Continuing working on Apostol 1.11 exercises.
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@ -22,7 +22,7 @@ The properties of area in this set of exercises are to be deduced from the
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Prove that each of the following sets is measurable and has zero area:
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\subsection*{\proceeding{Exercise 1a}}%
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\subsection*{\unverified{Exercise 1a}}%
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\label{sub:exercise-1a}
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A set consisting of a single point.
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@ -38,7 +38,7 @@ A set consisting of a single point.
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\end{proof}
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\subsection*{\proceeding{Exercise 1b}}%
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\subsection*{\unverified{Exercise 1b}}%
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\label{sub:exercise-1b}
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A set consisting of a finite number of points in a plane.
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@ -97,7 +97,7 @@ A set consisting of a finite number of points in a plane.
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\end{proof}
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\subsection*{\proceeding{Exercise 1c}}%
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\subsection*{\unverified{Exercise 1c}}%
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\label{sub:exercise-1c}
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The union of a finite collection of line segments in a plane.
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@ -382,24 +382,19 @@ Prove that the formula is valid for right triangles and parallelograms.
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Likewise,
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\begin{equation}
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\label{sub:exercise-4b-eq2}
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I_T = \frac{1}{2}(I_R - H_L + 2).
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I_T = \frac{1}{2}(I_R - (H_L - 2)).
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\end{equation}
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The following shows the lattice point area formula is in agreement with
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the expected result:
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\begin{align*}
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I_T + \frac{1}{2}B_T - 1
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& = \frac{1}{2}(I_R - H_L + 2) + \frac{1}{2}B_T - 1
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& = \frac{1}{2}(I_R - (H_L - 2)) + \frac{1}{2}B_T - 1
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& \eqref{sub:exercise-4b-eq2} \\
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& = \frac{1}{2}\left[ I_R - H_L + 2 + B_T - 2 \right] \\
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& = \frac{1}{2}\left[ I_R - H_L + B_T \right] \\
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& = \frac{1}{2}\left[ I_R - H_L + \frac{1}{2}B_R - 1 + H_L \right]
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& \eqref{sub:exercise-4b-eq1} \\
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& = \frac{1}{2}\left[ I_R + \frac{1}{2}B_R - 1 \right] \\
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& = \frac{1}{2}\left[ (w - 1)(h - 1) + \frac{1}{2}(2(w + h)) - 1 \right]
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& \text{\nameref{sub:exercise-4a}} \\
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& = \frac{1}{2}\left[ (w - 1)(h - 1) + w + h - 1 \right] \\
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& = \frac{1}{2}\left[ wh - w - h + 1 + w + h - 1 \right] \\
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& = \frac{wh}{2}.
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& = \frac{1}{2}\left[ wh \right] & \text{\nameref{sub:exercise-4a}}.
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\end{align*}
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We do not prove this formula is valid for parallelograms here.
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@ -1,12 +1,14 @@
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import Mathlib.Algebra.BigOperators.Basic
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import Mathlib.Data.Real.Basic
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import Mathlib.Data.Finset.Basic
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import Mathlib.Tactic.LibrarySearch
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import Common.Real.Floor
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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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@ -76,49 +78,14 @@ theorem exercise_4c (x y : ℝ)
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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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namespace Hermite
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/--
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Constructs a partition of `[0, 1)` that looks as follows:
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```
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[0, 1/n), [1/n, 2/n), ..., [(n-1)/n, 1)
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```
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-/
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def partition (n : ℕ) (i : ℕ) : Set ℝ := Set.Ico (i / n) ((i + 1) / n)
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/--
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The indexed union of the family of sets of a `partition` is equal to `[0, 1)`.
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-/
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theorem partition_eq_Ico_zero_one
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: (⋃ i ∈ Finset.range n, partition n i) = Set.Ico 0 1 := by
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sorry
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/--
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The fractional portion of any real number is always in `[0, 1)`.
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-/
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theorem fract_mem_Ico_zero_one (x : ℝ)
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: Int.fract x ∈ Set.Ico 0 1 := ⟨Int.fract_nonneg x, Int.fract_lt_one x⟩
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/--
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The fractional portion of any real number always exists in some member of the
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indexed family of sets formed by any `partition`.
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-/
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theorem fract_mem_partition (r : ℝ) (hr : r ∈ Set.Ico 0 1)
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: ∀ n : ℕ, ∃ j : ℕ, r ∈ Set.Ico ↑(j / n) ↑((j + 1) / n) := by
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sorry
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end Hermite
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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⌋) := by
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let r := Int.fract x
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have hx : x = ⌊x⌋ + r := Eq.symm (add_eq_of_eq_sub' rfl)
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sorry
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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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@ -157,6 +124,9 @@ 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 : True := sorry
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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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end Apostol.Chapter_1_11
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@ -2,6 +2,8 @@
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\input{../../preamble}
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\externaldocument[C:1:07:]{Chapter_1_07}[Chapter_1_07.pdf]
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\newcommand{\lean}[1]{\leanref
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{./Chapter\_1\_11.html\#Apostol.Chapter\_1\_11.#1}
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{Apostol.Chapter\_1\_11.#1}}
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@ -15,7 +17,7 @@
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Prove that the greatest-integer function has the properties indicated:
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\subsection*{\proceeding{Exercise 4a}}%
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\subsection*{\verified{Exercise 4a}}%
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\label{sub:exercise-4a}
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$\floor{x + n} = \floor{x} + n$ for every integer $n$.
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@ -24,9 +26,19 @@ $\floor{x + n} = \floor{x} + n$ for every integer $n$.
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\lean{exercise\_4a}
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\divider
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Let $x$ be a real number and $n$ an integer.
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Let $m = \floor{x + n}$.
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By definition of the floor function, $m$ is the unique integer such that
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$m \leq x + n < m + 1$.
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Then $m - n \leq x < (m - n) + 1$.
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That is, $m - n = \floor{x}$.
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Rearranging terms we see that $m = \floor{x} + n$ as expected.
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\end{proof}
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\subsection*{\proceeding{Exercise 4b}}%
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\subsection*{\verified{Exercise 4b}}%
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\label{sub:exercise-4b}
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$\floor{-x} =
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@ -37,16 +49,45 @@ $\floor{-x} =
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\begin{proof}
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\ % Force space prior to *Proof.*
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\ \vspace{6pt}
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\begin{enumerate}[(a)]
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\item \lean{exercise\_4b\_1}
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\item \lean{exercise\_4b\_2}
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\end{enumerate}
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\lean{exercise\_4b\_1}
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\lean{exercise\_4b\_2}
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\divider
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There are two cases to consider:
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\paragraph{Case 1}%
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Suppose $x$ is an integer.
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Then $x = \floor{x}$ and $-x = \floor{-x}$.
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It immediately follows that $$\floor{-x} = -x = -\floor{x}.$$
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\paragraph{Case 2}%
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Suppose $x$ is not an integer.
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Let $m = \floor{-x}$.
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By definition of the floor function, $m$ is the unique integer such that
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$m \leq -x < m + 1$.
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Equivalently, $-m - 1 < x \leq -m$.
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Since $x$ is not an integer, it follows $-m - 1 \leq x < -m$.
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Then, by definition of the floor function, $\floor{x} = -m - 1$.
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Solving for $m$ yields $$\floor{-x} = m = -\floor{x} - 1.$$
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\paragraph{Conclusion}%
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The above two cases are exhaustive. Thus
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$$\floor{-x} =
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\begin{cases}
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-\floor{x} & \text{if } x \text{ is an integer}, \\
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-\floor{x} - 1 & \text{otherwise}.
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\end{cases}$$
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\end{proof}
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\subsection*{\proceeding{Exercise 4c}}%
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\subsection*{\verified{Exercise 4c}}%
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\label{sub:exercise-4c}
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$\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$.
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@ -55,6 +96,40 @@ $\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$.
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\lean{exercise\_4c}
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\divider
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Rewrite $x$ and $y$ as the sum of their floor and fractional components:
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$x = \floor{x} + \{x\}$ and $y = \floor{y} + \{y\}$.
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Now
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\begin{align}
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\floor{x + y}
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& = \floor{\floor{x} + \{x\} + \floor{y} + \{y\}} \nonumber \\
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& = \floor{\floor{x} + \floor{y} + \{x\} + \{y\}} \nonumber \\
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& = \floor{x} + \floor{y} + \floor{\{x\} + \{y\}}
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& \text{\nameref{sub:exercise-4a}} \label{sub:exercise-4c-eq1}
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\end{align}
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There are two cases to consider:
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\paragraph{Case 1}%
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Suppose $\{x\} + \{y\} < 1$.
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Then $\floor{\{x\} + \{y\}} = 0$.
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Substituting this value into \eqref{sub:exercise-4c-eq1} yields
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$$\floor{x + y} = \floor{x} + \floor{y}.$$
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\paragraph{Case 2}%
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Suppose $\{x\} + \{y\} \geq 1$.
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Because $\{x\}$ and $\{y\}$ are both less than $1$, $\{x\} + \{y\} < 2$.
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Thus $\floor{\{x\} + \{y\}} = 1$.
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Substituting this value into \eqref{sub:exercise-4c-eq1} yields
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$$\floor{x + y} = \floor{x} + \floor{y} + 1.$$
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\paragraph{Conclusion}%
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Since the above two cases are exhaustive, it follows
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$\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$.
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\end{proof}
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\subsection*{\proceeding{Exercise 4d}}%
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@ -66,6 +141,11 @@ $\floor{2x} = \floor{x} + \floor{x + \frac{1}{2}}.$
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\lean{exercise\_4d}
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\divider
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This is immediately proven by applying Hermite's Identity as shown in
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\nameref{sec:exercise-5}.
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\end{proof}
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\subsection*{\proceeding{Exercise 4e}}%
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@ -77,6 +157,11 @@ $\floor{3x} = \floor{x} + \floor{x + \frac{1}{3}} + \floor{x + \frac{2}{3}}.$
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\lean{exercise\_4e}
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\divider
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This is immediately proven by applying Hermite's Identity as shown in
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\nameref{sec:exercise-5}.
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\end{proof}
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\section*{\proceeding{Exercise 5}}%
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@ -116,7 +201,7 @@ State and prove such a generalization.
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\paragraph{Left-Hand Side}%
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Consider the left-hande side of identity \eqref{sec:exercise-5-eq1}
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Consider the left-hand side of identity \eqref{sec:exercise-5-eq1}.
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By \eqref{sec:exercise-5-eq2}, $nr \in \ico{j}{j + 1}$.
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Therefore $\floor{nr} = j$.
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Thus
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@ -182,13 +267,32 @@ Prove that the number of lattice points in $S$ is equal to the sum
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\begin{proof}
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Define $S_i = \mathbb{Z} \cap \ioc{0}{f(i)}$ for all $i \in \mathbb{Z}$.
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By definition, the set of lattice points of $S$ is given by
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$$L = \{ (i, j) : i = a, \ldots, b \land j \in S_i \}.$$
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By construction, it follows $$\sum_{j \in S_i} 1 = \floor{f(i)}.$$
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Therefore $$\abs{L}
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= \sum_{i=a}^b \sum_{j \in S_i} 1
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= \sum_{i=1}^b \floor{f(i)}.$$
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Let $i = a, \ldots, b$ and define $S_i = \mathbb{N} \cap \ioc{0}{f(i)}$.
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By construction, the number of lattice points in $S$ is
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\begin{equation}
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\label{sec:exercise-6-eq1}
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\sum_{n = a}^b \abs{S_n}.
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\end{equation}
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All that remains is to show $\abs{S_i} = \floor{f(i)}$.
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There are two cases to consider:
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\paragraph{Case 1}%
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Suppose $f(i)$ is an integer.
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Then the number of integers in $\ioc{0}{f(i)}$ is $f(i) = \floor{f(i)}$.
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\paragraph{Case 2}%
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Suppose $f(i)$ is not an integer.
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Then the number of integers in $\ioc{0}{f(i)}$ is the same as that of
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$\ioc{0}{\floor{f(i)}}$.
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Once again, that number is $\floor{f(i)}$.
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\paragraph{Conclusion}%
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By cases 1 and 2, $\abs{S_i} = \floor{f(i)}$.
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Substituting this identity into \eqref{sec:exercise-6-eq1} finishes the
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proof.
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\end{proof}
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@ -199,6 +303,9 @@ If $a$ and $b$ are positive integers with no common factor, we have the formula
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$$\sum_{n=1}^{b-1} \floor{\frac{na}{b}} = \frac{(a - 1)(b - 1)}{2}.$$
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When $b = 1$, the sum on the left is understood to be $0$.
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\note{When $b = 1$, the proofs of (a) and (b) are trivial. We continue under the
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assumption $b > 1$.}
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\subsection*{\unverified{Exercise 7a}}%
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\label{sub:exercise-7a}
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@ -207,7 +314,78 @@ Derive this result by a geometric argument, counting lattice points in a right
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\begin{proof}
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TODO
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Let $f \colon [1, b - 1] \rightarrow \mathbb{R}$ be given by $f(x) = ax / b$.
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Let $S$ denote the set of points $(x, y)$ satisfying $1 \leq x \leq b - 1$,
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$0 < y \leq f(x)$.
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By \nameref{sec:exercise-6}, the number of lattice points of $S$ is equal to
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the sum
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\begin{equation}
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\label{sub:exercise-7a-eq1}
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\sum_{n=1}^{b-1} \floor{f(n)} = \sum_{n=1}^{b-1} \floor{\frac{na}{b}}.
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\end{equation}
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Define $T$ to be the triangle of width $w = b$ and height $h = f(b) = a$
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as $$T = \{ (x, y) : 0 < x < b, 0 < y \leq f(x) \}.$$
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By construction, $T$ does not introduce any additional lattice points.
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Thus $S$ and $T$ have the same number of lattice points.
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Let $H_L$ denote the number of boundary points on $T$'s hypotenuse.
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We prove that (i) $H_L = 2$ and (ii) that $T$ has $\frac{(a - 1)(b - 1)}{2}$
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lattice points.
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\paragraph{(i)}%
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\label{par:exercise-7a-i}
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Consider the line $L$ overlapping the hypotenuse of $T$.
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By construction, $T$'s hypotenuse has endpoints $(0, 0)$ and $(b, a)$.
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By hypothesis, $a$ and $b$ are positive, excluding the possibility of $L$
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being vertical.
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Define the slope of $L$ as $$m = \frac{a}{b}.$$
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$H_L$ coincides with the number of indices $i = 0, \ldots, b$ such that
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$(i, i * m)$ is a lattice point.
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But $a$ and $b$ are coprime by hypothesis and $i \leq b$.
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Thus $i * m$ is an integer if and only if $i = 0$ or $i = b$.
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Thus $H_L = 2$.
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\paragraph{(ii)}%
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Next we count the number of lattice points in $T$.
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Let $R$ be the overlapping retangle of width $w$ and height $h$, situated
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with bottom-left corner at $(0, 0)$.
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Let $I_R$ denote the number of interior lattice points of $R$.
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Let $I_T$ and $B_T$ denote the interior and boundary lattice points of $T$
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respectively.
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By \nameref{C:1:07:sub:exercise-4b-eq2},
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\begin{align}
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I_T
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& = \frac{1}{2}(I_R - (H_L - 2)) \nonumber \\
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& = \frac{1}{2}(I_R - (2 - 2))
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& \text{\nameref{par:exercise-7a-i}} \nonumber \\
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& = \frac{1}{2}I_R. & \label{sub:exercise-7a-eq2}
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\end{align}
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Furthermore, since both the adjacent and opposite side of $T$ are not
|
||||
included in $T$ and there exist no lattice points on $T$'s hypotenuse
|
||||
besides the endpoints, it follows
|
||||
\begin{equation}
|
||||
\label{sub:exercise-7a-eq3}
|
||||
B_T = 0.
|
||||
\end{equation}
|
||||
Thus the number of lattice points of $T$ equals
|
||||
\begin{align}
|
||||
I_T + B_T
|
||||
& = I_T & \eqref{sub:exercise-7a-eq3} \nonumber \\
|
||||
& = \frac{1}{2}I_R & \eqref{sub:exercise-7a-eq2} \nonumber \\
|
||||
& = \frac{(b - 1)(a - 1)}{2}.
|
||||
& \text{\nameref{C:1:07:sub:exercise-4a}}
|
||||
\label{sub:exercise-7a-eq4}
|
||||
\end{align}
|
||||
|
||||
\paragraph{Conclusion}%
|
||||
|
||||
By \eqref{sub:exercise-7a-eq1} the number of lattice points of $S$ is equal
|
||||
to the sum $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}}.$$
|
||||
But the number of lattice points of $S$ is the same as that of $T$.
|
||||
By \eqref{sub:exercise-7a-eq4}, the number of lattice points in $T$ is equal
|
||||
to $$\frac{(b - 1)(a - 1)}{2}.$$
|
||||
Thus $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}} = \frac{(a - 1)(b - 1)}{2}.$$
|
||||
|
||||
\end{proof}
|
||||
|
||||
|
|
@ -223,6 +401,35 @@ Now apply Exercises 4(a) and (b) to the bracket on the right.
|
|||
|
||||
\lean{exercise\_7b}
|
||||
|
||||
\divider
|
||||
|
||||
Let $n = 1, \ldots, b - 1$.
|
||||
By hypothesis, $a$ and $b$ are coprime.
|
||||
Furthermore, $n < b$ for all values of $n$.
|
||||
Thus $an / b$ is not an integer.
|
||||
By \nameref{sub:exercise-4b},
|
||||
\begin{equation}
|
||||
\label{sub:exercise-7b-eq1}
|
||||
\floor{-\frac{an}{b}} = -\floor{\frac{an}{b}} - 1.
|
||||
\end{equation}
|
||||
Consider the following:
|
||||
\begin{align*}
|
||||
\sum_{n=1}^{b-1} \floor{\frac{na}{b}}
|
||||
& = \sum_{n=1}^{b-1} \floor{\frac{a(b - n)}{b}} \\
|
||||
& = \sum_{n=1}^{b-1} \floor{\frac{ab - an}{b}} \\
|
||||
& = \sum_{n=1}^{b-1} \floor{-\frac{an}{b} + a} \\
|
||||
& = \sum_{n=1}^{b-1} \floor{-\frac{an}{b}} + a.
|
||||
& \text{\nameref{sub:exercise-4a}} \\
|
||||
& = \sum_{n=1}^{b-1} -\floor{\frac{an}{b}} - 1 + a
|
||||
& \eqref{sub:exercise-7b-eq1} \\
|
||||
& = -\sum_{n=1}^{b-1} \floor{\frac{an}{b}} - \sum_{n=1}^{b-1} 1 +
|
||||
\sum_{n=1}^{b-1} a \\
|
||||
& = -\sum_{n=1}^{b-1} \floor{\frac{an}{b}} - (b - 1) + a(b - 1).
|
||||
\end{align*}
|
||||
Rearranging the above yields
|
||||
$$2\sum_{n=1}^{b-1} \floor{\frac{an}{b}} = (a - 1)(b - 1).$$
|
||||
Dividing both sides of the above identity concludes the proof.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\unverified{Exercise 8}}%
|
||||
|
|
|
|||
|
|
@ -0,0 +1,149 @@
|
|||
import Mathlib.Algebra.BigOperators.Basic
|
||||
import Mathlib.Data.Real.Basic
|
||||
import Mathlib.Data.Finset.Basic
|
||||
import Mathlib.Tactic.LibrarySearch
|
||||
|
||||
/-! # Common.Real.Floor
|
||||
|
||||
A collection of useful definitions and theorems around the floor function.
|
||||
-/
|
||||
|
||||
namespace Real.Floor
|
||||
|
||||
/--
|
||||
The fractional portion of any real number is always in `[0, 1)`.
|
||||
-/
|
||||
theorem fract_mem_Ico_zero_one (x : ℝ)
|
||||
: Int.fract x ∈ Set.Ico 0 1 :=
|
||||
⟨Int.fract_nonneg x, Int.fract_lt_one x⟩
|
||||
|
||||
/-! ## Hermite's Identity
|
||||
|
||||
Definitions and theorems in support of proving Hermite's identity.
|
||||
-/
|
||||
|
||||
namespace Hermite
|
||||
|
||||
/--
|
||||
A partition of `[0, 1)` that looks as follows:
|
||||
|
||||
```
|
||||
[0, 1/n), [1/n, 2/n), ..., [(n-1)/n, 1)
|
||||
```
|
||||
|
||||
This is expected to be used as an indexing function of a union of sets, e.g.
|
||||
`⋃ i ∈ Finset.range n, partition n i`.
|
||||
-/
|
||||
def partition (n : ℕ) (i : ℕ) : Set ℝ := Set.Ico (↑i / n) ((↑i + 1) / n)
|
||||
|
||||
/--
|
||||
The fractional portion of any real number always exists in some member of the
|
||||
indexed family of sets formed by any `partition`.
|
||||
-/
|
||||
theorem fract_mem_partition (r : ℝ) (hr : r ∈ Set.Ico 0 1)
|
||||
: ∀ n : ℕ, ∃ j : ℕ,
|
||||
j < n ∧ r ∈ Set.Ico (((j : ℕ) : ℝ) / n) ((↑j + 1) / n) := by
|
||||
sorry
|
||||
|
||||
/--
|
||||
The indexed union of the family of sets of a `partition` is a subset of `[0, 1)`.
|
||||
-/
|
||||
theorem partition_subset_Ico_zero_one
|
||||
: (⋃ i ∈ Finset.range n, partition n i) ⊆ Set.Ico 0 1 := by
|
||||
simp only [
|
||||
Finset.mem_range,
|
||||
gt_iff_lt,
|
||||
zero_lt_one,
|
||||
not_true,
|
||||
ge_iff_le,
|
||||
Set.unionᵢ_subset_iff
|
||||
]
|
||||
intro i hi x hx
|
||||
have hn : (0 : ℝ) < n := calc (0 : ℝ)
|
||||
_ ≤ i := Nat.cast_nonneg i
|
||||
_ < n := Nat.cast_lt.mpr hi
|
||||
apply And.intro
|
||||
· have h_zero_le_i_div_n : (0 : ℝ) ≤ i / n := by
|
||||
rw [← mul_le_mul_right hn, zero_mul, div_mul, div_self, div_one]
|
||||
· exact Nat.cast_nonneg i
|
||||
· exact ne_iff_lt_or_gt.mpr (Or.inr hn)
|
||||
calc (0 : ℝ)
|
||||
_ ≤ i / n := h_zero_le_i_div_n
|
||||
_ ≤ x := hx.left
|
||||
· have h_succ_div_n_le_one : (i + 1) / n ≤ (1 : ℝ) := by
|
||||
rw [div_le_one_iff]
|
||||
refine Or.inl ?_
|
||||
exact ⟨hn, by norm_cast⟩
|
||||
calc x
|
||||
_ < (i + 1) / n := hx.right
|
||||
_ ≤ 1 := h_succ_div_n_le_one
|
||||
|
||||
/--
|
||||
`[0, 1)` is a subset of the indexed union of the family of sets of a `partition`.
|
||||
-/
|
||||
theorem Ico_zero_one_subset_partition
|
||||
: Set.Ico 0 1 ⊆ (⋃ i ∈ Finset.range n, partition n i) := by
|
||||
intro x hx
|
||||
simp only [Finset.mem_range, Set.mem_unionᵢ, exists_prop]
|
||||
unfold partition
|
||||
exact fract_mem_partition x hx n
|
||||
|
||||
/--
|
||||
The indexed union of the family of sets of a `partition` is equal to `[0, 1)`.
|
||||
-/
|
||||
theorem partition_eq_Ico_zero_one
|
||||
: (⋃ i ∈ Finset.range n, partition n i) = Set.Ico 0 1 :=
|
||||
Set.Subset.antisymm_iff.mpr
|
||||
⟨partition_subset_Ico_zero_one, Ico_zero_one_subset_partition⟩
|
||||
|
||||
end Hermite
|
||||
|
||||
open BigOperators
|
||||
|
||||
/-- ### Hermite's Identity
|
||||
|
||||
The following decomposes the floor of a multiplication into a sum of floors.
|
||||
-/
|
||||
theorem floor_mul_eq_sum_range_floor_add_index_div (n : ℕ) (x : ℝ)
|
||||
: ⌊n * x⌋ = ∑ i in Finset.range n, ⌊x + i / n⌋ := by
|
||||
let r := Int.fract x
|
||||
|
||||
-- Here we see there exists some `j` such that `r ∈ [j / n, (j + 1) / n]`.
|
||||
have hx : x = ⌊x⌋ + r := Eq.symm (add_eq_of_eq_sub' rfl)
|
||||
have ⟨j, ⟨hj, hr⟩⟩ :=
|
||||
Hermite.fract_mem_partition r (fract_mem_Ico_zero_one x) n
|
||||
|
||||
-- With the above definitions established, we now show the left- and
|
||||
-- right-hand sides of the goal evaluate to the same number.
|
||||
|
||||
have hlhs : ⌊n * x⌋ = n * ⌊x⌋ + j := by
|
||||
have hn : (0 : ℝ) < n := calc (0 : ℝ)
|
||||
_ ≤ j := Nat.cast_nonneg j
|
||||
_ < n := Nat.cast_lt.mpr hj
|
||||
-- We prove that `nr ∈ [j, j + 1)`. It must then follow `⌊nr⌋ = j`.
|
||||
have hnr : n * r ∈ Set.Ico ((j : ℕ) : ℝ) (j + 1) := by
|
||||
apply And.intro
|
||||
· have := hr.left
|
||||
rw [← mul_le_mul_right hn, div_mul, div_self, div_one] at this
|
||||
· rwa [mul_comm]
|
||||
· exact ne_of_gt hn
|
||||
· have := hr.right
|
||||
rw [← mul_lt_mul_right hn, div_mul, div_self, div_one] at this
|
||||
· rwa [mul_comm]
|
||||
· exact ne_of_gt hn
|
||||
have hnr_eq_j : ⌊n * r⌋ = j := by
|
||||
have := Int.floor_eq_on_Ico' j (n * r) hnr
|
||||
norm_cast at this
|
||||
conv =>
|
||||
lhs
|
||||
rw [hx, mul_add, add_comm]
|
||||
norm_cast
|
||||
rw [Int.floor_add_int, hnr_eq_j, add_comm]
|
||||
|
||||
have hrhs : ∑ i in Finset.range n, ⌊x + i / n⌋ = n * ⌊x⌋ + j := by
|
||||
sorry
|
||||
|
||||
-- Close out goal by showing left- and right-hand side equal a common value.
|
||||
rw [hlhs, hrhs]
|
||||
|
||||
end Real.Floor
|
||||
|
|
@ -37,9 +37,9 @@ def index : BaseHtmlM Html := do templateExtends (baseHtml "Index") <|
|
|||
status:
|
||||
<ul>
|
||||
<li>
|
||||
<span style="color:cyan">Cyan statements </span> indicate axioms and
|
||||
definitions. There must exist a corresponding <code>axiom</code> or
|
||||
<code>def</code> in Lean.
|
||||
<span style="color:darkgray">Dark gray statements </span> indicate
|
||||
axioms and definitions. There must exist a corresponding
|
||||
<code>axiom</code> or <code>def</code> in Lean.
|
||||
</li>
|
||||
<li>
|
||||
<span style="color:teal">Teal statements </span> indicate those
|
||||
|
|
|
|||
|
|
@ -50,7 +50,7 @@
|
|||
% Indicates a statement corresponds to an axiom or definition. There must exist
|
||||
% a corresponding `axiom` or `def` in Lean.
|
||||
\DeclareRobustCommand{\defined}[1]{%
|
||||
\texorpdfstring{\color{cyan}\faParagraph\ #1}{#1}}
|
||||
\texorpdfstring{\color{darkgray}\faParagraph\ #1}{#1}}
|
||||
|
||||
% Indicates a statement has a complete proof in both LaTeX *and* Lean.
|
||||
\DeclareRobustCommand{\verified}[1]{%
|
||||
|
|
|
|||
Loading…
Reference in New Issue