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(* Title: HOL/ex/NatSum.thy
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ID: $Id$
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Author: Tobias Nipkow
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*)
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header {* Summing natural numbers *}
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theory NatSum imports Main begin
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text {*
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Summing natural numbers, squares, cubes, etc.
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Thanks to Sloane's On-Line Encyclopedia of Integer Sequences,
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\url{http://www.research.att.com/~njas/sequences/}.
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*}
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lemmas [simp] =
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left_distrib right_distrib
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left_diff_distrib right_diff_distrib --{*for true subtraction*}
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diff_mult_distrib diff_mult_distrib2 --{*for type nat*}
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text {*
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\medskip The sum of the first @{text n} odd numbers equals @{text n}
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squared.
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*}
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lemma sum_of_odds: "(\<Sum>i=0..<n. Suc (i + i)) = n * n"
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by (induct n) auto
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text {*
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\medskip The sum of the first @{text n} odd squares.
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*}
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lemma sum_of_odd_squares:
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"3 * (\<Sum>i=0..<n. Suc(2*i) * Suc(2*i)) = n * (4 * n * n - 1)"
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by (induct n) auto
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text {*
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\medskip The sum of the first @{text n} odd cubes
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*}
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lemma sum_of_odd_cubes:
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"(\<Sum>i=0..<n. Suc (2*i) * Suc (2*i) * Suc (2*i)) =
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n * n * (2 * n * n - 1)"
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by (induct n) auto
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text {*
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\medskip The sum of the first @{text n} positive integers equals
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@{text "n (n + 1) / 2"}.*}
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lemma sum_of_naturals:
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"2 * (\<Sum>i=0..n. i) = n * Suc n"
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by (induct n) auto
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lemma sum_of_squares:
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"6 * (\<Sum>i=0..n. i * i) = n * Suc n * Suc (2 * n)"
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by (induct n) auto
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lemma sum_of_cubes:
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"4 * (\<Sum>i=0..n. i * i * i) = n * n * Suc n * Suc n"
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by (induct n) auto
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text{* \medskip A cute identity: *}
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lemma sum_squared: "(\<Sum>i=0..n. i)^2 = (\<Sum>i=0..n::nat. i^3)"
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proof(induct n)
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case 0 show ?case by simp
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next
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case (Suc n)
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have "(\<Sum>i = 0..Suc n. i)^2 =
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(\<Sum>i = 0..n. i^3) + (2*(\<Sum>i = 0..n. i)*(n+1) + (n+1)^2)"
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(is "_ = ?A + ?B")
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using Suc by(simp add:nat_number)
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also have "?B = (n+1)^3"
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using sum_of_naturals by(simp add:nat_number)
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also have "?A + (n+1)^3 = (\<Sum>i=0..Suc n. i^3)" by simp
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finally show ?case .
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qed
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text {*
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\medskip Sum of fourth powers: three versions.
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*}
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lemma sum_of_fourth_powers:
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"30 * (\<Sum>i=0..n. i * i * i * i) =
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n * Suc n * Suc (2 * n) * (3 * n * n + 3 * n - 1)"
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apply (induct n)
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apply simp_all
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apply (case_tac n) -- {* eliminates the subtraction *}
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apply (simp_all (no_asm_simp))
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done
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text {*
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Two alternative proofs, with a change of variables and much more
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subtraction, performed using the integers. *}
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lemma int_sum_of_fourth_powers:
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"30 * int (\<Sum>i=0..<m. i * i * i * i) =
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int m * (int m - 1) * (int(2 * m) - 1) *
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(int(3 * m * m) - int(3 * m) - 1)"
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by (induct m) (simp_all add: int_mult)
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lemma of_nat_sum_of_fourth_powers:
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"30 * of_nat (\<Sum>i=0..<m. i * i * i * i) =
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of_nat m * (of_nat m - 1) * (of_nat (2 * m) - 1) *
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(of_nat (3 * m * m) - of_nat (3 * m) - (1::int))"
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by (induct m) simp_all
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text {*
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\medskip Sums of geometric series: @{text 2}, @{text 3} and the
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general case.
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*}
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lemma sum_of_2_powers: "(\<Sum>i=0..<n. 2^i) = 2^n - (1::nat)"
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by (induct n) (auto split: nat_diff_split)
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lemma sum_of_3_powers: "2 * (\<Sum>i=0..<n. 3^i) = 3^n - (1::nat)"
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by (induct n) auto
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lemma sum_of_powers: "0 < k ==> (k - 1) * (\<Sum>i=0..<n. k^i) = k^n - (1::nat)"
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by (induct n) auto
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end
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