(* Title: HOL/ex/NatSum.thy
ID: $Id$
Author: Tobias Nipkow
*)
header {* Summing natural numbers *}
theory NatSum imports Main 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] =
left_distrib right_distrib
left_diff_distrib right_diff_distrib --{*for true subtraction*}
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: "(\<Sum>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 * (\<Sum>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:
"(\<Sum>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 * (\<Sum>i=0..n. i) = n * Suc n"
by (induct n) auto
lemma sum_of_squares:
"6 * (\<Sum>i=0..n. i * i) = n * Suc n * Suc (2 * n)"
by (induct n) auto
lemma sum_of_cubes:
"4 * (\<Sum>i=0..n. i * i * i) = n * n * Suc n * Suc n"
by (induct n) auto
text {*
\medskip Sum of fourth powers: three versions.
*}
lemma sum_of_fourth_powers:
"30 * (\<Sum>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 (\<Sum>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 (\<Sum>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
text {*
\medskip Sums of geometric series: @{text 2}, @{text 3} and the
general case.
*}
lemma sum_of_2_powers: "(\<Sum>i=0..<n. 2^i) = 2^n - (1::nat)"
by (induct n) (auto split: nat_diff_split)
lemma sum_of_3_powers: "2 * (\<Sum>i=0..<n. 3^i) = 3^n - (1::nat)"
by (induct n) auto
lemma sum_of_powers: "0 < k ==> (k - 1) * (\<Sum>i=0..<n. k^i) = k^n - (1::nat)"
by (induct n) auto
end