Post on 28-Jun-2020
Polynomial Families with Interesting Zeros
Robert BoyerDrexel University
April 30, 2010
Taylor Expansions with Finite Radius of Convergence
Work of Robert Jentzsch, 1917
Partial Sums of Geometric Series
pn(z) =n
!
k=0
zk = 1 + z + z2 + · · · + zn =zn+1 ! 1
z ! 1
pn(z) has zeros at all the (n + 1)-st roots of unity except at z = 1
Zeros all lie on the unit circle and fill it up as n " #
Taylor Polynomials for f (z)
pn(z) =n
!
k=0
f (k)(0)
k!zk
No simple formula to get the zeros of pn(z)
Assume that the Taylor series!!
k=0
f (k)(0)
k!zk converges for |z | < R
Question: What happens to the zeros of pn(z) as n " #?
Figure: Taylor polynomial for sec(x) – Degree 100
Figure: Taylor polynomial for arcsin(x) – Degree 100
Figure: !(x) – Degree 100
2
2
1
10
-1
0
-2
-1-2
Figure:"
xk/sin(k) – Degree 400
-1 1.5-0.5
1
0.5
-0.5
0.50
-1
01
Figure:"
xk/sin(k) – Degree 7,000
Taylor Polynomials for Exponential and Cosine
Work of Szego
Figure:"
(70x)k/k! – Degree 70
Figure: Szego Curve |ze1!z | = 1
Figure:"
(70x)k/k! – Degree 70 – with limiting curve
Figure: Exponential with zero attractor - degree 1000
Figure: Scaled Cosine Polynomial with limiting curve – Degree 70
Figure: Cosine with zero attractor - degree 1000
Taylor Polynomials for Linear Combinations of Exponentials
Work of Bleher and Mallison, 2006
3 e(8+2 i)z ! (9 ! 12 i) e(4+7 i)z + (2 + i) e("7+4 i)z
+ (30 + 30 i) e(6"7 i)z + (8 ! 5 i) e(6"4 i)z
+ (3 ! 9 i) e(4+4 i)z + 2 ie("2"4 i)z
Figure: Linear Combination of exp’s - degree 250
Figure: Linear Combination of exp’s - degree 1000
Bernoulli Polynomials
Work of Boyer and Goh, 2007
Bernoulli Polynomials
Generating Function
text
et ! 1=
!!
n=0
Bn(x)tn
Formula for Sums of Powers of Integers
m!
k=1
kn =Bn+1(m) ! Bn+1(1)
m + 1
Table of Bernoulli Polynomials
B1(x) = x ! 1/2
B2(x) = x2 ! x + 1/6
B3(x) = x3 ! 3/2 x2 + 1/2 x
B4(x) = x4 ! 2 x3 + x2 ! 1/30
B5(x) = x5 ! 5/2 x4 + 5/3 x3 ! 1/6 x
B6(x) = x6 ! 3 x5 + 5/2 x4 ! 1/2 x2 + 1/42
B7(x) = x7 ! 7/2 x6 + 7/2 x5 ! 7/6 x3 + 1/6 x
B8(x) = x8 ! 4 x7 + 14/3 x6 ! 7/3 x4 + 2/3 x2 ! 1/30
Figure: Bernoulli Polynomial Degree 70
Figure: Bernoulli Polynomial Degree 70 with limiting curve
Figure: Bernoulli Polynomial Degree 500
Figure: Bernoulli Polynomial Degree 1000
Figure: Bernoulli Polynomial Degree 500 with limiting curve
Figure: Bernoulli Polynomial Degree 1,000 with limiting curve
Appell Polynomials
Work of Boyer and Goh, 2010
Appell Polynomials
ext
g(t)=
!!
n=0
Pn(x)tn, P #
n(x) = Pn"1(x)
Basic Example: Pn(x) =xn
n!with g(t) = 1
Example:
n!
k=0
xn
n!, with generating function g(t) = 1 ! t
Another Method: Given any sequence {an}, the polynomialfamily below is Appell:
Pn(x) =n
!
k=0
an"k
k!xk
Figure: Appell Polynomial – Degree 100 – g(t) = (t ! 1)(t2 + 2)
Figure: Two Szego Curves
Figure: Degree 100, g(t) = (t ! 1)(t2 + 2)
Figure: Degree 400, g(t) = (t ! 1)(t2 + 2)
Figure: Appell Polynomial – Degree 400 –g(t) = (t ! 1/(1.2e i3!/16))(t ! 1/(1.3e i7!/16))(t ! 1/1.5)
Figure: Appell Polynomial –g(t) = (t ! 1/(1.2e i3!/16))(t ! 1/(1.3e i7!/16))(t ! 1/1.5)
Figure: Appell Polynomials
Figure: Appell Polynomials
Polynomials Satisfying a Linear Recurrence
pn+1(z) =k
!
j=1
qj(z)pn"j(z)
Example
pn+1(z) = (z + 1 ! i) pn + (z + 1) (z ! i) pn"1 +#
z3 + 10$
pn"2
Example
pn+1(z) = [(z + 1 ! i) + (z + 1) (z ! i)]pn"1 +#
z3 + 10$
pn"2
Fibonacci Type Polynomials
Fibonacci numbers: Fn+2 = Fn+1 + Fn have polynomial version:
Fn+1(x) = xFn(x) + Fn"1(x), F1(x) = 1,F2(x) = x .
Their zeros are all purely imaginary.
More general versions “Tribonacci”:
Tn+3(x) = x2Tn+2(x) + xTn+1 + Tn(x),
T0(x) = 0,T1(x) = 1,T2(x) = x2
The following two examples have the following recurrences.Both have the same initial conditions:
p0(z) = z6!z4+i , p1(z) = z!i+2, p2(z) = (2+i)2 (z2!8)
Example One
pn+1(z) = [(z +1+ i) + (z +1)] (z ! i) pn"1(z)+ (z3 +10) pn"2(z)
Example Two
pn+1(z) = (z+1+i) pn(x)+(z+1) (z!i) pn"1(z)+(z3+10) pn"2(z)
Figure: Tribonacci Polynomial - Degree 238
Example One
Figure: Generalized Fibonacci - Degree 76
Figure: Generalized Fibonacci - Degree 506
Figure: Generalized Fiboncaci - Degree 1006
Example Two
Figure: Another Example: Generalized Fibonaaci - Degree 506
Jacobi Polynomials
K Driver and P. Duren, 1999
P(!,")n (z) =
1
n!
!(" + n + 1)
!(" + # + n + 1)
n!
m=0
%
n
m
&
!(" + # + n + m + 1)
!(" + m + 1)
%
z ! 1
2
&m
" = kn + 1, # = !n ! 1, with k = 2
Lemniscate: |z ! 1|k |z + 1| =
%
2
k + 1
&k+1
kk
Figure: Jacobi Polynomial - Degree 50
Figure: Jacobi Polynomial - Degree 500
Figure: Jacobi Polynomial - Degree 700
Mandelbrot Polynomials with Fractal Zeros
pn+1(x) = xpn(x)2 + 1, p0(x) = 1
Figure: Mandelbot Polynomial - Degree 210 ! 1 = 1023
Figure: Mandelbot Polynomial - Degree 211 ! 1 = 2, 048
Figure: Mandelbot Polynomial - Degree 212 ! 1 = 4, 095
Polynomials Associated with Painleve Equations
Peter A Clarkson and Elizabeth L Mansfield, 2003
Painleve Di!erential Equations and VorobevYablonskii
Polynomials
Suppose that Qn(z) satisfies the recursion relation
Qn+1Qn"1 = zQ2n ! 4[QnQ
##
n ! (Q #
n)2], Q0(z) = 1,Q1(z) = z .
Then the rational function
w(z ; n) =d
dzln
Qn"1(z)
Qn(z)
satisfies PII
w ## = 2w3 + zw + ", " = n $ Z+.
Further, w(z ; 0) = 0 and w(z ;!n) = !w(z ; n).The VorobevYablonskii polynomials are monic with degreen(n + 1)/2.
Figure: Painleve Polynomial - Degree 325
Figure: Painleve Polynomial - Degree 1, 275
Richard Stanley Examples from Combinatorics, 2001
Chromatic Polynomial of a Graph
A complete bipartite graph G has its vertices broken into twodisjoint subsets A and B so that every vertex in A is connected byan edge with every vertex in B .
A coloring of a graph with r colors is an assignment that uses allthe possible colors so that if vertices v and w are connected by anedge they must have di"erent colors.
The chromatic polynomial pG (x) of a graph G is determined by itsvalues on the positive integers:
pG (n) = # all colorings of G using n colors
Figure: Chromatic Polynomial for a Complete Bipartite Graph – Degree1,000
q-Catalan Numbers
Catalan numbers: Cn = 1n+1
#2nn
$
with recurrence
Cn+1 =n
!
i=0
CiCn"1
Counts properly parathenized expressions or nondecreasing binarypaths
q-Catalan Polynomials Cn(q)
Cn+1(q) =n
!
i=0
Ci (q)Cn"i (q)q(i+1)(n"i), C0(q) = 1,
deg(Cn) =
%
n
2
&
, Cn(1) = Cn
Geometric-Combinatorial Meaning
Cn(q) =!
P:path
qarea(P)
where P is any lattice path from (0, 0) to (n, n) with step either(1, 0) or (0, 1) satisfies the additional condition that the path P
never rises above the line y = x .area(P) means the area underneath the path.
Figure: q-catalan Polynomial - Degree 190
Figure: q-catalan Polynomial - Degree 11,175
Partition Polynomials
Work of Boyer and Goh, 2007
Polynomial Partition Polynomials
Partition Numbers
4 = 4, 4 = 3+1, 4 = 2+2, 4 = 2+1+1, 4 = 1+1+1+1
p1(4) = 1, p2(4) = 2, p3(4) = 1, p4(4) = 1
Hardy-Ramanujan Asymptotics
p(n) %1
4n&
3e#&
2n/3.
Partition Polynomials
Fn(x) =n
!
k=1
pk(n)xk
1
1
0.5
0.50
-0.5
0
-1
-0.5-1
Figure: Partition Polynomial - degree 200
1000 400
16
300200
4
500
12
0
8
Figure: Digits of Partition Polynomial - degree 500
250
200
150
50
0
300
800000 20000
100
6000040000
Figure: Digits of Partition Polynomial - degree 80,000
10
1
0.5-0.5
-1
-0.5
0.5
0-1
Figure: All Zeros of Partition Polynomial - degree 10,000
1
1
0.5
0.50
-0.5
0
-1
-0.5-1
Figure: Partition Polynomial Attractor
1
0.8
0.6
0.4
0.2
010.50-0.5-1
Figure: Partition Polynomial Attractor in Upper Half Plane
0.4
-0.4
0.2
0-0.6-0.8-1
1
0
0.8
-0.2
0.6
Figure: Partition Polynomial Attractor in Second Quadrant
0.7
-0.6
0.65
0.6
-0.640.55
-0.68-0.72
0.8
-0.56
0.75
Figure: Triple Point for Partition Polynomial - degree 400
0.76
-0.6
0.72
0.68
-0.64
0.64
-0.68-0.72
0.8
-0.56
Figure: Triple Point for Partition Polynomial - degree 5,000
0.76
-0.6
0.72
0.68
-0.64
0.64
-0.68-0.72
0.8
-0.56
Figure: Triple Point for Partition Polynomial - degree 50,000
1
0.8
0.6
0.4
0.2
0
10.5-0.5 0-1
Figure: Region I for Partition Polynomial
0.5
0.1
0.6
0.4
0
10.50-0.5-1
0.3
0.7
0.2
Figure: Region II for Partition Polynomial
-0.5-0.6-0.7-0.8
0.9
0.8
0.7
0.6
-0.3
0.5
0.4
-0.4
Figure: Region III for Partition Polynomial
0.4
0.2
1
0.8
0
10 0.5
-0.5
0.6
-1