Tácticas  matemáticas. Mathematical  ploys.⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀...

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Enunciado

x + 1/x = √ 3 , 1 qué vale x ²⁰²⁴ + ―― x ²⁰²⁴

Este tipo de problemas pertenecen al ámbito de las funciones simétricas y aunque es posible solucionarlos de una forma algebraica , vamos a seguir la senda del análisis combinatorio .

Preliminares

x = α , 1/x = β

-a₁= (1) = ∑ α = α + β = √ 3

a₂= (1²) = ∑ α β = α ∙ β = 1

Nos facilitan los valores de las funciones simétricas elementales (1), (1²).
Nos piden determinar el valor de la suma de potencias de índice 2024.

s₂₀₂₄ = (2024) = ∑α ²⁰²⁴= α ²⁰²⁴ + β ²⁰²⁴

Desarrollo

La relación entre las funciones simétricas elementales y la suma de potencias es la razón de ser de las identidades de Newton ,

s_n − a₁s_n-1 + a₂s_n-2 − ... + (−1)ⁿna_n = 0

Esta ecuación nos permite establecer una relación de recurrencia en s_n

s₁ = a₁
...
s_n = a₁s_n-1 − a₂s_n-2 + ... + (−1)^{n + 1}na_n

s₁= − √ 3
s₂= (− √ 3 ∙ − √ 3 ) − (2 ∙ 1) = 1

a_n= 0 , n > 2

s_n+2= − √ 3 ∙ s_n+1 − s_n

Hemos obtenido una recurrencia de orden dos, su ecuación característica determina el término general de la sucesión,

p² + √ 3 p + 1 = 0

Con soluciones,

p = e^{± i ∙5/6∙π}

De la naturaleza de las soluciones de la ecuación característica inferimos que la sucesión es periódica,

5/6 ∙ n = 2k
5n = 12k
n = 12

∴ s_n+12= s_n
s_n+6 = -s_n

Resolución

Solución

2024 ≡ 8 mod 12

s₂₀₂₄ = s₈= - s₂

s₂₀₂₄ = - 1

∎

English

Español

Statement

x + 1/x = √ 3 , 1 calculate x ²⁰²⁴ + ―― x ²⁰²⁴

This is a problem on symmetric functions , solvable by pure algebraic methods, but here we are taking the combinatorial route and seeing...

Prelude

x = α , 1/x = β

-a₁= (1) = ∑ α = α + β = √ 3

a₂= (1²) = ∑ α β = α ∙ β = 1

We know value of the elementary symmetric functions (1), (1²).
We are asked for the value of the powers sum with 2024 index .

s₂₀₂₄ = (2024) = ∑α ²⁰²⁴= α ²⁰²⁴ + β ²⁰²⁴

Development

Symmetric elementary functions related to powers sum ones by means of the Newtonian identities ,

s_n − a₁s_n-1 + a₂s_n-2 − ... + (−1)ⁿna_n = 0

That is a recurrence relation on s_n,

s₁ = a₁
...
s_n = a₁s_n-1 − a₂s_n-2 + ... + (−1)^{n + 1}na_n

s₁= − √ 3
s₂= (− √ 3 ∙ − √ 3 ) − (2 ∙ 1) = 1

a_n= 0 , n > 2

s_n+2= − √ 3 ∙ s_n+1 − s_n

We have a recurrence of second order, the general term is formed with solutions of its characteristic equation ,

p² + √ 3 p + 1 = 0

p = e^{± i ∙5/6∙π}

We can infer that the sequence is periodic as the solutions of its charaterisitic equation .

Tácticas  matemáticas. Mathematical  ploys.⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Enunciado

x + 1/x = √ 3 ,

1
qué vale x ²⁰²⁴ + ――
x ²⁰²⁴

Preliminares

x = α , 1/x = β

-a₁= (1) = ∑ α = α + β = √ 3

a₂= (1²) = ∑ α β = α ∙ β = 1

s₂₀₂₄ = (2024) = ∑α ²⁰²⁴= α ²⁰²⁴ + β ²⁰²⁴

Desarrollo

s_n − a₁s_n-1 + a₂s_n-2 − ... + (−1)ⁿna_n = 0

s₁ = a₁
...
s_n = a₁s_n-1 − a₂s_n-2 + ... + (−1)^{n + 1}na_n

s₁= − √ 3
s₂= (− √ 3 ∙ − √ 3 ) − (2 ∙ 1) = 1

a_n= 0 , n > 2

s_n+2= − √ 3 ∙ s_n+1 − s_n

p² + √ 3 p + 1 = 0

p = e^{± i ∙5/6∙π}

5/6 ∙ n = 2k
5n = 12k
n = 12

∴ s_n+12= s_n
s_n+6 = -s_n

Resolución

Solución

2024 ≡ 8 mod 12

s₂₀₂₄ = s₈= - s₂

s₂₀₂₄ = - 1

Statement

x + 1/x = √ 3 ,

1
calculate x ²⁰²⁴ + ――
x ²⁰²⁴

Prelude

x = α , 1/x = β

-a₁= (1) = ∑ α = α + β = √ 3

a₂= (1²) = ∑ α β = α ∙ β = 1

s₂₀₂₄ = (2024) = ∑α ²⁰²⁴= α ²⁰²⁴ + β ²⁰²⁴

Development

s_n − a₁s_n-1 + a₂s_n-2 − ... + (−1)ⁿna_n = 0

s₁ = a₁
...
s_n = a₁s_n-1 − a₂s_n-2 + ... + (−1)^{n + 1}na_n

s₁= − √ 3
s₂= (− √ 3 ∙ − √ 3 ) − (2 ∙ 1) = 1

a_n= 0 , n > 2

s_n+2= − √ 3 ∙ s_n+1 − s_n

p² + √ 3 p + 1 = 0

p = e^{± i ∙5/6∙π}

5/6 ∙ n = 2k
5n = 12k
n = 12

∴ s_n+12= s_n
s_n+6 = -s_n

Solution

Answer

2024 ≡ 8 mod 12

s₂₀₂₄ = s₈= - s₂

s₂₀₂₄ = - 1

Tácticas matemáticas. Mathematical ploys.⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Enunciado

x + 1/x = √ 3 , 1qué vale x 2024 + ―― x 2024

Preliminares

x = α , 1/x = β-a1 = (1) = ∑ α = α + β = √ 3 a2 = (12 ) = ∑ α β = α ∙ β = 1

s2024 = (2024) = ∑α 2024 = α 2024 + β 2024

Desarrollo

sn − a1sn-1 + a2sn-2 − ... + (−1)nnan = 0

s1 = a1 ... sn = a1sn-1 − a2sn-2 + ... + (−1)n + 1nan

s1 = − √ 3 s2 = (− √ 3 ∙ − √ 3 ) − (2 ∙ 1) = 1 an = 0 , n > 2 sn+2 = − √ 3 ∙ sn+1 − sn

p2 + √ 3 p + 1 = 0

p = e ± i ∙5/6∙π

5/6 ∙ n = 2k 5n = 12k n = 12 ∴ sn+12 = sn sn+6 = -sn

Resolución

Solución

2024 ≡ 8 mod 12s2024 = s8 = - s2s2024 = - 1

Statement

x + 1/x = √ 3 , 1calculate x 2024 + ―― x 2024

Prelude

x = α , 1/x = β-a1 = (1) = ∑ α = α + β = √ 3 a2 = (12 ) = ∑ α β = α ∙ β = 1

s2024 = (2024) = ∑α 2024 = α 2024 + β 2024

Development

sn − a1sn-1 + a2sn-2 − ... + (−1)nnan = 0

s1 = a1 ... sn = a1sn-1 − a2sn-2 + ... + (−1)n + 1nan

s1 = − √ 3 s2 = (− √ 3 ∙ − √ 3 ) − (2 ∙ 1) = 1 an = 0 , n > 2 sn+2 = − √ 3 ∙ sn+1 − sn

p2 + √ 3 p + 1 = 0

p = e ± i ∙5/6∙π

5/6 ∙ n = 2k 5n = 12k n = 12 ∴ sn+12 = sn sn+6 = -sn

Solution

Answer

2024 ≡ 8 mod 12s2024 = s8 = - s2s2024 = - 1

Tácticas  matemáticas. Mathematical  ploys.⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

x + 1/x = √ 3 ,

1
qué vale x ²⁰²⁴ + ――
x ²⁰²⁴

x = α , 1/x = β

-a₁= (1) = ∑ α = α + β = √ 3

a₂= (1²) = ∑ α β = α ∙ β = 1

s₂₀₂₄ = (2024) = ∑α ²⁰²⁴= α ²⁰²⁴ + β ²⁰²⁴

s_n − a₁s_n-1 + a₂s_n-2 − ... + (−1)ⁿna_n = 0

s₁ = a₁
...
s_n = a₁s_n-1 − a₂s_n-2 + ... + (−1)^{n + 1}na_n

s₁= − √ 3
s₂= (− √ 3 ∙ − √ 3 ) − (2 ∙ 1) = 1

a_n= 0 , n > 2

s_n+2= − √ 3 ∙ s_n+1 − s_n

p² + √ 3 p + 1 = 0

p = e^{± i ∙5/6∙π}

5/6 ∙ n = 2k
5n = 12k
n = 12

∴ s_n+12= s_n
s_n+6 = -s_n

2024 ≡ 8 mod 12

s₂₀₂₄ = s₈= - s₂

s₂₀₂₄ = - 1

x + 1/x = √ 3 ,

1
calculate x ²⁰²⁴ + ――
x ²⁰²⁴

x = α , 1/x = β

-a₁= (1) = ∑ α = α + β = √ 3

a₂= (1²) = ∑ α β = α ∙ β = 1

s₂₀₂₄ = (2024) = ∑α ²⁰²⁴= α ²⁰²⁴ + β ²⁰²⁴

s_n − a₁s_n-1 + a₂s_n-2 − ... + (−1)ⁿna_n = 0

s₁ = a₁
...
s_n = a₁s_n-1 − a₂s_n-2 + ... + (−1)^{n + 1}na_n

s₁= − √ 3
s₂= (− √ 3 ∙ − √ 3 ) − (2 ∙ 1) = 1

a_n= 0 , n > 2

s_n+2= − √ 3 ∙ s_n+1 − s_n

p² + √ 3 p + 1 = 0

p = e^{± i ∙5/6∙π}

5/6 ∙ n = 2k
5n = 12k
n = 12

∴ s_n+12= s_n
s_n+6 = -s_n

2024 ≡ 8 mod 12

s₂₀₂₄ = s₈= - s₂

s₂₀₂₄ = - 1