(* Title: HOL/ex/NatSum.thy ID: $Id: NatSum.thy,v 1.27 2007/06/23 17:33:23 nipkow Exp $ 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:nat_number) also have "?B = (n+1)^3" using sum_of_naturals by(simp add:nat_number) 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
lemma
a * (b + c) = a * b + a * c
(a + b) * c = a * c + b * c
(a - b) * c = a * c - b * c
a * (b - c) = a * b - a * c
(m - n) * k = m * k - n * k
k * (m - n) = k * m - k * n
lemma sum_of_odds:
(∑i = 0..<n. Suc (i + i)) = n * n
lemma sum_of_odd_squares:
3 * (∑i = 0..<n. Suc (2 * i) * Suc (2 * i)) = n * (4 * n * n - 1)
lemma sum_of_odd_cubes:
(∑i = 0..<n. Suc (2 * i) * Suc (2 * i) * Suc (2 * i)) = n * n * (2 * n * n - 1)
lemma sum_of_naturals:
2 * ∑{0..n} = n * Suc n
lemma sum_of_squares:
6 * (∑i = 0..n. i * i) = n * Suc n * Suc (2 * n)
lemma sum_of_cubes:
4 * (∑i = 0..n. i * i * i) = n * n * Suc n * Suc n
lemma sum_squared:
∑{0..n} ^ 2 = (∑i = 0..n. i ^ 3)
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)
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)
lemma of_nat_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)
lemma sum_of_2_powers:
setsum (op ^ 2) {0..<n} = 2 ^ n - 1
lemma sum_of_3_powers:
2 * setsum (op ^ 3) {0..<n} = 3 ^ n - 1
lemma sum_of_powers:
0 < k ==> (k - 1) * setsum (op ^ k) {0..<n} = k ^ n - 1