Roots of Polynomials Under Repeated Differentiation Stefan - - PowerPoint PPT Presentation

Roots of Polynomials Under Repeated Differentiation

Stefan Steinerberger

UCLA/Caltech, October 2020

Outline of the Talk

• 1. Roots of Polynomials

Outline of the Talk

• 1. Roots of Polynomials
• 2. A Nonlinear PDE

Outline of the Talk

• 1. Roots of Polynomials
• 2. A Nonlinear PDE
• 3. Hermite Polynomials out of thin air (with Jeremy Hoskins)

Outline of the Talk

• 1. Roots of Polynomials
• 2. A Nonlinear PDE
• 3. Hermite Polynomials out of thin air (with Jeremy Hoskins)
• 4. Free Probability, Random Matrices, ...

Outline of the Talk

• 1. Roots of Polynomials
• 2. A Nonlinear PDE
• 3. Hermite Polynomials out of thin air (with Jeremy Hoskins)
• 4. Free Probability, Random Matrices, ...
• 5. A Nonlinear PDE in the Complex Plane (with Sean O’Rourke)

pn will be a polynomial of degree n having n distinct roots.

pn will be a polynomial of degree n having n distinct roots. The Gauss-Lucas theorem (1830s). The roots of p′

n are

contained in the convex hull of the roots of pn.

pn will be a polynomial of degree n having n distinct roots. The Gauss-Lucas theorem (1830s). The roots of p′

n are

contained in the convex hull of the roots of pn.

• Proof. The ‘electrostatic interpretation’:

p′

n(z)

pn(z) =

n

• k=1

1 z − zk .

pn will be a polynomial of degree n having n distinct roots. The Gauss-Lucas theorem (1830s). The roots of p′

n are

contained in the convex hull of the roots of pn.

• Proof. The ‘electrostatic interpretation’:

p′

n(z)

pn(z) =

n

• k=1

1 z − zk . If you are outside the convex hull, the charges ‘push you away’.

p′

n(z)

pn(z) =

n

• k=1

1 z − zk .

p′

n(z)

pn(z) =

n

• k=1

1 z − zk .

Suppose µ is a probability measure on C and suppose pn(z) =

n

• k=1

(z − zk), where z1, . . . , zn ∼ µ are i.i.d. random variables distributed according to µ.

Suppose µ is a probability measure on C and suppose pn(z) =

n

• k=1

(z − zk), where z1, . . . , zn ∼ µ are i.i.d. random variables distributed according to µ. What is the distribution of the critical points of pn (the roots of p′

n)?

Suppose µ is a probability measure on C and suppose pn(z) =

n

• k=1

(z − zk), where z1, . . . , zn ∼ µ are i.i.d. random variables distributed according to µ. What is the distribution of the critical points of pn (the roots of p′

n)?

Conjecture (Rivin-Pemantle), Theorem (Kabluchko, 2015)

The critical points are also distributed according to µ.

Suppose µ is a probability measure on C and suppose pn(z) =

n

• k=1

(z − zk), where z1, . . . , zn ∼ µ are i.i.d. random variables distributed according to µ. What is the distribution of the critical points of pn (the roots of p′

n)?

Conjecture (Rivin-Pemantle), Theorem (Kabluchko, 2015)

The critical points are also distributed according to µ. Note: The critical points are w1, . . . , wn−1 ∈ C. The statement really says that 1 n − 1

n−1

• k=1

δwk ⇀ µ.

Here’s a heuristic why this is not too surprising.

Here’s a heuristic why this is not too surprising. Suppose µ is absolutely continuous and compactly supported. Then the roots look a bit like this

Here’s a heuristic why this is not too surprising. Suppose µ is absolutely continuous and compactly supported. Then the roots look a bit like this roots of pn

Here’s a heuristic why this is not too surprising. Suppose µ is absolutely continuous and compactly supported typically ∼ n−1/2

typically ∼ n−1/2 Every root of the derivative satisfies p′

n(z) = 0 which means

1 z − zℓ = −

n

• k=1

k=ℓ

1 z − zk .

typically ∼ n−1/2 Every root of the derivative satisfies p′

n(z) = 0 which means

1 z − zℓ = −

n

• k=1

k=ℓ

1 z − zk . The right-hand side is typically size ∼ n.

typically ∼ n−1/2 Every root of the derivative satisfies p′

n(z) = 0 which means

1 z − zℓ = −

n

• k=1

k=ℓ

1 z − zk . The right-hand side is typically size ∼ n. So the root of the derivative has to be distance ∼ n−1 from one of the existing roots

typically ∼ n−1/2 Every root of the derivative satisfies p′

n(z) = 0 which means

1 z − zℓ = −

n

• k=1

k=ℓ

1 z − zk . The right-hand side is typically size ∼ n. So the root of the derivative has to be distance ∼ n−1 from one of the existing roots and the existing roots are ∼ n−1/2 separated, so no two of them are very close.

typically ∼ n−1/2 What this argument tells us is roughly the following:

typically ∼ n−1/2 What this argument tells us is roughly the following: if µ is a sufficiently nice measure, then for each (random) root pn(zk) = 0, we would expect that there is a critical point p′

n(z) = 0 that is at

most distance ∼ n−1 nearby.

typically ∼ n−1/2 What this argument tells us is roughly the following: if µ is a sufficiently nice measure, then for each (random) root pn(zk) = 0, we would expect that there is a critical point p′

n(z) = 0 that is at

most distance ∼ n−1 nearby. This is roughly correct and there are recent papers by Sean O’Rourke and Noah Williams in this direction.

picture from O’Rourke and Williams (2018)

Theorem (O’Rourke and Williams)

Under reasonable assumptions on the measure W1(µn, µ′

n) (log n)10

n .

picture from O’Rourke and Williams (2018)

?

picture from O’Rourke and Williams (2018) n

k=1 k=ℓ

1 z−zk vanishes

In fact, the bijective relationship has to fail somewhere: there are n roots and n − 1 critical points.

In fact, the bijective relationship has to fail somewhere: there are n roots and n − 1 critical points. The unpaired root is frequently close to the root of V (z) =

n

• k=1

1 z − zk .

picture from O’Rourke and Williams (2018) This looks almost like a flow of particles captured at nearby times.

Let’s now return to the one-dimensional setting.

Let’s now return to the one-dimensional setting. Things are slightly different here pn(x) =

n

• k=1

(x − xk).

Let’s now return to the one-dimensional setting. Things are slightly different here pn(x) =

n

• k=1

(x − xk). Between any two roots there is a maximum or a minimum, thus a root of p′

n.

Let’s now return to the one-dimensional setting. Things are slightly different here pn(x) =

n

• k=1

(x − xk). Between any two roots there is a maximum or a minimum, thus a root of p′

• n. Moreover, there are n − 1 intervals between the n

roots, so each interval has exactly one root. Thus the roots of pn and the roots of p′

n interlace.

Let’s now return to the one-dimensional setting. Things are slightly different here pn(x) =

n

• k=1

(x − xk). Between any two roots there is a maximum or a minimum, thus a root of p′

• n. Moreover, there are n − 1 intervals between the n

roots, so each interval has exactly one root. Thus the roots of pn and the roots of p′

n interlace.

Let’s now return to the one-dimensional setting. Things are slightly different here pn(x) =

n

• k=1

(x − xk). Between any two roots there is a maximum or a minimum, thus a root of p′

• n. Moreover, there are n − 1 intervals between the n

roots, so each interval has exactly one root. Thus the roots of pn and the roots of p′

n interlace.

Conjecture (Rivin-Pemantle), Theorem (Kabluchko, 2015)

The critical points are also distributed according to µ.

Conjecture (Rivin-Pemantle), Theorem (Kabluchko, 2015)

The critical points are also distributed according to µ.

Fact for real-rooted polynomials

The roots of p

(

n log n )

n

also distributed according to µ as n → ∞.

Conjecture (Rivin-Pemantle), Theorem (Kabluchko, 2015)

The critical points are also distributed according to µ.

Fact for real-rooted polynomials

The roots of p

(

n log n )

n

also distributed according to µ as n → ∞.

• Sketch. Each root moves roughly ±n−1 under one step of

differentiation.

Conjecture (Rivin-Pemantle), Theorem (Kabluchko, 2015)

The critical points are also distributed according to µ.

Fact for real-rooted polynomials

The roots of p

(

n log n )

n

also distributed according to µ as n → ∞.

• Sketch. Each root moves roughly ±n−1 under one step of

differentiation.

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1?

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? t = 0 t = 0.5 t = 0.99

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? t = 0 t = 0.5 t = 0.99

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? t = 0 t = 0.5 t = 0.99

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1?

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? Some History.

• 1. The question hasn’t been studied very much.

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? Some History.

• 1. The question hasn’t been studied very much.
• 2. Polya asked a whole number of questions in the setting of real

entire functions.

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? Some History.

• 1. The question hasn’t been studied very much.
• 2. Polya asked a whole number of questions in the setting of real

entire functions.

• 3. The smallest gap grows under differentiation.

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? Some History.

• 1. The question hasn’t been studied very much.
• 2. Polya asked a whole number of questions in the setting of real

entire functions.

• 3. The smallest gap grows under differentiation. Denoting the

smallest gap of a polynomial pn having n real roots {x1, . . . , xn} by G(pn) = min

i=j |xi − xj|,

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? Some History.

• 1. The question hasn’t been studied very much.
• 2. Polya asked a whole number of questions in the setting of real

entire functions.

• 3. The smallest gap grows under differentiation. Denoting the

smallest gap of a polynomial pn having n real roots {x1, . . . , xn} by G(pn) = min

i=j |xi − xj|,

we have (Riesz, Sz-Nagy, Walker, 1920s) G(p′

n) ≥ G(pn).

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? Let us denote the answer by u(t, x). Here, the idea is that u(t, x) is the limiting behavior as n → ∞.

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? Let us denote the answer by u(t, x). Here, the idea is that u(t, x) is the limiting behavior as n → ∞. In particular µ = u(0, x)dx

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? Let us denote the answer by u(t, x). Here, the idea is that u(t, x) is the limiting behavior as n → ∞. In particular µ = u(0, x)dx and

• R

u(t, x)dx = 1 − t.

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? Let us denote the answer by u(t, x). Here, the idea is that u(t, x) is the limiting behavior as n → ∞. In particular µ = u(0, x)dx and

• R

u(t, x)dx = 1 − t. What can one say about u(t, x)?

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? Let us denote the answer by u(t, x). 1.

• R u(t, x)dx = 1 − t.

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? Let us denote the answer by u(t, x). 1.

• R u(t, x)dx = 1 − t.

2.

• R u(t, x)x dx = (1 − t)
• R u(0, x)x dx

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? Let us denote the answer by u(t, x). 1.

• R u(t, x)dx = 1 − t.

2.

• R u(t, x)x dx = (1 − t)
• R u(0, x)x dx

3.

• R
• R u(t, x)(x − y)2u(t, y) dxdy =

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? Let us denote the answer by u(t, x). 1.

• R u(t, x)dx = 1 − t.

2.

• R u(t, x)x dx = (1 − t)
• R u(0, x)x dx

3.

• R
• R u(t, x)(x − y)2u(t, y) dxdy =

(1 − t)3

R

• R u(0, x)(x − y)2u(0, y) dxdy

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? Let us denote the answer by u(t, x). 1.

• R u(t, x)dx = 1 − t.

2.

• R u(t, x)x dx = (1 − t)
• R u(0, x)x dx

3.

• R
• R u(t, x)(x − y)2u(t, y) dxdy =

(1 − t)3

R

• R u(0, x)(x − y)2u(0, y) dxdy

This means: the distribution shrinks linearly in mass,

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? Let us denote the answer by u(t, x). 1.

• R u(t, x)dx = 1 − t.

2.

• R u(t, x)x dx = (1 − t)
• R u(0, x)x dx

3.

• R
• R u(t, x)(x − y)2u(t, y) dxdy =

(1 − t)3

R

• R u(0, x)(x − y)2u(0, y) dxdy

This means: the distribution shrinks linearly in mass, its mean is preserved and

Main Question

What about the roots of p(t·n)

n

where 0 < t < 1? Let us denote the answer by u(t, x). 1.

• R u(t, x)dx = 1 − t.

2.

• R u(t, x)x dx = (1 − t)
• R u(0, x)x dx

3.

• R
• R u(t, x)(x − y)2u(t, y) dxdy =

(1 − t)3

R

• R u(0, x)(x − y)2u(0, y) dxdy

This means: the distribution shrinks linearly in mass, its mean is preserved and the mass is distributed over area ∼ √1 − t.

An Equation (S. 2018)

There’s some good heuristic reasoning for ∂u ∂t + 1 π ∂ ∂x arctan Hu u

• = 0
• n supp(u)

An Equation (S. 2018)

There’s some good heuristic reasoning for ∂u ∂t + 1 π ∂ ∂x arctan Hu u

• = 0
• n supp(u)

where Hf (x) = p.v. 1 π

• R

f (y) x − y dy is the Hilbert transform.

An Equation (S. 2018)

There’s some good heuristic reasoning for ∂u ∂t + 1 π ∂ ∂x arctan Hu u

• = 0
• n supp(u)

where Hf (x) = p.v. 1 π

• R

f (y) x − y dy is the Hilbert transform. The argument is actually fun and I can give it in full. But before, let’s explore this strange equation.

A nice way to understand a PDE is through explicit closed-form solutions (if they exist).

A nice way to understand a PDE is through explicit closed-form solutions (if they exist). So the relevant question is: are there nice special solutions that we can construct? For this we need polynomials pn whose roots have a nice distribution

A nice way to understand a PDE is through explicit closed-form solutions (if they exist). So the relevant question is: are there nice special solutions that we can construct? For this we need polynomials pn whose roots have a nice distribution and whose derivatives p(k)

n

also have a nice distribution?

• 1. Hermite polynomials

A nice way to understand a PDE is through explicit closed-form solutions (if they exist). So the relevant question is: are there nice special solutions that we can construct? For this we need polynomials pn whose roots have a nice distribution and whose derivatives p(k)

n

also have a nice distribution?

• 1. Hermite polynomials
• 2. (associated) Laguerre polynomials

A nice way to understand a PDE is through explicit closed-form solutions (if they exist). So the relevant question is: are there nice special solutions that we can construct? For this we need polynomials pn whose roots have a nice distribution and whose derivatives p(k)

n

also have a nice distribution?

• 1. Hermite polynomials
• 2. (associated) Laguerre polynomials

Presumably there are many others(?)

Hermite Polynomials

Hermite polynomials Hn : R → R satisfy a nice recurrence relation dm dxm Hn(x) = 2nn! (n − m)!Hn−m(x).

Hermite Polynomials

Hermite polynomials Hn : R → R satisfy a nice recurrence relation dm dxm Hn(x) = 2nn! (n − m)!Hn−m(x). Moreover, the roots of Hn converge, in a suitable sense, to µ = 1 π

• 2n − x2dx.

Hermite Polynomials

Hermite polynomials Hn : R → R satisfy a nice recurrence relation dm dxm Hn(x) = 2nn! (n − m)!Hn−m(x). Moreover, the roots of Hn converge, in a suitable sense, to µ = 1 π

• 2n − x2dx.

This suggests that u(t, x) = 2 π

• 1 − t − x2 · χ|x|≤√1−t

for t ≤ 1 should be a solution of the PDE (and it is).

Hermite Polynomials

u(t, x) = 2 π

• 1 − t − x2 · χ|x|≤√1−t

for t ≤ 1

Laguerre Polynomials

(Associated) Laguerre polynomials Hn : R → R satisfy the recurrence relation dk dxk L(α)

n (x) = (−1)kL(α+k) n−k (x).

Laguerre Polynomials

(Associated) Laguerre polynomials Hn : R → R satisfy the recurrence relation dk dxk L(α)

n (x) = (−1)kL(α+k) n−k (x).

The roots converge in distribution to the Marchenko-Pastur distribution v(c, x) =

• (x+ − x)(x − x−)

2πx χ(x−,x+)dx where x± = ( √ c + 1 ± 1)2.

Laguerre Polynomials

(Associated) Laguerre polynomials Hn : R → R satisfy the recurrence relation dk dxk L(α)

n (x) = (−1)kL(α+k) n−k (x).

The roots converge in distribution to the Marchenko-Pastur distribution v(c, x) =

• (x+ − x)(x − x−)

2πx χ(x−,x+)dx where x± = ( √ c + 1 ± 1)2. Indeed, uc(t, x) = v c + t 1 − t , x 1 − t

• is a solution of the PDE.

Laguerre Polynomials

uc(t, x) = v c + t 1 − t , x 1 − t

• .

Figure: Marchenko-Pastur solutions uc(t, x): c = 1 (left) and c = 15 (right) shown for t ∈ {0, 0.2, 0.4, 0.6, 0.8, 0.9, 0.95, 0.99}.

A Bonus Solution

There are several classical orthogonal polynomials on [−1, 1] (Gegenbauer, Jacobi, ...).

A Bonus Solution

There are several classical orthogonal polynomials on [−1, 1] (Gegenbauer, Jacobi, ...). For fairly general classes (Erd˝

• s-Freud

theorem) of such polynomials, the distribution of roots is asymptotically given by µ = 1 π dx √ 1 − x2 .

A Bonus Solution

There are several classical orthogonal polynomials on [−1, 1] (Gegenbauer, Jacobi, ...). For fairly general classes (Erd˝

• s-Freud

theorem) of such polynomials, the distribution of roots is asymptotically given by µ = 1 π dx √ 1 − x2 . As it turns out, u(t, x) = c √ 1 − x2 is indeed a stationary solution of the equation.

A Bonus Solution

There are several classical orthogonal polynomials on [−1, 1] (Gegenbauer, Jacobi, ...). For fairly general classes (Erd˝

• s-Freud

theorem) of such polynomials, the distribution of roots is asymptotically given by µ = 1 π dx √ 1 − x2 . As it turns out, u(t, x) = c √ 1 − x2 is indeed a stationary solution of the equation.

Theorem (Tricomi?)

Let f : (−1, 1) → R≥0. If Hf ≡ 0 in (−1, 1), then f = c √ 1 − x2 .

∂u ∂t + 1 π ∂ ∂x arctan Hu u

• = 0
• n supp(u)

∂u ∂t + 1 π ∂ ∂x arctan Hu u

• = 0
• n supp(u)

Sketch of the Derivation. Crystallization as key assumption.

∂u ∂t + 1 π ∂ ∂x arctan Hu u

• = 0
• n supp(u)

Sketch of the Derivation. Crystallization as key assumption. u(t, x)

∂u ∂t + 1 π ∂ ∂x arctan Hu u

• = 0
• n supp(u)

Sketch of the Derivation. Crystallization as key assumption. u(t, x)

∂u ∂t + 1 π ∂ ∂x arctan Hu u

• = 0
• n supp(u)

Sketch of the Derivation. Crystallization as key assumption. u(t, x)

xk

n

k=1 1 x−xk = 0

xk

n

k=1 1 x−xk = 0

n

• k=1

1 x − xk =

• |xk−x| large

1 x − xk +

• |xk−x| small

1 x − xk

xk

n

k=1 1 x−xk = 0

n

• k=1

1 x − xk =

• |xk−x| large

1 x − xk +

• |xk−x| small

1 x − xk

• |xk−x| large

1 x − xk ∼ n

• R

1 x − y · u(t, y)dy = n · [Hu](t, x).

xk

n

k=1 1 x−xk = 0

n

• k=1

1 x − xk =

• |xk−x| large

1 x − xk +

• |xk−x| small

1 x − xk

• |xk−x| large

1 x − xk ∼ n

• R

1 x − y · u(t, y)dy = n · [Hu](t, x). It thus remains to understand the behavior of the local term.

The local term is

• |xk−x| small

1 x − xk .

The local term is

• |xk−x| small

1 x − xk . Crystallization means that the roots form, locally, an arithmetic progressions

The local term is

• |xk−x| small

1 x − xk . Crystallization means that the roots form, locally, an arithmetic progressions and thus

• |xk−x| small

1 x − xk ∼

• ℓ∈Z

1 x −

• xk +

ℓ u(t,x)n

.

The local term is

• |xk−x| small

1 x − xk . Crystallization means that the roots form, locally, an arithmetic progressions and thus

• |xk−x| small

1 x − xk ∼

• ℓ∈Z

1 x −

• xk +

ℓ u(t,x)n

. We are in luck: this sum has a closed-form expression due to Euler π cot πx = 1 x +

∞

• n=1
• 1

x + n + 1 x − n

• for x ∈ R \ Z.

The Local Field

The Local Field

We can then predict the behavior of the roots of the derivative: they are in places where the local (near) field and the global (far) field cancel out. This leads to the desired equation.

A Fast Numerical Algorithm

Jeremy Hoskins (U Chicago) used the electrostatic interpretation to produce an algorithm that can compute all derivatives of polynomials up to degree ∼ 100.000.

A Fast Numerical Algorithm

Jeremy Hoskins (U Chicago) used the electrostatic interpretation to produce an algorithm that can compute all derivatives of polynomials up to degree ∼ 100.000. Semicircles.

Theorem (J. Hoskins and S, 2020)

Let X be a random variable on R such that all moments are finite and EX = 0 as well as VX = 1.

Theorem (J. Hoskins and S, 2020)

Let X be a random variable on R such that all moments are finite and EX = 0 as well as VX = 1. Let pn be a random polynomial whose roots are i.i.d. copies of X and fix ℓ ∈ N.

Theorem (J. Hoskins and S, 2020)

Let X be a random variable on R such that all moments are finite and EX = 0 as well as VX = 1. Let pn be a random polynomial whose roots are i.i.d. copies of X and fix ℓ ∈ N. Then, as n → ∞, nℓ/2 ℓ! n! · p(n−ℓ)

n

x √n

• ∼ (1 + o(1)) · Heℓ(x + γn),

Theorem (J. Hoskins and S, 2020)

Let X be a random variable on R such that all moments are finite and EX = 0 as well as VX = 1. Let pn be a random polynomial whose roots are i.i.d. copies of X and fix ℓ ∈ N. Then, as n → ∞, nℓ/2 ℓ! n! · p(n−ℓ)

n

x √n

• ∼ (1 + o(1)) · Heℓ(x + γn),

where γn ∼ N(0, 1) and Heℓ is the ℓ−th Hermite polynomial.

Theorem (J. Hoskins and S, 2020)

Let X be a random variable on R such that all moments are finite and EX = 0 as well as VX = 1. Let pn be a random polynomial whose roots are i.i.d. copies of X and fix ℓ ∈ N. Then, as n → ∞, nℓ/2 ℓ! n! · p(n−ℓ)

n

x √n

• ∼ (1 + o(1)) · Heℓ(x + γn),

where γn ∼ N(0, 1) and Heℓ is the ℓ−th Hermite polynomial. Remarks.

• 1. The roots of the Hermite polynomial have a semicircle density.

Theorem (J. Hoskins and S, 2020)

Let X be a random variable on R such that all moments are finite and EX = 0 as well as VX = 1. Let pn be a random polynomial whose roots are i.i.d. copies of X and fix ℓ ∈ N. Then, as n → ∞, nℓ/2 ℓ! n! · p(n−ℓ)

n

x √n

• ∼ (1 + o(1)) · Heℓ(x + γn),

where γn ∼ N(0, 1) and Heℓ is the ℓ−th Hermite polynomial. Remarks.

• 1. The roots of the Hermite polynomial have a semicircle density.
• 2. If x1, x2, . . . , xn ∼ X, then

x1 + · · · + xn √n ∼ N(0, 1) and the mean of the roots is preserved under differentiation (hence the random shift).

Ideas behind the proof

pn(x) = xn + an−1xn−1 + · · · + a0

Ideas behind the proof

pn(x) = xn + an−1xn−1 + · · · + a0 Coefficients are preserved under differentiation, they are simply multiplied with degrees.

Ideas behind the proof

pn(x) = xn + an−1xn−1 + · · · + a0 Coefficients are preserved under differentiation, they are simply multiplied with degrees.

n

• i=1

(x − xi) =

n

• k=0

(−1)kek(x1, . . . , xn)xn−k.

Ideas behind the proof

pn(x) = xn + an−1xn−1 + · · · + a0 Coefficients are preserved under differentiation, they are simply multiplied with degrees.

n

• i=1

(x − xi) =

n

• k=0

(−1)kek(x1, . . . , xn)xn−k. We need elementary symmetric polynomials e0(x1, . . . , xn) = 1 e1(x1, . . . , xn) = x1 + · · · + xn e2(x1, . . . , xn) =

• i<j

xixj e3(x1, . . . , xn) =

• i<j<k

xixjxk

Ideas behind the proof

Given x1, . . . , xn ∼ X, what do we know about e0(x1, . . . , xn) = 1 e1(x1, . . . , xn) = x1 + · · · + xn e2(x1, . . . , xn) =

• i<j

xixj e3(x1, . . . , xn) =

• i<j<k

xixjxk

Ideas behind the proof

Given x1, . . . , xn ∼ X, what do we know about e0(x1, . . . , xn) = 1 e1(x1, . . . , xn) = x1 + · · · + xn e2(x1, . . . , xn) =

• i<j

xixj e3(x1, . . . , xn) =

• i<j<k

xixjxk As it turns out: e1 determines everything else.

Ideas behind the proof

e3(x1, . . . , xn) =

• i<j<k

xixjxk ek has ∼ nk terms which means we expect it to be size nk/2.

Ideas behind the proof

e3(x1, . . . , xn) =

• i<j<k

xixjxk ek has ∼ nk terms which means we expect it to be size nk/2.

Lemma

Let m ∈ N and let x1, . . . , xn be i.i.d. random variables sampled from a distribution on R with EX = 0, EX 2 = 1 and E|X|m < ∞. Then, as n → ∞, E

• em −

⌊m/2⌋

• k=0

(−1)k 1 k!(m − 2k)!2k · em−2k

1

nk

• X n

m−1 2

Sep 3, 2020

Sep 3, 2020

Sep 3, 2020

Sep 3, 2020

The same PDE in a supposedly different context is presumably not a coincidence.

Sep 3, 2020

is concerned with free convolution of measures µ ⊞ ν as introduced by Voiculescu in the 1980s.

Sep 3, 2020

is concerned with free convolution of measures µ ⊞ ν as introduced by Voiculescu in the 1980s. It is an analogue of classical convolution in the non-commutative setting.

Sep 3, 2020

is concerned with free convolution of measures µ ⊞ ν as introduced by Voiculescu in the 1980s. It is an analogue of classical convolution in the non-commutative setting. In particular, we expect that u(t, x) is given by a fractional free convolution µ⊞k with k ≥ 1.

Sep 3, 2020

is concerned with free convolution of measures µ ⊞ ν as introduced by Voiculescu in the 1980s. It is an analogue of classical convolution in the non-commutative setting. In particular, we expect that u(t, x) is given by a fractional free convolution µ⊞k with k ≥ 1. As t → 1, we have k → ∞.

Sep 3, 2020

is concerned with free convolution of measures µ ⊞ ν as introduced by Voiculescu in the 1980s. It is an analogue of classical convolution in the non-commutative setting. In particular, we expect that u(t, x) is given by a fractional free convolution µ⊞k with k ≥ 1. As t → 1, we have k → ∞.

An Optimistic Conjecture

Under some reasonable assumptions µ⊞k = u

• 1 − 1

k , x k

• dx.

An Optimistic Conjecture

Under some reasonable assumptions µ⊞k = u

• 1 − 1

k , x k

• dx.

An Optimistic Conjecture

Under some reasonable assumptions µ⊞k = u

• 1 − 1

k , x k

• dx.

This would have a large number of implications.

An Optimistic Conjecture

Under some reasonable assumptions µ⊞k = u

• 1 − 1

k , x k

• dx.

This would have a large number of implications. ◮ Fractional Free Convolution preserves free cumulants κ1(µ) =

• R

xdµ κ2(µ) =

• R

x2dµ −

• R

xdµ 2 . . .

An Optimistic Conjecture

Under some reasonable assumptions µ⊞k = u

• 1 − 1

k , x k

• dx.

This would have a large number of implications. ◮ Fractional Free Convolution preserves free cumulants κ1(µ) =

• R

xdµ κ2(µ) =

• R

x2dµ −

• R

xdµ 2 . . . since κn(µ⊞k) = k · κn(µ).

An Optimistic Conjecture

Under some reasonable assumptions µ⊞k = u

• 1 − 1

k , x k

• dx.

This would have a large number of implications. ◮ Fractional Free Convolution preserves free cumulants κ1(µ) =

• R

xdµ κ2(µ) =

• R

x2dµ −

• R

xdµ 2 . . . since κn(µ⊞k) = k · κn(µ). Infinitely many conserved quantities.

Conjecture

µ⊞k = u

• 1 − 1

k , x k

• dx.

would have a large number of implications.

Conjecture

µ⊞k = u

• 1 − 1

k , x k

• dx.

would have a large number of implications.

Voiculescu’s Free Central Limit Theorem

µ ⊞ µ ⊞ · · · ⊞ µ → semicircle.

Conjecture

µ⊞k = u

• 1 − 1

k , x k

• dx.

would have a large number of implications.

Voiculescu’s Free Central Limit Theorem

µ ⊞ µ ⊞ · · · ⊞ µ → semicircle. This would then imply that u(t, x) should be a semicircle for t close to 1.

Sep 3, 2020

Sep 4, 2020

Sep 4, 2020

Sep 4, 2020

which proves that, in a certain setting, the crystallization assumption for roots is justified in the bulk

Sep 4, 2020

which proves that, in a certain setting, the crystallization assumption for roots is justified in the bulk and a couple of weeks later

Sep 4, 2020

which proves that, in a certain setting, the crystallization assumption for roots is justified in the bulk and a couple of weeks later which establishes a connection between Bessel processes and free convolution.

Sep 4, 2020

which proves that, in a certain setting, the crystallization assumption for roots is justified in the bulk and a couple of weeks later which establishes a connection between Bessel processes and free

• convolution. So I think we are pretty close to having completely

rigorous arguments for most things.

What’s left to do?

What’s left to do?

◮ Can the PDE be useful? Linearization seems really nice?

What’s left to do?

◮ Can the PDE be useful? Linearization seems really nice? ◮ Is Jeremy Hoskins’ algorithm a useful method to compute µ⊞k?

What’s left to do?

◮ Can the PDE be useful? Linearization seems really nice? ◮ Is Jeremy Hoskins’ algorithm a useful method to compute µ⊞k? ◮ What about the complex case?

What’s left to do?

picture from O’Rourke and Williams (2018)

The Complex Case

One can derive the same sort of PDE in the complex case. The derivation is actually simpler

The Complex Case

One can derive the same sort of PDE in the complex case. The derivation is actually simpler typically ∼ n−1/2 1 z − zℓ = −

n

• k=1

k=ℓ

1 z − zk .

A Nonlocal Transport Equation

Sean O’Rourke and I tried to see whether the equation simplifies if we assume that the initial distribution is radial around the origin.

A Nonlocal Transport Equation

Sean O’Rourke and I tried to see whether the equation simplifies if we assume that the initial distribution is radial around the origin. If the density is ψ(t, x), then ∂ψ ∂t = ∂ ∂x 1 x x ψ(s)ds −1 ψ(x)

• .

A Nonlocal Transport Equation

Sean O’Rourke and I tried to see whether the equation simplifies if we assume that the initial distribution is radial around the origin. If the density is ψ(t, x), then ∂ψ ∂t = ∂ ∂x 1 x x ψ(s)ds −1 ψ(x)

• .

r

∂ψ ∂t = ∂ ∂x 1 x x ψ(s)ds −1 ψ(x)

• has a nice closed form solution

u(t, x) = χ0≤x≤1−t.

∂ψ ∂t = ∂ ∂x 1 x x ψ(s)ds −1 ψ(x)

• has a nice closed form solution

u(t, x) = χ0≤x≤1−t. This corresponds to Random Taylor Polynomials.

1000 500 500 1000 1000 500 500 1000

1000 500 500 1000 1000 500 500 1000

Random Taylor polynomials are defined by pn =

n

• k=0

γk zk k! , where γk ∼ N(0, 1).

1000 500 500 1000 1000 500 500 1000

Random Taylor polynomials are defined by pn =

n

• k=0

γk zk k! , where γk ∼ N(0, 1). They are preserved under differentiation.

1000 500 500 1000 1000 500 500 1000

Random Taylor polynomials are defined by pn =

n

• k=0

γk zk k! , where γk ∼ N(0, 1). They are preserved under differentiation.

Theorem ( Kabluchko & Zaporozhets)

1 n

n

• k=1

δzkn−1 → χ|z|≤1 2π|z| as n → ∞.

A final pretty fact: when trying to study L2−stability of the solution, one runs into the following beautiful inequality.

Lemma

For f : (0, ∞) → R≥0 ∞ f (x) x2 x f (y)dy

• dx ≤

∞ f (x)2 x dx,

A final pretty fact: when trying to study L2−stability of the solution, one runs into the following beautiful inequality.

Lemma

For f : (0, ∞) → R≥0 ∞ f (x) x2 x f (y)dy

• dx ≤

∞ f (x)2 x dx,

• Proof. follows easily from a general Hardy inequality.

Thank you!