Problemilla  con  las  raíces  de  una  ecuación  cúbica. Tricky  roots  of  cubics  problem.⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

5 months ago (Edited)

Español	English

Enunciado

α , β , γ raíces de la ecuación cúbica, x ³ + 2x + 1 = 0 qué vale α ⁵ + β ⁵ + γ ⁵

Es de sobra conocida la relación entre los coeficientes de una ecuación algebraica y sus raíces, en la forma de las funciones simétricas elementales de las mismas.

Este problema nos propone determinar el valor de la suma de las quintas potencias de las raíces de un polinomio, nos están pidiendo que expresemos el valor de la función simétrica de las quintas potencias en función de las funciones simétricas elementales.

①

⠀⠀⠀⠀ _\\|/_ ⠀⠀⠀⠀ (o  o) ----oOO-{_}-OOo--- La relación entre la suma de potencias y las funciones simétricas unitarias es un problema en análisis combinatorio que encuentra su solución con las celebradas Identidades de Newton

Utilizando la relación de Newton ,

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

Donde,

			a_n = (1ⁿ) = ∑α₁α₂ ... α_n , funciones elementales s_n = (n) = ∑α ⁿ, suma de potencias

Solución

Tenemos una ecuación cúbica de forma que conocemos el valor de las funciones elementales (1³), (1²), (1)

			x ³ + 2x + 1 , x ³ - a₁x² + a₂x - a₃ a₁ = 0, a₂ = 2, a₃ = -1

La ecuación ①, es una ecuación recurrente, resolviendo los valores de s_n,

			s_n = a₁s_n-1 - a₂s_n-2 + ... + (-1)^{n + 1}na_n s₁ = a₁ = 0 s₂ = a₁s₁- 2a₂ = 0 - 4 = -4 s₃ = a₁s₂- a₂s₁+ 3a₃ = 0 - 0 + (-3) = -3 s₄ = a₁s₃- a₂s₂+ a₃s₁- 4a₄ = 0 - (-8) + 0 - 0 = 8 s₅ = a₁s₄- a₂s₃+ a₃s₂- a₄s₁+ 5a₅ = 0 - (-6) + 4 + 0 + 0 = 10

Por lo tanto,

α ⁵ + β ⁵ + γ ⁵ = 10

Fácil y sencillo a la par que simpático y agradable.

∎

English	Español

Statement

α , β , γ roots of the cubic, x ³ + 2x + 1 = 0 calculate α ⁵ + β ⁵ + γ ⁵

The roots of an algebraic equation and coefficients are related by means of the elementary symmetric functions, as well known.

This problem poses the question on how fifth power sums and elementary symmetric functions related, asking to put fifth power sums symmetric function as elementary ones.

①

⠀⠀⠀⠀ _\\|/_ ⠀⠀⠀⠀ (o  o) ----oOO-{_}-OOo--- The relations between power sums and unitary symmetric functions are problems on combinatorial analysis , solved by the celebrated Newtonian Identities

By means of the Newtonian relation,

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

Where,

			a_n = (1ⁿ) = ∑α₁α₂ ... α_n , elementary functions s_n = (n) = ∑α ², power sums

Answer

Our equation is a cubic one, so we know value of the elementary symmetric functions (1³), (1²), (1)

			x ³ + 2x + 1 , x ³ - a₁x² + a₂x - a₃ a₁ = 0, a₂ = 2, a₃ = -1

Relation ①, is a recurrence relation, solving for s_n,

			s_n = a₁s_n-1 - a₂s_n-2 + ... + (-1)^{n + 1}na_n s₁ = a₁ = 0 s₂ = a₁s₁- 2a₂ = 0 - 4 = -4 s₃ = a₁s₂- a₂s₁+ 3a₃ = 0 - 0 + (-3) = -3 s₄ = a₁s₃- a₂s₂+ a₃s₁- 4a₄ = 0 - (-8) + 0 - 0 = 8 s₅ = a₁s₄- a₂s₃+ a₃s₂- a₄s₁+ 5a₅ = 0 - (-6) + 4 + 0 + 0 = 10

Hence,

α ⁵ + β ⁵ + γ ⁵ = 10

Not so tricky.

∎

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Problemilla  con  las  raíces  de  una  ecuación  cúbica. Tricky  roots  of  cubics  problem.⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Enunciado

α , β , γ
raíces de la ecuación cúbica, x ³ + 2x + 1 = 0

qué vale α ⁵ + β ⁵ + γ ⁵

①

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

a_n = (1ⁿ) = ∑α₁α₂ ... α_n , funciones elementales

s_n = (n) = ∑α ⁿ, suma de potencias

Solución

x ³ + 2x + 1 , x ³ - a₁x² + a₂x - a₃

a₁ = 0, a₂ = 2, a₃ = -1

α ⁵ + β ⁵ + γ ⁵ = 10

Statement

α , β , γ
roots of the cubic, x ³ + 2x + 1 = 0

calculate α ⁵ + β ⁵ + γ ⁵

①

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

a_n = (1ⁿ) = ∑α₁α₂ ... α_n , elementary functions

s_n = (n) = ∑α ², power sums

Answer

x ³ + 2x + 1 , x ³ - a₁x² + a₂x - a₃

a₁ = 0, a₂ = 2, a₃ = -1

α ⁵ + β ⁵ + γ ⁵ = 10

Keep up the good work. 👏🎵

What's next?

Expose local music from your area!

Join our community 🐬

Bad news: Saying see you later to Hive! 👋

Curious about the future of Earth + ET relations? New economic systems?

Want to build tools like I used to share this?

Problemilla con las raíces de una ecuación cúbica. Tricky roots of cubics problem.⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Enunciado

α , β , γ raíces de la ecuación cúbica, x 3 + 2x + 1 = 0qué vale α 5 + β 5 + γ 5

①

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

an = (1n) = ∑α1α2 ... αn , funciones elementalessn = (n) = ∑α n, suma de potencias

Solución

x 3 + 2x + 1 , x 3 - a1x2 + a2x - a3 a1 = 0, a2 = 2, a3 = -1

sn = a1sn-1 - a2sn-2 + ... + (-1)n + 1nan s1 = a1 = 0 s2 = a1s1 - 2a2 = 0 - 4 = -4 s3 = a1s2 - a2s1 + 3a3 = 0 - 0 + (-3) = -3 s4 = a1s3 - a2s2 + a3s1 - 4a4 = 0 - (-8) + 0 - 0 = 8 s5 = a1s4 - a2s3 + a3s2 - a4s1 + 5a5 = 0 - (-6) + 4 + 0 + 0 = 10

α 5 + β 5 + γ 5 = 10

Statement

α , β , γ roots of the cubic, x 3 + 2x + 1 = 0calculate α 5 + β 5 + γ 5

①

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

an = (1n) = ∑α1α2 ... αn , elementary functionssn = (n) = ∑α 2, power sums

Answer

x 3 + 2x + 1 , x 3 - a1x2 + a2x - a3 a1 = 0, a2 = 2, a3 = -1

sn = a1sn-1 - a2sn-2 + ... + (-1)n + 1nan s1 = a1 = 0 s2 = a1s1 - 2a2 = 0 - 4 = -4 s3 = a1s2 - a2s1 + 3a3 = 0 - 0 + (-3) = -3 s4 = a1s3 - a2s2 + a3s1 - 4a4 = 0 - (-8) + 0 - 0 = 8 s5 = a1s4 - a2s3 + a3s2 - a4s1 + 5a5 = 0 - (-6) + 4 + 0 + 0 = 10

α 5 + β 5 + γ 5 = 10

Keep up the good work. 👏🎵

What's next?

Expose local music from your area!

Join our community 🐬

Bad news: Saying see you later to Hive! 👋

Curious about the future of Earth + ET relations? New economic systems?

Want to build tools like I used to share this?

Problemilla  con  las  raíces  de  una  ecuación  cúbica. Tricky  roots  of  cubics  problem.⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

α , β , γ
raíces de la ecuación cúbica, x ³ + 2x + 1 = 0

qué vale α ⁵ + β ⁵ + γ ⁵

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

a_n = (1ⁿ) = ∑α₁α₂ ... α_n , funciones elementales

s_n = (n) = ∑α ⁿ, suma de potencias

x ³ + 2x + 1 , x ³ - a₁x² + a₂x - a₃

a₁ = 0, a₂ = 2, a₃ = -1

α ⁵ + β ⁵ + γ ⁵ = 10

α , β , γ
roots of the cubic, x ³ + 2x + 1 = 0

calculate α ⁵ + β ⁵ + γ ⁵

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

a_n = (1ⁿ) = ∑α₁α₂ ... α_n , elementary functions

s_n = (n) = ∑α ², power sums

x ³ + 2x + 1 , x ³ - a₁x² + a₂x - a₃

a₁ = 0, a₂ = 2, a₃ = -1

α ⁵ + β ⁵ + γ ⁵ = 10