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Casos particulares delProducto Triple de Jacobi
Jesus A. CorralChiara ForacePiera Galber
Luis J. Salmeron ContrerasMaria Soler Facundo
Universitat de Valencia
16 Enero 2014
Producto Triple de Jacobi
Producto Triple de Jacobi
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
A partir de esta identidad podemos deducir:I El Teorema de Jacobi de los Numeros TriangularesI El Teorema de Euler de los Numeros Pentagonales
Producto Triple de Jacobi
Producto Triple de Jacobi
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
A partir de esta identidad podemos deducir:I El Teorema de Jacobi de los Numeros TriangularesI El Teorema de Euler de los Numeros Pentagonales
Deduccion del Teorema de Jacobi
Producto Triple de Jacobi
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
⇓
Teorema de los numeros triangulares
( ∞∏k=1
(1− qk )
)3
=∞∑
n=0
(−1)n(2n + 1)qn(n+1)
2
Deduccion del Teorema de Jacobi
Producto Triple de Jacobi
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
⇓
Teorema de los numeros triangulares
( ∞∏k=1
(1− qk )
)3
=∞∑
n=0
(−1)n(2n + 1)qn(n+1)
2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q12 e y = ia
12 q−
14
1− x2k → 1−(
q12
)2k= 1− qk
1+x2k−1y2 → 1+(
q12
)2k−1 (ia
12 q−
14
)2= 1−qk− 1
2 aq−12 = 1− aqk−1
1+x2k−1y−2 → 1+(
q12
)2k−1 (ia
12 q−
14
)−2= 1−qk− 1
2 a−1q12 = 1− a−1qk
xn2y2n →
(q
12
)n2 (ia
12 q−
14
)2n= q
n22 (−1)nanq−
n2 = (−1)nanq
n(n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q12 e y = ia
12 q−
14
1− x2k
→ 1−(
q12
)2k= 1− qk
1+x2k−1y2 → 1+(
q12
)2k−1 (ia
12 q−
14
)2= 1−qk− 1
2 aq−12 = 1− aqk−1
1+x2k−1y−2 → 1+(
q12
)2k−1 (ia
12 q−
14
)−2= 1−qk− 1
2 a−1q12 = 1− a−1qk
xn2y2n →
(q
12
)n2 (ia
12 q−
14
)2n= q
n22 (−1)nanq−
n2 = (−1)nanq
n(n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q12 e y = ia
12 q−
14
1− x2k → 1−(
q12
)2k
= 1− qk
1+x2k−1y2 → 1+(
q12
)2k−1 (ia
12 q−
14
)2= 1−qk− 1
2 aq−12 = 1− aqk−1
1+x2k−1y−2 → 1+(
q12
)2k−1 (ia
12 q−
14
)−2= 1−qk− 1
2 a−1q12 = 1− a−1qk
xn2y2n →
(q
12
)n2 (ia
12 q−
14
)2n= q
n22 (−1)nanq−
n2 = (−1)nanq
n(n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q12 e y = ia
12 q−
14
1− x2k → 1−(
q12
)2k= 1− qk
1+x2k−1y2 → 1+(
q12
)2k−1 (ia
12 q−
14
)2= 1−qk− 1
2 aq−12 = 1− aqk−1
1+x2k−1y−2 → 1+(
q12
)2k−1 (ia
12 q−
14
)−2= 1−qk− 1
2 a−1q12 = 1− a−1qk
xn2y2n →
(q
12
)n2 (ia
12 q−
14
)2n= q
n22 (−1)nanq−
n2 = (−1)nanq
n(n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q12 e y = ia
12 q−
14
1− x2k → 1−(
q12
)2k= 1− qk
1+x2k−1y2
→ 1+(
q12
)2k−1 (ia
12 q−
14
)2= 1−qk− 1
2 aq−12 = 1− aqk−1
1+x2k−1y−2 → 1+(
q12
)2k−1 (ia
12 q−
14
)−2= 1−qk− 1
2 a−1q12 = 1− a−1qk
xn2y2n →
(q
12
)n2 (ia
12 q−
14
)2n= q
n22 (−1)nanq−
n2 = (−1)nanq
n(n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q12 e y = ia
12 q−
14
1− x2k → 1−(
q12
)2k= 1− qk
1+x2k−1y2 → 1+(
q12
)2k−1 (ia
12 q−
14
)2
= 1−qk− 12 aq−
12 = 1− aqk−1
1+x2k−1y−2 → 1+(
q12
)2k−1 (ia
12 q−
14
)−2= 1−qk− 1
2 a−1q12 = 1− a−1qk
xn2y2n →
(q
12
)n2 (ia
12 q−
14
)2n= q
n22 (−1)nanq−
n2 = (−1)nanq
n(n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q12 e y = ia
12 q−
14
1− x2k → 1−(
q12
)2k= 1− qk
1+x2k−1y2 → 1+(
q12
)2k−1 (ia
12 q−
14
)2= 1−qk− 1
2 aq−12 = 1− aqk−1
1+x2k−1y−2 → 1+(
q12
)2k−1 (ia
12 q−
14
)−2= 1−qk− 1
2 a−1q12 = 1− a−1qk
xn2y2n →
(q
12
)n2 (ia
12 q−
14
)2n= q
n22 (−1)nanq−
n2 = (−1)nanq
n(n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q12 e y = ia
12 q−
14
1− x2k → 1−(
q12
)2k= 1− qk
1+x2k−1y2 → 1+(
q12
)2k−1 (ia
12 q−
14
)2= 1−qk− 1
2 aq−12 = 1− aqk−1
1+x2k−1y−2
→ 1+(
q12
)2k−1 (ia
12 q−
14
)−2= 1−qk− 1
2 a−1q12 = 1− a−1qk
xn2y2n →
(q
12
)n2 (ia
12 q−
14
)2n= q
n22 (−1)nanq−
n2 = (−1)nanq
n(n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q12 e y = ia
12 q−
14
1− x2k → 1−(
q12
)2k= 1− qk
1+x2k−1y2 → 1+(
q12
)2k−1 (ia
12 q−
14
)2= 1−qk− 1
2 aq−12 = 1− aqk−1
1+x2k−1y−2 → 1+(
q12
)2k−1 (ia
12 q−
14
)−2
= 1−qk− 12 a−1q
12 = 1− a−1qk
xn2y2n →
(q
12
)n2 (ia
12 q−
14
)2n= q
n22 (−1)nanq−
n2 = (−1)nanq
n(n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q12 e y = ia
12 q−
14
1− x2k → 1−(
q12
)2k= 1− qk
1+x2k−1y2 → 1+(
q12
)2k−1 (ia
12 q−
14
)2= 1−qk− 1
2 aq−12 = 1− aqk−1
1+x2k−1y−2 → 1+(
q12
)2k−1 (ia
12 q−
14
)−2= 1−qk− 1
2 a−1q12 = 1− a−1qk
xn2y2n →
(q
12
)n2 (ia
12 q−
14
)2n= q
n22 (−1)nanq−
n2 = (−1)nanq
n(n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q12 e y = ia
12 q−
14
1− x2k → 1−(
q12
)2k= 1− qk
1+x2k−1y2 → 1+(
q12
)2k−1 (ia
12 q−
14
)2= 1−qk− 1
2 aq−12 = 1− aqk−1
1+x2k−1y−2 → 1+(
q12
)2k−1 (ia
12 q−
14
)−2= 1−qk− 1
2 a−1q12 = 1− a−1qk
xn2y2n
→(
q12
)n2 (ia
12 q−
14
)2n= q
n22 (−1)nanq−
n2 = (−1)nanq
n(n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q12 e y = ia
12 q−
14
1− x2k → 1−(
q12
)2k= 1− qk
1+x2k−1y2 → 1+(
q12
)2k−1 (ia
12 q−
14
)2= 1−qk− 1
2 aq−12 = 1− aqk−1
1+x2k−1y−2 → 1+(
q12
)2k−1 (ia
12 q−
14
)−2= 1−qk− 1
2 a−1q12 = 1− a−1qk
xn2y2n →
(q
12
)n2 (ia
12 q−
14
)2n
= qn22 (−1)nanq−
n2 = (−1)nanq
n(n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q12 e y = ia
12 q−
14
1− x2k → 1−(
q12
)2k= 1− qk
1+x2k−1y2 → 1+(
q12
)2k−1 (ia
12 q−
14
)2= 1−qk− 1
2 aq−12 = 1− aqk−1
1+x2k−1y−2 → 1+(
q12
)2k−1 (ia
12 q−
14
)−2= 1−qk− 1
2 a−1q12 = 1− a−1qk
xn2y2n →
(q
12
)n2 (ia
12 q−
14
)2n= q
n22 (−1)nanq−
n2 = (−1)nanq
n(n−1)2
Sustitucion
Por tanto obtenemos:∞∏
k=1
(1− qk )(1− aqk−1)(1− a−1qk ) =∞∑
n=−∞(−1)nanq
n(n−1)2
De los tres factores que aparecen en el productorio, elsegundo, cuando k = 1 queda simplemente (1− a).
Sustitucion
Por tanto obtenemos:∞∏
k=1
(1− qk )(1− aqk−1)(1− a−1qk ) =∞∑
n=−∞(−1)nanq
n(n−1)2
De los tres factores que aparecen en el productorio, elsegundo, cuando k = 1 queda simplemente (1− a).
Deduccion del Teorema de Jacobi
Luego ese productorio se puede escribir como:
∞∏k=1
(1− qk )(1− aqk−1)(1− a−1qk )
=
( ∞∏k=1
(1− qk )(1− a−1qk )
)( ∞∏k=1
(1− aqk−1)
)
=
( ∞∏k=1
(1− qk )(1− a−1qk )
)(1− a)
( ∞∏k=2
(1− aqk−1)
)
= (1− a)
( ∞∏k=1
(1− qk )(1− a−1qk )
)( ∞∏k=1
(1− aqk )
)
= (1− a)
( ∞∏k=1
(1− qk )(1− a−1qk )(1− aqk )
)
Deduccion del Teorema de Jacobi
Luego ese productorio se puede escribir como:
∞∏k=1
(1− qk )(1− aqk−1)(1− a−1qk )
=
( ∞∏k=1
(1− qk )(1− a−1qk )
)( ∞∏k=1
(1− aqk−1)
)
=
( ∞∏k=1
(1− qk )(1− a−1qk )
)(1− a)
( ∞∏k=2
(1− aqk−1)
)
= (1− a)
( ∞∏k=1
(1− qk )(1− a−1qk )
)( ∞∏k=1
(1− aqk )
)
= (1− a)
( ∞∏k=1
(1− qk )(1− a−1qk )(1− aqk )
)
Deduccion del Teorema de Jacobi
Luego ese productorio se puede escribir como:
∞∏k=1
(1− qk )(1− aqk−1)(1− a−1qk )
=
( ∞∏k=1
(1− qk )(1− a−1qk )
)( ∞∏k=1
(1− aqk−1)
)
=
( ∞∏k=1
(1− qk )(1− a−1qk )
)(1− a)
( ∞∏k=2
(1− aqk−1)
)
= (1− a)
( ∞∏k=1
(1− qk )(1− a−1qk )
)( ∞∏k=1
(1− aqk )
)
= (1− a)
( ∞∏k=1
(1− qk )(1− a−1qk )(1− aqk )
)
Deduccion del Teorema de Jacobi
Luego ese productorio se puede escribir como:
∞∏k=1
(1− qk )(1− aqk−1)(1− a−1qk )
=
( ∞∏k=1
(1− qk )(1− a−1qk )
)( ∞∏k=1
(1− aqk−1)
)
=
( ∞∏k=1
(1− qk )(1− a−1qk )
)(1− a)
( ∞∏k=2
(1− aqk−1)
)
= (1− a)
( ∞∏k=1
(1− qk )(1− a−1qk )
)( ∞∏k=1
(1− aqk )
)
= (1− a)
( ∞∏k=1
(1− qk )(1− a−1qk )(1− aqk )
)
Deduccion del Teorema de Jacobi
Veamos ahora el sumatorio de la derecha:∞∑
n=−∞(−1)nanq
n(n−1)2
=0∑
n=−∞(−1)nanq
n(n−1)2 +
∞∑n=1
(−1)nanqn(n−1)
2
=∞∑
n=0
(−1)−na−nq(−n)(−n−1)
2 +∞∑
n=0
(−1)n+1an+1q(n+1)n
2
=∞∑
n=0
(−1)na−nqn(n+1)
2 +∞∑
n=0
−(−1)nan+1qn(n+1)
2
=∞∑
n=0
(−1)n(a−n − an+1)qn(n+1)
2
Deduccion del Teorema de Jacobi
Veamos ahora el sumatorio de la derecha:∞∑
n=−∞(−1)nanq
n(n−1)2
=0∑
n=−∞(−1)nanq
n(n−1)2 +
∞∑n=1
(−1)nanqn(n−1)
2
=∞∑
n=0
(−1)−na−nq(−n)(−n−1)
2 +∞∑
n=0
(−1)n+1an+1q(n+1)n
2
=∞∑
n=0
(−1)na−nqn(n+1)
2 +∞∑
n=0
−(−1)nan+1qn(n+1)
2
=∞∑
n=0
(−1)n(a−n − an+1)qn(n+1)
2
Deduccion del Teorema de Jacobi
Veamos ahora el sumatorio de la derecha:∞∑
n=−∞(−1)nanq
n(n−1)2
=0∑
n=−∞(−1)nanq
n(n−1)2 +
∞∑n=1
(−1)nanqn(n−1)
2
=∞∑
n=0
(−1)−na−nq(−n)(−n−1)
2 +∞∑
n=0
(−1)n+1an+1q(n+1)n
2
=∞∑
n=0
(−1)na−nqn(n+1)
2 +∞∑
n=0
−(−1)nan+1qn(n+1)
2
=∞∑
n=0
(−1)n(a−n − an+1)qn(n+1)
2
Deduccion del Teorema de Jacobi
Veamos ahora el sumatorio de la derecha:∞∑
n=−∞(−1)nanq
n(n−1)2
=0∑
n=−∞(−1)nanq
n(n−1)2 +
∞∑n=1
(−1)nanqn(n−1)
2
=∞∑
n=0
(−1)−na−nq(−n)(−n−1)
2 +∞∑
n=0
(−1)n+1an+1q(n+1)n
2
=∞∑
n=0
(−1)na−nqn(n+1)
2 +∞∑
n=0
−(−1)nan+1qn(n+1)
2
=∞∑
n=0
(−1)n(a−n − an+1)qn(n+1)
2
Deduccion del Teorema de Jacobi
El factor (a−n − an+1) se puede escribir como:
(a−n − an+1) =1an (1− a2n+1)
=(1− a)
an(1− a2n+1)
(1− a)
Como en general es cierto que:
m∑k=0
ak =(1− am+1)
(1− a)
Deduccion del Teorema de Jacobi
El factor (a−n − an+1) se puede escribir como:
(a−n − an+1) =1an (1− a2n+1) =
(1− a)an
(1− a2n+1)
(1− a)
Como en general es cierto que:
m∑k=0
ak =(1− am+1)
(1− a)
Deduccion del Teorema de Jacobi
El factor (a−n − an+1) se puede escribir como:
(a−n − an+1) =1an (1− a2n+1) =
(1− a)an
(1− a2n+1)
(1− a)
Como en general es cierto que:
m∑k=0
ak =(1− am+1)
(1− a)
Deduccion del Teorema de Jacobi
El factor (a−n − an+1) se puede escribir como:
(a−n − an+1) =1an (1− a2n+1) =
(1− a)an
(1− a2n+1)
(1− a)
Como en general es cierto que:
m∑k=0
ak =1− am+1
1− a
Podemos escribir que
(1− a2n+1)
(1− a)=
2n∑k=0
ak = (1 + a + a2 + a3 + . . .+ a2n)
Deduccion del Teorema de Jacobi
Entonces
(a−n−an+1) =(1− a)
an(1− a2n+1)
(1− a)
=(1− a)
an (1+a+a2+a3+. . .+a2n)
Sustituyendo este valor en el sumatorio de la derecha obtenemos:
∞∑n=0
(−1)n(a−n − an+1)qn(n+1)
2
= (1− a)∞∑
n=0
(−1)n
an (1 + a + a2 + a3 + . . .+ a2n)qn(n+1)
2
Deduccion del Teorema de Jacobi
Entonces
(a−n−an+1) =(1− a)
an(1− a2n+1)
(1− a)=
(1− a)an (1+a+a2+a3+. . .+a2n)
Sustituyendo este valor en el sumatorio de la derecha obtenemos:
∞∑n=0
(−1)n(a−n − an+1)qn(n+1)
2
= (1− a)∞∑
n=0
(−1)n
an (1 + a + a2 + a3 + . . .+ a2n)qn(n+1)
2
Deduccion del Teorema de Jacobi
Entonces
(a−n−an+1) =(1− a)
an(1− a2n+1)
(1− a)=
(1− a)an (1+a+a2+a3+. . .+a2n)
Sustituyendo este valor en el sumatorio de la derecha obtenemos:
∞∑n=0
(−1)n(a−n − an+1)qn(n+1)
2
= (1− a)∞∑
n=0
(−1)n
an (1 + a + a2 + a3 + . . .+ a2n)qn(n+1)
2
Conclusion
Ası hemos visto que tanto en el miembro de la izquierda de laecuacion, como en el de la derecha, aparece el factor (1− a).
(1− a)
( ∞∏k=1
(1− qk )(1− a−1qk )(1− aqk )
)
= (1− a)∞∑
n=0
(−1)n
an (1 + a + a2 + a3 + . . .+ a2n)qn(n+1)
2
Si eliminamos el factor (1-a) y substituimos a por 1, se obtiene:( ∞∏k=1
(1− qk )
)3
=∞∑
n=0
(−1)n(2n + 1)qn(n+1)
2
Conclusion
Ası hemos visto que tanto en el miembro de la izquierda de laecuacion, como en el de la derecha, aparece el factor (1− a).
(1− a)
( ∞∏k=1
(1− qk )(1− a−1qk )(1− aqk )
)
= (1− a)∞∑
n=0
(−1)n
an (1 + a + a2 + a3 + . . .+ a2n)qn(n+1)
2
Si eliminamos el factor (1-a) y substituimos a por 1, se obtiene:( ∞∏k=1
(1− qk )
)3
=∞∑
n=0
(−1)n(2n + 1)qn(n+1)
2
Conclusion
Ası hemos visto que tanto en el miembro de la izquierda de laecuacion, como en el de la derecha, aparece el factor (1− a).
(1− a)
( ∞∏k=1
(1− qk )(1− a−1qk )(1− aqk )
)
= (1− a)∞∑
n=0
(−1)n
an (1 + a + a2 + a3 + . . .+ a2n)qn(n+1)
2
Si eliminamos el factor (1-a) y substituimos a por 1, se obtiene:( ∞∏k=1
(1− qk )
)3
=∞∑
n=0
(−1)n(2n + 1)qn(n+1)
2
Conclusion
Entonces a partir de
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
⇓∞∏
k=1
(1− qk )(1− aqk−1)(1− a−1qk ) =∞∑
n=−∞(−1)nanq
n(n−1)2
⇓
Teorema de los numeros triangulares
( ∞∏k=1
(1− qk )
)3
=∞∑
n=0
(−1)n(2n + 1)qn(n+1)/2
Conclusion
Entonces a partir de
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
⇓∞∏
k=1
(1− qk )(1− aqk−1)(1− a−1qk ) =∞∑
n=−∞(−1)nanq
n(n−1)2
⇓
Teorema de los numeros triangulares
( ∞∏k=1
(1− qk )
)3
=∞∑
n=0
(−1)n(2n + 1)qn(n+1)/2
Conclusion
Entonces a partir de
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
⇓∞∏
k=1
(1− qk )(1− aqk−1)(1− a−1qk ) =∞∑
n=−∞(−1)nanq
n(n−1)2
⇓
Teorema de los numeros triangulares
( ∞∏k=1
(1− qk )
)3
=∞∑
n=0
(−1)n(2n + 1)qn(n+1)/2
Deduccion del Teorema de Euler
Producto triple de Jacobi
∞∏k=1
(1− x2k )(1 + x2ky2)(1 + x2ky−2) =∞∑
n=−∞xn2
y2n
⇓
Teorema del numero pentagonal
∞∏k=1
(1− qk ) =∞∑
n=−∞(−1)nq
(3n−1)n2
Deduccion del Teorema de Euler
Producto triple de Jacobi
∞∏k=1
(1− x2k )(1 + x2ky2)(1 + x2ky−2) =∞∑
n=−∞xn2
y2n
⇓
Teorema del numero pentagonal
∞∏k=1
(1− qk ) =∞∑
n=−∞(−1)nq
(3n−1)n2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q32 e y = iq−
14
1− x2k → 1−(
q32
)2k= 1− q3k
1+x2k−1y2 → 1+(
q32
)2k−1 (iq−
14
)2= 1−q3k− 3
2 q−12 = 1− q3k−2
1+x2k−1y−2 → 1+(
q32
)2k−1 (iq−
14
)−2= 1−q3k− 3
2 q12 = 1− q3k−1
xn2y2n →
(q
32
)n2 (iq−
14
)2n= q
3n22 (−1)nq−
n2 = (−1)nq
n(3n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q32 e y = iq−
14
1− x2k
→ 1−(
q32
)2k= 1− q3k
1+x2k−1y2 → 1+(
q32
)2k−1 (iq−
14
)2= 1−q3k− 3
2 q−12 = 1− q3k−2
1+x2k−1y−2 → 1+(
q32
)2k−1 (iq−
14
)−2= 1−q3k− 3
2 q12 = 1− q3k−1
xn2y2n →
(q
32
)n2 (iq−
14
)2n= q
3n22 (−1)nq−
n2 = (−1)nq
n(3n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q32 e y = iq−
14
1− x2k → 1−(
q32
)2k
= 1− q3k
1+x2k−1y2 → 1+(
q32
)2k−1 (iq−
14
)2= 1−q3k− 3
2 q−12 = 1− q3k−2
1+x2k−1y−2 → 1+(
q32
)2k−1 (iq−
14
)−2= 1−q3k− 3
2 q12 = 1− q3k−1
xn2y2n →
(q
32
)n2 (iq−
14
)2n= q
3n22 (−1)nq−
n2 = (−1)nq
n(3n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q32 e y = iq−
14
1− x2k → 1−(
q32
)2k= 1− q3k
1+x2k−1y2 → 1+(
q32
)2k−1 (iq−
14
)2= 1−q3k− 3
2 q−12 = 1− q3k−2
1+x2k−1y−2 → 1+(
q32
)2k−1 (iq−
14
)−2= 1−q3k− 3
2 q12 = 1− q3k−1
xn2y2n →
(q
32
)n2 (iq−
14
)2n= q
3n22 (−1)nq−
n2 = (−1)nq
n(3n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q32 e y = iq−
14
1− x2k → 1−(
q32
)2k= 1− q3k
1+x2k−1y2
→ 1+(
q32
)2k−1 (iq−
14
)2= 1−q3k− 3
2 q−12 = 1− q3k−2
1+x2k−1y−2 → 1+(
q32
)2k−1 (iq−
14
)−2= 1−q3k− 3
2 q12 = 1− q3k−1
xn2y2n →
(q
32
)n2 (iq−
14
)2n= q
3n22 (−1)nq−
n2 = (−1)nq
n(3n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q32 e y = iq−
14
1− x2k → 1−(
q32
)2k= 1− q3k
1+x2k−1y2 → 1+(
q32
)2k−1 (iq−
14
)2
= 1−q3k− 32 q−
12 = 1− q3k−2
1+x2k−1y−2 → 1+(
q32
)2k−1 (iq−
14
)−2= 1−q3k− 3
2 q12 = 1− q3k−1
xn2y2n →
(q
32
)n2 (iq−
14
)2n= q
3n22 (−1)nq−
n2 = (−1)nq
n(3n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q32 e y = iq−
14
1− x2k → 1−(
q32
)2k= 1− q3k
1+x2k−1y2 → 1+(
q32
)2k−1 (iq−
14
)2= 1−q3k− 3
2 q−12 = 1− q3k−2
1+x2k−1y−2 → 1+(
q32
)2k−1 (iq−
14
)−2= 1−q3k− 3
2 q12 = 1− q3k−1
xn2y2n →
(q
32
)n2 (iq−
14
)2n= q
3n22 (−1)nq−
n2 = (−1)nq
n(3n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q32 e y = iq−
14
1− x2k → 1−(
q32
)2k= 1− q3k
1+x2k−1y2 → 1+(
q32
)2k−1 (iq−
14
)2= 1−q3k− 3
2 q−12 = 1− q3k−2
1+x2k−1y−2
→ 1+(
q32
)2k−1 (iq−
14
)−2= 1−q3k− 3
2 q12 = 1− q3k−1
xn2y2n →
(q
32
)n2 (iq−
14
)2n= q
3n22 (−1)nq−
n2 = (−1)nq
n(3n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q32 e y = iq−
14
1− x2k → 1−(
q32
)2k= 1− q3k
1+x2k−1y2 → 1+(
q32
)2k−1 (iq−
14
)2= 1−q3k− 3
2 q−12 = 1− q3k−2
1+x2k−1y−2 → 1+(
q32
)2k−1 (iq−
14
)−2
= 1−q3k− 32 q
12 = 1− q3k−1
xn2y2n →
(q
32
)n2 (iq−
14
)2n= q
3n22 (−1)nq−
n2 = (−1)nq
n(3n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q32 e y = iq−
14
1− x2k → 1−(
q32
)2k= 1− q3k
1+x2k−1y2 → 1+(
q32
)2k−1 (iq−
14
)2= 1−q3k− 3
2 q−12 = 1− q3k−2
1+x2k−1y−2 → 1+(
q32
)2k−1 (iq−
14
)−2= 1−q3k− 3
2 q12 = 1− q3k−1
xn2y2n →
(q
32
)n2 (iq−
14
)2n= q
3n22 (−1)nq−
n2 = (−1)nq
n(3n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q32 e y = iq−
14
1− x2k → 1−(
q32
)2k= 1− q3k
1+x2k−1y2 → 1+(
q32
)2k−1 (iq−
14
)2= 1−q3k− 3
2 q−12 = 1− q3k−2
1+x2k−1y−2 → 1+(
q32
)2k−1 (iq−
14
)−2= 1−q3k− 3
2 q12 = 1− q3k−1
xn2y2n
→(
q32
)n2 (iq−
14
)2n= q
3n22 (−1)nq−
n2 = (−1)nq
n(3n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q32 e y = iq−
14
1− x2k → 1−(
q32
)2k= 1− q3k
1+x2k−1y2 → 1+(
q32
)2k−1 (iq−
14
)2= 1−q3k− 3
2 q−12 = 1− q3k−2
1+x2k−1y−2 → 1+(
q32
)2k−1 (iq−
14
)−2= 1−q3k− 3
2 q12 = 1− q3k−1
xn2y2n →
(q
32
)n2 (iq−
14
)2n
= q3n2
2 (−1)nq−n2 = (−1)nq
n(3n−1)2
Sustitucion
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
En la ecuacion ponemos x = q32 e y = iq−
14
1− x2k → 1−(
q32
)2k= 1− q3k
1+x2k−1y2 → 1+(
q32
)2k−1 (iq−
14
)2= 1−q3k− 3
2 q−12 = 1− q3k−2
1+x2k−1y−2 → 1+(
q32
)2k−1 (iq−
14
)−2= 1−q3k− 3
2 q12 = 1− q3k−1
xn2y2n →
(q
32
)n2 (iq−
14
)2n= q
3n22 (−1)nq−
n2 = (−1)nq
n(3n−1)2
Sustitucion
Por tanto obtenemos:∞∏
k=1
(1− q3k )(1− q3k−1)(1− q3k−2) =∞∑
n=−∞(−1)nq
n(3n−1)2
Podemos ver que el miembro de la derecha es igual que elTeorema del Numero Pentagonal. Nos falta demonstrar que elde la izquierda es igual tambien.
Sustitucion
Por tanto obtenemos:∞∏
k=1
(1− q3k )(1− q3k−1)(1− q3k−2) =∞∑
n=−∞(−1)nq
n(3n−1)2
Podemos ver que el miembro de la derecha es igual que elTeorema del Numero Pentagonal. Nos falta demonstrar que elde la izquierda es igual tambien.
Deduccion del Teorema de Euler
Desarrollando el productorio:
∞∏k=1
(1− q3k )(1− q3k−1)(1− q3k−2)
= (1− q3) · (1− q2) · (1− q1)︸ ︷︷ ︸k=1
· (1− q6) · (1− q5) · (1− q4)︸ ︷︷ ︸k=2
· (1− q9) · (1− q8) · (1− q7)︸ ︷︷ ︸k=3
. . .
Reordenando los terminos:
= (1− q1) · (1− q2) · (1− q3) · (1− q4) · (1− q5) · (1− q6)
· (1− q7) · (1− q8) · (1− q9) . . .
Deduccion del Teorema de Euler
Desarrollando el productorio:
∞∏k=1
(1− q3k )(1− q3k−1)(1− q3k−2)
= (1− q3) · (1− q2) · (1− q1)︸ ︷︷ ︸k=1
· (1− q6) · (1− q5) · (1− q4)︸ ︷︷ ︸k=2
· (1− q9) · (1− q8) · (1− q7)︸ ︷︷ ︸k=3
. . .
Reordenando los terminos:
= (1− q1) · (1− q2) · (1− q3) · (1− q4) · (1− q5) · (1− q6)
· (1− q7) · (1− q8) · (1− q9) . . .
Deduccion del Teorema de Euler
Desarrollando el productorio:
∞∏k=1
(1− q3k )(1− q3k−1)(1− q3k−2)
= (1− q3) · (1− q2) · (1− q1)︸ ︷︷ ︸k=1
· (1− q6) · (1− q5) · (1− q4)︸ ︷︷ ︸k=2
· (1− q9) · (1− q8) · (1− q7)︸ ︷︷ ︸k=3
. . .
Reordenando los terminos:
= (1− q1) · (1− q2) · (1− q3) · (1− q4) · (1− q5) · (1− q6)
· (1− q7) · (1− q8) · (1− q9) . . .
Deduccion del Teorema de Euler
Por tanto:∞∏
k=1
(1− q3k )(1− q3k−1)(1− q3k−2) =∞∏
k=1
(1− qk )
que es igual que el miembro de la izquierda del Teorema delNumero Pentagonal.
Conclusion
Entonces a partir de
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
⇓∞∏
k=1
(1− q3k )(1− q3k−1)(1− q3k−2) =∞∑
n=−∞(−1)nq
n(3n−1)2
⇓
Teorema del Numero Pentagonal
∞∏k=1
(1− qk ) =∞∑
n=−∞(−1)nq
(3n−1)n2
Conclusion
Entonces a partir de
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
⇓∞∏
k=1
(1− q3k )(1− q3k−1)(1− q3k−2) =∞∑
n=−∞(−1)nq
n(3n−1)2
⇓
Teorema del Numero Pentagonal
∞∏k=1
(1− qk ) =∞∑
n=−∞(−1)nq
(3n−1)n2
Conclusion
Entonces a partir de
∞∏k=1
(1− x2k )(1 + x2k−1y2)(1 + x2k−1y−2) =∞∑
n=−∞xn2
y2n
⇓∞∏
k=1
(1− q3k )(1− q3k−1)(1− q3k−2) =∞∑
n=−∞(−1)nq
n(3n−1)2
⇓
Teorema del Numero Pentagonal
∞∏k=1
(1− qk ) =∞∑
n=−∞(−1)nq
(3n−1)n2
Casos particulares delProducto Triple de Jacobi
Jesus A. CorralChiara ForacePiera Galber
Luis J. Salmeron ContrerasMaria Soler Facundo
Universitat de Valencia
16 Enero 2014