Improve evaluation (Horner methods) of Taylor1s with intervals (#249)

* Improve evaluation of Taylor1 with intervals

* Bump patch version

Project.toml

@@ -1,7 +1,7 @@
 name = "TaylorSeries"
 uuid = "6aa5eb33-94cf-58f4-a9d0-e4b2c4fc25ea"
 repo = "https://github.com/JuliaDiff/TaylorSeries.jl.git"
-version = "0.10.6"
+version = "0.10.7"
 
 [deps]
 InteractiveUtils = "b77e0a4c-d291-57a0-90e8-8db25a27a240"
@@ -11,7 +11,7 @@ Requires = "ae029012-a4dd-5104-9daa-d747884805df"
 SparseArrays = "2f01184e-e22b-5df5-ae63-d93ebab69eaf"
 
 [compat]
-IntervalArithmetic = "0.15, 0.16"
+IntervalArithmetic = "0.15, 0.16, 0.17"
 Requires = "0.5.2, 1.0"
 julia = "1"
 

src/intervals.jl

@@ -1,27 +1,15 @@
 using .IntervalArithmetic
 
 function evaluate(a::Taylor1, dx::Interval)
-    dx == (-1..1) && return _evaluate(a, dx, Val(true))
-    # Usual Horner rule
+    order = a.order
     uno = one(dx)
-    @inbounds begin
-        suma = a[end]*uno
-        for k in a.order-1:-1:0
-            suma = suma*dx + a[k]
-        end
-    end
-    return suma
-end
-
-function _evaluate(a::Taylor1, dx::Interval{T}, ::Val{true}) where {T}
-    uno = Interval{T}(1, 1)
-    dx2 = Interval{T}(0, 1)
-    if iseven(a.order)
-        kend = a.order-2
+    dx2 = dx^2
+    if iseven(order)
+        kend = order-2
         @inbounds sum_even = a[end]*uno
-        @inbounds sum_odd = a[end-1]*Interval{T}(0, 0)
+        @inbounds sum_odd = a[end-1]*zero(dx)
     else
-        kend = a.order-3
+        kend = order-3
         @inbounds sum_odd = a[end]*uno
         @inbounds sum_even = a[end-1]*uno
     end
@@ -32,8 +20,6 @@ function _evaluate(a::Taylor1, dx::Interval{T}, ::Val{true}) where {T}
     return sum_even + sum_odd*dx
 end
 
-_evaluate(a::Taylor1, dx::Interval, ::Val{false}) = a(dx)
-
 
 function evaluate(a::TaylorN, dx::IntervalBox{N,T}) where {N,T}
 

test/intervals.jl

@@ -53,6 +53,12 @@ eeuler = Base.MathConstants.e
     @test evaluate(x*y^2, (-1..1)×(-1..1)) == (-1..1)
     @test evaluate(x^2*y^2, (-1..1)×(-1..1)) == (0..1)
 
+    ii = -1..1
+    t = Taylor1(1)
+    @test 0..2 ⊆ (1+t)(ii)
+    t = Taylor1(2)
+    @test 0..4 ⊆ ((1+t)^2)(ii)
+
     ii = 0..6
     t = Taylor1(4)
     f(x) = 0.1 * x^3 - 0.5*x^2 + 1