Theory NatSum

theory NatSum
imports Parity
(*  Title:  HOL/ex/NatSum.thy
    Author: Tobias Nipkow
*)

header {* Summing natural numbers *}

theory NatSum imports Main Parity begin

text {*
  Summing natural numbers, squares, cubes, etc.

  Thanks to Sloane's On-Line Encyclopedia of Integer Sequences,
  @{url "http://www.research.att.com/~njas/sequences/"}.
*}

lemmas [simp] =
  ring_distribs
  diff_mult_distrib diff_mult_distrib2 --{*for type nat*}

text {*
  \medskip The sum of the first @{text n} odd numbers equals @{text n}
  squared.
*}

lemma sum_of_odds: "(∑i=0..<n. Suc (i + i)) = n * n"
  by (induct n) auto


text {*
  \medskip The sum of the first @{text n} odd squares.
*}

lemma sum_of_odd_squares:
  "3 * (∑i=0..<n. Suc(2*i) * Suc(2*i)) = n * (4 * n * n - 1)"
  by (induct n) auto


text {*
  \medskip The sum of the first @{text n} odd cubes
*}

lemma sum_of_odd_cubes:
  "(∑i=0..<n. Suc (2*i) * Suc (2*i) * Suc (2*i)) =
    n * n * (2 * n * n - 1)"
  by (induct n) auto

text {*
  \medskip The sum of the first @{text n} positive integers equals
  @{text "n (n + 1) / 2"}.*}

lemma sum_of_naturals:
    "2 * (∑i=0..n. i) = n * Suc n"
  by (induct n) auto

lemma sum_of_squares:
    "6 * (∑i=0..n. i * i) = n * Suc n * Suc (2 * n)"
  by (induct n) auto

lemma sum_of_cubes:
    "4 * (∑i=0..n. i * i * i) = n * n * Suc n * Suc n"
  by (induct n) auto

text{* \medskip A cute identity: *}

lemma sum_squared: "(∑i=0..n. i)^2 = (∑i=0..n::nat. i^3)"
proof(induct n)
  case 0 show ?case by simp
next
  case (Suc n)
  have "(∑i = 0..Suc n. i)^2 =
        (∑i = 0..n. i^3) + (2*(∑i = 0..n. i)*(n+1) + (n+1)^2)"
    (is "_ = ?A + ?B")
    using Suc by(simp add:eval_nat_numeral)
  also have "?B = (n+1)^3"
    using sum_of_naturals by(simp add:eval_nat_numeral)
  also have "?A + (n+1)^3 = (∑i=0..Suc n. i^3)" by simp
  finally show ?case .
qed

text {*
  \medskip Sum of fourth powers: three versions.
*}

lemma sum_of_fourth_powers:
  "30 * (∑i=0..n. i * i * i * i) =
    n * Suc n * Suc (2 * n) * (3 * n * n + 3 * n - 1)"
  apply (induct n)
   apply simp_all
  apply (case_tac n)  -- {* eliminates the subtraction *} 
   apply (simp_all (no_asm_simp))
  done

text {*
  Two alternative proofs, with a change of variables and much more
  subtraction, performed using the integers. *}

lemma int_sum_of_fourth_powers:
  "30 * int (∑i=0..<m. i * i * i * i) =
    int m * (int m - 1) * (int(2 * m) - 1) *
    (int(3 * m * m) - int(3 * m) - 1)"
  by (induct m) (simp_all add: int_mult)

lemma of_nat_sum_of_fourth_powers:
  "30 * of_nat (∑i=0..<m. i * i * i * i) =
    of_nat m * (of_nat m - 1) * (of_nat (2 * m) - 1) *
    (of_nat (3 * m * m) - of_nat (3 * m) - (1::int))"
  by (induct m) (simp_all add: of_nat_mult)


text {*
  \medskip Sums of geometric series: @{text 2}, @{text 3} and the
  general case.
*}

lemma sum_of_2_powers: "(∑i=0..<n. 2^i) = 2^n - (1::nat)"
  by (induct n) (auto split: nat_diff_split)

lemma sum_of_3_powers: "2 * (∑i=0..<n. 3^i) = 3^n - (1::nat)"
  by (induct n) auto

lemma sum_of_powers: "0 < k ==> (k - 1) * (∑i=0..<n. k^i) = k^n - (1::nat)"
  by (induct n) auto

end