# Theory Sum_of_Powers

theory Sum_of_Powers
imports Complex_Main
```(*  Author:     Lukas Bulwahn <lukas.bulwahn-at-gmail.com> *)
section ‹Sum of Powers›

theory Sum_of_Powers
imports Complex_Main
begin

subsection ‹Additions to @{theory HOL.Binomial} Theory›

lemma (in field_char_0) one_plus_of_nat_neq_zero [simp]:
"1 + of_nat n ≠ 0"
proof -
have "of_nat (Suc n) ≠ of_nat 0"
unfolding of_nat_eq_iff by simp
then show ?thesis by simp
qed

lemma of_nat_binomial_eq_mult_binomial_Suc:
assumes "k ≤ n"
shows "(of_nat :: (nat ⇒ ('a :: field_char_0))) (n choose k) = of_nat (n + 1 - k) / of_nat (n + 1) * of_nat (Suc n choose k)"
proof (cases k)
case 0 then show ?thesis by simp
next
case (Suc l)
have "of_nat (n + 1) * (∏i=0..<k. of_nat (n - i)) = (of_nat :: (nat ⇒ 'a)) (n + 1 - k) * (∏i=0..<k. of_nat (Suc n - i))"
using prod.atLeast0_lessThan_Suc [where ?'a = 'a, symmetric, of "λi. of_nat (Suc n - i)" k]
also have "... = (of_nat :: (nat ⇒ 'a)) (Suc n - k) * (∏i=0..<k. of_nat (Suc n - i))"
by (simp add: Suc atLeast0_atMost_Suc atLeastLessThanSuc_atLeastAtMost)
also have "... = (of_nat :: (nat ⇒ 'a)) (n + 1 - k) * (∏i=0..<k. of_nat (Suc n - i))"
by (simp only: Suc_eq_plus1)
finally have "(∏i=0..<k. of_nat (n - i)) = (of_nat :: (nat ⇒ 'a)) (n + 1 - k) / of_nat (n + 1) * (∏i=0..<k. of_nat (Suc n - i))"
with assms show ?thesis
qed

lemma real_binomial_eq_mult_binomial_Suc:
assumes "k ≤ n"
shows "(n choose k) = (n + 1 - k) / (n + 1) * (Suc n choose k)"
by (metis Suc_eq_plus1 add.commute assms le_SucI of_nat_Suc of_nat_binomial_eq_mult_binomial_Suc of_nat_diff)

subsection ‹Preliminaries›

lemma integrals_eq:
assumes "f 0 = g 0"
assumes "⋀ x. ((λx. f x - g x) has_real_derivative 0) (at x)"
shows "f x = g x"
proof -
show "f x = g x"
proof (cases "x ≠ 0")
case True
from assms DERIV_const_ratio_const[OF this, of "λx. f x - g x" 0]
show ?thesis by auto
qed

lemma sum_diff: "((∑i≤n::nat. f (i + 1) - f i)::'a::field) = f (n + 1) - f 0"
by (induct n) (auto simp add: field_simps)

declare One_nat_def [simp del]

subsection ‹Bernoulli Numbers and Bernoulli Polynomials›

declare sum.cong [fundef_cong]

fun bernoulli :: "nat ⇒ real"
where
"bernoulli 0 = (1::real)"
| "bernoulli (Suc n) =  (-1 / (n + 2)) * (∑k ≤ n. ((n + 2 choose k) * bernoulli k))"

declare bernoulli.simps[simp del]

definition
"bernpoly n = (λx. ∑k ≤ n. (n choose k) * bernoulli k * x ^ (n - k))"

subsection ‹Basic Observations on Bernoulli Polynomials›

lemma bernpoly_0: "bernpoly n 0 = bernoulli n"
proof (cases n)
case 0
then show "bernpoly n 0 = bernoulli n"
unfolding bernpoly_def bernoulli.simps by auto
next
case (Suc n')
have "(∑k≤n'. real (Suc n' choose k) * bernoulli k * 0 ^ (Suc n' - k)) = 0"
by (rule sum.neutral) auto
with Suc show ?thesis
unfolding bernpoly_def by simp
qed

lemma sum_binomial_times_bernoulli:
"(∑k≤n. ((Suc n) choose k) * bernoulli k) = (if n = 0 then 1 else 0)"
proof (cases n)
case 0
then show ?thesis by (simp add: bernoulli.simps)
next
case Suc
then show ?thesis
qed

subsection ‹Sum of Powers with Bernoulli Polynomials›

lemma bernpoly_derivative [derivative_intros]:
"(bernpoly (Suc n) has_real_derivative ((n + 1) * bernpoly n x)) (at x)"
proof -
have "(bernpoly (Suc n) has_real_derivative (∑k≤n. real (Suc n - k) * x ^ (n - k) * (real (Suc n choose k) * bernoulli k))) (at x)"
unfolding bernpoly_def by (rule DERIV_cong) (fast intro!: derivative_intros, simp)
moreover have "(∑k≤n. real (Suc n - k) * x ^ (n - k) * (real (Suc n choose k) * bernoulli k)) = (n + 1) * bernpoly n x"
unfolding bernpoly_def
by (auto intro: sum.cong simp add: sum_distrib_left real_binomial_eq_mult_binomial_Suc[of _ n] Suc_eq_plus1 of_nat_diff)
ultimately show ?thesis by auto
qed

lemma diff_bernpoly:
"bernpoly n (x + 1) - bernpoly n x = n * x ^ (n - 1)"
proof (induct n arbitrary: x)
case 0
show ?case unfolding bernpoly_def by auto
next
case (Suc n)
have "bernpoly (Suc n) (0 + 1) - bernpoly (Suc n) 0 = (Suc n) * 0 ^ n"
unfolding bernpoly_0 unfolding bernpoly_def by (simp add: sum_binomial_times_bernoulli zero_power)
then have const: "bernpoly (Suc n) (0 + 1) - bernpoly (Suc n) 0 = real (Suc n) * 0 ^ n" by (simp add: power_0_left)
have hyps': "⋀x. (real n + 1) * bernpoly n (x + 1) - (real n + 1) * bernpoly n x = real n * x ^ (n - Suc 0) * real (Suc n)"
unfolding right_diff_distrib[symmetric] by (simp add: Suc.hyps One_nat_def)
note [derivative_intros] = DERIV_chain'[where f = "λx::real. x + 1" and g = "bernpoly (Suc n)" and s="UNIV"]
have derivative: "⋀x. ((%x. bernpoly (Suc n) (x + 1) - bernpoly (Suc n) x - real (Suc n) * x ^ n) has_real_derivative 0) (at x)"
by (rule DERIV_cong) (fast intro!: derivative_intros, simp add: hyps')
from integrals_eq[OF const derivative] show ?case by simp
qed

lemma sum_of_powers: "(∑k≤n::nat. (real k) ^ m) = (bernpoly (Suc m) (n + 1) - bernpoly (Suc m) 0) / (m + 1)"
proof -
from diff_bernpoly[of "Suc m", simplified] have "(m + (1::real)) * (∑k≤n. (real k) ^ m) = (∑k≤n. bernpoly (Suc m) (real k + 1) - bernpoly (Suc m) (real k))"
by (auto simp add: sum_distrib_left intro!: sum.cong)
also have "... = (∑k≤n. bernpoly (Suc m) (real (k + 1)) - bernpoly (Suc m) (real k))"
by simp
also have "... = bernpoly (Suc m) (n + 1) - bernpoly (Suc m) 0"
by (simp only: sum_diff[where f="λk. bernpoly (Suc m) (real k)"]) simp
finally show ?thesis by (auto simp add: field_simps intro!: eq_divide_imp)
qed

subsection ‹Instances for Square And Cubic Numbers›

lemma binomial_unroll:
"n > 0 ⟹ (n choose k) = (if k = 0 then 1 else (n - 1) choose (k - 1) + ((n - 1) choose k))"

lemma sum_unroll:
"(∑k≤n::nat. f k) = (if n = 0 then f 0 else f n + (∑k≤n - 1. f k))"
by auto (metis One_nat_def Suc_pred add.commute sum_atMost_Suc)

lemma bernoulli_unroll:
"n > 0 ⟹ bernoulli n = - 1 / (real n + 1) * (∑k≤n - 1. real (n + 1 choose k) * bernoulli k)"
by (cases n) (simp add: bernoulli.simps One_nat_def)+

lemmas unroll = binomial_unroll
bernoulli.simps(1) bernoulli_unroll sum_unroll bernpoly_def

lemma sum_of_squares: "(∑k≤n::nat. k ^ 2) = (2 * n ^ 3 + 3 * n ^ 2 + n) / 6"
proof -
have "real (∑k≤n::nat. k ^ 2) = (∑k≤n::nat. (real k) ^ 2)" by simp
also have "... = (bernpoly 3 (real (n + 1)) - bernpoly 3 0) / real (3 :: nat)"
also have "... = (2 * n ^ 3 + 3 * n ^ 2 + n) / 6"
by (simp add: unroll algebra_simps power2_eq_square power3_eq_cube One_nat_def[symmetric])
finally show ?thesis by simp
qed

lemma sum_of_squares_nat: "(∑k≤n::nat. k ^ 2) = (2 * n ^ 3 + 3 * n ^ 2 + n) div 6"
proof -
from sum_of_squares have "real (6 * (∑k≤n. k ^ 2)) = real (2 * n ^ 3 + 3 * n ^ 2 + n)"
then have "6 * (∑k≤n. k ^ 2) = 2 * n ^ 3 + 3 * n ^ 2 + n"
using of_nat_eq_iff by blast
then show ?thesis by auto
qed

lemma sum_of_cubes: "(∑k≤n::nat. k ^ 3) = (n ^ 2 + n) ^ 2 / 4"
proof -
have two_plus_two: "2 + 2 = 4" by simp
have power4_eq: "⋀x::real. x ^ 4 = x * x * x * x"
by (simp only: two_plus_two[symmetric] power_add power2_eq_square)
have "real (∑k≤n::nat. k ^ 3) = (∑k≤n::nat. (real k) ^ 3)" by simp
also have "... = ((bernpoly 4 (n + 1) - bernpoly 4 0)) / (real (4 :: nat))"
also have "... = ((n ^ 2 + n) / 2) ^ 2"
by (simp add: unroll algebra_simps power2_eq_square power4_eq power3_eq_cube)
finally show ?thesis by (simp add: power_divide)
qed

lemma sum_of_cubes_nat: "(∑k≤n::nat. k ^ 3) = (n ^ 2 + n) ^ 2 div 4"
proof -
from sum_of_cubes have "real (4 * (∑k≤n. k ^ 3)) = real ((n ^ 2 + n) ^ 2)"