154 lines
4.9 KiB
Plaintext
154 lines
4.9 KiB
Plaintext
import Common.Real.Floor
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import Common.Set.Basic
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import Mathlib.Data.Real.Basic
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/-! # Apostol.Chapter_1_11 -/
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namespace Apostol.Chapter_1_11
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open BigOperators
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/-! ## Exercise 4
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Prove that the greatest-integer function has the properties indicated.
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-/
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/-- ### Exercise 4a
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`⌊x + n⌋ = ⌊x⌋ + n` for every integer `n`.
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-/
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theorem exercise_4a (x : ℝ) (n : ℤ) : ⌊x + n⌋ = ⌊x⌋ + n :=
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Int.floor_add_int x n
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/-- ### Exercise 4b.1
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`⌊-x⌋ = -⌊x⌋` if `x` is an integer.
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-/
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theorem exercise_4b_1 (x : ℤ) : ⌊-x⌋ = -⌊x⌋ := by
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simp only [Int.floor_int, id_eq]
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/-- ### Exercise 4b.2
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`⌊-x⌋ = -⌊x⌋ - 1` otherwise.
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-/
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theorem exercise_4b_2 (x : ℝ) (h : ∃ n : ℤ, x ∈ Set.Ioo ↑n (↑n + (1 : ℝ)))
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: ⌊-x⌋ = -⌊x⌋ - 1 := by
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rw [Int.floor_neg]
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suffices ⌈x⌉ = ⌊x⌋ + 1 by
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have := congrArg (HMul.hMul (-1)) this
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simp only [neg_mul, one_mul, neg_add_rev, add_comm] at this
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exact this
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have ⟨n, hn⟩ := h
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have hn' : x ∈ Set.Ico ↑n (↑n + (1 : ℝ)) :=
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Set.mem_of_subset_of_mem Set.Ioo_subset_Ico_self hn
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rw [Int.ceil_eq_iff, Int.floor_eq_on_Ico n x hn']
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simp only [Int.cast_add, Int.cast_one, add_sub_cancel]
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apply And.intro
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· exact (Set.mem_Ioo.mp hn).left
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· exact le_of_lt (Set.mem_Ico.mp hn').right
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/-- ### Exercise 4c
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`⌊x + y⌋ = ⌊x⌋ + ⌊y⌋` or `⌊x⌋ + ⌊y⌋ + 1`.
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-/
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theorem exercise_4c (x y : ℝ)
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: ⌊x + y⌋ = ⌊x⌋ + ⌊y⌋ ∨ ⌊x + y⌋ = ⌊x⌋ + ⌊y⌋ + 1 := by
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have hx : x = Int.floor x + Int.fract x := Eq.symm (add_eq_of_eq_sub' rfl)
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have hy : y = Int.floor y + Int.fract y := Eq.symm (add_eq_of_eq_sub' rfl)
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by_cases Int.fract x + Int.fract y < 1
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· refine Or.inl ?_
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rw [Int.floor_eq_iff]
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simp only [Int.cast_add]
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apply And.intro
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· exact add_le_add (Int.floor_le x) (Int.floor_le y)
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· conv => lhs; rw [hx, hy, add_add_add_comm]; arg 1; rw [add_comm]
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rwa [add_comm, ← add_assoc, ← sub_lt_iff_lt_add', ← sub_sub, add_sub_cancel, add_sub_cancel]
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· refine Or.inr ?_
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rw [Int.floor_eq_iff]
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simp only [Int.cast_add, Int.cast_one]
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have h := le_of_not_lt h
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apply And.intro
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· conv => lhs; rw [← add_rotate]
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conv => rhs; rw [hx, hy, add_add_add_comm]; arg 1; rw [add_comm]
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rwa [← sub_le_iff_le_add', ← sub_sub, add_sub_cancel, add_sub_cancel]
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· conv => lhs; rw [hx, hy, add_add_add_comm]; arg 1; rw [add_comm]
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conv => lhs; rw [add_comm, ← add_assoc]
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conv => rhs; rw [add_assoc]
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rw [← sub_lt_iff_lt_add', ← sub_sub, add_sub_cancel, add_sub_cancel]
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exact add_lt_add (Int.fract_lt_one x) (Int.fract_lt_one y)
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/-- ### Exercise 5
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The formulas in Exercises 4(d) and 4(e) suggest a generalization for `⌊nx⌋`.
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State and prove such a generalization.
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-/
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theorem exercise_5 (n : ℕ) (x : ℝ)
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: ⌊n * x⌋ = Finset.sum (Finset.range n) (fun i => ⌊x + i/n⌋) :=
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Real.Floor.floor_mul_eq_sum_range_floor_add_index_div n x
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/-- ### Exercise 4d
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`⌊2x⌋ = ⌊x⌋ + ⌊x + 1/2⌋`
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-/
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theorem exercise_4d (x : ℝ)
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: ⌊2 * x⌋ = ⌊x⌋ + ⌊x + 1/2⌋ := by
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suffices ⌊x⌋ + ⌊x + 1/2⌋ = Finset.sum (Finset.range 2) (fun i => ⌊x + i/2⌋) by
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rw [this]
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exact exercise_5 2 x
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unfold Finset.sum
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simp
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rw [add_comm]
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/-- ### Exercise 4e
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`⌊3x⌋ = ⌊x⌋ + ⌊x + 1/3⌋ + ⌊x + 2/3⌋`
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-/
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theorem exercise_4e (x : ℝ)
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: ⌊3 * x⌋ = ⌊x⌋ + ⌊x + 1/3⌋ + ⌊x + 2/3⌋ := by
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suffices ⌊x⌋ + ⌊x + 1/3⌋ + ⌊x + 2/3⌋ = Finset.sum (Finset.range 3) (fun i => ⌊x + i/3⌋) by
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rw [this]
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exact exercise_5 3 x
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unfold Finset.sum
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simp
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conv => rhs; rw [← add_rotate']; arg 2; rw [add_comm]
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rw [← add_assoc]
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/-- ### Exercise 7b
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If `a` and `b` are positive integers with no common factor, we have the formula
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`∑_{n=1}^{b-1} ⌊na / b⌋ = ((a - 1)(b - 1)) / 2`. When `b = 1`, the sum on the
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left is understood to be `0`.
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Derive the result analytically as follows: By changing the index of summation,
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note that `Σ_{n=1}^{b-1} ⌊na / b⌋ = Σ_{n=1}^{b-1} ⌊a(b - n) / b⌋`. Now apply
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Exercises 4(a) and (b) to the bracket on the right.
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-/
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theorem exercise_7b (ha : a > 0) (hb : b > 0) (hp : Nat.coprime a b)
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: ∑ n in (Finset.range b).filter (· > 0), ⌊n * ((a : ℕ) : ℝ) / b⌋ =
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((a - 1) * (b - 1)) / 2 := by
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sorry
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section
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/-- ### Exercise 8
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Let `S` be a set of points on the real line. The *characteristic function* of
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`S` is, by definition, the function `Χ` such that `Χₛ(x) = 1` for every `x` in
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`S`, and `Χₛ(x) = 0` for those `x` not in `S`. Let `f` be a step function which
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takes the constant value `cₖ` on the `k`th open subinterval `Iₖ` of some
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partition of an interval `[a, b]`. Prove that for each `x` in the union
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`I₁ ∪ I₂ ∪ ⋯ ∪ Iₙ` we have
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```
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f(x) = ∑_{k=1}^n cₖΧ_{Iₖ}(x).
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```
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This property is described by saying that every step function is a linear
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combination of characteristic functions of intervals.
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-/
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theorem exercise_8 : True := sorry
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end
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end Apostol.Chapter_1_11
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