luis angel gonzalez serrano´ en base de estudios conjuntos...

Pol. sim. completos Formula recursiva Funcion generadora Formula progr. geom.

Polinomios simetricos completos

Luis Angel Gonzalez SerranoEn base de estudios conjuntos con:

Egor Maximenko y Mario Alberto Moctezuma Salazar

ESFM - IPN

Seminario “Matrices y operadores”14 Octubre de 2020

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1 Polinomios simetricos completos

2 Formula recursiva

3 Funcion generadora

4 Formula con progresiones geometricas

2 / 23

Ejemplos de polinomios completos, n = 2

h0(x1, x2) = 1,

h1(x1, x2) = x1 + x2,

h2(x1, x2) = x21 + x2

2 + x1x2,

h3(x1, x2) = x31 + x3

2 + x21x2 + x1x

h4(x1, x2) = x41 + x4

2 + x31x2 + x1x

32 + x2

h5(x1, x2) = x51 + x5

2 + x41x2 + x1x

42 + x3

1x22 + x2

Se define hm(x1, x2) = 0 si m < 0.

3 / 23

h0(x1, x2) = 1,

h1(x1, x2) = x1 + x2,

h2(x1, x2) = x21 + x2

2 + x1x2,

h3(x1, x2) = x31 + x3

2 + x21x2 + x1x

h4(x1, x2) = x41 + x4

2 + x31x2 + x1x

32 + x2

h5(x1, x2) = x51 + x5

2 + x41x2 + x1x

42 + x3

1x22 + x2

3 / 23

h0(x1, x2) = 1,

h1(x1, x2) = x1 + x2,

h2(x1, x2) = x21 + x2

2 + x1x2,

h3(x1, x2) = x31 + x3

2 + x21x2 + x1x

h4(x1, x2) = x41 + x4

2 + x31x2 + x1x

32 + x2

h5(x1, x2) = x51 + x5

2 + x41x2 + x1x

42 + x3

1x22 + x2

3 / 23

h0(x1, x2) = 1,

h1(x1, x2) = x1 + x2,

h2(x1, x2) = x21 + x2

2 + x1x2,

h3(x1, x2) = x31 + x3

2 + x21x2 + x1x

h4(x1, x2) = x41 + x4

2 + x31x2 + x1x

32 + x2

h5(x1, x2) = x51 + x5

2 + x41x2 + x1x

42 + x3

1x22 + x2

3 / 23

h0(x1, x2, x3) = 1,

h1(x1, x2, x3) = x1 + x2 + x3,

h2(x1, x2, x3) = x21 + x2

2 + x23 + x1x2 + x1x3 + x2x3,

h3(x1, x2, x3) = x31 + x3

2 + x33 + x2

1x2 + x21x3 + x2

2x1 + x22x3

+ x23x1 + x2

3x2 + x1x2x3,

h4(x1, x2, x3) = x41 + x4

2 + x43 + x3

1x2 + x31x3 + x3

2x3 + x1x33

+ x1x32 + x2x

33 + x2

1x22 + x2

1x23 + x2

+ x21x2x3 + x1x

22x3 + x1x2x

4 / 23

h0(x1, x2, x3) = 1,

h1(x1, x2, x3) = x1 + x2 + x3,

h2(x1, x2, x3) = x21 + x2

2 + x23 + x1x2 + x1x3 + x2x3,

h3(x1, x2, x3) = x31 + x3

2 + x33 + x2

1x2 + x21x3 + x2

2x1 + x22x3

+ x23x1 + x2

3x2 + x1x2x3,

h4(x1, x2, x3) = x41 + x4

2 + x43 + x3

1x2 + x31x3 + x3

2x3 + x1x33

+ x1x32 + x2x

33 + x2

1x22 + x2

1x23 + x2

+ x21x2x3 + x1x

22x3 + x1x2x

4 / 23

Definicion combinatoriaEl polinomio simetrico completo de orden m en n variablesx1, . . . , xn es la suma de todos los monomios de grado total m:

hm(x1, . . . , xn) =∑

k1+...+kn=mk1,...,kn≥0

xk11 · · ·x

Otra forma equivalente de la definicion:

hm(x1, . . . , xn) =∑

1≤i1≤...≤im≤n

xi1xi2 · · ·xim .

Ejemplo de las definiciones equivalentes:

h2(x1, x2) = x21x

02 + x0

1x22 + x1x2,

h2(x1, x2) = x1x1 + x1x2 + x2x2.

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hm(x1, . . . , xn) =∑

k1+...+kn=mk1,...,kn≥0

xk11 · · ·x

hm(x1, . . . , xn) =∑

1≤i1≤...≤im≤n

h2(x1, x2) = x21x

02 + x0

1x22 + x1x2,

h2(x1, x2) = x1x1 + x1x2 + x2x2.

5 / 23

hm(x1, . . . , xn) =∑

k1+...+kn=mk1,...,kn≥0

xk11 · · ·x

hm(x1, . . . , xn) =∑

1≤i1≤...≤im≤n

h2(x1, x2) = x21x

02 + x0

1x22 + x1x2,

h2(x1, x2) = x1x1 + x1x2 + x2x2.

5 / 23

Ejercicio:¿Cuantos sumandos hay en hm(x1, . . . , xn)?

Por ejemplo: Calculemos el numero de sumando deh2(x1, x2, x3),

−→ x21

−→ x1x2

−→ x22

−→ x1x3

−→ x2x3

−→ x23

Respuesta:(

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Ejercicio:¿Cuantos sumandos hay en hm(x1, . . . , xn)?Por ejemplo: Calculemos el numero de sumando deh2(x1, x2, x3),

−→ x21

−→ x1x2

−→ x22

−→ x1x3

−→ x2x3

−→ x23

Respuesta:(

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−→ x21

−→ x1x2

−→ x22

−→ x1x3

−→ x2x3

−→ x23

Respuesta:(

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−→ x21

−→ x1x2

−→ x22

−→ x1x3

−→ x2x3

−→ x23

Respuesta:(

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−→ x21

−→ x1x2

−→ x22

−→ x1x3

−→ x2x3

−→ x23

Respuesta:(

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−→ x21

−→ x1x2

−→ x22

−→ x1x3

−→ x2x3

−→ x23

Respuesta:(

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−→ x21

−→ x1x2

−→ x22

−→ x1x3

−→ x2x3

−→ x23

Respuesta:(

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−→ x21

−→ x1x2

−→ x22

−→ x1x3

−→ x2x3

−→ x23

Respuesta:(

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−→ x21

−→ x1x2

−→ x22

−→ x1x3

−→ x2x3

−→ x23

Respuesta:(

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−→ x21

−→ x1x2

−→ x22

−→ x1x3

−→ x2x3

−→ x23

Respuesta:(

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−→ x21

−→ x1x2

−→ x22

−→ x1x3

−→ x2x3

−→ x23

Respuesta:(

6 / 23

−→ x21

−→ x1x2

−→ x22

−→ x1x3

−→ x2x3

−→ x23

Respuesta:(

)6 / 23

Otro ejemplo:

Veamos solo dos sumandos de h7(x1, x2, x3, x4)

−→ x21x

−→ x1x32x3x

¿Cuantos sumandos hay?

Respuesta:(

¿Cuantos sumandos hay en hm(x1, . . . , xn)?

7 / 23

Otro ejemplo: Veamos solo dos sumandos de h7(x1, x2, x3, x4)

−→ x21x

−→ x1x32x3x

Respuesta:(

7 / 23

−→ x21x

−→ x1x32x3x

Respuesta:(

7 / 23

−→ x21x

−→ x1x32x3x

Respuesta:(

7 / 23

−→ x21x

−→ x1x32x3x

Respuesta:(

7 / 23

−→ x21x

−→ x1x32x3x

Respuesta:(

7 / 23

−→ x21x

−→ x1x32x3x

Respuesta:(

7 / 23

Hacia una formula recursiva

Recordemos el polinomio completo de grado tres para tresvariables:

h3(x1, x2, x3) = x31 + x3

2 + x33 + x2

1x2 + x21x3 + x2

2x1 + x22x3

+ x23x1 + x2

3x2 + x1x2x3

1 + x32 + x2

1x2 + x22x1)

+ x3(x2

1 + x22 + x2

3 + x1x2 + x1x3 + x2x3)

= h3(x1, x2) + x3 h2(x1, x2, x3).

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h3(x1, x2, x3) = x31 + x3

2 + x33 + x2

1x2 + x21x3 + x2

2x1 + x22x3

+ x23x1 + x2

3x2 + x1x2x3

1 + x32 + x2

1x2 + x22x1)

+ x3(x2

1 + x22 + x2

3 + x1x2 + x1x3 + x2x3)

= h3(x1, x2) + x3 h2(x1, x2, x3).

8 / 23

h3(x1, x2, x3) = x31 + x3

2 + x33 + x2

1x2 + x21x3 + x2

2x1 + x22x3

+ x23x1 + x2

3x2 + x1x2x3

1 + x32 + x2

1x2 + x22x1)

+ x3(x2

1 + x22 + x2

3 + x1x2 + x1x3 + x2x3)

= h3(x1, x2) + x3 h2(x1, x2, x3).

8 / 23

h3(x1, x2, x3) = x31 + x3

2 + x33 + x2

1x2 + x21x3 + x2

2x1 + x22x3

+ x23x1 + x2

3x2 + x1x2x3

1 + x32 + x2

1x2 + x22x1)

+ x3(x2

1 + x22 + x2

3 + x1x2 + x1x3 + x2x3)

= h3(x1, x2) + x3 h2(x1, x2, x3).

8 / 23

Nota:Denotemos a la lista x1, . . . , xn por x.

Proposicion (Formula recursiva de los polinomios completos)

hm+1(x, xn+1) = hm+1(x) + xn+1 hm(x, xn+1).

Demostracion:

hm+1(x, xn+1) =∑

k1+...+kn+1=m+1k1,...,kn+1≥0

xk11 · · ·x

kn+1n+1

k1+...+kn=m+1kn+1=0

xk11 x

k22 · · ·x

k1+...+kn+1=m+1kn+1>0

xk11 x

k22 · · ·x

kn+1n+1

9 / 23

Demostracion:

hm+1(x, xn+1) =∑

k1+...+kn+1=m+1k1,...,kn+1≥0

xk11 · · ·x

kn+1n+1

k1+...+kn=m+1kn+1=0

xk11 x

k22 · · ·x

k1+...+kn+1=m+1kn+1>0

xk11 x

k22 · · ·x

kn+1n+1

9 / 23

Demostracion:

hm+1(x, xn+1) =∑

k1+...+kn+1=m+1k1,...,kn+1≥0

xk11 · · ·x

kn+1n+1

k1+...+kn=m+1kn+1=0

xk11 x

k22 · · ·x

k1+...+kn+1=m+1kn+1>0

xk11 x

k22 · · ·x

kn+1n+1

9 / 23

Demostracion:

hm+1(x, xn+1) =∑

k1+...+kn+1=m+1k1,...,kn+1≥0

xk11 · · ·x

kn+1n+1

k1+...+kn=m+1kn+1=0

xk11 x

k22 · · ·x

k1+...+kn+1=m+1kn+1>0

xk11 x

k22 · · ·x

kn+1n+1

9 / 23

hm+1(x, xn+1) =∑

k1+...+kn=m+1kn+1=0

xk11 x

k22 · · ·x

+ xn+1∑

k1+...+kn+kn+1−1=mkn+1−1≥0

xk11 x

k22 · · ·x

kn+1−1n+1

k1+...+kn=m+1kn+1=0

xk11 x

k22 · · ·x

+ xn+1∑

j1+...+jn+1=mjn+1≥0

xj11 x

j22 · · ·x

jn+1n+1

= hm+1(x) + xn+1 hm(x, xn+1).

10 / 23

hm+1(x, xn+1) =∑

k1+...+kn=m+1kn+1=0

xk11 x

k22 · · ·x

+ xn+1∑

k1+...+kn+kn+1−1=mkn+1−1≥0

xk11 x

k22 · · ·x

kn+1−1n+1

Denotemos j1 = k1, . . . , jn = kn, jn+1 = kn+1 − 1

k1+...+kn=m+1kn+1=0

xk11 x

k22 · · ·x

+ xn+1∑

j1+...+jn+1=mjn+1≥0

xj11 x

j22 · · ·x

jn+1n+1

= hm+1(x) + xn+1 hm(x, xn+1).

10 / 23

hm+1(x, xn+1) =∑

k1+...+kn=m+1kn+1=0

xk11 x

k22 · · ·x

+ xn+1∑

k1+...+kn+kn+1−1=mkn+1−1≥0

xk11 x

k22 · · ·x

kn+1−1n+1

k1+...+kn=m+1kn+1=0

xk11 x

k22 · · ·x

+ xn+1∑

j1+...+jn+1=mjn+1≥0

xj11 x

j22 · · ·x

jn+1n+1

= hm+1(x) + xn+1 hm(x, xn+1).

10 / 23

hm+1(x, xn+1) =∑

k1+...+kn=m+1kn+1=0

xk11 x

k22 · · ·x

+ xn+1∑

k1+...+kn+kn+1−1=mkn+1−1≥0

xk11 x

k22 · · ·x

kn+1−1n+1

k1+...+kn=m+1kn+1=0

xk11 x

k22 · · ·x

+ xn+1∑

j1+...+jn+1=mjn+1≥0

xj11 x

j22 · · ·x

jn+1n+1

= hm+1(x) + xn+1 hm(x, xn+1).

10 / 23

Diagrama para la formula recursiva

h0(x1) h1(x1) h2(x1) h3(x1)

h0(x1, x2) h1(x1, x2) h2(x1, x2) h3(x1, x2)

h0(x1, x2, x3) h1(x1, x2, x3) h2(x1, x2, x3) h3(x1, x2, x3)

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h0(x1) h1(x1) h2(x1) h3(x1)

h0(x1, x2) h1(x1, x2) h2(x1, x2) h3(x1, x2)

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h0(x1) h1(x1) h2(x1) h3(x1)

h0(x1, x2) h1(x1, x2) h2(x1, x2) h3(x1, x2)

11 / 23

h0(x1) h1(x1) h2(x1) h3(x1)

h0(x1, x2) h1(x1, x2) h2(x1, x2) h3(x1, x2)

11 / 23

Algoritmo en SageMath

12 / 23

Ejercicio

Proposicionhm+1(x, xn+1)−hm+1(x, xn+2) = (xn+1−xn+2) hm(x, xn+1, xn+2).

Indicacion al ejercicio: Usar la formula recursiva.

13 / 23

Ejercicio

Proposicionhm+1(x, xn+1)−hm+1(x, xn+2) = (xn+1−xn+2) hm(x, xn+1, xn+2).

Indicacion al ejercicio: Usar la formula recursiva.

13 / 23

Funcion generadora

Dada una sucesion (αk)∞k=0 se define su funcion generadoracomo ∞∑

k=0αkt

Funcion generadora de la sucesion (hm)∞m=0

H(x1, . . . , xn)(t) =∞∑

m=0hm(x1, . . . , xn)tm.

14 / 23

Funcion generadora

Dada una sucesion (αk)∞k=0 se define su funcion generadoracomo ∞∑

k=0αkt

Funcion generadora de la sucesion (hm)∞m=0

H(x1, . . . , xn)(t) =∞∑

m=0hm(x1, . . . , xn)tm.

14 / 23

H(x)(t) = (1− xn+1t) H(x, xn+1)(t)

Demostracion:

H(x, xn+1)(t) =∞∑

j=0hj(x, xn+1)tj = 1 +

∞∑j=0

hj+1(x, xn+1)tj+1

= 1 +∞∑

j=0(hj+1(x) + xn+1 hj(x, xn+1)) tj+1

1 +∞∑

j=1hj(x)tj

+∞∑

j=0(xn+1 hj(x, xn+1)) tj+1

=∞∑

j=0hj(x)tj + xn+1t

∞∑j=0

hj(x, xn+1)tj

= H(x)(t) + xn+1tH(x, xn+1)(t).

15 / 23

H(x)(t) = (1− xn+1t) H(x, xn+1)(t)

Demostracion:

H(x, xn+1)(t) =∞∑

j=0hj(x, xn+1)tj

= 1 +∞∑

j=0hj+1(x, xn+1)tj+1

= 1 +∞∑

j=0(hj+1(x) + xn+1 hj(x, xn+1)) tj+1

1 +∞∑

j=1hj(x)tj

+∞∑

j=0(xn+1 hj(x, xn+1)) tj+1

=∞∑

j=0hj(x)tj + xn+1t

∞∑j=0

hj(x, xn+1)tj

= H(x)(t) + xn+1tH(x, xn+1)(t).

15 / 23

H(x)(t) = (1− xn+1t) H(x, xn+1)(t)

Demostracion:

H(x, xn+1)(t) =∞∑

j=0hj(x, xn+1)tj = 1 +

∞∑j=0

hj+1(x, xn+1)tj+1

= 1 +∞∑

j=0(hj+1(x) + xn+1 hj(x, xn+1)) tj+1

1 +∞∑

j=1hj(x)tj

+∞∑

j=0(xn+1 hj(x, xn+1)) tj+1

=∞∑

j=0hj(x)tj + xn+1t

∞∑j=0

hj(x, xn+1)tj

= H(x)(t) + xn+1tH(x, xn+1)(t).

15 / 23

H(x)(t) = (1− xn+1t) H(x, xn+1)(t)

Demostracion:

H(x, xn+1)(t) =∞∑

j=0hj(x, xn+1)tj = 1 +

∞∑j=0

hj+1(x, xn+1)tj+1

= 1 +∞∑

j=0(hj+1(x) + xn+1 hj(x, xn+1)) tj+1

1 +∞∑

j=1hj(x)tj

+∞∑

j=0(xn+1 hj(x, xn+1)) tj+1

=∞∑

j=0hj(x)tj + xn+1t

∞∑j=0

hj(x, xn+1)tj

= H(x)(t) + xn+1tH(x, xn+1)(t).

15 / 23

H(x)(t) = (1− xn+1t) H(x, xn+1)(t)

Demostracion:

H(x, xn+1)(t) =∞∑

j=0hj(x, xn+1)tj = 1 +

∞∑j=0

hj+1(x, xn+1)tj+1

= 1 +∞∑

j=0(hj+1(x) + xn+1 hj(x, xn+1)) tj+1

1 +∞∑

j=1hj(x)tj

+∞∑

j=0(xn+1 hj(x, xn+1)) tj+1

=∞∑

j=0hj(x)tj + xn+1t

∞∑j=0

hj(x, xn+1)tj

= H(x)(t) + xn+1tH(x, xn+1)(t).

15 / 23

H(x)(t) = (1− xn+1t) H(x, xn+1)(t)

Demostracion:

H(x, xn+1)(t) =∞∑

j=0hj(x, xn+1)tj = 1 +

∞∑j=0

hj+1(x, xn+1)tj+1

= 1 +∞∑

j=0(hj+1(x) + xn+1 hj(x, xn+1)) tj+1

1 +∞∑

j=1hj(x)tj

+∞∑

j=0(xn+1 hj(x, xn+1)) tj+1

=∞∑

j=0hj(x)tj + xn+1t

∞∑j=0

hj(x, xn+1)tj

= H(x)(t) + xn+1tH(x, xn+1)(t).

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H(x)(t) = (1− xn+1t) H(x, xn+1)(t)

Demostracion:

H(x, xn+1)(t) =∞∑

j=0hj(x, xn+1)tj = 1 +

∞∑j=0

hj+1(x, xn+1)tj+1

= 1 +∞∑

j=0(hj+1(x) + xn+1 hj(x, xn+1)) tj+1

1 +∞∑

j=1hj(x)tj

+∞∑

j=0(xn+1 hj(x, xn+1)) tj+1

=∞∑

j=0hj(x)tj + xn+1t

∞∑j=0

hj(x, xn+1)tj

= H(x)(t) + xn+1tH(x, xn+1)(t).

15 / 23

Otra forma equivalente:

H(x, xn+1)(t) = H(x)(t)(1− xn+1t)

16 / 23

Proposicion

H(x)(t) =n∏

i=1(1− xit)−1.

Demostracion: Por Induccion sobre n

H(x1) =∞∑

j=0hj(x1)tj =

∞∑j=0

(x1t)j = 11− x1t

Suponiendo para n, demostremos para n+ 1

H(x, xn+1)(t) Lem==== H(x)(t)1− xn+1t

H.I.==== (1− xn+1t)−1n∏

i=1(1− xit)−1

=n+1∏i=1

(1− xit)−1.

17 / 23

Proposicion

H(x)(t) =n∏

i=1(1− xit)−1.

H(x1) =∞∑

j=0hj(x1)tj

=∞∑

j=0(x1t)j = 1

1− x1t.

H(x, xn+1)(t) Lem==== H(x)(t)1− xn+1t

H.I.==== (1− xn+1t)−1n∏

i=1(1− xit)−1

=n+1∏i=1

(1− xit)−1.

17 / 23

Proposicion

H(x)(t) =n∏

i=1(1− xit)−1.

H(x1) =∞∑

j=0hj(x1)tj =

∞∑j=0

(x1t)j

= 11− x1t

H(x, xn+1)(t) Lem==== H(x)(t)1− xn+1t

H.I.==== (1− xn+1t)−1n∏

i=1(1− xit)−1

=n+1∏i=1

(1− xit)−1.

17 / 23

Proposicion

H(x)(t) =n∏

i=1(1− xit)−1.

H(x1) =∞∑

j=0hj(x1)tj =

∞∑j=0

(x1t)j = 11− x1t

H(x, xn+1)(t) Lem==== H(x)(t)1− xn+1t

H.I.==== (1− xn+1t)−1n∏

i=1(1− xit)−1

=n+1∏i=1

(1− xit)−1.

17 / 23

Proposicion

H(x)(t) =n∏

i=1(1− xit)−1.

H(x1) =∞∑

j=0hj(x1)tj =

∞∑j=0

(x1t)j = 11− x1t

H(x, xn+1)(t) Lem==== H(x)(t)1− xn+1t

H.I.==== (1− xn+1t)−1n∏

i=1(1− xit)−1

=n+1∏i=1

(1− xit)−1.

17 / 23

Proposicion

H(x)(t) =n∏

i=1(1− xit)−1.

H(x1) =∞∑

j=0hj(x1)tj =

∞∑j=0

(x1t)j = 11− x1t

H(x, xn+1)(t) Lem==== H(x)(t)1− xn+1t

H.I.==== (1− xn+1t)−1n∏

i=1(1− xit)−1

=n+1∏i=1

(1− xit)−1.

17 / 23

Recordemos que

H(x)(t) =n∏

11− xit

H(x)(t) =n∑

1− xit,

dondeCi = xn−1

i∏j 6=i

(xi − xj).

18 / 23

Demostracion: Supongamos que existe Ci que cumple laigualdad, entonces para n = 2

1(1− x1t)(1− x2t)

= C11− x1t

+ C21− x2t

multiplicando por 1− x1t en ambos lados, resulta

11− x2t

= C1 + C2(1− x1t)1− x2t

Entonces tomando t = x−11

C1 = x1x1 − x2

De manera analoga se obtiene que

C2 = x2x2 − x1

19 / 23

1(1− x1t)(1− x2t)

= C11− x1t

+ C21− x2t

11− x2t

= C1 + C2(1− x1t)1− x2t

C1 = x1x1 − x2

C2 = x2x2 − x1

19 / 23

1(1− x1t)(1− x2t)

= C11− x1t

+ C21− x2t

11− x2t

= C1 + C2(1− x1t)1− x2t

C1 = x1x1 − x2

C2 = x2x2 − x1

19 / 23

1(1− x1t)(1− x2t)

= C11− x1t

+ C21− x2t

11− x2t

= C1 + C2(1− x1t)1− x2t

C1 = x1x1 − x2

C2 = x2x2 − x1

19 / 23

Para n = 31

(1− x1t)(1− x2t)(1− x3t)

C1(1− x2t)(1− x3t) + C2(1− x1t)(1− x3t) + C3(1− x1t)(1− x2t)(1− x1t)(1− x2t)(1− x3t)

Multiplicando en ambos lados por, por ejemplo, 1− x1t

1(1− x2t)(1− x3t)

C1 + C2(1− x1t)(1− x3t) + C3(1− x1t)(1− x2t)(1− x2t)(1− x3t)

Tomando, t = x−11

C1 = 1(1− x2x

) (1− x3x

) = x21

(x1 − x2)(x1 − x3) .

De manera analoga, se obtiene la igualdad para C2 y C3

20 / 23

Para n = 31

(1− x1t)(1− x2t)(1− x3t)=

1(1− x2t)(1− x3t)

Tomando, t = x−11

C1 = 1(1− x2x

) (1− x3x

) = x21

(x1 − x2)(x1 − x3) .

20 / 23

Para n = 31

(1− x1t)(1− x2t)(1− x3t)=

1(1− x2t)(1− x3t)

Tomando, t = x−11

C1 = 1(1− x2x

) (1− x3x

) = x21

(x1 − x2)(x1 − x3) .

20 / 23

Para n = 31

(1− x1t)(1− x2t)(1− x3t)=

1(1− x2t)(1− x3t)

Tomando, t = x−11

C1 = 1(1− x2x

) (1− x3x

) = x21

(x1 − x2)(x1 − x3) .

20 / 23

Para n = 31

(1− x1t)(1− x2t)(1− x3t)=

1(1− x2t)(1− x3t)

Tomando, t = x−11

C1 = 1(1− x2x

) (1− x3x

) = x21

(x1 − x2)(x1 − x3) .

20 / 23

Para n = 31

(1− x1t)(1− x2t)(1− x3t)=

1(1− x2t)(1− x3t)

Tomando, t = x−11

C1 = 1(1− x2x

) (1− x3x

) = x21

(x1 − x2)(x1 − x3) .

De manera analoga, se obtiene la igualdad para C2 y C320 / 23

Entonces para n, se obtiene que∑1≤i≤n

∏j 6=i

Ci(1− xjt)∏1≤j≤n

(1− xjt)=

∏1≤j≤n

1(1− xjt)

Multiplicando por 1− xit

∑r 6=i

∏j 6=r

Cr(1− xjt)∏1≤j≤n

(1− xjt)=

∏1≤j≤n

1(1− xjt)

Haciendo t = x−1i

Ci =∏

1≤j≤nj 6=i

xn−1i

xi − xj.

21 / 23

∏j 6=i

(1− xjt)=

∏1≤j≤n

1(1− xjt)

∑r 6=i

∏j 6=r

(1− xjt)=

∏1≤j≤n

1(1− xjt)

Haciendo t = x−1i

Ci =∏

1≤j≤nj 6=i

xn−1i

xi − xj.

21 / 23

∏j 6=i

(1− xjt)=

∏1≤j≤n

1(1− xjt)

∑r 6=i

∏j 6=r

(1− xjt)=

∏1≤j≤n

1(1− xjt)

Haciendo t = x−1i

Ci =∏

1≤j≤nj 6=i

xn−1i

xi − xj.

21 / 23

Polinomios completos como combinacion lineal de progresionesgeometricas

hm(x) =n∑

xn+m−1j∏

k 6=j(xj − xk) .

Demostracion:∞∑

m=0hm(x)tm = H(x)(t) =

n∏i=1

11− xit

1− xit

xn−1i

(1− xit)∏

k 6=i(xi − xk)

=∞∑

n∑i=1

xn+m−1i∏

k 6=i(xi − xk) tm.

Por ultimo igualamos los coeficientes.

22 / 23

hm(x) =n∑

xn+m−1j∏

k 6=j(xj − xk) .

Demostracion:∞∑

m=0hm(x)tm = H(x)(t)

11− xit

1− xit

xn−1i

(1− xit)∏

k 6=i(xi − xk)

=∞∑

n∑i=1

xn+m−1i∏

22 / 23

hm(x) =n∑

xn+m−1j∏

k 6=j(xj − xk) .

Demostracion:∞∑

n∏i=1

11− xit

1− xit

xn−1i

(1− xit)∏

k 6=i(xi − xk)

=∞∑

n∑i=1

xn+m−1i∏

22 / 23

hm(x) =n∑

xn+m−1j∏

k 6=j(xj − xk) .

Demostracion:∞∑

n∏i=1

11− xit

1− xit

xn−1i

(1− xit)∏

k 6=i(xi − xk)

=∞∑

n∑i=1

xn+m−1i∏

22 / 23

hm(x) =n∑

xn+m−1j∏

k 6=j(xj − xk) .

Demostracion:∞∑

n∏i=1

11− xit

1− xit

xn−1i

(1− xit)∏

k 6=i(xi − xk)

=∞∑

n∑i=1

xn+m−1i∏

22 / 23

hm(x) =n∑

xn+m−1j∏

k 6=j(xj − xk) .

Demostracion:∞∑

n∏i=1

11− xit

1− xit

xn−1i

(1− xit)∏

k 6=i(xi − xk)

=∞∑

n∑i=1

xn+m−1i∏

22 / 23

hm(x) =n∑

xn+m−1j∏

k 6=j(xj − xk) .

Demostracion:∞∑

n∏i=1

11− xit

1− xit

xn−1i

(1− xit)∏

k 6=i(xi − xk)

=∞∑

n∑i=1

xn+m−1i∏

Por ultimo igualamos los coeficientes.22 / 23

¿Que pasa si xj = xk para algun j 6= k en la forma anterior?

Veamos un ejemplo: Por un lado

hm(x1, x1) =m∑

j=0xm−j

1 xj1 = (m+ 1)xm

Por otro lado, recordando que para dos variables se tiene que

hm(x1, x2) = xm+11 − xm+1

2x1 − x2

Entonces, cuando las dos variables coinciden, por la regla deL’Hopital

hm(x1, x1) = limx2→x1

xm+12 − xm+1

1x2 − x1

= (m+ 1)xm1 .

23 / 23

¿Que pasa si xj = xk para algun j 6= k en la forma anterior?Veamos un ejemplo: Por un lado

hm(x1, x1) =m∑

j=0xm−j

1 xj1 = (m+ 1)xm

hm(x1, x2) = xm+11 − xm+1

2x1 − x2

xm+12 − xm+1

1x2 − x1

= (m+ 1)xm1 .

23 / 23

hm(x1, x1) =m∑

j=0xm−j

1 xj1 = (m+ 1)xm

hm(x1, x2) = xm+11 − xm+1

2x1 − x2

xm+12 − xm+1

1x2 − x1

= (m+ 1)xm1 .

23 / 23

hm(x1, x1) =m∑

j=0xm−j

1 xj1 = (m+ 1)xm

hm(x1, x2) = xm+11 − xm+1

2x1 − x2

xm+12 − xm+1

1x2 − x1

= (m+ 1)xm1 .

23 / 23

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