Polynomial Families with Interesting Zeros Robert Boyer Drexel University April 30, 2010
Taylor Expansions with Finite Radius of Convergence Work of Robert Jentzsch, 1917
Partial Sums of Geometric Series p n (z) = n k=0 z k = 1 + z + z 2 + + z n = z n+1 1 z 1 p n (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) p n (z) = n k=0 f (k) (0) k! z k No simple formula to get the zeros of p n (z) Assume that the Taylor series k=0 f (k) (0) k! z k converges for z < R Question: What happens to the zeros of p n (z) as n?
Figure: Taylor polynomial for sec(x) Degree 100
Figure: Taylor polynomial for arcsin(x) Degree 100
Figure: ζ(x) Degree 100
2 1-2 -1 0 0 1 2-1 -2 Figure: x k /sin(k) Degree 400
1 0.5 0-1 -0.5 0 0.5 1 1.5-0.5-1 Figure: x k /sin(k) Degree 7,000
Taylor Polynomials for Exponential and Cosine Work of Szegö
Figure: (70x) k /k! Degree 70
Figure: Szegö Curve ze 1 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 ( 7+4 i)z + (2 + i) e + (30 + 30 i) e (6 7 i)z (6 4 i)z + (8 5 i) e + (3 9 i) e (4+4 i)z ( 2 4 i)z +2ie
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 te xt e t 1 = n=0 B n (x)t n Formula for Sums of Powers of Integers m k=1 k n = B n+1(m) B n+1 (1) m +1
Table of Bernoulli Polynomials B 1 (x) = x 1/2 B 2 (x) = x 2 x +1/6 B 3 (x) = x 3 3/2 x 2 +1/2x B 4 (x) = x 4 2 x 3 + x 2 1/30 B 5 (x) = x 5 5/2 x 4 +5/3x 3 1/6 x B 6 (x) = x 6 3 x 5 +5/2x 4 1/2 x 2 +1/42 B 7 (x) = x 7 7/2 x 6 +7/2x 5 7/6 x 3 +1/6x B 8 (x) = x 8 4 x 7 + 14/3 x 6 7/3 x 4 +2/3x 2 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 Basic Example: Example: e xt g(t) = n k=0 n=0 P n (x)t n, P n (x) = xn n! P n (x) =P n 1(x) with g(t) = 1 x n, with generating function g(t) = 1 t n! Another Method: Given any sequence {a n }, the polynomial family below is Appell: P n (x) = n k=0 a n k x k k!
Figure: Appell Polynomial Degree 100 g(t) = (t 1)(t 2 + 2)
Figure: Two Szegö Curves
Figure: Degree 100, g(t) = (t 1)(t 2 + 2)
Figure: Degree 400, g(t) = (t 1)(t 2 + 2)
Figure: Appell Polynomial Degree 400 g(t) = (t 1/(1.2e i3π/16 ))(t 1/(1.3e i7π/16 ))(t 1/1.5)
Figure: Appell Polynomial g(t) = (t 1/(1.2e i3π/16 ))(t 1/(1.3e i7π/16 ))(t 1/1.5)
Figure: Appell Polynomials
Figure: Appell Polynomials
Polynomials Satisfying a Linear Recurrence k p n+1 (z) = q j (z)p n j (z) j=1 Example p n+1 (z) = (z +1 i) p n +(z + 1) (z i) p n 1 + ( z 3 + 10 ) p n 2 Example p n+1 (z) = [(z +1 i) + (z + 1) (z i)]p n 1 + ( z 3 + 10 ) p n 2
Fibonacci Type Polynomials Fibonacci numbers: F n+2 = F n+1 + F n have polynomial version: F n+1 (x) =xf n (x)+f n 1 (x), Their zeros are all purely imaginary. F 1 (x) = 1, F 2 (x) =x. More general versions Tribonacci : T n+3 (x) = x 2 T n+2 (x)+xt n+1 + T n (x), T 0 (x) = 0, T 1 (x) = 1, T 2 (x) =x 2
The following two examples have the following recurrences. Both have the same initial conditions: p 0 (z) =z 6 z 4 +i, p 1 (z) =z i +2, p 2 (z) = (2+i) 2 (z 2 8) Example One p n+1 (z) = [(z + 1 + i) + (z + 1)] (z i) p n 1 (z) + (z 3 + 10) p n 2 (z) Example Two p n+1 (z) = (z+1+i) p n (x)+(z+1) (z i) p n 1 (z)+(z 3 +10) p n 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 (α,β) n (z) = 1 n! Γ(α + n + 1) Γ(α + β + n + 1) n m=0 ( ) n Γ(α + β + n + m + 1) m Γ(α + m + 1) α = kn +1, β = n 1, with k =2 ( z 1 2 ) m Lemniscate: z 1 k z +1 = ( ) 2 k+1 k k k +1
Figure: Jacobi Polynomial - Degree 50
Figure: Jacobi Polynomial - Degree 500
Figure: Jacobi Polynomial - Degree 700
Mandelbrot Polynomials with Fractal Zeros p n+1 (x) =xp n (x) 2 +1, p 0 (x) = 1
Figure: Mandelbot Polynomial - Degree 2 10 1 = 1023
Figure: Mandelbot Polynomial - Degree 2 11 1 = 2, 048
Figure: Mandelbot Polynomial - Degree 2 12 1 = 4, 095
Polynomials Associated with Painlevé Equations Peter A Clarkson and Elizabeth L Mansfield, 2003
Painlevé Differential Equations and VorobevYablonskii Polynomials Suppose that Q n (z) satisfies the recursion relation Q n+1 Q n 1 = zq 2 n 4[Q nq n (Q n )2 ], Q 0 (z) = 1, Q 1 (z) =z. Then the rational function satisfies P II w(z; n) = d dz ln Q n 1(z) Q n (z) w =2w 3 + zw + α, α = n Z +. Further, w(z; 0) = 0 and w(z; n) = w(z; n). The VorobevYablonskii polynomials are monic with degree n(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 two disjoint subsets A and B so that every vertex in A is connected by an edge with every vertex in B. A coloring of a graph with r colors is an assignment that uses all the possible colors so that if vertices v and w are connected by an edge they must have different colors. The chromatic polynomial p G (x) of a graph G is determined by its values on the positive integers: p G (n) = # all colorings of G using n colors
Figure: Chromatic Polynomial for a Complete Bipartite Graph Degree 1,000
q-catalan Numbers Catalan numbers: C n = 1 n+1( 2n n ) with recurrence C n+1 = n C i C n 1 i=0 Counts properly parathenized expressions or nondecreasing binary paths
q-catalan Polynomials C n (q) C n+1 (q) = deg(c n ) = n C i (q)c n i (q)q (i+1)(n i), C 0 (q) = 1, i=0 ( ) n, C n (1) = C n 2 Geometric-Combinatorial Meaning C n (q) = P:path q area(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 p 1 (4) = 1, p 2 (4) = 2, p 3 (4) = 1, p 4 (4) = 1 Hardy-Ramanujan Asymptotics Partition Polynomials p(n) 1 4n 3 eπ 2n/3. F n (x) = n p k (n)x k k=1
1 0.5-1 -0.5 0 0 0.5 1-0.5-1 Figure: Partition Polynomial - degree 200
16 12 8 4 0 0 100 200 300 400 500 Figure: Digits of Partition Polynomial - degree 500
300 250 200 150 100 50 0 0 20000 40000 60000 80000 Figure: Digits of Partition Polynomial - degree 80,000
1 0.5-1 -0.5 0 0 0.5 1-0.5-1 Figure: All Zeros of Partition Polynomial - degree 10,000
1 0.5-1 -0.5 0 0 0.5 1-0.5-1 Figure: Partition Polynomial Attractor
1 0.8 0.6 0.4 0.2-1 -0.5 0 0 0.5 1 Figure: Partition Polynomial Attractor in Upper Half Plane
1 0.8 0.6 0.4 0.2-1 -0.8-0.6-0.4-0.2 0 0 Figure: Partition Polynomial Attractor in Second Quadrant
0.8 0.75 0.7 0.65 0.6-0.72-0.68-0.64-0.6 0.55-0.56 Figure: Triple Point for Partition Polynomial - degree 400
0.8 0.76 0.72 0.68 0.64-0.72-0.68-0.64-0.6-0.56 Figure: Triple Point for Partition Polynomial - degree 5,000
0.8 0.76 0.72 0.68 0.64-0.72-0.68-0.64-0.6-0.56 Figure: Triple Point for Partition Polynomial - degree 50,000
1 0.8 0.6 0.4 0.2 0-1 -0.5 0 0.5 1 Figure: Region I for Partition Polynomial
0.5 0.7 0.6 0.4 0.3 0.2 0.1 0-1 -0.5 0 0.5 1 Figure: Region II for Partition Polynomial
-0.5 0.9 0.8 0.7 0.6 0.5 0.4-0.8-0.7-0.6-0.4-0.3 Figure: Region III for Partition Polynomial
0.4 1 0.8 0.6 0.2 0-1 -0.5 0 0.5 1