L AST L ECTURE OF M ATH 788F
L AST L ECTURE
OF
M ATH 788F
Final Exam:
Final Exam: Thursday, December 12, 2:00 p.m.
Final Exam: Thursday, December 12, 2:00 p.m.The Final will be in this room.
Final Exam: Thursday, December 12, 2:00 p.m.The Final will be in this room.The Final is optional.
Final Exam: Thursday, December 12, 2:00 p.m.The Final will be in this room.The Final is optional.It can only help your grade.
Final Exam: Thursday, December 12, 2:00 p.m.The Final will be in this room.The Final is optional.It can only help your grade.
Material to Know:
Final Exam: Thursday, December 12, 2:00 p.m.The Final will be in this room.The Final is optional.It can only help your grade.
Material to Know: Same as what you needed to know forthe test.
Final Exam: Thursday, December 12, 2:00 p.m.The Final will be in this room.The Final is optional.It can only help your grade.
Material to Know: Same as what you needed to know forthe test.
Will I be Around?
Final Exam: Thursday, December 12, 2:00 p.m.The Final will be in this room.The Final is optional.It can only help your grade.
Material to Know: Same as what you needed to know forthe test.
Wil l I beAround? Can be
Final Exam: Thursday, December 12, 2:00 p.m.The Final will be in this room.The Final is optional.It can only help your grade.
Material to Know: Same as what you needed to know forthe test.
Wil l I beAround? Can be (maybe wil l be).
Final Exam: Thursday, December 12, 2:00 p.m.The Final will be in this room.The Final is optional.It can only help your grade.
Material to Know: Same as what you needed to know forthe test.
Will I be Around? Can be (maybe will be). Please sendme email if you would like to get to-gether (or if you have questions).
WHAT WERE WE DISCUSSING BEFORE OUR TEST ?
WHAT WERE WE DISCUSSING BEFORE OUR TEST ?
Laguerre Polynomials
WHAT WERE WE DISCUSSING BEFORE OUR TEST ?
Laguerre Polynomials
REALLY ? WHAT ARE THEY ?
WHAT WERE WE DISCUSSING BEFORE OUR TEST ?
Laguerre Polynomials
REALLY ? WHAT ARE THEY ?
m∑j=0
(m+α)(m−1+α) · · · (j+1+α)(−x)j
(m − j)!j!
m∑j=0
(m+α)(m−1+α) · · · (j+1+α)(−x)j
(m − j)!j!
m∑j=0
(m+α)(m−1+α) · · · (j+1+α)(−x)j
(m − j)!j!
We denote thisL(α)m (x).
m∑j=0
(m+α)(m−1+α) · · · (j+1+α)(−x)j
(m − j)!j!
We denote thisL(α)m (x).
THEOREM 7.7.2. ∀α ∈ Q − Z−, ∃ finitely many
m ∈ Z+ such thatL(α)m (x) is reducible.
α =u
v6∈ Z−, v > 0, gcd(u, v) = 1
α =u
v6∈ Z−, v > 0, gcd(u, v) = 1
bj =
(m
j
)(m + α)(m − 1 + α) · · · (j + 1 + α)
α =u
v6∈ Z−, v > 0, gcd(u, v) = 1
bj =
(m
j
)(m + α)(m − 1 + α) · · · (j + 1 + α)
=
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
α =u
v6∈ Z−, v > 0, gcd(u, v) = 1
bj =
(m
j
)(m + α)(m − 1 + α) · · · (j + 1 + α)
=
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
Let g(x) =m∑
j=0
bjxj andf(x) =
m∑j=0
ajbjxj.
α =u
v6∈ Z−, v > 0, gcd(u, v) = 1
bj =
(m
j
)(m + α)(m − 1 + α) · · · (j + 1 + α)
=
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
Let g(x) =m∑
j=0
bjxj andf(x) =
m∑j=0
ajbjxj. Then
g(x) is monic.
α =u
v6∈ Z−, v > 0, gcd(u, v) = 1
bj =
(m
j
)(m + α)(m − 1 + α) · · · (j + 1 + α)
=
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
Let g(x) =m∑
j=0
bjxj andf(x) =
m∑j=0
ajbjxj. Then
g(x) is monic. Assumef(x) has a factor of degreek in[1, m/2].
α =u
v6∈ Z−, v > 0, gcd(u, v) = 1
bj =
(m
j
)(m + α)(m − 1 + α) · · · (j + 1 + α)
=
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
Let g(x) =m∑
j=0
bjxj andf(x) =
m∑j=0
ajbjxj. Then
g(x) is monic. Assumef(x) has a factor of degreek in[1, m/2]. Use Lemma 2.4.2 to obtain a contradiction.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
Let g(x) =m∑
j=0
bjxj andf(x) =
m∑j=0
ajbjxj. Then
g(x) is monic. Assumef(x) has a factor of degreek in[1, m/2]. Use Lemma 2.4.2 to obtain a contradiction.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
Let g(x) =m∑
j=0
bjxj andf(x) =
m∑j=0
ajbjxj. Then
g(x) is monic. Assumef(x) has a factor of degreek in[1, m/2]. Use Lemma 2.4.2 to obtain a contradiction.
We want a primep that satisfies certain conditions withg(x).
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
Let g(x) =m∑
j=0
bjxj andf(x) =
m∑j=0
ajbjxj. Then
g(x) is monic. Assumef(x) has a factor of degreek in[1, m/2]. Use Lemma 2.4.2 to obtain a contradiction.
We want a primep that satisfies certain conditions withg(x). One of them is thatp does not divide the leadingcoefficient of g(x).
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
Let g(x) =m∑
j=0
bjxj andf(x) =
m∑j=0
ajbjxj. Then
g(x) is monic. Assumef(x) has a factor of degreek in[1, m/2]. Use Lemma 2.4.2 to obtain a contradiction.
We want a primep that satisfies certain conditions withg(x). One of them is thatp does not divide the leadingcoefficient ofg(x). This is clear.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p prime
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p prime
CASES:
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p prime
CASES: • k > m/ log2 m
• k0 ≤ k ≤ m/ log2 m
• 2 ≤ k < k0
• k = 1
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p prime
CASES: • k > m/ log2 m
• k0 ≤ k ≤ m/ log2 m
• 2 ≤ k < k0
• k = 1
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p prime
BASIC I DEA IN EACH CASE:
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p prime
BASIC I DEA IN EACH CASE:
• Wantp|(v(m−j)+u) for somej ∈{0, 1,..., k−1}.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p prime
BASIC I DEA IN EACH CASE:
• Wantp|(v(m−j)+u) for somej ∈{0, 1,..., k−1}.
• Wantp > v.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p prime
BASIC I DEA IN EACH CASE:
• Wantp|(v(m−j)+u) for somej ∈{0, 1,..., k−1}.
• Wantp > v.
• Thenνp(bj) ≥ 1 for all j ∈{0, 1,..., m−k}.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p prime
BASIC I DEA IN EACH CASE:
• Wantp|(v(m−j)+u) for somej ∈{0, 1,..., k−1}.
• Wantp > v.
• Thenνp(bj) ≥ 1 for all j ∈{0, 1,..., m−k}.
• Show slope of right-most edge of N. P. ofg(x) is < 1k.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p prime
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
Slope of right-most edge ismax1≤j≤m
{ν(b0)−ν(bj)
j
}.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
Slope of right-most edge ismax1≤j≤m
{ν(b0)−ν(bj)
j
}.
ν(b0)−ν(bj)
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
Slope of right-most edge ismax1≤j≤m
{ν(b0)−ν(bj)
j
}.
ν(b0)−ν(bj) ≤ ν((vj+u)(v(j−1)+u)···(v+u))
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
Slope of right-most edge ismax1≤j≤m
{ν(b0)−ν(bj)
j
}.
ν(b0)−ν(bj) ≤ ν((vj+u)(v(j−1)+u)···(v+u))
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
Slope of right-most edge ismax1≤j≤m
{ν(b0)−ν(bj)
j
}.
ν(b0)−ν(bj) ≤ ν((vj+u)(v(j−1)+u)···(v+u))
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
Slope of right-most edge ismax1≤j≤m
{ν(b0)−ν(bj)
j
}.
ν(b0)−ν(bj) ≤ ν((vj+u)(v(j−1)+u)···(v+u))
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
Slope of right-most edge ismax1≤j≤m
{ν(b0)−ν(bj)
j
}.
ν(b0)−ν(bj) ≤ ν((vj+u)(v(j−1)+u)···(v+u))
≤ ν((vj+|u|)!)
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
Slope of right-most edge ismax1≤j≤m
{ν(b0)−ν(bj)
j
}.
ν(b0)−ν(bj) ≤ ν((vj+u)(v(j−1)+u)···(v+u))
≤ ν((vj+|u|)!) <vj+|u|p−1
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
Slope of right-most edge ismax1≤j≤m
{ν(b0)−ν(bj)
j
}.
ν(b0)−ν(bj) ≤ ν((vj+u)(v(j−1)+u)···(v+u))
≤ ν((vj+|u|)!) <vj+|u|p−1
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
Slope of right-most edge ismax1≤j≤m
{ν(b0)−ν(bj)
j
}.
ν(b0)−ν(bj) ≤ ν((vj+u)(v(j−1)+u)···(v+u))
≤ ν((vj+|u|)!) <vj+|u|
(v + |u|)k
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
Slope of right-most edge ismax1≤j≤m
{ν(b0)−ν(bj)
j
}.
ν(b0)−ν(bj) ≤ ν((vj+u)(v(j−1)+u)···(v+u))
≤ ν((vj+|u|)!) <(v + |u|)j(v + |u|)k
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
Slope of right-most edge ismax1≤j≤m
{ν(b0)−ν(bj)
j
}.
ν(b0)−ν(bj) ≤ ν((vj+u)(v(j−1)+u)···(v+u))
≤ ν((vj+|u|)!) <j
k
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
Slope of right-most edge ismax1≤j≤m
{ν(b0)−ν(bj)
j
}.
ν(b0)−ν(bj) ≤ ν((vj+u)(v(j−1)+u)···(v+u))
≤ ν((vj+|u|)!) <j
k
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
CASES: • k > m/ log2 m
• k0 ≤ k ≤ m/ log2 m
• 2 ≤ k < k0
• k = 1
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
CASES: • 2 ≤ k < k0
• k = 1
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
CASES: • 2 ≤ k < k0
• k = 1
BASIC I DEA IN EACH CASE:
• Wantp|(v(m−j)+u) for somej ∈{0, 1,..., k−1}.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
CASES: • 2 ≤ k < k0, p|(vm+u)(v(m−1)+u)
• k = 1
BASIC I DEA IN EACH CASE:
• Wantp|(v(m−j)+u) for somej ∈{0, 1,..., k−1}.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
CASES: • 2 ≤ k < k0, p|(vm+u)(v(m−1)+u)
• k = 1, p|(vm+u)
BASIC I DEA IN EACH CASE:
• Wantp|(v(m−j)+u) for somej ∈{0, 1,..., k−1}.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
CASES: • 2 ≤ k < k0, p|(vm+u)(v(m−1)+u)
• k = 1, p|(vm+u)
L EMMA 7.7.7.If a, b, c, d ∈ Z with ad−bc 6= 0, thenthe largest prime factor of(am + b)(cm + d) tends toinfinity with m.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
CASES: • 2 ≤ k < k0, p|(vm+u)(v(m−1)+u)
• k = 1, p|(vm+u)
L EMMA 7.7.7.If a, b, c, d ∈ Z with ad−bc 6= 0, thenthe largest prime factor of(am + b)(cm + d) tends toinfinity with m.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
CASES: • k = 1, p|(vm+u)
L EMMA 7.7.7.If a, b, c, d ∈ Z with ad−bc 6= 0, thenthe largest prime factor of(am + b)(cm + d) tends toinfinity with m.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
CASES: • k = 1, p|(vm+u)
L EMMA 7.7.7.If a, b, c, d ∈ Z with ad−bc 6= 0, thenthe largest prime factor of(am + b)(cm + d) tends toinfinity with m.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
CASES: • k = 1, p|m(vm+u)
L EMMA 7.7.7.If a, b, c, d ∈ Z with ad−bc 6= 0, thenthe largest prime factor of(am + b)(cm + d) tends toinfinity with m.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
CASES: • k = 1, p|m(vm+u)
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
CASES: • k = 1, p|m(vm+u)
BASIC I DEA IN EACH CASE:
• Wantp|(v(m−j)+u) for somej ∈{0, 1,..., k−1}.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
CASES: • k = 1, p|m(vm+u)
BASIC I DEA IN EACH CASE:
• Wantp|(v(m−j)+u) for somej ∈{0, 1,..., k−1}.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
CASES: • k = 1, p|m(vm+u)
BASIC I DEA IN EACH CASE:
• Wantp|(v(m−j)+u) for somej ∈{0, 1,..., k−1}.
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
CASES: • k = 1, p|m(vm+u)
BASIC I DEA IN CASE k = 1:
bj =
(m
j
)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)
vm−j
g(x) =m∑
j=0
bjxj, k ∈ [1, m/2], p > (v + |u|)k
CASES: • k = 1, p|m(vm+u)
BASIC I DEA IN CASE k = 1:
• Use that ifp|m andp is large, thenp|(m
j
)for smallj
and the numerator above for largej.