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Zeros and irreducibility of some classes of special polynomials - - PowerPoint PPT Presentation

Zeros and irreducibility of some classes of special polynomials Karl Dilcher Dalhousie Number Theory Seminar January 21, 2019 Karl Dilcher Zeros and irreducibility of some classes of special polynomials Part I: Chebyshev-like polynomials


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Zeros and irreducibility of some classes of special polynomials

Karl Dilcher Dalhousie Number Theory Seminar January 21, 2019

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Part I: Chebyshev-like polynomials

Pafnutiy L ’vovich Chebyshev 1821 – 1894

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Joint work with Kenneth B. Stolarsky University of Illinois, Urbana-Champaign

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  • 1. Introduction

The Chebyshev polynomials Tn(x) are among the most important and interesting classical orthogonal polynomials.

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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  • 1. Introduction

The Chebyshev polynomials Tn(x) are among the most important and interesting classical orthogonal polynomials. Numerous applications, e.g., in Approximation Theory.

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  • 1. Introduction

The Chebyshev polynomials Tn(x) are among the most important and interesting classical orthogonal polynomials. Numerous applications, e.g., in Approximation Theory. They can be defined by T0(x) = 1, T1(x) = x, and Tn+1(x) = 2xTn(x) − Tn−1(x) (n ≥ 1).

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  • 1. Introduction

The Chebyshev polynomials Tn(x) are among the most important and interesting classical orthogonal polynomials. Numerous applications, e.g., in Approximation Theory. They can be defined by T0(x) = 1, T1(x) = x, and Tn+1(x) = 2xTn(x) − Tn−1(x) (n ≥ 1).

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Here is the definition again: T0(x) = 1, T1(x) = x, and Tn+1(x) = 2xTn(x) − Tn−1(x) (n ≥ 1).

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Here is the definition again: T0(x) = 1, T1(x) = x, and Tn+1(x) = 2xTn(x) − Tn−1(x) (n ≥ 1). We compute: T2(x) = 2x2 − 1, T3(x) = 4x3 − 3x, T4(x) = 8x4 − 8x2 + 1, . . .

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Here is the definition again: T0(x) = 1, T1(x) = x, and Tn+1(x) = 2xTn(x) − Tn−1(x) (n ≥ 1). We compute: T2(x) = 2x2 − 1, T3(x) = 4x3 − 3x, T4(x) = 8x4 − 8x2 + 1, . . . Now consider a slight variant: V0(x) = 1, V1(x) = x, and

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Here is the definition again: T0(x) = 1, T1(x) = x, and Tn+1(x) = 2xTn(x) − Tn−1(x) (n ≥ 1). We compute: T2(x) = 2x2 − 1, T3(x) = 4x3 − 3x, T4(x) = 8x4 − 8x2 + 1, . . . Now consider a slight variant: V0(x) = 1, V1(x) = x, and Vn+1(x) = 2xVn(x) − Vn−1(x) − xn+1 (n ≥ 1).

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Here is the definition again: T0(x) = 1, T1(x) = x, and Tn+1(x) = 2xTn(x) − Tn−1(x) (n ≥ 1). We compute: T2(x) = 2x2 − 1, T3(x) = 4x3 − 3x, T4(x) = 8x4 − 8x2 + 1, . . . Now consider a slight variant: V0(x) = 1, V1(x) = x, and Vn+1(x) = 2xVn(x) − Vn−1(x) − xn+1 (n ≥ 1). Do we get anything sensible?

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Let’s look at a table:

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Let’s look at a table:

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n Vn(x) 1 x 2 x2 − 1 3 x3 − 3x 4 x4 − 7x2 + 1 5 x5 − 15x3 + 5x 6 x6 − 31x4 + 17x2 − 1 7 x7 − 63x5 + 49x3 − 7x 8 x8 − 127x6 + 129x4 − 31x2 + 1 9 x9 − 255x7 + 321x5 − 111x3 + 9x 10 x10 − 511x8 + 769x6 − 351x4 + 49x2 − 1 11 x11 − 1023x9 + 1793x7 − 1023x5 + 209x3 − 11x 12 x12 − 2047x10 + 4097x8 − 2815x6 + 769x4 − 71x2 + 1

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V0(x) = 1, V1(x) = x, and Vn+1(x) = 2xVn(x) − Vn−1(x) − xn+1 (n ≥ 1).

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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V0(x) = 1, V1(x) = x, and Vn+1(x) = 2xVn(x) − Vn−1(x) − xn+1 (n ≥ 1). Some properties: Vn(x) = xn+2 − Tn(x) x2 − 1 ; (1)

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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V0(x) = 1, V1(x) = x, and Vn+1(x) = 2xVn(x) − Vn−1(x) − xn+1 (n ≥ 1). Some properties: Vn(x) = xn+2 − Tn(x) x2 − 1 ; (1) Vn(x) = xn −

⌊ n 2 ⌋

  • k=1

n 2k

  • (x2 − 1)k−1xn−2k.

(2)

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V0(x) = 1, V1(x) = x, and Vn+1(x) = 2xVn(x) − Vn−1(x) − xn+1 (n ≥ 1). Some properties: Vn(x) = xn+2 − Tn(x) x2 − 1 ; (1) Vn(x) = xn −

⌊ n 2 ⌋

  • k=1

n 2k

  • (x2 − 1)k−1xn−2k.

(2) Compare with Tn(x) = xn +

⌊ n 2 ⌋

  • k=1

n 2k

  • (x2 − 1)kxn−2k,

from which (2) is derived, by way of (1).

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Some special values: Vn(1) = 1 − n 2

  • ,

Vn(−1) = (−1)n

  • 1 −

n 2

  • .

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Some special values: Vn(1) = 1 − n 2

  • ,

Vn(−1) = (−1)n

  • 1 −

n 2

  • .

Generating function: 1 − 2tx (1 − tx)(1 − 2tx + t2) =

  • n=0

Vn(x)tn. (3)

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Some special values: Vn(1) = 1 − n 2

  • ,

Vn(−1) = (−1)n

  • 1 −

n 2

  • .

Generating function: 1 − 2tx (1 − tx)(1 − 2tx + t2) =

  • n=0

Vn(x)tn. (3) Compare with 1 − tx 1 − 2tx + t2 =

  • n=0

Tn(x)tn, from which (3) is derived.

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  • 2. Irreducibility and Zeros

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  • 2. Irreducibility and Zeros

The Chebyshev polynomial Tn(x)

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  • 2. Irreducibility and Zeros

The Chebyshev polynomial Tn(x)

  • has a well-known factorization over Q in terms of cyclotomic

polynomials

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  • 2. Irreducibility and Zeros

The Chebyshev polynomial Tn(x)

  • has a well-known factorization over Q in terms of cyclotomic

polynomials

  • is irreducible over Q iff n = 2k, k = 0, 1, 2, . . ..

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  • 2. Irreducibility and Zeros

The Chebyshev polynomial Tn(x)

  • has a well-known factorization over Q in terms of cyclotomic

polynomials

  • is irreducible over Q iff n = 2k, k = 0, 1, 2, . . ..

How about the Vn(x)?

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  • 2. Irreducibility and Zeros

The Chebyshev polynomial Tn(x)

  • has a well-known factorization over Q in terms of cyclotomic

polynomials

  • is irreducible over Q iff n = 2k, k = 0, 1, 2, . . ..

How about the Vn(x)? Easy to see: V2(x) = (x − 1)(x + 1), V4(x) = (x2 − 3x + 1)(x2 + 3x + 1)

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  • 2. Irreducibility and Zeros

The Chebyshev polynomial Tn(x)

  • has a well-known factorization over Q in terms of cyclotomic

polynomials

  • is irreducible over Q iff n = 2k, k = 0, 1, 2, . . ..

How about the Vn(x)? Easy to see: V2(x) = (x − 1)(x + 1), V4(x) = (x2 − 3x + 1)(x2 + 3x + 1) However, all other V2k(x) and

1 x V2k+1(x) appear to be

irreducible.

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  • 2. Irreducibility and Zeros

The Chebyshev polynomial Tn(x)

  • has a well-known factorization over Q in terms of cyclotomic

polynomials

  • is irreducible over Q iff n = 2k, k = 0, 1, 2, . . ..

How about the Vn(x)? Easy to see: V2(x) = (x − 1)(x + 1), V4(x) = (x2 − 3x + 1)(x2 + 3x + 1) However, all other V2k(x) and

1 x V2k+1(x) appear to be

irreducible. We can prove a partial result:

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Proposition The following are irreducible over Q: (a) V2k−2(x) for all k ≥ 3;

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Proposition The following are irreducible over Q: (a) V2k−2(x) for all k ≥ 3; (b)

1 x Vp(x) for all odd primes p.

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Proposition The following are irreducible over Q: (a) V2k−2(x) for all k ≥ 3; (b)

1 x Vp(x) for all odd primes p.

Sketch of Proof: Using the explicit expansion Vn(x) = xn −

⌊ n 2 ⌋−1

  • r=0

(−1)r   

⌊ n 2 ⌋

  • k=r+1

n 2k k − 1 r

  xn−2−2r, it can be shown that the polynomials in (a) and (b) are 2-Eisenstein.

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Proposition The following are irreducible over Q: (a) V2k−2(x) for all k ≥ 3; (b)

1 x Vp(x) for all odd primes p.

Sketch of Proof: Using the explicit expansion Vn(x) = xn −

⌊ n 2 ⌋−1

  • r=0

(−1)r   

⌊ n 2 ⌋

  • k=r+1

n 2k k − 1 r

  xn−2−2r, it can be shown that the polynomials in (a) and (b) are 2-Eisenstein. (No other V2k(x) or

1 x V2k+1(x) is Eisenstein).

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Recall: All zeros of Tn(x) lie in the interval (−1, 1).

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Recall: All zeros of Tn(x) lie in the interval (−1, 1). The zeros of Vn(x) are also all real. However:

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Recall: All zeros of Tn(x) lie in the interval (−1, 1). The zeros of Vn(x) are also all real. However: n rn n rn 1 11 31.956928 2 1 12 45.221645 3 1.7320508 13 63.974591 4 2.6180339 14 90.490325 5 3.8286956 15 127.98534 6 5.5174860 16 181.00828 7 7.8875983 17 255.99169 8 11.223990 18 362.03245 9 15.929112 19 511.99536 10 22.571929 20 724.07389 Table 2: The largest zeros rn of Vn(x), 2 ≤ n ≤ 20.

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Recall: All zeros of Tn(x) lie in the interval (−1, 1). The zeros of Vn(x) are also all real. However: n rn 2(n−1)/2 n rn 2(n−1)/2 1 1 11 31.956928 32 2 1 1.4142135 12 45.221645 45.254833 3 1.7320508 2 13 63.974591 64 4 2.6180339 2.8284271 14 90.490325 90.509667 5 3.8286956 4 15 127.98534 128 6 5.5174860 5.6568542 16 181.00828 181.01933 7 7.8875983 8 17 255.99169 256 8 11.223990 11.313708 18 362.03245 362.03867 9 15.929112 16 19 511.99536 512 10 22.571929 22.627416 20 724.07389 724.07734 Table 2: The largest zeros rn of Vn(x), 2 ≤ n ≤ 20.

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Proposition Let n ≥ 2, and ±rn be the largest zeros in absolute value of Vn(x). Then (a) n − 2 zeros of Vn(x) lie in the interval (−1, 1);

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Proposition Let n ≥ 2, and ±rn be the largest zeros in absolute value of Vn(x). Then (a) n − 2 zeros of Vn(x) lie in the interval (−1, 1); (b) ( √ 2)n−1 −

n ( √ 2)n−1 < rn < (

√ 2)n−1.

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Proposition Let n ≥ 2, and ±rn be the largest zeros in absolute value of Vn(x). Then (a) n − 2 zeros of Vn(x) lie in the interval (−1, 1); (b) ( √ 2)n−1 −

n ( √ 2)n−1 < rn < (

√ 2)n−1. Idea of proof: For (a), use (x2 − 1)Vn(x) = xn+2 − Tn(x).

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Proposition Let n ≥ 2, and ±rn be the largest zeros in absolute value of Vn(x). Then (a) n − 2 zeros of Vn(x) lie in the interval (−1, 1); (b) ( √ 2)n−1 −

n ( √ 2)n−1 < rn < (

√ 2)n−1. Idea of proof: For (a), use (x2 − 1)Vn(x) = xn+2 − Tn(x). Consider graph of y = Tn(x); count intersections with y = xn+2.

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Proposition Let n ≥ 2, and ±rn be the largest zeros in absolute value of Vn(x). Then (a) n − 2 zeros of Vn(x) lie in the interval (−1, 1); (b) ( √ 2)n−1 −

n ( √ 2)n−1 < rn < (

√ 2)n−1. Idea of proof: For (a), use (x2 − 1)Vn(x) = xn+2 − Tn(x). Consider graph of y = Tn(x); count intersections with y = xn+2. (b): Evaluate Vn(x) at the two boundary points of the interval. T20(x)

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  • 3. A Related Polynomial

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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  • 3. A Related Polynomial

The Chebyshev polynomials Tn(x) satisfy the (2 × 2 Hankel determinant) identity Tn+1(x)2 − Tn(x)Tn+2(x) = 1 − x2 (n ≥ 0).

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  • 3. A Related Polynomial

The Chebyshev polynomials Tn(x) satisfy the (2 × 2 Hankel determinant) identity Tn+1(x)2 − Tn(x)Tn+2(x) = 1 − x2 (n ≥ 0). How about the analogue for {Vn(x)} ?

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  • 3. A Related Polynomial

The Chebyshev polynomials Tn(x) satisfy the (2 × 2 Hankel determinant) identity Tn+1(x)2 − Tn(x)Tn+2(x) = 1 − x2 (n ≥ 0). How about the analogue for {Vn(x)} ? Define Wn(x) := Vn+1(x)2 − Vn(x)Vn+2(x) (n ≥ 0).

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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  • 3. A Related Polynomial

The Chebyshev polynomials Tn(x) satisfy the (2 × 2 Hankel determinant) identity Tn+1(x)2 − Tn(x)Tn+2(x) = 1 − x2 (n ≥ 0). How about the analogue for {Vn(x)} ? Define Wn(x) := Vn+1(x)2 − Vn(x)Vn+2(x) (n ≥ 0). We’ll see: These polynomials have some interesting properties.

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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n Wn(x) 1 1 x2 + 1 2 2x4 + x2 + 1 3 4x6 + x4 + x2 + 1 4 8x8 + x4 + x2 + 1 5 16x10 − 4x8 + x6 + x4 + x2 + 1 6 32x12 − 16x10 + 2x8 + x6 + x4 + x2 + 1 7 64x14 − 48x12 + 8x10 + x8 + x6 + x4 + x2 + 1 8 128x16 − 128x14 + 32x12 + x8 + x6 + x4 + x2 + 1 9 256x18 − 320x16 + 112x14 − 8x12 + x10 + x8 + x6 +x4 + x2 + 1 10 512x20 − 768x18 + 352x16 − 48x14 + 2x12 + x10 +x8 + x6 + x4 + x2 + 1

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Some properties: Wn(x) = 1 − xn+2Tn(x) 1 − x2 .

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Some properties: Wn(x) = 1 − xn+2Tn(x) 1 − x2 . Compare: Vn(x) = Tn(x) − xn+2 1 − x2 .

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Some properties: Wn(x) = 1 − xn+2Tn(x) 1 − x2 . Compare: Vn(x) = Tn(x) − xn+2 1 − x2 . Recurrence: W0(x) = 1, W1(x) = x2 + 1, and for n ≥ 1, Wn+1(x) = x2 (2Wn(x) − Wn−1(x)) + 1.

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Some properties: Wn(x) = 1 − xn+2Tn(x) 1 − x2 . Compare: Vn(x) = Tn(x) − xn+2 1 − x2 . Recurrence: W0(x) = 1, W1(x) = x2 + 1, and for n ≥ 1, Wn+1(x) = x2 (2Wn(x) − Wn−1(x)) + 1. Generating function: 1 − tx2 + t2x2 (1 − t)(1 − 2tx2 + t2x2) =

  • n=0

Wn(x)tn.

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Let’s look at the table again:

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Let’s look at the table again: n Wn(x) 1 1 x2 + 1 2 2x4 + x2 + 1 3 4x6 + x4 + x2 + 1 4 8x8 + x4 + x2 + 1 5 16x10 − 4x8 + x6 + x4 + x2 + 1 6 32x12 − 16x10 + 2x8 + x6 + x4 + x2 + 1 7 64x14 − 48x12 + 8x10 + x8 + x6 + x4 + x2 + 1 8 128x16 − 128x14 + 32x12 + x8 + x6 + x4 + x2 + 1 9 256x18 − 320x16 + 112x14 − 8x12 + x10 + x8 + x6 +x4 + x2 + 1 10 512x20 − 768x18 + 352x16 − 48x14 + 2x12 + x10 +x8 + x6 + x4 + x2 + 1

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Let’s look at the table again: n Wn(x) 1 1 x2 + 1 2 2x4 + x2 + 1 3 4x6 + x4 + x2 + 1 4 8x8 + x4 + x2 + 1 5 16x10 − 4x8 + x6 + x4 + x2 + 1 6 32x12 − 16x10 + 2x8 + x6 + x4 + x2 + 1 7 64x14 − 48x12 + 8x10 + x8 + x6 + x4 + x2 + 1 8 128x16 − 128x14 + 32x12 + x8 + x6 + x4 + x2 + 1 9 256x18 − 320x16 + 112x14 − 8x12 + x10 + x8 + x6 +x4 + x2 + 1 10 512x20 − 768x18 + 352x16 − 48x14 + 2x12 + x10 +x8 + x6 + x4 + x2 + 1 Do we get anything sensible if we cut the Wn(x) into two halves?

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Define the lower and upper parts, respectively, of Wn(x) by W ℓ

n(x) := ⌊ n+1

2 ⌋

  • j=0

x2j, W u

n (x) :=

1 xn+2

  • Wn(x) − W ℓ

n(x)

  • .

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Define the lower and upper parts, respectively, of Wn(x) by W ℓ

n(x) := ⌊ n+1

2 ⌋

  • j=0

x2j, W u

n (x) :=

1 xn+2

  • Wn(x) − W ℓ

n(x)

  • .

Easy to establish generating functions for both, and with these we get W u

n (x) = 2 ⌊ n−2

2 ⌋

  • k=0

Un−2−2k(x) where the Un(x) are the Chebyshev polynomials of the second kind, which can be defined by the generating function 1 1 − 2tx + t2 =

  • n=0

Un(x)tn.

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Using known identities: W u

2k(x) = 1 − T2k(x)

1 − x2 = 2Uk−1(x)2, W u

2k+1(x) = x − T2k+1(x)

1 − x2 = 2Uk−1(x)Uk(x).

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Using known identities: W u

2k(x) = 1 − T2k(x)

1 − x2 = 2Uk−1(x)2, W u

2k+1(x) = x − T2k+1(x)

1 − x2 = 2Uk−1(x)Uk(x). This, together with the definition of the W ℓ

n(z), gives

Proposition For all n ≥ 1, the zeros (a) of W ℓ

n(z) lie on the unit circle;

(b) of W u

n (z) lie in the open interval (−1, 1).

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Using known identities: W u

2k(x) = 1 − T2k(x)

1 − x2 = 2Uk−1(x)2, W u

2k+1(x) = x − T2k+1(x)

1 − x2 = 2Uk−1(x)Uk(x). This, together with the definition of the W ℓ

n(z), gives

Proposition For all n ≥ 1, the zeros (a) of W ℓ

n(z) lie on the unit circle;

(b) of W u

n (z) lie in the open interval (−1, 1).

What can we say about the zeros of Wn(z) as a whole?

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Plot of the zeros of W50(z) (degree 100):

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Plot of the zeros of W50(z) (degree 100):

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Plot of the zeros of W50(z) (degree 100): Do they lie on (or near) an identifiable curve?

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Proposition The zeros of Wn(z), as n → ∞, lie arbitrarily close to the curve 3r 8 − 8r 6 cos(2θ) + 6r 4 − 1 = 0, z = reiθ, 0 ≤ θ ≤ 2π. (4) Furthermore, they all lie outside the closed region defined by this curve.

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Proposition The zeros of Wn(z), as n → ∞, lie arbitrarily close to the curve 3r 8 − 8r 6 cos(2θ) + 6r 4 − 1 = 0, z = reiθ, 0 ≤ θ ≤ 2π. (4) Furthermore, they all lie outside the closed region defined by this curve.

Figure: The zeros of W50(z) and the curve (4).

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Proof:

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Proof:

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Ingredients in the proof:

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Ingredients in the proof:

  • The identity

Wn(x) = 1 − xn+2Tn(x) 1 − x2 .

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Ingredients in the proof:

  • The identity

Wn(x) = 1 − xn+2Tn(x) 1 − x2 .

  • The Binet-type expression

Tn(x) = 1 2

  • (x −
  • x2 − 1)n + (x +
  • x2 − 1)n

.

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Ingredients in the proof:

  • The identity

Wn(x) = 1 − xn+2Tn(x) 1 − x2 .

  • The Binet-type expression

Tn(x) = 1 2

  • (x −
  • x2 − 1)n + (x +
  • x2 − 1)n

.

  • Concentrate on the larger of the two summands.

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Ingredients in the proof:

  • The identity

Wn(x) = 1 − xn+2Tn(x) 1 − x2 .

  • The Binet-type expression

Tn(x) = 1 2

  • (x −
  • x2 − 1)n + (x +
  • x2 − 1)n

.

  • Concentrate on the larger of the two summands.
  • A chain of tricky estimates.

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An older result of a similar flavour: Let Lp(x), Up(x) be the lower and upper sections of an even-degree polynomial p(x).

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An older result of a similar flavour: Let Lp(x), Up(x) be the lower and upper sections of an even-degree polynomial p(x). Proposition (D. & Stolarsky, 1992) There is a sequence of polynomials {Qn(x)} such that (a) the zeros of Qn(x) lie on the oval |x(x − 1)| = 1/2;

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An older result of a similar flavour: Let Lp(x), Up(x) be the lower and upper sections of an even-degree polynomial p(x). Proposition (D. & Stolarsky, 1992) There is a sequence of polynomials {Qn(x)} such that (a) the zeros of Qn(x) lie on the oval |x(x − 1)| = 1/2; (b) the zeros of LQn(x) lie on the circle of radius 1/ √ 2 centered at the origin;

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An older result of a similar flavour: Let Lp(x), Up(x) be the lower and upper sections of an even-degree polynomial p(x). Proposition (D. & Stolarsky, 1992) There is a sequence of polynomials {Qn(x)} such that (a) the zeros of Qn(x) lie on the oval |x(x − 1)| = 1/2; (b) the zeros of LQn(x) lie on the circle of radius 1/ √ 2 centered at the origin; (c) the zeros of UQn(x) lie on the circle of radius 1/ √ 2 centered at x = 1. Remarks: (i) The centers of the circles in (b), (c) are the foci of the oval (an oval of Cassini) in (a).

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An older result of a similar flavour: Let Lp(x), Up(x) be the lower and upper sections of an even-degree polynomial p(x). Proposition (D. & Stolarsky, 1992) There is a sequence of polynomials {Qn(x)} such that (a) the zeros of Qn(x) lie on the oval |x(x − 1)| = 1/2; (b) the zeros of LQn(x) lie on the circle of radius 1/ √ 2 centered at the origin; (c) the zeros of UQn(x) lie on the circle of radius 1/ √ 2 centered at x = 1. Remarks: (i) The centers of the circles in (b), (c) are the foci of the oval (an oval of Cassini) in (a). (ii) The polynomials can be given explicitly and are also related to Chebyshev polynomials.

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Part II: Zeros and irreducibility of gcd-polynomials

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Joint work with Sinai Robins University of São Paulo, Brazil

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  • 1. Introduction

Some classes of polynomials with special number theoretic sequences as coefficients:

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  • 1. Introduction

Some classes of polynomials with special number theoretic sequences as coefficients:

  • 1. Fekete polynomials:

fp(z) :=

p−1

  • j=0

j p

  • zj

(p prime), where ( a

p) is the Legendre symbol.

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  • 1. Introduction

Some classes of polynomials with special number theoretic sequences as coefficients:

  • 1. Fekete polynomials:

fp(z) :=

p−1

  • j=0

j p

  • zj

(p prime), where ( a

p) is the Legendre symbol.

Conrey, Granville, Poonen, and Soundararajan (2000) showed: For each p, at least half of the zeros of fp(z) lie on the unit circle.

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  • 1. Introduction

Some classes of polynomials with special number theoretic sequences as coefficients:

  • 1. Fekete polynomials:

fp(z) :=

p−1

  • j=0

j p

  • zj

(p prime), where ( a

p) is the Legendre symbol.

Conrey, Granville, Poonen, and Soundararajan (2000) showed: For each p, at least half of the zeros of fp(z) lie on the unit circle. Deep connections with the distribution of primes.

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  • 2. Ramanujan polynomials:

R2k+1(z) :=

k+1

  • j=0
  • B2jB2k+2−2j

(2j)!(2k + 2 − 2j)!

  • z2j,

where Bn is the nth Bernoulli number.

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  • 2. Ramanujan polynomials:

R2k+1(z) :=

k+1

  • j=0
  • B2jB2k+2−2j

(2j)!(2k + 2 − 2j)!

  • z2j,

where Bn is the nth Bernoulli number. Murty, Smyth, and Wang (2011) showed: With the exception of four real zeros, all others zeros lie on the unit circle and have uniform angular distribution.

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  • 2. Ramanujan polynomials:

R2k+1(z) :=

k+1

  • j=0
  • B2jB2k+2−2j

(2j)!(2k + 2 − 2j)!

  • z2j,

where Bn is the nth Bernoulli number. Murty, Smyth, and Wang (2011) showed: With the exception of four real zeros, all others zeros lie on the unit circle and have uniform angular distribution. Applications to the theory of the Riemann zeta function.

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  • 2. Ramanujan polynomials:

R2k+1(z) :=

k+1

  • j=0
  • B2jB2k+2−2j

(2j)!(2k + 2 − 2j)!

  • z2j,

where Bn is the nth Bernoulli number. Murty, Smyth, and Wang (2011) showed: With the exception of four real zeros, all others zeros lie on the unit circle and have uniform angular distribution. Applications to the theory of the Riemann zeta function. Later extended by other authors to similar polynomials (Lalín & Smyth, 2013; Berndt & Straub, 2017).

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  • 3. Dedekind polynomials:

pk(z) :=

k−1

  • j=0

s(j, k)zj, where s(d, c) is the Dedekind sum

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  • 3. Dedekind polynomials:

pk(z) :=

k−1

  • j=0

s(j, k)zj, where s(d, c) is the Dedekind sum defined by s(d, c) =

c

  • j=1

j c dj c

  • ,

with ((x)) denoting the “sawtooth function" ((x)) =

  • 0,

if x ∈ Z, x − [x] − 1

2,

  • therwise.

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  • 3. Dedekind polynomials:

pk(z) :=

k−1

  • j=0

s(j, k)zj, where s(d, c) is the Dedekind sum defined by s(d, c) =

c

  • j=1

j c dj c

  • ,

with ((x)) denoting the “sawtooth function" ((x)) =

  • 0,

if x ∈ Z, x − [x] − 1

2,

  • therwise.

Observation: For each k, most of the zeros of pk(z) lies on the unit circle.

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  • 3. Dedekind polynomials:

pk(z) :=

k−1

  • j=0

s(j, k)zj, where s(d, c) is the Dedekind sum defined by s(d, c) =

c

  • j=1

j c dj c

  • ,

with ((x)) denoting the “sawtooth function" ((x)) =

  • 0,

if x ∈ Z, x − [x] − 1

2,

  • therwise.

Observation: For each k, most of the zeros of pk(z) lies on the unit circle. In an effort to prove this, we were led to studying the following class of polynomials.

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  • 2. GCD Polynomials

What can we say about the polynomials

n

  • j=0

gcd(n, j)zj?

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  • 2. GCD Polynomials

What can we say about the polynomials

n

  • j=0

gcd(n, j)zj? It turns out: A more general class has basically the same properties.

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  • 2. GCD Polynomials

What can we say about the polynomials

n

  • j=0

gcd(n, j)zj? It turns out: A more general class has basically the same

  • properties. For k ≥ 0 and n ≥ 1, let

g(k)

n (z) := n

  • j=0

gcd(n, j)kzj.

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  • 2. GCD Polynomials

What can we say about the polynomials

n

  • j=0

gcd(n, j)zj? It turns out: A more general class has basically the same

  • properties. For k ≥ 0 and n ≥ 1, let

g(k)

n (z) := n

  • j=0

gcd(n, j)kzj. For k = 0, obviously g(0)

n (z) = zn+1 − 1

z − 1 , so all the zeros are roots of unity and thus lie on the unit circle.

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  • 2. GCD Polynomials

What can we say about the polynomials

n

  • j=0

gcd(n, j)zj? It turns out: A more general class has basically the same

  • properties. For k ≥ 0 and n ≥ 1, let

g(k)

n (z) := n

  • j=0

gcd(n, j)kzj. For k = 0, obviously g(0)

n (z) = zn+1 − 1

z − 1 , so all the zeros are roots of unity and thus lie on the unit circle. For n = p − 1 (p a prime), these are cyclotomic polynomials; hence irreducible.

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From now on: Disregard the case k = 0.

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From now on: Disregard the case k = 0. However, we will see: g(k)

n (z) for k ≥ 1 have properties similar to the case k = 0.

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From now on: Disregard the case k = 0. However, we will see: g(k)

n (z) for k ≥ 1 have properties similar to the case k = 0.

Theorem For all k ≥ 1 and all n ≥ 1, all the zeros of g(k)

n (z) lie on the unit

circle and have uniform angular distribution.

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From now on: Disregard the case k = 0. However, we will see: g(k)

n (z) for k ≥ 1 have properties similar to the case k = 0.

Theorem For all k ≥ 1 and all n ≥ 1, all the zeros of g(k)

n (z) lie on the unit

circle and have uniform angular distribution. Idea of proof: Consider g(k)

n (e2πix)

and show it has n real zeros for 0 < x < 1.

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  • 3. Zeros; proof of the Theorem

Since gcd(j, n) = gcd(n − j, n) for 0 ≤ j ≤ n, the g(k)

n (z) are self-inversive (or reciprocal):

g(k)

n (z) = zng(k) n ( 1 z ).

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  • 3. Zeros; proof of the Theorem

Since gcd(j, n) = gcd(n − j, n) for 0 ≤ j ≤ n, the g(k)

n (z) are self-inversive (or reciprocal):

g(k)

n (z) = zng(k) n ( 1 z ).

Set z = e2πix for a real variable x. Then e−πinxg(k)

n (e2πix) = eπinxg(k) n (e−2πix).

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  • 3. Zeros; proof of the Theorem

Since gcd(j, n) = gcd(n − j, n) for 0 ≤ j ≤ n, the g(k)

n (z) are self-inversive (or reciprocal):

g(k)

n (z) = zng(k) n ( 1 z ).

Set z = e2πix for a real variable x. Then e−πinxg(k)

n (e2πix) = eπinxg(k) n (e−2πix).

If we define h(k)

n (x) := e−πinxg(k) n (e2πix),

then h(k)

n (x) = h(k) n (x) for x ∈ R.

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  • 3. Zeros; proof of the Theorem

Since gcd(j, n) = gcd(n − j, n) for 0 ≤ j ≤ n, the g(k)

n (z) are self-inversive (or reciprocal):

g(k)

n (z) = zng(k) n ( 1 z ).

Set z = e2πix for a real variable x. Then e−πinxg(k)

n (e2πix) = eπinxg(k) n (e−2πix).

If we define h(k)

n (x) := e−πinxg(k) n (e2πix),

then h(k)

n (x) = h(k) n (x) for x ∈ R.

Hence h(k)

n (x) is real-valued.

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h(k)

n (x) := e−πinxg(k) n (e2πix).

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h(k)

n (x) := e−πinxg(k) n (e2πix).

For m = 0, 1, . . . , n, consider h(k)

n ( m n ) = e−πimg(k) n (e2πim/n) = (−1)m n

  • j=0

gcd(j, n)ke2πijm/n.

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h(k)

n (x) := e−πinxg(k) n (e2πix).

For m = 0, 1, . . . , n, consider h(k)

n ( m n ) = e−πimg(k) n (e2πim/n) = (−1)m n

  • j=0

gcd(j, n)ke2πijm/n. Last sum is, essentially, discrete Fourier transform of gcd(j, n)k.

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h(k)

n (x) := e−πinxg(k) n (e2πix).

For m = 0, 1, . . . , n, consider h(k)

n ( m n ) = e−πimg(k) n (e2πim/n) = (−1)m n

  • j=0

gcd(j, n)ke2πijm/n. Last sum is, essentially, discrete Fourier transform of gcd(j, n)k. Denote it here by S(k)(m, n) :=

n

  • j=1

gcd(j, n)ke2πijm/n.

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h(k)

n (x) := e−πinxg(k) n (e2πix).

For m = 0, 1, . . . , n, consider h(k)

n ( m n ) = e−πimg(k) n (e2πim/n) = (−1)m n

  • j=0

gcd(j, n)ke2πijm/n. Last sum is, essentially, discrete Fourier transform of gcd(j, n)k. Denote it here by S(k)(m, n) :=

n

  • j=1

gcd(j, n)ke2πijm/n. So we have h(k)

n ( m n ) = (−1)m

S(k)(m, n) + nk .

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h(k)

n ( m n ) = (−1)m

S(k)(m, n) + nk .

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h(k)

n ( m n ) = (−1)m

S(k)(m, n) + nk . Thus, if we can show S(k)(m, n) > 0, (5) then for fixed k and n, h(k)

n ( m n ) is alternating positive and

negative.

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h(k)

n ( m n ) = (−1)m

S(k)(m, n) + nk . Thus, if we can show S(k)(m, n) > 0, (5) then for fixed k and n, h(k)

n ( m n ) is alternating positive and

negative. This means that h(k)

n (x) has n real zeros between the n + 1

points 0, 1/n, 2/n, . . . , 1.

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h(k)

n ( m n ) = (−1)m

S(k)(m, n) + nk . Thus, if we can show S(k)(m, n) > 0, (5) then for fixed k and n, h(k)

n ( m n ) is alternating positive and

negative. This means that h(k)

n (x) has n real zeros between the n + 1

points 0, 1/n, 2/n, . . . , 1. This in turn implies that g(k)

n (z)

  • has all its n zeros on the unit circle, and

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h(k)

n ( m n ) = (−1)m

S(k)(m, n) + nk . Thus, if we can show S(k)(m, n) > 0, (5) then for fixed k and n, h(k)

n ( m n ) is alternating positive and

negative. This means that h(k)

n (x) has n real zeros between the n + 1

points 0, 1/n, 2/n, . . . , 1. This in turn implies that g(k)

n (z)

  • has all its n zeros on the unit circle, and
  • one each in adjacent sectors of angle 2π/n.

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h(k)

n ( m n ) = (−1)m

S(k)(m, n) + nk . Thus, if we can show S(k)(m, n) > 0, (5) then for fixed k and n, h(k)

n ( m n ) is alternating positive and

negative. This means that h(k)

n (x) has n real zeros between the n + 1

points 0, 1/n, 2/n, . . . , 1. This in turn implies that g(k)

n (z)

  • has all its n zeros on the unit circle, and
  • one each in adjacent sectors of angle 2π/n.

This proves Theorem 1, provided we can prove (5).

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h(k)

n ( m n ) = (−1)m

S(k)(m, n) + nk . Thus, if we can show S(k)(m, n) > 0, (5) then for fixed k and n, h(k)

n ( m n ) is alternating positive and

negative. This means that h(k)

n (x) has n real zeros between the n + 1

points 0, 1/n, 2/n, . . . , 1. This in turn implies that g(k)

n (z)

  • has all its n zeros on the unit circle, and
  • one each in adjacent sectors of angle 2π/n.

This proves Theorem 1, provided we can prove (5).

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h(k)

n ( m n ) = (−1)m

S(k)(m, n) + nk .

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h(k)

n ( m n ) = (−1)m

S(k)(m, n) + nk . DFTs have recently been studied for

  • arithmetic, especially multiplicative, functions in general;

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h(k)

n ( m n ) = (−1)m

S(k)(m, n) + nk . DFTs have recently been studied for

  • arithmetic, especially multiplicative, functions in general;
  • the gcd and its powers as special cases.

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h(k)

n ( m n ) = (−1)m

S(k)(m, n) + nk . DFTs have recently been studied for

  • arithmetic, especially multiplicative, functions in general;
  • the gcd and its powers as special cases.

For instance: Theorem (L. Tóth, 2011) For all m ∈ Z and n ∈ N, S(1)(m, n) =

  • d| gcd(m,n)

dϕ( n

d ).

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h(k)

n ( m n ) = (−1)m

S(k)(m, n) + nk . DFTs have recently been studied for

  • arithmetic, especially multiplicative, functions in general;
  • the gcd and its powers as special cases.

For instance: Theorem (L. Tóth, 2011) For all m ∈ Z and n ∈ N, S(1)(m, n) =

  • d| gcd(m,n)

dϕ( n

d ).

This proves our theorem for k = 1.

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h(k)

n ( m n ) = (−1)m

S(k)(m, n) + nk . DFTs have recently been studied for

  • arithmetic, especially multiplicative, functions in general;
  • the gcd and its powers as special cases.

For instance: Theorem (L. Tóth, 2011) For all m ∈ Z and n ∈ N, S(1)(m, n) =

  • d| gcd(m,n)

dϕ( n

d ).

This proves our theorem for k = 1. Can this be extended to general k ≥ 1?

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  • 4. Jordan’s totient function

We need a generalization of Euler’s ϕ-function. Definition Jordan’s totient function is defined by Jk(n) = nk

p|n

  • 1 − 1

pk

  • ,

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  • 4. Jordan’s totient function

We need a generalization of Euler’s ϕ-function. Definition Jordan’s totient function is defined by Jk(n) = nk

p|n

  • 1 − 1

pk

  • ,
  • r equivalently as the number of different sets of k (equal or

distinct) positive integers ≤ n whose gcd is relatively prime to n.

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  • 4. Jordan’s totient function

We need a generalization of Euler’s ϕ-function. Definition Jordan’s totient function is defined by Jk(n) = nk

p|n

  • 1 − 1

pk

  • ,
  • r equivalently as the number of different sets of k (equal or

distinct) positive integers ≤ n whose gcd is relatively prime to n. This equivalence was first established by Camille Jordan in 1870.

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  • 4. Jordan’s totient function

We need a generalization of Euler’s ϕ-function. Definition Jordan’s totient function is defined by Jk(n) = nk

p|n

  • 1 − 1

pk

  • ,
  • r equivalently as the number of different sets of k (equal or

distinct) positive integers ≤ n whose gcd is relatively prime to n. This equivalence was first established by Camille Jordan in 1870. Clearly, J1(n) = ϕ(n).

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Camille Jordan (1838–1922)

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Other properties are similar to those of Euler’s ϕ-function; e.g., mk =

  • d|m

Jk(d).

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Other properties are similar to those of Euler’s ϕ-function; e.g., mk =

  • d|m

Jk(d).

  • W. Schramm (2008) showed;

S(k)(1, n) = Jk(n) (n ≥ 1).

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Other properties are similar to those of Euler’s ϕ-function; e.g., mk =

  • d|m

Jk(d).

  • W. Schramm (2008) showed;

S(k)(1, n) = Jk(n) (n ≥ 1). This can be extended:

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Proposition For all k, n ∈ N and all m ∈ Z we have S(k)(m, n) =

  • d| gcd(m,n)

dJk( n

d ).

In particular, S(k)(m, n) is always a positive integer.

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Proposition For all k, n ∈ N and all m ∈ Z we have S(k)(m, n) =

  • d| gcd(m,n)

dJk( n

d ).

In particular, S(k)(m, n) is always a positive integer. Since the summands on the right are positive, this proves the Theorem.

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Proposition For all k, n ∈ N and all m ∈ Z we have S(k)(m, n) =

  • d| gcd(m,n)

dJk( n

d ).

In particular, S(k)(m, n) is always a positive integer. Since the summands on the right are positive, this proves the Theorem. Compare with Tóth’s result: S(1)(m, n) =

  • d| gcd(m,n)

dϕ( n

d ).

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Proof of Proposition. Using gcd(j, n)k =

  • d| gcd(j,n)

Jk(d), we have S(k)(m, n) =

n

  • j=1
  • ℓ| gcd(n,j)

Jk(ℓ)e2πijm/n =

  • ℓ|n

Jk(ℓ)

n/ℓ

  • j=1

e2πijm/(n/ℓ).

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Proof of Proposition. Using gcd(j, n)k =

  • d| gcd(j,n)

Jk(d), we have S(k)(m, n) =

n

  • j=1
  • ℓ| gcd(n,j)

Jk(ℓ)e2πijm/n =

  • ℓ|n

Jk(ℓ)

n/ℓ

  • j=1

e2πijm/(n/ℓ). Inner sum in the last term is

  • n/ℓ if n/ℓ divides m;
  • 0 otherwise.

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Proof of Proposition. Using gcd(j, n)k =

  • d| gcd(j,n)

Jk(d), we have S(k)(m, n) =

n

  • j=1
  • ℓ| gcd(n,j)

Jk(ℓ)e2πijm/n =

  • ℓ|n

Jk(ℓ)

n/ℓ

  • j=1

e2πijm/(n/ℓ). Inner sum in the last term is

  • n/ℓ if n/ℓ divides m;
  • 0 otherwise.

Hence, setting d = n/ℓ, we get the desired identity.

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An interesting consequence: Recall S(k)(m, n) :=

n

  • j=1

gcd(j, n)ke2πijm/n

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An interesting consequence: Recall S(k)(m, n) :=

n

  • j=1

gcd(j, n)ke2πijm/n and S(k)(m, n) =

  • d| gcd(m,n)

dJk( n

d ).

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An interesting consequence: Recall S(k)(m, n) :=

n

  • j=1

gcd(j, n)ke2πijm/n and S(k)(m, n) =

  • d| gcd(m,n)

dJk( n

d ).

Set m = n; then Corollary For all k, n ∈ N we have

  • d|n

dJk( n

d ) = n

  • j=1

gcd(j, n)k.

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An interesting consequence: Recall S(k)(m, n) :=

n

  • j=1

gcd(j, n)ke2πijm/n and S(k)(m, n) =

  • d| gcd(m,n)

dJk( n

d ).

Set m = n; then Corollary For all k, n ∈ N we have

  • d|n

dJk( n

d ) = n

  • j=1

gcd(j, n)k. This was published by K. Alladi (1975) when he was 19 years

  • ld, and with a different goal in mind.

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  • 5. Irreducibility

Recall: g(k)

n (z) := n

  • j=0

gcd(n, j)kzj.

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  • 5. Irreducibility

Recall: g(k)

n (z) := n

  • j=0

gcd(n, j)kzj. Observation: When n is odd then by symmetry, g(k)

n (−1) = 0,

so z + 1 is always a factor of g(k)

n (z) in that case.

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  • 5. Irreducibility

Recall: g(k)

n (z) := n

  • j=0

gcd(n, j)kzj. Observation: When n is odd then by symmetry, g(k)

n (−1) = 0,

so z + 1 is always a factor of g(k)

n (z) in that case.

However, it appears that this is the only factor. In fact:

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  • 5. Irreducibility

Recall: g(k)

n (z) := n

  • j=0

gcd(n, j)kzj. Observation: When n is odd then by symmetry, g(k)

n (−1) = 0,

so z + 1 is always a factor of g(k)

n (z) in that case.

However, it appears that this is the only factor. In fact: Theorem For α, k ∈ N and odd primes p, g(k)

2α (z)

and g(k)

pα (z)

z + 1 are irreducible over Q.

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  • Proof. (Sketch).

Part 1: We begin with the smallest cases: g(k)

2 (z) = 2k +z+2kz2,

1 z + 1g(k)

3 (z) = 3k +(1−3k)z+3kz2.

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  • Proof. (Sketch).

Part 1: We begin with the smallest cases: g(k)

2 (z) = 2k +z+2kz2,

1 z + 1g(k)

3 (z) = 3k +(1−3k)z+3kz2.

The only self-reciprocal reducible quadratics are z2 ± 2z + 1 and their integer multiples.

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  • Proof. (Sketch).

Part 1: We begin with the smallest cases: g(k)

2 (z) = 2k +z+2kz2,

1 z + 1g(k)

3 (z) = 3k +(1−3k)z+3kz2.

The only self-reciprocal reducible quadratics are z2 ± 2z + 1 and their integer multiples. But none of the polynomials above are of this form. This proves the Theorem for p = 2, p = 3 and α = 1.

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  • Proof. (Sketch).

Part 1: We begin with the smallest cases: g(k)

2 (z) = 2k +z+2kz2,

1 z + 1g(k)

3 (z) = 3k +(1−3k)z+3kz2.

The only self-reciprocal reducible quadratics are z2 ± 2z + 1 and their integer multiples. But none of the polynomials above are of this form. This proves the Theorem for p = 2, p = 3 and α = 1. For the remaining cases, let p ≥ 2 be any prime, and α, k ∈ N. Set g(k)

n (z) =

  • g(k)

n (z)

when n is even,

1 z+1g(k) n (z)

when n is odd.

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Part 2: Assume that g(k)

n (z) is reducible for n ≥ 4.

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Part 2: Assume that g(k)

n (z) is reducible for n ≥ 4.

Then it’s a product of r ≥ 2 irreducible polynomials with integer coefficients.

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Part 2: Assume that g(k)

n (z) is reducible for n ≥ 4.

Then it’s a product of r ≥ 2 irreducible polynomials with integer coefficients. These are themselves self-inversive and thus have even degrees since all their zeros are conjugate pairs of complex numbers with modulus 1.

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Part 2: Assume that g(k)

n (z) is reducible for n ≥ 4.

Then it’s a product of r ≥ 2 irreducible polynomials with integer coefficients. These are themselves self-inversive and thus have even degrees since all their zeros are conjugate pairs of complex numbers with modulus 1. So we can write, for any n ≥ 4, g(k)

n (z) = (a1 + b1z + . . . )(a2 + b2z + . . . ) . . . (ar + brz + . . . )

= a1a2 . . . ar + a1a2 . . . ar  

r

  • j=1

bj aj   z + . . .

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On the other hand, it is clear from the definition that g(k)

pα (z) =

  • pαk + (1 − pαk)z + . . .

when p ≥ 3, pαk + z + . . . when p = 2.

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On the other hand, it is clear from the definition that g(k)

pα (z) =

  • pαk + (1 − pαk)z + . . .

when p ≥ 3, pαk + z + . . . when p = 2. Equating coefficients, we therefore have a1a2 . . . ar = pαk, (6) b1a2 . . . ar + a1b2 . . . ar + . . . + a1a2 . . . br = 1 − [p ≥ 3]pαk, (7)

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On the other hand, it is clear from the definition that g(k)

pα (z) =

  • pαk + (1 − pαk)z + . . .

when p ≥ 3, pαk + z + . . . when p = 2. Equating coefficients, we therefore have a1a2 . . . ar = pαk, (6) b1a2 . . . ar + a1b2 . . . ar + . . . + a1a2 . . . br = 1 − [p ≥ 3]pαk, (7)

  • By (6): the aj can only be powers of p;

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On the other hand, it is clear from the definition that g(k)

pα (z) =

  • pαk + (1 − pαk)z + . . .

when p ≥ 3, pαk + z + . . . when p = 2. Equating coefficients, we therefore have a1a2 . . . ar = pαk, (6) b1a2 . . . ar + a1b2 . . . ar + . . . + a1a2 . . . br = 1 − [p ≥ 3]pαk, (7)

  • By (6): the aj can only be powers of p;
  • by (7): at least one of them has to be 1

(otherwise p would divide LHS of (7) — contradiction.)

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This means: at least one of the r irreducible factors (which are self-inversive) is monic, with all its zeros on the unit circle.

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This means: at least one of the r irreducible factors (which are self-inversive) is monic, with all its zeros on the unit circle. We now use a classical theorem of Kronecker (1857):

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This means: at least one of the r irreducible factors (which are self-inversive) is monic, with all its zeros on the unit circle. We now use a classical theorem of Kronecker (1857): Leopold Kronecker 1823 – 1891

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These polynomials have to be cyclotomic, i.e., of the form Φn(z) =

n

  • j=1

(j,n)=1

  • z − e2πij/n

.

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These polynomials have to be cyclotomic, i.e., of the form Φn(z) =

n

  • j=1

(j,n)=1

  • z − e2πij/n

. Our proof is complete if we can show that this cannot happen.

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These polynomials have to be cyclotomic, i.e., of the form Φn(z) =

n

  • j=1

(j,n)=1

  • z − e2πij/n

. Our proof is complete if we can show that this cannot happen. Proof requires a detailed analysis using resultants of polynomials. We skip this.

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SLIDE 164

Thank you

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Part 3: Given f(z) = amzm + · · · + a1z + a0, g(z) = bnzn + · · · + b1z + b0,

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Part 3: Given f(z) = amzm + · · · + a1z + a0, g(z) = bnzn + · · · + b1z + b0, the resultant of f and g is usually defined by the Sylvester determinant,

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Part 3: Given f(z) = amzm + · · · + a1z + a0, g(z) = bnzn + · · · + b1z + b0, the resultant of f and g is usually defined by the Sylvester determinant, i.e., the determinant of a certain (m + n) × (m + n) matrix which has the coefficients of f and g as entries.

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Part 3: Given f(z) = amzm + · · · + a1z + a0, g(z) = bnzn + · · · + b1z + b0, the resultant of f and g is usually defined by the Sylvester determinant, i.e., the determinant of a certain (m + n) × (m + n) matrix which has the coefficients of f and g as entries. In particular, this means:

  • the resultant of two integer polynomials is a rational integer;

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Part 3: Given f(z) = amzm + · · · + a1z + a0, g(z) = bnzn + · · · + b1z + b0, the resultant of f and g is usually defined by the Sylvester determinant, i.e., the determinant of a certain (m + n) × (m + n) matrix which has the coefficients of f and g as entries. In particular, this means:

  • the resultant of two integer polynomials is a rational integer;
  • reducing the coefficients of f and g modulo some integer will

carry through to their resultant.

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Part 3: Given f(z) = amzm + · · · + a1z + a0, g(z) = bnzn + · · · + b1z + b0, the resultant of f and g is usually defined by the Sylvester determinant, i.e., the determinant of a certain (m + n) × (m + n) matrix which has the coefficients of f and g as entries. In particular, this means:

  • the resultant of two integer polynomials is a rational integer;
  • reducing the coefficients of f and g modulo some integer will

carry through to their resultant. We denote the resultant of f and g by Res(f, g) if there is no ambiguity as to the variable z.

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Suppose that the zeros of f and g are α1, . . . , αm and β1, . . . , βn, respectively. Then the most important property is Res(f, g) = an

mbm n m

  • i=1

n

  • j=1

(αi − βj), an alternative definition.

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Suppose that the zeros of f and g are α1, . . . , αm and β1, . . . , βn, respectively. Then the most important property is Res(f, g) = an

mbm n m

  • i=1

n

  • j=1

(αi − βj), an alternative definition. Some consequences: Res(f, g) = an

m m

  • i=1

g(αi), Res(f, g) = (−1)nmRes(g, f), Res(f, g1g2) = Res(f, g1)Res(f, g2).

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Suppose that the zeros of f and g are α1, . . . , αm and β1, . . . , βn, respectively. Then the most important property is Res(f, g) = an

mbm n m

  • i=1

n

  • j=1

(αi − βj), an alternative definition. Some consequences: Res(f, g) = an

m m

  • i=1

g(αi), Res(f, g) = (−1)nmRes(g, f), Res(f, g1g2) = Res(f, g1)Res(f, g2). The first identity shows that Res(f, g) = 0 iff f and g have a factor in common.

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Important for us: Lemma (Apostol (1970)) For m > n > 1 we have Res(Φm(z), Φn(z)) =

  • pϕ(n)

if m

n is a power of a prime p,

1

  • therwise.

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Important for us: Lemma (Apostol (1970)) For m > n > 1 we have Res(Φm(z), Φn(z)) =

  • pϕ(n)

if m

n is a power of a prime p,

1

  • therwise.

With this we will prove Lemma Let p be any prime and α, k be positive integers. Then Res(g(k)

pα (z), Φn(z)) = 0

for any n ≥ 3.

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Important for us: Lemma (Apostol (1970)) For m > n > 1 we have Res(Φm(z), Φn(z)) =

  • pϕ(n)

if m

n is a power of a prime p,

1

  • therwise.

With this we will prove Lemma Let p be any prime and α, k be positive integers. Then Res(g(k)

pα (z), Φn(z)) = 0

for any n ≥ 3. Hence no cyclotomic polynomial of degree ≥ 2 can divide any g(k)

pα (z).

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Important for us: Lemma (Apostol (1970)) For m > n > 1 we have Res(Φm(z), Φn(z)) =

  • pϕ(n)

if m

n is a power of a prime p,

1

  • therwise.

With this we will prove Lemma Let p be any prime and α, k be positive integers. Then Res(g(k)

pα (z), Φn(z)) = 0

for any n ≥ 3. Hence no cyclotomic polynomial of degree ≥ 2 can divide any g(k)

pα (z). This completes the proof of the Theorem.

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Part 4: Proof of the Lemma. Case 1: p is odd. We’ll prove the Lemma by showing: Resultant cannot be simultaneously 0 (mod 2) and 0 (mod p).

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Part 4: Proof of the Lemma. Case 1: p is odd. We’ll prove the Lemma by showing: Resultant cannot be simultaneously 0 (mod 2) and 0 (mod p). (a) The gcd’s are all odd, and therefore g(k)

pα (z) ≡ 1 + z + · · · + zpα =

  • d|pα+1

d=1

Φd(z) (mod 2),

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Part 4: Proof of the Lemma. Case 1: p is odd. We’ll prove the Lemma by showing: Resultant cannot be simultaneously 0 (mod 2) and 0 (mod p). (a) The gcd’s are all odd, and therefore g(k)

pα (z) ≡ 1 + z + · · · + zpα =

  • d|pα+1

d=1

Φd(z) (mod 2), so by multiplicativity of resultants, Res(g(k)

pα (z), Φn(z)) ≡

  • d|pα+1

d=1

Res(Φd(z), Φn(z)) (mod 2).

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Part 4: Proof of the Lemma. Case 1: p is odd. We’ll prove the Lemma by showing: Resultant cannot be simultaneously 0 (mod 2) and 0 (mod p). (a) The gcd’s are all odd, and therefore g(k)

pα (z) ≡ 1 + z + · · · + zpα =

  • d|pα+1

d=1

Φd(z) (mod 2), so by multiplicativity of resultants, Res(g(k)

pα (z), Φn(z)) ≡

  • d|pα+1

d=1

Res(Φd(z), Φn(z)) (mod 2). By Apostol’s result and commutativity (up to sign) of resultants: Res(g(k)

pα (z), Φn(z)) ≡ 1

(mod 2)

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Part 4: Proof of the Lemma. Case 1: p is odd. We’ll prove the Lemma by showing: Resultant cannot be simultaneously 0 (mod 2) and 0 (mod p). (a) The gcd’s are all odd, and therefore g(k)

pα (z) ≡ 1 + z + · · · + zpα =

  • d|pα+1

d=1

Φd(z) (mod 2), so by multiplicativity of resultants, Res(g(k)

pα (z), Φn(z)) ≡

  • d|pα+1

d=1

Res(Φd(z), Φn(z)) (mod 2). By Apostol’s result and commutativity (up to sign) of resultants: Res(g(k)

pα (z), Φn(z)) ≡ 1

(mod 2) unless n = 2jd for some nonzero j and d > 1 where d | pα + 1 (j may be positive or negative).

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(b) On the other hand, g(k)

pα (z) ≡ (z + · · · + zp−1) + (zp+1 + · · · + z2p−1)

+ · · · + (zpα−p+1 + · · · + zpα−1) (mod p) = z

  • 1 + z + · · · + zp−2

1 + zp + · · · + z(pα−1−1)p = z · zp−1 − 1 z − 1 · zpα − 1 zp − 1 = z

  • d|p−1

d=1

Φd(z)

α

  • j=2

Φpj(z).

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By properties of resultants, Res(z, Φn(z)) = 1 for n ≥ 3, and so Res(g(k)

pα (z), Φn(z)) ≡ ±

  • d|p−1

d=1

Res(Φd(z), Φn(z)) ×

α

  • j=2

Res(Φpj(z), Φn(z)) (mod p).

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By properties of resultants, Res(z, Φn(z)) = 1 for n ≥ 3, and so Res(g(k)

pα (z), Φn(z)) ≡ ±

  • d|p−1

d=1

Res(Φd(z), Φn(z)) ×

α

  • j=2

Res(Φpj(z), Φn(z)) (mod p). By Apostol’s result: Res(g(k)

pα (z), Φn(z)) ≡ ±1

(mod p)

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By properties of resultants, Res(z, Φn(z)) = 1 for n ≥ 3, and so Res(g(k)

pα (z), Φn(z)) ≡ ±

  • d|p−1

d=1

Res(Φd(z), Φn(z)) ×

α

  • j=2

Res(Φpj(z), Φn(z)) (mod p). By Apostol’s result: Res(g(k)

pα (z), Φn(z)) ≡ ±1

(mod p) unless n = pℓd for some ℓ ≥ 1 and d ≥ 1 with d | p − 1.

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Combining the conditions: The above congruences (mod 2) and (mod p) fail simultaneously only if 2jd1 = pℓd2, where d1 | pα + 1, d2 | p − 1.

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Combining the conditions: The above congruences (mod 2) and (mod p) fail simultaneously only if 2jd1 = pℓd2, where d1 | pα + 1, d2 | p − 1. Impossible for an odd prime p since ℓ ≥ 1 and p ∤ d1.

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Combining the conditions: The above congruences (mod 2) and (mod p) fail simultaneously only if 2jd1 = pℓd2, where d1 | pα + 1, d2 | p − 1. Impossible for an odd prime p since ℓ ≥ 1 and p ∤ d1. Hence at least one of the congruences holds, which means that the resultant is nonzero.

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Combining the conditions: The above congruences (mod 2) and (mod p) fail simultaneously only if 2jd1 = pℓd2, where d1 | pα + 1, d2 | p − 1. Impossible for an odd prime p since ℓ ≥ 1 and p ∤ d1. Hence at least one of the congruences holds, which means that the resultant is nonzero. Case 2: p = 2 — Similar.

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Combining the conditions: The above congruences (mod 2) and (mod p) fail simultaneously only if 2jd1 = pℓd2, where d1 | pα + 1, d2 | p − 1. Impossible for an odd prime p since ℓ ≥ 1 and p ∤ d1. Hence at least one of the congruences holds, which means that the resultant is nonzero. Case 2: p = 2 — Similar. This completes the proof of the resultant lemma, and thus of the irreducibility theorem.

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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  • 6. Further Remarks
  • 1. Irreducibility proof fails when n has ≥ 2 prime divisors.

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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  • 6. Further Remarks
  • 1. Irreducibility proof fails when n has ≥ 2 prime divisors.

Still, we propose Conjecture For any integers n ≥ 2 and k ≥ 1, the polynomial g(k)

n (z) is

irreducible, apart from the obvious factor z + 1 when n is odd.

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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  • 6. Further Remarks
  • 1. Irreducibility proof fails when n has ≥ 2 prime divisors.

Still, we propose Conjecture For any integers n ≥ 2 and k ≥ 1, the polynomial g(k)

n (z) is

irreducible, apart from the obvious factor z + 1 when n is odd. Verified by computation for all n ≤ 1000 and 1 ≤ k ≤ 10.

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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  • 6. Further Remarks
  • 1. Irreducibility proof fails when n has ≥ 2 prime divisors.

Still, we propose Conjecture For any integers n ≥ 2 and k ≥ 1, the polynomial g(k)

n (z) is

irreducible, apart from the obvious factor z + 1 when n is odd. Verified by computation for all n ≤ 1000 and 1 ≤ k ≤ 10.

  • 2. Our results give a large supply of algebraic numbers on the

unit circle that are not roots of unity.

Karl Dilcher Zeros and irreducibility of some classes of special polynomials

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Thank you

Karl Dilcher Zeros and irreducibility of some classes of special polynomials