The Beauty of Roots. John Baez, Dan Christensen and Sam Derbyshire with lots of help from Greg Egan
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1 The Beauty of Roots John Baez, Dan Christensen and Sam Derbyshire with lots of help from Greg Egan
2 Definition. A Littlewood polynomial is a polynomial whose coefficients are all 1 and -1. Let s draw all roots of all Littlewood polynomials!
3
4 Certain regions seem particularly interesting:
5 The hole at 1: Note the line along the real axis: more Littlewood polynomials have real roots than nearly real roots.
6 The holes at i and e iπ/4 :
7 This plot is centered at the point 54 :
8 This is centered at the point 45 i:
9 This is centered at 1 2 e i/5 :
10 Can we understand these pictures? Let D = {z C : z is the root of some Littlewood polynomial}
11 Can we understand these pictures? Let D = {z C : z is the root of some Littlewood polynomial} Theorem 1. D {1/2 < z < 2}
12 Can we understand these pictures? Let D = {z C : z is the root of some Littlewood polynomial} Theorem 1. D {1/2 < z < 2} Proof. Suppose z is a root of a Littlewood polynomial. Then If z < 1 then 1 = ±z ± z 2 ± ± z n 1 z + z z n < z 1 z so z > 1/2. Since z is the root of a Littlewood polynomial if and only if z 1 is, D is contained in the annulus 1 2 < z < 2.
13 Theorem 2. {2 1/4 z 2 1/4 } D.
14 Theorem 2. {2 1/4 z 2 1/4 } D. Proof. This was proved by Thierry Bousch in We won t prove it here.
15 The closure D is easier to study than D. For example: Theorem 3. D is connected.
16 The closure D is easier to study than D. For example: Theorem 3. D is connected. Proof. This was proved by Bousch in Let s sketch how the proof works. It s enough to show D { z < r} is connected where r is slightly less than 1.
17 Definition. A Littlewood series is a power series all of whose coefficients are 1 or 1.
18 Definition. A Littlewood series is a power series all of whose coefficients are 1 or 1. Littlewood series converge for z < 1.
19 Definition. A Littlewood series is a power series all of whose coefficients are 1 or 1. Littlewood series converge for z < 1. Lemma 1. A point z C with z < 1 lies in D if and only if some Littlewood series vanishes at this point.
20 A Littlewood polynomial is not a Littlewood series! But any Littlewood polynomial, say P(z) = a a d z d gives a Littlewood series having the same roots with z < 1: P(z) 1 z d+1 = a 0+ +a d z d +a 0 z d+1 + +a d z 2d+1 +a 0 z 2d+2 +
21 A Littlewood polynomial is not a Littlewood series! But any Littlewood polynomial, say P(z) = a a d z d gives a Littlewood series having the same roots with z < 1: P(z) 1 z d+1 = a 0+ +a d z d +a 0 z d+1 + +a d z 2d+1 +a 0 z 2d+2 + Thus D R, where R is the set of roots of Littlewood series.
22 A Littlewood polynomial is not a Littlewood series! But any Littlewood polynomial, say P(z) = a a d z d gives a Littlewood series having the same roots with z < 1: P(z) 1 z d+1 = a 0+ +a d z d +a 0 z d+1 + +a d z 2d+1 +a 0 z 2d+2 + Thus D R, where R is the set of roots of Littlewood series. Our job is to show D = R.
23 A Littlewood polynomial is not a Littlewood series! But any Littlewood polynomial, say P(z) = a a d z d gives a Littlewood series having the same roots with z < 1: P(z) 1 z d+1 = a 0+ +a d z d +a 0 z d+1 + +a d z 2d+1 +a 0 z 2d+2 + Thus D R, where R is the set of roots of Littlewood series. Our job is to show D = R. To do this, let s show that R is closed and D is dense in R.
24 Let L be the set of Littlewood series. L = { 1, 1} N, so with the product topology it s homeomorphic to the Cantor set.
25 Let L be the set of Littlewood series. L = { 1, 1} N, so with the product topology it s homeomorphic to the Cantor set. Choose 0 < r < 1. Let M be the space of finite multisets of points with z r, modulo the equivalence relation generated by S S {p} when p = r.
26 Let L be the set of Littlewood series. L = { 1, 1} N, so with the product topology it s homeomorphic to the Cantor set. Choose 0 < r < 1. Let M be the space of finite multisets of points with z r, modulo the equivalence relation generated by S S {p} when p = r. Lemma 2. Any Littlewood series has finitely many roots in the disc { z r}. The map ρ : L M sending a Littlewood series to its multiset of roots in this disc is continuous.
27 Let L be the set of Littlewood series. L = { 1, 1} N, so with the product topology it s homeomorphic to the Cantor set. Choose 0 < r < 1. Let M be the space of finite multisets of points with z r, modulo the equivalence relation generated by S S {p} when p = r. Lemma 2. Any Littlewood series has finitely many roots in the disc { z r}. The map ρ : L M sending a Littlewood series to its multiset of roots in this disc is continuous. Since L is compact, the image of ρ is closed. From this we can show that R, the set of roots of Littlewood series, is closed.
28 Let L be the set of Littlewood series. L = { 1, 1} N, so with the product topology it s homeomorphic to the Cantor set. Choose 0 < r < 1. Let M be the space of finite multisets of points with z r, modulo the equivalence relation generated by S S {p} when p = r. Lemma 2. Any Littlewood series has finitely many roots in the disc { z r}. The map ρ : L M sending a Littlewood series to its multiset of roots in this disc is continuous. Since L is compact, the image of ρ is closed. From this we can show that R, the set of roots of Littlewood series, is closed. Since Littlewood polynomials are densely included in L and ρ is continuous, we can show that D, the set of roots of Littlewood polynomials, is dense in R.
29 Let L be the set of Littlewood series. L = { 1, 1} N, so with the product topology it s homeomorphic to the Cantor set. Choose 0 < r < 1. Let M be the space of finite multisets of points with z r, modulo the equivalence relation generated by S S {p} when p = r. Lemma 2. Any Littlewood series has finitely many roots in the disc { z r}. The map ρ : L M sending a Littlewood series to its multiset of roots in this disc is continuous. Since L is compact, the image of ρ is closed. From this we can show that R, the set of roots of Littlewood series, is closed. Since Littlewood polynomials are densely included in L and ρ is continuous, we can show that D, the set of roots of Littlewood polynomials, is dense in R. It follows that D = R.
30 Lemma 3. The set R is connected.
31 Lemma 3. The set R is connected. Proof. It s enough to show R { z < r} is connected, where 2 1/4 < r < 1.
32 Lemma 3. The set R is connected. Proof. It s enough to show R { z < r} is connected, where 2 1/4 < r < 1. Suppose U R { z < r} is closed and open in the relative topology. We want to show U is empty.
33 Lemma 3. The set R is connected. Proof. It s enough to show R { z < r} is connected, where 2 1/4 < r < 1. Suppose U R { z < r} is closed and open in the relative topology. We want to show U is empty. Let L U be the set of Littlewood series with a root in U. L U is a closed and open subset of L. Thus, we can determine whether f L lies in L U by looking at its first d terms: f (z) = a 0 + a 1 z + + a d 1 z d 1 + a d z d + for some d. Choose the smallest d with this property.
34 Lemma 3. The set R is connected. Proof. It s enough to show R { z < r} is connected, where 2 1/4 < r < 1. Suppose U R { z < r} is closed and open in the relative topology. We want to show U is empty. Let L U be the set of Littlewood series with a root in U. L U is a closed and open subset of L. Thus, we can determine whether f L lies in L U by looking at its first d terms: f (z) = a 0 + a 1 z + + a d 1 z d 1 + a d z d + for some d. Choose the smallest d with this property. We will get a contradiction if U, and thus L U, is nonempty! We ll show d 1 has the same property.
35 Suppose f L U : f (z) = a 0 + a 1 z + + a d 1 z d 1 + a d z d +
36 Suppose f L U : f (z) = a 0 + a 1 z + + a d 1 z d 1 + a d z d + Suppose g has the same first d 1 terms: g(z) = a 0 + a 1 z + + a d 1 z d 1 + b d z d + We ll show g L U too.
37 Suppose f L U : f (z) = a 0 + a 1 z + + a d 1 z d 1 + a d z d + Suppose g has the same first d 1 terms: g(z) = a 0 + a 1 z + + a d 1 z d 1 + b d z d + We ll show g L U too. Multiplying by 1 if necessary, we can assume a 0 = 1: f (z) = 1 + a 1 z + + a d 1 z d 1 + a d z d +
38 Suppose f L U : f (z) = a 0 + a 1 z + + a d 1 z d 1 + a d z d + Suppose g has the same first d 1 terms: g(z) = a 0 + a 1 z + + a d 1 z d 1 + b d z d + We ll show g L U too. Multiplying by 1 if necessary, we can assume a 0 = 1: f (z) = 1 + a 1 z + + a d 1 z d 1 + a d z d + There are two cases, a d = 1 and a d = 1. We ll just do the first, since the second is similar. So: f (z) = 1 + a 1 z + + a d 1 z d 1 + z d +
39 f (z) = 1 + a 1 z + + a d 1 z d 1 + z d + a d+1 z d+1 + has a root in U, and we want to show the same for any g L with the same first d 1 terms, say g(z) = 1 + a 1 z + + a d 1 z d 1 + b d z d + b d+1 z d+1 +
40 f (z) = 1 + a 1 z + + a d 1 z d 1 + z d + a d+1 z d+1 + has a root in U, and we want to show the same for any g L with the same first d 1 terms, say g(z) = 1 + a 1 z + + a d 1 z d 1 + b d z d + b d+1 z d+1 + We may assume g differs from f in its dth term: g(z) = 1 + a 1 z + + a d 1 z d 1 z d + b d+1 z d +
41 f (z) = 1 + a 1 z + + a d 1 z d 1 + z d + a d+1 z d+1 + has a root in U, and we want to show the same for any g L with the same first d 1 terms, say g(z) = 1 + a 1 z + + a d 1 z d 1 + b d z d + b d+1 z d+1 + We may assume g differs from f in its dth term: g(z) = 1 + a 1 z + + a d 1 z d 1 z d + b d+1 z d + It suffices to show that g has a root in U: ( g(z) = 1 + a 1 z + + a d 1 z d 1) / (1 + z d) since this has the same first d terms as g.
42 f (z) = 1 + a 1 z + + a d 1 z d 1 + z d + a d+1 z d+1 + has a root in U, and we want to show the same for any g L with the same first d 1 terms, say g(z) = 1 + a 1 z + + a d 1 z d 1 + b d z d + b d+1 z d+1 + We may assume g differs from f in its dth term: g(z) = 1 + a 1 z + + a d 1 z d 1 z d + b d+1 z d + It suffices to show that g has a root in U: ( g(z) = 1 + a 1 z + + a d 1 z d 1) / (1 + z d) since this has the same first d terms as g. Since f has a root in U, so does f (z) = (1 + a 1 z + + a d 1 z d 1) / (1 z d) since this has the same first d terms as f.
43 ( g(z) = 1 + a 1 z + + a d 1 z d 1) / (1 + z d) and f (z) = (1 + a 1 z + + a d 1 z d 1) / (1 z d) so ( ) 1 z d g(z) = f (z) 1 + z d Since f has a root in U, so does g. QED!
44 Here is the key to understanding the beautiful patterns in the set D. Define two functions from the complex plane to itself, depending on a complex parameter q: f +q (z) = 1 + qz f q (z) = 1 qz
45 Here is the key to understanding the beautiful patterns in the set D. Define two functions from the complex plane to itself, depending on a complex parameter q: f +q (z) = 1 + qz f q (z) = 1 qz When q < 1 these are both contraction mappings, so by Hutchinson s theorem on iterated function systems there s a unique nonempty compact set D q C with We call this set a dragon. D q = f +q (D q ) f q (D q )
46 Here is the key to understanding the beautiful patterns in the set D. Define two functions from the complex plane to itself, depending on a complex parameter q: f +q (z) = 1 + qz f q (z) = 1 qz When q < 1 these are both contraction mappings, so by Hutchinson s theorem on iterated function systems there s a unique nonempty compact set D q C with We call this set a dragon. D q = f +q (D q ) f q (D q ) Here s the marvelous fact: the portion of D in a small neighborhood of q C tends to look like D q.
47 For example, here s the set D near q = i:
48 And here s the dragon D q for q = i:
49 Let s zoom in on the set of roots of Littlewood polynomials of degree 20. When we zoom in enough, we ll see it s a discrete set!
50 Let s zoom in on the set of roots of Littlewood polynomials of degree 20. When we zoom in enough, we ll see it s a discrete set! Then we ll increase the degree and see how the set fills in.
51 Let s zoom in on the set of roots of Littlewood polynomials of degree 20. When we zoom in enough, we ll see it s a discrete set! Then we ll increase the degree and see how the set fills in. Then we ll switch to a zoomed-in view of the corresponding dragon, and then zoom out.
52 center i, height.62508, degree 20
53 center i, height.31304, degree 20
54 center i, height.15652, degree 20
55 center i, height.07826, degree 20
56 center i, height.03913, degree 20
57 center i, height , degree 20
58 center i, height , degree 20
59 center i, height , degree 20
60 center i, height , degree 20
61 center i, height , degree 21
62 center i, height , degree 22
63 center i, height , degree 23
64 center i, height , degree 24
65 center i, height , degree 25
66 center i, height , degree 26
67 center i, height , degree 27
68 dragon for i, height
69 dragon for i, height
70 dragon for i, height
71 dragon for i, height
72 dragon for i, height.03913
73 dragon for i, height.07826
74 dragon i, height.15652
75 dragon for i, height.31304
76 dragon for i, height.62508
77 The set D is the set of roots of all Littlewood series. The set D q is the set of values of all Littlewood series at the point q: Theorem 4. For q < 1, D q = {f (q): f L}.
78 The set D is the set of roots of all Littlewood series. The set D q is the set of values of all Littlewood series at the point q: Theorem 4. For q < 1, D q = {f (q): f L}. This is easy to show.
79 The set D is the set of roots of all Littlewood series. The set D q is the set of values of all Littlewood series at the point q: Theorem 4. For q < 1, D q = {f (q): f L}. This is easy to show. But why does D near q tend to resemble D q?
80 Each Littlewood series f maps q to a point f (q) D q. For p near q, f (p) f (q) + f (q) (p q)
81 Each Littlewood series f maps q to a point f (q) D q. For p near q, f (p) f (q) + f (q) (p q) Thus, we expect f (p) = 0 when p q f (q) f (q) If this reasoning is good, this formula approximately gives points p D near q from points f (q) D q.
82 Each Littlewood series f maps q to a point f (q) D q. For p near q, f (p) f (q) + f (q) (p q) Thus, we expect f (p) = 0 when p q f (q) f (q) If this reasoning is good, this formula approximately gives points p D near q from points f (q) D q. So, we should expect that near q, the set D will approximately look like a somewhat distorted copy of the dragon D q, or sometimes a union of such copies.
83 Each Littlewood series f maps q to a point f (q) D q. For p near q, f (p) f (q) + f (q) (p q) Thus, we expect f (p) = 0 when p q f (q) f (q) If this reasoning is good, this formula approximately gives points p D near q from points f (q) D q. So, we should expect that near q, the set D will approximately look like a somewhat distorted copy of the dragon D q, or sometimes a union of such copies. We re working on stating this precisely and proving it.
84 D near q = 45 i:
85 union of distorted dragons for q = 45 i:
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