrack1, 0\lt x\bigr\rbrack\bigr\rbrack, 0)$

$\operatorname{Clip}(x,\{-1,1\})$ expanded into its three branches

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{Clip}(x, \bigl\lbrack-1, 1\bigr\rbrack), x∈ℝ)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack-1, x\lt-1\bigr\rbrack, \bigl\lbrack1, 1\lt x\bigr\rbrack\bigr\rbrack, x)$

$\max(x,y)$ as a two-branch Piecewise

live ↗

In$\mathrm{PiecewiseExpand}(\max(x, y), x∈ℝ∧y∈ℝ)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack x, 0≤x-y\bigr\rbrack\bigr\rbrack, y)$

A fractional modulus: the values over a common denominator, in Wolfram's order of them

live ↗

In$\mathrm{PiecewiseExpand}(x\bmod\frac{3}{2}, 0≤x≤4)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack x-3, 3≤x\bigr\rbrack, \bigl\lbrack x, x\lt3/2\bigr\rbrack\bigr\rbrack, \frac{2x-3}{2})$

Quotient by $3/2$: a branch per multiple of it

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{Quotient}(x, \frac{3}{2}), 0≤x≤4)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 3/2≤x∧x\lt3\bigr\rbrack, \bigl\lbrack2, 3≤x\bigr\rbrack\bigr\rbrack, 0)$

A constant modulus keeps its multiples in the bounds

live ↗

In$\mathrm{PiecewiseExpand}(x\bmod π, 0\lt x\lt10)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack x, x\lt π\bigr\rbrack, \bigl\lbrack x-3π, 3π≤x\bigr\rbrack, \bigl\lbrack x-2π, 2π≤x∧x\lt3π\bigr\rbrack\bigr\rbrack, x-π)$

A symbolic modulus: the conditions are on $k/m$

live ↗

In$\mathrm{PiecewiseExpand}(k\bmod m, 0≤k≤3m)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack k, k/m\lt1\bigr\rbrack, \bigl\lbrack k-3m, 3≤k/m\bigr\rbrack, \bigl\lbrack k-2m, 2≤k/m∧k/m\lt3\bigr\rbrack\bigr\rbrack, k-m)$

Quotient by a symbol, over $-m \le k \le 2m$

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{Quotient}(k, m), -m≤k≤2m)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack-1, k/m\lt0\bigr\rbrack, \bigl\lbrack1, 1≤k/m∧k/m\lt2\bigr\rbrack, \bigl\lbrack2, 2≤k/m\bigr\rbrack\bigr\rbrack, 0)$

The argument is solved for $x$: $\lfloor 2x\rfloor$ steps at the halves

live ↗

In$\mathrm{PiecewiseExpand}(\lfloor2x\rfloor, 0\lt x\lt2)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 1/2≤x∧x\lt1\bigr\rbrack, \bigl\lbrack2, 1≤x∧x\lt3/2\bigr\rbrack, \bigl\lbrack3, 3/2≤x\bigr\rbrack\bigr\rbrack, 0)$

A falling argument flips which end of each stretch the condition names

live ↗

In$\mathrm{PiecewiseExpand}(\lfloor-x\rfloor, 0\lt x\lt3)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack-3, 2\lt x\bigr\rbrack, \bigl\lbrack-2, 1\lt x∧x≤2\bigr\rbrack\bigr\rbrack, -1)$

The values carry their common factor out front, as Wolfram's Together writes them

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{FractionalPart}(2x), 0\lt x\lt2)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack2(x-1), 1≤x∧x\lt3/2\bigr\rbrack, \bigl\lbrack2x, x\lt1/2\bigr\rbrack, \bigl\lbrack2x-3, 3/2≤x\bigr\rbrack\bigr\rbrack, 2x-1)$

A triangle wave of $2x$ over half a unit: four stretches, each rising or falling

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{TriangleWave}(2x), 0≤x≤\frac{1}{2})$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack8x, x\lt1/8\bigr\rbrack, \bigl\lbrack4(2x-1), 3/8≤x\bigr\rbrack\bigr\rbrack, -2(4x-1))$

$\lfloor x^2\rfloor$: the stretches are bounded by square roots

live ↗

In$\mathrm{PiecewiseExpand}(\lfloor x^2\rfloor, 0\lt x\lt2)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 1≤x∧x\lt2^{1/2}\bigr\rbrack, \bigl\lbrack2, 2^{1/2}≤x∧x\lt3^{1/2}\bigr\rbrack, \bigl\lbrack3, 3^{1/2}≤x\bigr\rbrack\bigr\rbrack, 0)$

$\lfloor\sqrt{x}\rfloor$ is compared as $\sqrt{x}$, over its range

live ↗

In$\mathrm{PiecewiseExpand}(\lfloor\sqrt{x}\rfloor, 0\lt x\lt5)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 1≤x^{1/2}∧x^{1/2}\lt2\bigr\rbrack, \bigl\lbrack2, 2≤x^{1/2}\bigr\rbrack\bigr\rbrack, 0)$

The bound $x^2 < 3$ is $-\sqrt{3} < x < \sqrt{3}$

live ↗

In$\mathrm{PiecewiseExpand}(\lfloor x\rfloor, x^2\lt3)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack-2, x\lt-1\bigr\rbrack, \bigl\lbrack-1, -1≤x∧x\lt0\bigr\rbrack, \bigl\lbrack1, 1≤x\bigr\rbrack\bigr\rbrack, 0)$

$x < 3y$ with $0 < y < 1$ bounds $x$ above by 3

live ↗

In$\mathrm{PiecewiseExpand}(\lfloor x\rfloor, 0\lt x∧x\lt3y∧0\lt y\lt1)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 1≤x∧x\lt2\bigr\rbrack, \bigl\lbrack2, 2≤x\bigr\rbrack\bigr\rbrack, 0)$

A branch the assumption rules out goes, and the last takes the default

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{Clip}(x, \bigl\lbrack0, 1\bigr\rbrack), 0\lt x\lt3)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 1\lt x\bigr\rbrack\bigr\rbrack, x)$

With $x > 3$ the constant never wins, and $2 < x$ needs no saying

live ↗

In$\mathrm{PiecewiseExpand}(\max(x, 2, y), x\gt3)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack x, 0≤x-y\bigr\rbrack\bigr\rbrack, y)$

The peak, where both branches give 1, goes to the branch left of it when it is right of 0

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{UnitTriangle}(x-1))$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack2-x, 1\lt x∧x≤2\bigr\rbrack, \bigl\lbrack x, 0≤x≤1\bigr\rbrack\bigr\rbrack, 0)$

A sum of Piecewise values is one Piecewise: a point is $x == 0$, and a zero value is the default

live ↗

In$\mathrm{PiecewiseExpand}(\max(x, 0)+\mathrm{UnitStep}(x))$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, x=0\bigr\rbrack, \bigl\lbrack x+1, 0\lt x\bigr\rbrack\bigr\rbrack, 0)$

With no zero value the last in order is the default

live ↗

In$\mathrm{PiecewiseExpand}(\vert x\vert+1, ℝ)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1-x, x\lt0\bigr\rbrack\bigr\rbrack, x+1)$

Branches with the same value share their conditions

live ↗

In$\mathrm{PiecewiseExpand}(\vert x\vert\vert y\vert, ℝ)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack-(xy), x\lt0∧0≤y∨0≤x∧y\lt0\bigr\rbrack\bigr\rbrack, xy)$

Independent unknowns: the sum counts how many are non-negative

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{UnitStep}(x)+\mathrm{UnitStep}(y))$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 0≤x∧y\lt0∨x\lt0∧0≤y\bigr\rbrack, \bigl\lbrack2, 0≤x∧0≤y\bigr\rbrack\bigr\rbrack, 0)$

Scope #

Four arguments: a branch wins over the earlier ones strictly, over the later ones on a tie

live ↗

In$\mathrm{PiecewiseExpand}(\max(a, b, c, d))$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack a, 0≤a-b∧0≤a-c∧0≤a-d\bigr\rbrack, \bigl\lbrack b, a-b\lt0∧0≤b-c∧0≤b-d\bigr\rbrack, \bigl\lbrack c, a-c\lt0∧b-c\lt0∧0≤c-d\bigr\rbrack\bigr\rbrack, d)$

Floor on a bounded interval: a branch per integer, the zero stretch as the default

live ↗

In$\mathrm{PiecewiseExpand}(\lfloor x\rfloor, 0\lt x\lt4)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 1≤x∧x\lt2\bigr\rbrack, \bigl\lbrack2, 2≤x∧x\lt3\bigr\rbrack, \bigl\lbrack3, 3≤x\bigr\rbrack\bigr\rbrack, 0)$

PrimePi on an interval stays

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{PrimePi}(x), 0\lt x\lt6)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 2≤x∧x\lt3\bigr\rbrack, \bigl\lbrack2, 3≤x∧x\lt5\bigr\rbrack, \bigl\lbrack3, 5≤x\bigr\rbrack\bigr\rbrack, 0)$

Inequalities with floor and round remain symbolic

live ↗

In$\mathrm{PiecewiseExpand}(0\lt\lfloor x+\mathrm{round}(x^2)\rfloor-x\lt4)$Out$0\lt\lfloor x+\mathrm{round}(x^2)\rfloor-x\lt4$

An enclosing Assuming bounds the symbol, as the second argument does

live ↗

In$\mathrm{Assuming}(0\lt x\lt3, \mathrm{PiecewiseExpand}(\lfloor x\rfloor))$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 1≤x∧x\lt2\bigr\rbrack, \bigl\lbrack2, 2≤x\bigr\rbrack\bigr\rbrack, 0)$

A bound on each unknown by its square: the floor and the round together are one Piecewise

live ↗

In$\mathrm{Assuming}(x^2\lt3, \mathrm{PiecewiseExpand}(\lfloor x\rfloor+\mathrm{round}(y), y^2\lt2))$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack-3, x\lt-1∧y\lt-1/2\bigr\rbrack, \bigl\lbrack-2, x\lt-1∧-1/2≤y≤1/2∨-1≤x∧x\lt0∧y\lt-1/2\bigr\rbrack, \bigl\lbrack-1, x\lt-1∧1/2\lt y∨-1≤x∧x\lt0∧-1/2≤y≤1/2∨0≤x∧x\lt1∧y\lt-1/2\bigr\rbrack, \bigl\lbrack1, 1≤x∧-1/2≤y≤1/2∨0≤x∧x\lt1∧1/2\lt y\bigr\rbrack, \bigl\lbrack2, 1≤x∧1/2\lt y\bigr\rbrack\bigr\rbrack, 0)$

A symbol is left alone

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{pw})$Out$\mathrm{pw}$

A chained inequality with max stays

live ↗

In$\mathrm{PiecewiseExpand}(-1≤\max(x, y)≤1)$Out$-1≤\max(x, y)≤1$

A max of absolute values is one Piecewise, each branch carrying the conditions of the pieces it came from

live ↗

In$\mathrm{PiecewiseExpand}(\max(\vert x\vert, \vert y\vert), ℝ)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack-x, x\lt0∧y\lt0∧x-y≤0∨x\lt0∧0≤y∧x+y≤0\bigr\rbrack, \bigl\lbrack x, 0≤x∧y\lt0∧0≤x+y∨0≤x∧0≤y∧0≤x-y\bigr\rbrack, \bigl\lbrack-y, x\lt0∧y\lt0∧0\lt x-y∨0≤x∧y\lt0∧x+y\lt0\bigr\rbrack\bigr\rbrack, y)$

live ↗

In$\mathrm{PiecewiseExpand}(\text{if }a\text{ then }b\text{ else }c)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack b, a\bigr\rbrack\bigr\rbrack, c)$

Min and Max expand over any symbols, with no assumption on them

live ↗

In$\mathrm{PiecewiseExpand}(\min(x, y))$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack x, x-y≤0\bigr\rbrack\bigr\rbrack, y)$

A constant bound is written as $x \le 2$, not as a difference

live ↗

In$\mathrm{PiecewiseExpand}(\max(x, 2, y))$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack2, x≤2∧y≤2\bigr\rbrack, \bigl\lbrack x, 2\lt x∧0≤x-y\bigr\rbrack\bigr\rbrack, y)$

A zero-valued branch is the Piecewise default

live ↗

In$\mathrm{PiecewiseExpand}(\max(x, y, 0))$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack x, 0\lt x∧0≤x-y\bigr\rbrack, \bigl\lbrack y, 0\lt y∧x-y\lt0\bigr\rbrack\bigr\rbrack, 0)$

UnitStep of several arguments is the AND of their steps

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{UnitStep}(x, y-2))$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 0≤x∧2≤y\bigr\rbrack\bigr\rbrack, 0)$

Clip to symbolic limits

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{Clip}(x, \bigl\lbrack a, b\bigr\rbrack))$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack a, 0\lt a-x\bigr\rbrack, \bigl\lbrack b, b-x\lt0∧a-x≤0\bigr\rbrack\bigr\rbrack, x)$

Clip to $[0, 1]$: the zero-valued branch is the default

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{Clip}(x, \bigl\lbrack0, 1\bigr\rbrack))$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 1\lt x\bigr\rbrack, \bigl\lbrack x, 0≤x≤1\bigr\rbrack\bigr\rbrack, 0)$

UnitBox of an affine argument solves for the variable

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{UnitBox}(2x-1))$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 1/4≤x≤3/4\bigr\rbrack\bigr\rbrack, 0)$

Floor on $-2 \le x \le 2$: the zero-valued stretch is the default

live ↗

In$\mathrm{PiecewiseExpand}(\lfloor x\rfloor, -2≤x≤2)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack-2, x\lt-1\bigr\rbrack, \bigl\lbrack-1, -1≤x∧x\lt0\bigr\rbrack, \bigl\lbrack1, 1≤x∧x\lt2\bigr\rbrack, \bigl\lbrack2, 2≤x\bigr\rbrack\bigr\rbrack, 0)$

Ceil on $0 < x < 3$: the last stretch is the default

live ↗

In$\mathrm{PiecewiseExpand}(\lceil x\rceil, 0\lt x\lt3)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, x≤1\bigr\rbrack, \bigl\lbrack2, 1\lt x∧x≤2\bigr\rbrack\bigr\rbrack, 3)$

Round on $0 \le x \le 3$: an even integer owns both ends of its stretch

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{round}(x), 0≤x≤3)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 1/2\lt x\lt3/2\bigr\rbrack, \bigl\lbrack2, 3/2≤x≤5/2\bigr\rbrack, \bigl\lbrack3, 5/2\lt x\bigr\rbrack\bigr\rbrack, 0)$

IntegerPart on $-2 < x < 2$ truncates toward zero

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{IntegerPart}(x), -2\lt x\lt2)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack-1, x≤-1\bigr\rbrack, \bigl\lbrack1, 1≤x\bigr\rbrack\bigr\rbrack, 0)$

FractionalPart on $-3 < x < 0$ is negative, with the lowest stretch as the default

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{FractionalPart}(x), -3\lt x\lt0)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack x, -1\lt x\bigr\rbrack, \bigl\lbrack x+1, -2\lt x∧x≤-1\bigr\rbrack\bigr\rbrack, x+2)$

Mod by 2 on $0 \le x \le 5$

live ↗

In$\mathrm{PiecewiseExpand}(x\bmod2, 0≤x≤5)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack x-4, 4≤x\bigr\rbrack, \bigl\lbrack x-2, 2≤x∧x\lt4\bigr\rbrack\bigr\rbrack, x)$

Quotient by 3 on $-4 \le x \le 7$

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{Quotient}(x, 3), -4≤x≤7)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack-2, x\lt-3\bigr\rbrack, \bigl\lbrack-1, -3≤x∧x\lt0\bigr\rbrack, \bigl\lbrack1, 3≤x∧x\lt6\bigr\rbrack, \bigl\lbrack2, 6≤x\bigr\rbrack\bigr\rbrack, 0)$

SquareWave on $0 < x < 2$: its low half-periods, joined

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{SquareWave}(x), 0\lt x\lt2)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack-1, 1/2≤x∧x\lt1∨3/2≤x\bigr\rbrack\bigr\rbrack, 1)$

TriangleWave on $0 \le x \le 1$: rising stretches first, then the falling one as default

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{TriangleWave}(x), 0≤x≤1)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack4(x-1), 3/4≤x\bigr\rbrack, \bigl\lbrack4x, x\lt1/4\bigr\rbrack\bigr\rbrack, -2(2x-1))$

SawtoothWave on $-1 < x < 1$

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{SawtoothWave}(x), -1\lt x\lt1)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack x, 0≤x\bigr\rbrack\bigr\rbrack, x+1)$

The real cube root is negative below zero

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{CubeRoot}(x))$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack-(-x)^{1/3}, x\lt0\bigr\rbrack\bigr\rbrack, \sqrt[3]{x})$

The conditions compare the cube root as it stands

live ↗

In$\mathrm{PiecewiseExpand}(\lfloor\mathrm{CubeRoot}(x)\rfloor, 0\lt x\lt10)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 1≤x^{1/3}∧x^{1/3}\lt2\bigr\rbrack, \bigl\lbrack2, 2≤x^{1/3}\bigr\rbrack\bigr\rbrack, 0)$

The prime-counting staircase on a bounded interval

live ↗

In$\mathrm{PiecewiseExpand}(\mathrm{PrimePi}(x), 1\lt x\lt20)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 2≤x∧x\lt3\bigr\rbrack, \bigl\lbrack2, 3≤x∧x\lt5\bigr\rbrack, \bigl\lbrack3, 5≤x∧x\lt7\bigr\rbrack, \bigl\lbrack4, 7≤x∧x\lt11\bigr\rbrack, \bigl\lbrack5, 11≤x∧x\lt13\bigr\rbrack, \bigl\lbrack6, 13≤x∧x\lt17\bigr\rbrack, \bigl\lbrack7, 17≤x∧x\lt19\bigr\rbrack, \bigl\lbrack8, 19≤x\bigr\rbrack\bigr\rbrack, 0)$

Properties #

live ↗

In$\mathrm{Map}(\mathrm{PiecewiseExpand}, \bigl\lbrack\min(x, y), \max(x, y, z), \mathrm{UnitStep}(x), \mathrm{Clamp}(x, a, b)\bigr\rbrack)$Out$\bigl\lbrack\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack x, x-y≤0\bigr\rbrack\bigr\rbrack, y), \mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack x, 0≤x-y∧0≤x-z\bigr\rbrack, \bigl\lbrack y, x-y\lt0∧0≤y-z\bigr\rbrack\bigr\rbrack, z), \mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 0≤x\bigr\rbrack\bigr\rbrack, 0), \mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack a, 0\lt a-x\bigr\rbrack, \bigl\lbrack b, b-x\lt0∧a-x≤0\bigr\rbrack\bigr\rbrack, x)\bigr\rbrack$

live ↗

In$\mathrm{Assuming}(\bigl\lbrack0≤x≤3\bigr\rbrack, \mathrm{Map}(\mathrm{PiecewiseExpand}, \bigl\lbrack\lfloor x\rfloor, \lceil x\rceil, \mathrm{round}(x), \mathrm{IntegerPart}(x), \mathrm{FractionalPart}(x)\bigr\rbrack))$Out$\bigl\lbrack\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 1≤x∧x\lt2\bigr\rbrack, \bigl\lbrack2, 2≤x∧x\lt3\bigr\rbrack, \bigl\lbrack3, 3≤x\bigr\rbrack\bigr\rbrack, 0), \mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 0\lt x∧x≤1\bigr\rbrack, \bigl\lbrack2, 1\lt x∧x≤2\bigr\rbrack, \bigl\lbrack3, 2\lt x\bigr\rbrack\bigr\rbrack, 0), \mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 1/2\lt x\lt3/2\bigr\rbrack, \bigl\lbrack2, 3/2≤x≤5/2\bigr\rbrack, \bigl\lbrack3, 5/2\lt x\bigr\rbrack\bigr\rbrack, 0), \mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 1≤x∧x\lt2\bigr\rbrack, \bigl\lbrack2, 2≤x∧x\lt3\bigr\rbrack, \bigl\lbrack3, 3≤x\bigr\rbrack\bigr\rbrack, 0), \mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack x-3, 3≤x\bigr\rbrack, \bigl\lbrack x-2, 2≤x∧x\lt3\bigr\rbrack, \bigl\lbrack x-1, 1≤x∧x\lt2\bigr\rbrack\bigr\rbrack, x)\bigr\rbrack$

live ↗

In$\mathrm{Map}(\mathrm{PiecewiseExpand}, \bigl\lbrack\mathrm{UnitBox}(x), \mathrm{UnitTriangle}(x)\bigr\rbrack)$Out$\bigl\lbrack\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, -1/2≤x≤1/2\bigr\rbrack\bigr\rbrack, 0), \mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1-x, 0≤x≤1\bigr\rbrack, \bigl\lbrack x+1, -1≤x∧x\lt0\bigr\rbrack\bigr\rbrack, 0)\bigr\rbrack$

live ↗

In$\mathrm{Assuming}(\bigl\lbrack0≤x≤3\bigr\rbrack, \mathrm{Map}(\mathrm{PiecewiseExpand}, \bigl\lbrack\mathrm{SquareWave}(x), \mathrm{TriangleWave}(x), \mathrm{SawtoothWave}(x)\bigr\rbrack))$Out$\bigl\lbrack\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack-1, 1/2≤x∧x\lt1∨3/2≤x∧x\lt2∨5/2≤x∧x\lt3\bigr\rbrack\bigr\rbrack, 1), \mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack4(x-3), 11/4≤x\bigr\rbrack, \bigl\lbrack4(x-2), 7/4≤x∧x\lt9/4\bigr\rbrack, \bigl\lbrack4(x-1), 3/4≤x∧x\lt5/4\bigr\rbrack, \bigl\lbrack4x, x\lt1/4\bigr\rbrack, \bigl\lbrack-2(2x-5), 9/4≤x∧x\lt11/4\bigr\rbrack, \bigl\lbrack-2(2x-3), 5/4≤x∧x\lt7/4\bigr\rbrack\bigr\rbrack, -2(2x-1)), \mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack x-3, 3≤x\bigr\rbrack, \bigl\lbrack x-2, 2≤x∧x\lt3\bigr\rbrack, \bigl\lbrack x-1, 1≤x∧x\lt2\bigr\rbrack\bigr\rbrack, x)\bigr\rbrack$

A symbolic modulus: the bound $0 \le k \le 3m$ is read as $0 \le k/m \le 3$

live ↗

In$\mathrm{Assuming}(0≤k≤3m, \mathrm{Map}(\mathrm{PiecewiseExpand}, \bigl\lbrack k\bmod m, \mathrm{Quotient}(k, m)\bigr\rbrack))$Out$\bigl\lbrack\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack k, k/m\lt1\bigr\rbrack, \bigl\lbrack k-3m, 3≤k/m\bigr\rbrack, \bigl\lbrack k-2m, 2≤k/m∧k/m\lt3\bigr\rbrack\bigr\rbrack, k-m), \mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, 1≤k/m∧k/m\lt2\bigr\rbrack, \bigl\lbrack2, 2≤k/m∧k/m\lt3\bigr\rbrack, \bigl\lbrack3, 3≤k/m\bigr\rbrack\bigr\rbrack, 0)\bigr\rbrack$

live ↗

In$\mathrm{Assuming}(x∈ℝ, \mathrm{Map}(\mathrm{PiecewiseExpand}, \bigl\lbrack\vert x\vert, \mathrm{sgn}(x), \arg(x)\bigr\rbrack))$Out$\bigl\lbrack\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack-x, x\lt0\bigr\rbrack\bigr\rbrack, x), \mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack-1, x\lt0\bigr\rbrack, \bigl\lbrack1, 0\lt x\bigr\rbrack\bigr\rbrack, 0), \mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack π, x\lt0\bigr\rbrack\bigr\rbrack, 0)\bigr\rbrack$

live ↗

In$\mathrm{Map}(\mathrm{PiecewiseExpand}, \bigl\lbrack δ_{xy}, \mathrm{DiscreteDelta}(x, y)\bigr\rbrack)$Out$\bigl\lbrack\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, x-y=0\bigr\rbrack\bigr\rbrack, 0), \mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, x=0∧y=0\bigr\rbrack\bigr\rbrack, 0)\bigr\rbrack$

live ↗

In$\mathrm{PiecewiseExpand}(\lbrack a\rbrack)$Out$\mathrm{Piecewise}(\bigl\lbrack\bigl\lbrack1, a\bigr\rbrack\bigr\rbrack, 0)$

Possible issues #

with no assumption that $x$ is real, Abs is left alone — same as Wolfram

live ↗

In$\mathrm{PiecewiseExpand}(\vert x\vert)$Out$\vert x\vert$

From a span of 100, Floor is left as it is, as in Wolfram

live ↗

In$\mathrm{PiecewiseExpand}(\lfloor x\rfloor, 0\lt x\lt100)$Out$\lfloor x\rfloor$

About

Examples	64
Checked against	Wolfram Language 60 agree2 differ
Crosswalk	Wolfram Language