Calculemus (Publicaciones sobre Induction)

Nicomachus’s theorem

José A. Alonso — Fri, 03 Jan 2025 04:00:00 GMT

Prove the Nicomachus's theorem which states that the sum of the cubes of the first \(n\) natural numbers is equal to the square of the sum of the first \(n\) natural numbers; that is, for any natural number n we have \[ 1^3 + 2^3 + ... + n^3 = (1 + 2 + ... + n)^2 \]

To do this, complete the following Lean4 theory:

import Mathlib.Data.Nat.Defs
import Mathlib.Tactic

open Nat

variable ( n : ℕ)

def sum : ℕ → ℕ
  | 0      => 0
  | succ n => sum n + (n+1)

def sumCubes : ℕ → ℕ
  | 0   => 0
  | n+1 => sumCubes n + (n+1)^3

example :
  sumCubes n = (sum n)^2 :=
by sorry

Leer más… (quedan 8 minutos de lectura)

Proofs of "If p > -1, then (1+p)ⁿ ≥ 1+np"

José A. Alonso — Thu, 12 Sep 2024 04:00:00 GMT

Let \(p ∈ ℝ\) and \(n ∈ ℕ\). Prove that if \(p > -1\), then \[ (1 + p)^n ≥ 1 + np \]

To do this, complete the following Lean4 theory:

import Mathlib.Data.Real.Basic
import Mathlib.Tactic

open Nat

variable (p : ℝ)
variable (n : ℕ)

example
  (h : p > -1)
  : (1 + p)^n ≥ 1 + n*p :=
by sorry

Leer más… (quedan 4 minutos de lectura)

Proofs of 0³+1³+2³+3³+···+n³ = (n(n+1)/2)²

José A. Alonso — Tue, 10 Sep 2024 04:00:00 GMT

Prove that the sum of the first cubes \[ 0^3 + 1^3 + 2^3 + 3^3 + ··· + n^3 \] is \[ \dfrac{n(n+1)}{2}^2 \]

To do this, complete the following Lean4 theory:

import Mathlib.Data.Nat.Defs
import Mathlib.Tactic

open Nat

variable (n : ℕ)

def sumCubes : ℕ → ℕ
  | 0   => 0
  | n+1 => sumCubes n + (n+1)^3

example :
  4 * sumCubes n = (n*(n+1))^2 :=
by sorry

Leer más… (quedan 3 minutos de lectura)

Proofs of a+aq+aq²+···+aqⁿ = a(1-qⁿ⁺¹)/(1-q)

José A. Alonso — Mon, 09 Sep 2024 04:00:00 GMT

Prove that if \(q ≠ 1\), then the sum of the terms of the geometric progression \[ a + aq + aq^2 + ··· + aq^n \] is \[ \dfrac{a(1 - q^{n+1})}{1 - q} \]

To do this, complete the following Lean4 theory:

import Mathlib.Data.Nat.Defs
import Mathlib.Tactic

open Nat

variable (n : ℕ)
variable (a q : ℝ)

def sumGP : ℝ → ℝ → ℕ → ℝ
  | a, _, 0     => a
  | a, q, n + 1 => sumGP a q n + (a * q^(n + 1))

example
  (h : q ≠ 1)
  : sumGP a q n = a * (1 - q^(n + 1)) / (1 - q) :=
by sorry

Leer más… (quedan 6 minutos de lectura)

Proofs of a+(a+d)+(a+2d)+···+(a+nd)=(n+1)(2a+nd)/2

José A. Alonso — Sat, 07 Sep 2024 04:00:00 GMT

Prove that the sum of the terms of the arithmetic progression \[ a + (a + d) + (a + 2d) + ··· + (a + nd) \] is \[ \dfrac{(n + 1)(2a + nd)}{2} \]

To do this, complete the following Lean4 theory:

import Mathlib.Data.Nat.Defs
import Mathlib.Tactic

open Nat

variable (n : ℕ)
variable (a d : ℝ)

def sumAP : ℝ → ℝ → ℕ → ℝ
  | a, _, 0     => a
  | a, d, n + 1 => sumAP a d n + (a + (n + 1) * d)

example :
  2 * sumAP a d n = (n + 1) * (2 * a + n * d) :=
by sorry

Leer más… (quedan 5 minutos de lectura)

Proofs of "0+1+2+3+···+n=n(n + 1)/2"

José A. Alonso — Thu, 05 Sep 2024 04:00:00 GMT

Prove that the sum of the natural numbers \[ 0 + 1 + 2 + 3 + ··· + n \] is \[\dfrac{n(n + 1)}{2}\]

To do this, complete the following Lean4 theory:

import Init.Data.Nat.Basic
import Mathlib.Tactic

open Nat

variable (n : Nat)

set_option pp.fieldNotation false

def sum : Nat → Nat
  | 0      => 0
  | succ n => sum n + (n+1)

example :
  2 * sum n = n * (n + 1) :=
by sorry

Leer más… (quedan 7 minutos de lectura)

Equivalence of definitions of the Fibonacci function

José A. Alonso — Thu, 29 Aug 2024 04:00:00 GMT

In Lean4, the Fibonacci function can be defined as

   def fibonacci : Nat → Nat
     | 0     => 0
     | 1     => 1
     | n + 2 => fibonacci n + fibonacci (n+1)

Another more efficient definition is

   def fib (n : Nat) : Nat :=
     (loop n).1
   where
     loop : Nat → Nat × Nat
       | 0   => (0, 1)
       | n + 1 =>
         let p := loop n
         (p.2, p.1 + p.2)

Prove that both definitions are equivalent; that is,

   fibonacci = fib

To do this, complete the following Lean4 theory:

open Nat

set_option pp.fieldNotation false

def fibonacci : Nat → Nat
  | 0     => 0
  | 1     => 1
  | n + 2 => fibonacci n + fibonacci (n+1)

def fib (n : Nat) : Nat :=
  (loop n).1
where
  loop : Nat → Nat × Nat
    | 0   => (0, 1)
    | n + 1 =>
      let p := loop n
      (p.2, p.1 + p.2)

example : fibonacci = fib :=
by sorry

Leer más… (quedan 5 minutos de lectura)