(* Title: HOL/ex/NatSum.thy
Author: Tobias Nipkow
*)
section \<open>Summing natural numbers\<close>
theory NatSum
imports Main
begin
text \<open>
Summing natural numbers, squares, cubes, etc.
Thanks to Sloane's On-Line Encyclopedia of Integer Sequences,
\<^url>\<open>https://oeis.org\<close>.
\<close>
lemmas [simp] =
ring_distribs
diff_mult_distrib diff_mult_distrib2 \<comment> \<open>for type nat\<close>
text \<open>\<^medskip> The sum of the first \<open>n\<close> odd numbers equals \<open>n\<close> squared.\<close>
lemma sum_of_odds: "(\i=0..
by (induct n) auto
text \<open>\<^medskip> The sum of the first \<open>n\<close> odd squares.\<close>
lemma sum_of_odd_squares:
"3 * (\i=0..
by (induct n) auto
text \<open>\<^medskip> The sum of the first \<open>n\<close> odd cubes.\<close>
lemma sum_of_odd_cubes:
"(\i=0..
n * n * (2 * n * n - 1)"
by (induct n) auto
text \<open>\<^medskip> The sum of the first \<open>n\<close> positive integers equals \<open>n (n + 1) / 2\<close>.\<close>
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 \<open>\<^medskip> A cute identity:\<close>
lemma sum_squared: "(\i=0..n. i)^2 = (\i=0..n. i^3)" for n :: nat
proof (induct n)
case 0
show ?case by simp
next
case (Suc n)
have "(\i = 0..Suc n. i)^2 =
(\<Sum>i = 0..n. i^3) + (2*(\<Sum>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 \<open>\<^medskip> Sum of fourth powers: three versions.\<close>
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)"
proof (induct n)
case 0
show ?case by simp
next
case (Suc n)
then show ?case
by (cases n) \<comment> \<open>eliminates the subtraction\<close>
simp_all
qed
text \<open>
Two alternative proofs, with a change of variables and much more
subtraction, performed using the integers.
\<close>
lemma int_sum_of_fourth_powers:
"30 * int (\i=0..
int m * (int m - 1) * (int(2 * m) - 1) *
(int(3 * m * m) - int(3 * m) - 1)"
by (induct m) simp_all
lemma of_nat_sum_of_fourth_powers:
"30 * of_nat (\i=0..
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 \<open>\<^medskip> Sums of geometric series: \<open>2\<close>, \<open>3\<close> and the general case.\<close>
lemma sum_of_2_powers: "(\i=0..
by (induct n) auto
lemma sum_of_3_powers: "2 * (\i=0..
by (induct n) auto
lemma sum_of_powers: "0 < k \ (k - 1) * (\i=0..
for k :: nat
by (induct n) auto
end
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