LAST LECTURE OFpeople.math.sc.edu/.../Math788FLastLecture.pdf · Final Exam: Thursday, December 12, 2:00 p.m. The Final will be in this room. The Final is optional. It can only help

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.


Material to Know:


Material to Know: Same as what you needed to know forthe test.



Will I be Around?



Wil l I beAround? Can be



Wil l I beAround? Can be (maybe wil l be).



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 ?


Laguerre Polynomials



REALLY ? WHAT ARE THEY ?



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


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


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


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


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


CASES:

bj =

(m

j

)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)

vm−j

g(x) =m∑

j=0


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



• 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


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



• 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




• Wantp > v.

bj =

(m

j

)(vm+u)(v(m−1)+u) ··· (v(j+1)+u)

vm−j

g(x) =m∑

j=0




• 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




• 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


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


{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


{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


{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


{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


{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


{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


{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


{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


{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


{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


{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


{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


• 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



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



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)



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)


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)


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)


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


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




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


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


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.

LAST LECTURE OFpeople.math.sc.edu/.../Math788FLastLecture.pdf · Final Exam: Thursday, December 12, 2:00 p.m. The Final will be in this room. The Final is optional. It can only help

Documents

LAST LECTURE OFpeople.math.sc.edu/.../Math788FLastLecture.pdf · Final Exam: Thursday, December 12, 2:00 p.m. The Final will be in this room. The Final is optional. It can only help