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

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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.

 

 

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⠀⠀⠀⠀ _\|/_ ⠀⠀⠀⠀ (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 ,

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

Donde,

			an = (1n) = ∑α1α2 ... αn ,    funciones elementales sn =  (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 ,

			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

Por lo tanto,

α 5 + β 5 + γ 5 = 10

Fácil y sencillo a la par que simpático y agradable.

∎

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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.

 

 

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⠀⠀⠀⠀ _\|/_ ⠀⠀⠀⠀ (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,

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

Where,

			an = (1n) = ∑α1α2 ... αn ,    elementary functions sn =  (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 ,

			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

Hence,

α 5 + β 5 + γ 5 = 10

Not so tricky.

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