Around Hilberts 13th Problem Ziqin Feng Miami University February - - PowerPoint PPT Presentation

Around Hilbert’s 13th Problem

Ziqin Feng

Miami University

February 21, 2012

Ziqin Feng Around Hilbert’s 13th Problem

1 Introduction — Hilbert’s 13th Problem

The 13th Problem. Motivation. Answers: Vitushkin, Kolmogorov, Arnold

2 Kolmogorov’s Superposition Theorem

The KST and its Proof New Results on the KST

3 Further Results

A New Cardinal Invariant, basic (X) ‘Real World’ Applications Strong Approximation and Universal PDEs

Ziqin Feng Around Hilbert’s 13th Problem

The 13th Problem

Question: Can every continuous function of 3 variables be written as a composition of continuous functions of 2 variables? Example f (x, y, z) = ex2 3

• sin(y + z) + 1

1 + x2 + z2 . Composition of: exp, squaring, ×,

3

√·, sin, +, /.

Ziqin Feng Around Hilbert’s 13th Problem

The 13th Problem

Question: Can every continuous function of 3 variables be written as a superposition of continuous functions of 2 variables? Example f (x, y, z) = ex2 3

• sin(y + z) + 1

1 + x2 + z2 . Superposition of: exp, squaring, ×,

3

√·, sin, +, /.

Ziqin Feng Around Hilbert’s 13th Problem

Solving Functions for Polynomials

Quadratic az2 + bz + c → z2 + pz + q Solution map: z(a, b, c) → z(p, q) = 1

2(−p +

• p2 − 4q),

p = b/a, q = c/a. Quintic General Quintic → z5 + pz + q Solution map: z(a, b, c, d, e, f ) → z(p, q) p, q arithmetic functions of coefficients. Degree 7 General Degree 7 → z7 + pz3 + qz2 + rz + 1 Solution map: → z(p, q, r), p, q, r arithmetic functions of coefficients.

Ziqin Feng Around Hilbert’s 13th Problem

Solving Functions for Polynomials

Quadratic az2 + bz + c → z2 + pz + q Solution map: z(a, b, c) → z(p, q) = 1

2(−p +

• p2 − 4q),

p = b/a, q = c/a. Quintic General Quintic → z5 + pz + q Solution map: z(a, b, c, d, e, f ) → z(p, q) p, q arithmetic functions of coefficients. Degree 7 General Degree 7 → z7 + pz3 + qz2 + rz + 1 Solution map: → z(p, q, r), p, q, r arithmetic functions of coefficients. Arithmetic functions are compositions of maps of 2 variables.

Ziqin Feng Around Hilbert’s 13th Problem

Solving Functions for Polynomials

Quadratic az2 + bz + c → z2 + pz + q Solution map: z(a, b, c) → z(p, q) = 1

2(−p +

• p2 − 4q),

p = b/a, q = c/a. Quintic General Quintic → z5 + pz + q Solution map: z(a, b, c, d, e, f ) → z(p, q) p, q arithmetic functions of coefficients. Degree 7 General Degree 7 → z7 + pz3 + qz2 + rz + 1 Solution map: → z(p, q, r), p, q, r arithmetic functions of coefficients.

Ziqin Feng Around Hilbert’s 13th Problem

Hilbert Thought No

Ziqin Feng Around Hilbert’s 13th Problem

Vitushkin’s Negative Answer

Theorem (Vitushkin, 1954) There is a function f : I n → R of n–variables which has continuous qth order derivatives but can not be expressed as a superposition

• f functions of n′–variables which have continuous q′th order

derivatives, if n/q > n′/q′. Example There are continuously differentiable functions of 3 variables which are not the superposition of continuously differentiable functions of 2 variables.

Ziqin Feng Around Hilbert’s 13th Problem

Vitushkin’s Negative Answer

Theorem (Vitushkin, 1954) There is a function f : I n → R of n–variables which has continuous qth order derivatives but can not be expressed as a superposition

• f functions of n′–variables which have continuous q′th order

derivatives, if n/q > n′/q′. Example There are continuously differentiable functions of 3 variables which are not the superposition of continuously differentiable functions of 2 variables.

Ziqin Feng Around Hilbert’s 13th Problem

Kolmogorov and Arnold

Theorem (Kolmogorov, 1956) Every f ∈ C(I n) for n ≥ 4 is the superposition of continuous functions with ≤ 3 variables. Theorem (Arnold, 1957) Every f ∈ C(I 3) is the superposition of continuous functions with ≤ 2 variables.

Ziqin Feng Around Hilbert’s 13th Problem

Solution to Hilbert 13

YES! Every continuous function of 3 variables, from I, can be written as a superposition of continuous functions of ≤ 2 variables. Applies to: Solution function z(p, q, r) to z7 + pz3 + qz2 + rz + 1 = 0.

Ziqin Feng Around Hilbert’s 13th Problem

New Results A space X satisfies Hilbert 13 every continuous function of n variables from X is a superposition of continuous functions of ≤ 2 variables Example Every continuous function of n real variables can be written as a superposition of continuous functions of ≤ 2 real variables.

Ziqin Feng Around Hilbert’s 13th Problem

New Results A space X satisfies Hilbert 13 every continuous function of n variables from X is a superposition of continuous functions of ≤ 2 variables if and only if X is locally compact, finite dimensional and separable metric (homeomorphic to a closed subspace of Euclidean space). Example Every continuous function of n real variables can be written as a superposition of continuous functions of ≤ 2 real variables.

Ziqin Feng Around Hilbert’s 13th Problem

New Results A space X satisfies Hilbert 13 every continuous function of n variables from X is a superposition of continuous functions of ≤ 2 variables if and only if X is locally compact, finite dimensional and separable metric (homeomorphic to a closed subspace of Euclidean space). Example Every continuous function of n real variables can be written as a superposition of continuous functions of ≤ 2 real variables.

Ziqin Feng Around Hilbert’s 13th Problem

Kolmogorov’s Superposition Theorem (KST)

Theorem (Kolmogorov, 1957) Step 1 There exist φ1 . . . , φ2n+1 in C(I n) such that ∀f ∈ C(I n) f =

2n+1

• i=1

gi◦φi, for some gi ∈ C(R). Step 2 Further, can choose the φ1, . . . , φ2n+1 such that: φi(x1, . . . xn) =

n

• j=1

ψij(xj)

Ziqin Feng Around Hilbert’s 13th Problem

Kolmogorov– Idea of Proof n = 2

Theorem (KST: n=2) Step 1 There exist φ1 . . . , φ5 in C(I 2) such that ∀f ∈ C(I 2) f =

5

• i=1

gi ◦φi, for some gi ∈ C(R). Step 2 Further, can choose the φ1, . . . , φ5 such that: φi(x1, x2) = ψi1(x1) + ψi2(x2)

Ziqin Feng Around Hilbert’s 13th Problem

Ideas in Proof of KST: The ψpq and φq

0.1 0.2 0.3 0.4 0.1 0.2 0.3 0.4 0.02 0.04 0.06 0.08 0.10

Ziqin Feng Around Hilbert’s 13th Problem

Proof of KST: Contour Map of φ

φ = φ(x1, x2)

0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 Ziqin Feng Around Hilbert’s 13th Problem

Proof of KST: Looking Deeper at ψ

0.0 0.1 0.2 0.3 0.4 0.5 0.00 0.02 0.04 0.06 0.08 0.10 Ziqin Feng Around Hilbert’s 13th Problem

Proof of KST: Looking Deeper at ψ

0.0 0.1 0.2 0.3 0.4 0.5 0.00 0.02 0.04 0.06 0.08 0.10 Ziqin Feng Around Hilbert’s 13th Problem

Proof of KST: Looking Deeper at ψ

0.20 0.22 0.24 0.26 0.28 0.30 0.040 0.042 0.044 0.046 0.048 0.050 0.052 0.054 0.056 Ziqin Feng Around Hilbert’s 13th Problem

Proof of KST: Looking Deeper at ψ

0.20 0.22 0.24 0.26 0.28 1.0 1.5 2.0 2.5 1e 8+3.999999e 2 Ziqin Feng Around Hilbert’s 13th Problem

Proof of KST: Looking Deeper at ψ

0.20 0.22 0.24 0.26 0.28 1.0 1.5 2.0 2.5 1e 8+3.999999e 2 Ziqin Feng Around Hilbert’s 13th Problem

Proof of KST: Approximate f

f

0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5

φ

0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5

Ziqin Feng Around Hilbert’s 13th Problem

Proof of KST: Approximate f

f

0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5

g ◦ φ, (g ◦ φ ≤ f )

0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5

g

0.00 0.02 0.04 0.06 0.08 0.10 0.12 3 2 1 1 2 3

Ziqin Feng Around Hilbert’s 13th Problem

Proof of KST: Approximate f

f

0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5

g ◦ φ, (g ◦ φ ≤ f )

0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5

g

0.00 0.02 0.04 0.06 0.08 0.10 0.12 3 2 1 1 2 3

Ziqin Feng Around Hilbert’s 13th Problem

Proof of KST: Replicate, Shift . . .

φ1 φ2 φ3 φ4 φ5

0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5

g1 ◦ φ1 g2 ◦ φ2 g3 ◦ φ3 g4 ◦ φ4 g5 ◦ φ5

0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5

Ziqin Feng Around Hilbert’s 13th Problem

Proof of KST: . . . and Average

f 5

q=1 1 3gq ◦ φq

0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5 0.0 0.1 0.2 0.3 0.4 0.5

Ziqin Feng Around Hilbert’s 13th Problem

Proof of KST: Correct and Try Again

Let f0 = f . Then f0 ∼

q 1 3g0 q ◦ φq.

Get f0 −

q 1 3g0 q ◦ φq ≤ 5 6f0.

Set f1 = f0 −

q 1 3g0 q ◦ φq.

And repeat f1 ∼

q 1 3g1 q ◦ φq.

Get f1 −

q 1 3g1 q ◦ φq ≤ 5 6f1.

And repeat . . . So that f =

q gq ◦ φq,

where gq =

k 1 3gk q .

Ziqin Feng Around Hilbert’s 13th Problem

Extending the KST

Smoothness Fridman Lipschitz inner functions Vitushkin & Khenkin Inner functions can not be C 1 Simplicity of Representation Lorentz Single outer function Sprecher Number of inner functions Domain Ostrand Compact, finite dimensional and metric Doss Step 1 (basic functions) for Rn

Ziqin Feng Around Hilbert’s 13th Problem

New Results on Hilbert 13 (I)

Theorem (KST for R) There exist ψpq ∈ C(R), for 1 ≤ q ≤ 2n + 1 and 1 ≤ p ≤ n, such that for any f ∈ C(Rn), there is a g in C(R) such that: f (x) =

2n+1

• q=1

g(φq(x)), where φq(x1, . . . , xn) =

n

• p=1

ψpq(xp). The ψpq can be taken to be Lipschitz.

Ziqin Feng Around Hilbert’s 13th Problem

New Results on Hilbert 13 (II)

Theorem (General KST) Let X be locally compact, finite dimensional and separable metrizable. There exist ψpq ∈ C(X), for 1 ≤ q ≤ m and 1 ≤ p ≤ n, such that for any f ∈ C(X n), there is a g in C(R) such that: f (x) =

m

• q=1

g(φq(x)), where φq(x1, . . . , xn) =

n

• p=1

ψpq(xp).

Ziqin Feng Around Hilbert’s 13th Problem

New Results on Hilbert 13 (III)

Theorem (Necessary Condition for KST for X) If a space X has a finite basic family then X is locally compact, finite dimensional and separable metrizable. Definition φ1, . . . , φm in C(X) are a finite basic family for X if for any f ∈ C(X), there are g1, . . . , gm in C(R) such that: f (x) =

m

• q=1

gq(φq(x)).

Ziqin Feng Around Hilbert’s 13th Problem

Difficulties of R versus I: Step 2

φ = φ(x1, x2) = ψ1(x1) + ψ2(x2).

Ziqin Feng Around Hilbert’s 13th Problem

Related Problems

• Minimal size of basic family on C(X)

Connections to set theory and Banach Algebra.

• Constructive proof of KST on R

Possible applcations in “real world”.

• ‘Fixed-formed’ Strong Analytic Approximation theorem

Applications to Universal PDE.

Ziqin Feng Around Hilbert’s 13th Problem

Related Problems

• Minimal size of basic family on C(X)

Connections to set theory and Banach Algebra.

• Constructive proof of KST on R

Possible applcations in “real world”.

• ‘Fixed-formed’ Strong Analytic Approximation theorem

Applications to Universal PDE.

Ziqin Feng Around Hilbert’s 13th Problem

Related Problems

• Minimal size of basic family on C(X)

Connections to set theory and Banach Algebra.

• Constructive proof of KST on R

Possible applcations in “real world”.

• ‘Fixed-formed’ Strong Analytic Approximation theorem

Applications to Universal PDE.

Ziqin Feng Around Hilbert’s 13th Problem

Basic Families, basic (X)

Definition A family Φ ⊆ C(X) is basic if for any f ∈ C(X), there are φ1, . . . , φm in Φ and g1, . . . , gm in C(R) such that: f (x) =

m

• q=1

gq(φq(x)). Definition basic (X) = min{|Φ| : Φ basic for X}.

Ziqin Feng Around Hilbert’s 13th Problem

Finite and Countable Basic Families

Theorem basic (X) ≤ 2n + 1 if and only if X is locally compact, separable metrizable and dim X ≤ n. Note: Sternfeld. Theorem basic (X) ≥ ℵ0 if and only if basic (X) ≥ ℵ1. Example basic (ω1) = ℵ1.

Ziqin Feng Around Hilbert’s 13th Problem

basic (X), X Separable Metric

Expected: Definition ck(X) = cof (K(X), ⊆) = min{|K| : K cofinal in K(X)}, where K(X) = all the compact subsets of X. Examples ck(R) = ℵ0, ck(P) = d = ck(Q).

Ziqin Feng Around Hilbert’s 13th Problem

basic (X), X Separable Metric

Actual: Theorem (Dichotomy) Let X be separable metrizable. Then either basic (X) finite which happens iff X locally compact, finite dimensional

• r basic (X) = c.

Ziqin Feng Around Hilbert’s 13th Problem

K Compact, PCF Theory

Lemma (Shelah’s Observation) κω = cof ([κ]ω, ⊆) × |P(ω)|. Theorem (Shelah) ℵω < cof ([ℵω]ℵ0, ⊆) < ℵω4. Making cof ([ℵω]ℵ0, ⊆) > ℵω+1 requires large cardinals.

Ziqin Feng Around Hilbert’s 13th Problem

basic (K), K Compact

Theorem For any finite dimensional K: basic (K) ≤ cof ([w(K)]ℵ0, ⊆). For any infinite dimensional K: basic (K) ≥ c. For any K with a discrete subspace D, |D| = w(K): basic (K) ≥ cof ([w(K)]ℵ0, ⊆). Hence, for such K: either K is finite dimensional, and basic (K) = cof ([w(K)]ℵ0, ⊆),

• r K is infinite dimensional, and

basic (K) = |C(K)| = w(K)ℵ0.

Ziqin Feng Around Hilbert’s 13th Problem

Computable KST for R

Theorem Let n be a natural number. Take any γ ≥ 2n + 2, and let D = {k/γℓ : k, ℓ ∈ Z} be the set of all rationals base γ. There are ψpq for 1 ≤ q ≤ 2n + 1 and 1 ≤ p ≤ n which are defined constructively on D, such that for any f ∈ C(Rn), there is a constructive algorithm for computing g1, . . . , g2n+1 in C(R) such that: f (x) =

2n+1

• q=1

gq(φq(x)), where φq(x1, . . . , xn) =

n

• p=1

ψpq(xp). to within a specified error ǫ > 0 on any specified compact subset K of Rn.

Ziqin Feng Around Hilbert’s 13th Problem

Possible Applications

Function reconstruction Statistical Pattern Recognition Image Compression Multidimensional Signal Processing Neural networks Analog computing devices etc...

Ziqin Feng Around Hilbert’s 13th Problem

The Strong Topology

The strong topology on C(Rn) has basic open sets: B(f , ǫ) = {g ∈ C(Rn) : |f (x) − g(x)| < ǫ(x) ∀x ∈ Rn}, where ǫ ∈ C(Rn, (0, ∞)). Example CS(Rn) is Baire, not first countable. The polynomials are not dense. But the analytic functions are dense. Frequently used in differential topology, geometric analysis.

Ziqin Feng Around Hilbert’s 13th Problem

The Strong Topology

The strong topology on C(Rn) has basic open sets: B(f , ǫ) = {g ∈ C(Rn) : |f (x) − g(x)| < ǫ(x) ∀x ∈ Rn}, where ǫ ∈ C(Rn, (0, ∞)). Example CS(Rn) is Baire, not first countable. The polynomials are not dense. But the analytic functions are dense. Frequently used in differential topology, geometric analysis.

Ziqin Feng Around Hilbert’s 13th Problem

Strong Approximation

Theorem (n = 2) Given f ∈ C(R2) and ǫ ∈ C(R2, (0, ∞)), ∃ analytic functions of 1–variable α, β, gi

q and

ψi

pq for i = 1, 2, p = 1, 2 and q = 1, . . . , 5

such that

• f (x, y) −

f (x, y)

• < ǫ(x, y),

where,

• f (x, y)

= a(x, y) f1(x, y) + (1 − a(x, y)) f2(x, y), a(x, y) = α ((β(x) + β(y)) y) , and

• fi(x, y)

=

5

• q=1
• gi

q

• ψi

1q((x + 1)2 + (y − (−1)i))2)

+ ψi

2q((x − 1)2 + (y − (−1)i)2

.

Ziqin Feng Around Hilbert’s 13th Problem

Strong Approximation

Theorem (n = 2) Given f ∈ C(R2) and ǫ ∈ C(R2, (0, ∞)), ∃ analytic functions of 1–variable α, β, gi

q and

ψi

pq for i = 1, 2, p = 1, 2 and q = 1, . . . , 5

such that

• f (x, y) −

f (x, y)

• < ǫ(x, y),

where,

• f (x, y)

= a(x, y) f1(x, y) + (1 − a(x, y)) f2(x, y), a(x, y) = α ((β(x) + β(y)) y) , and

• fi(x, y)

=

5

• q=1
• gi

q

• ψi

1q((x + 1)2 + (y − (−1)i))2)

+ ψi

2q((x − 1)2 + (y − (−1)i)2

.

Ziqin Feng Around Hilbert’s 13th Problem

Strong Approximation — Rn

Every f in C(Rn) can be strongly approximated by functions which are: the superposition of analytic functions of 1–variable, along with + and ×, in a Fixed form.

Ziqin Feng Around Hilbert’s 13th Problem

Strong Universal PDE

Theorem For every n ≥ 2, there is an algebraic n–variable PDE Q

• . . . , ∂αf

∂xα , . . .

• = 0,

whose analytic solutions are strongly dense in C(Rn). (Here Q is a polynomial with integer coefficients.)

Ziqin Feng Around Hilbert’s 13th Problem

Prior Work

(Rubel): there is an algebraic ODE whose C ∞ solutions are dense in CS(R). (Boshernitzan): there is an algebraic ODE whose analytic solutions are dense in Ck(R). (Buck): for every n ≥ 2, there is an algebraic PDE of functions of n variables whose C ∞ solutions are dense in C(I n).

Ziqin Feng Around Hilbert’s 13th Problem

The End

Thank you!

Ziqin Feng Around Hilbert’s 13th Problem

Aim Now...

We will show that: every function which is the superposition of analytic functions of 1–variable, along with + and ×, in a fixed form, satisfies an algebraic PDE (depending only on the form).

Ziqin Feng Around Hilbert’s 13th Problem

A Simplified Example

Consider functions of the form f (x, y) = g(ψ1(x) + ψ2(y)) where g, ψ1, ψ2 are analytic. We show that any such f satisfies a (fixed) algebraic PDE. Set u1 = g′(ψ1(x) + ψ2(y)), u2 = ψ′

1(x),

u3 = ψ′

2(y),

u4 = g′′(ψ1(x) + ψ2(y)), u5 = ψ′′

1(x),

u6 = ψ′′

2(y), and

u7 = g(3)(ψ1(x) + ψ2(y)). Differentiate f relentlessly! Simplify using the ui’s.

Ziqin Feng Around Hilbert’s 13th Problem

A Simplified Example

Consider functions of the form f (x, y) = g(ψ1(x) + ψ2(y)) where g, ψ1, ψ2 are analytic. We show that any such f satisfies a (fixed) algebraic PDE. Set u1 = g′(ψ1(x) + ψ2(y)), u2 = ψ′

1(x),

u3 = ψ′

2(y),

u4 = g′′(ψ1(x) + ψ2(y)), u5 = ψ′′

1(x),

u6 = ψ′′

2(y), and

u7 = g(3)(ψ1(x) + ψ2(y)). Differentiate f relentlessly! Simplify using the ui’s. P(1) = ∂f ∂x = ψ′

1(x) · g′(ψ1(x) + ψ2(y)) = u2u1.

Ziqin Feng Around Hilbert’s 13th Problem

A Simplified Example

Consider functions of the form f (x, y) = g(ψ1(x) + ψ2(y)) where g, ψ1, ψ2 are analytic. We show that any such f satisfies a (fixed) algebraic PDE. Set u1 = g′(ψ1(x) + ψ2(y)), u2 = ψ′

1(x),

u3 = ψ′

2(y),

u4 = g′′(ψ1(x) + ψ2(y)), u5 = ψ′′

1(x),

u6 = ψ′′

2(y), and

u7 = g(3)(ψ1(x) + ψ2(y)). Differentiate f relentlessly! Simplify using the ui’s. P(2) = ∂f ∂y = ψ′

2(y) · g′(ψ1(x) + ψ2(y)) = u3u1.

Ziqin Feng Around Hilbert’s 13th Problem

A Simplified Example

Consider functions of the form f (x, y) = g(ψ1(x) + ψ2(y)) where g, ψ1, ψ2 are analytic. We show that any such f satisfies a (fixed) algebraic PDE. Set u1 = g′(ψ1(x) + ψ2(y)), u2 = ψ′

1(x),

u3 = ψ′

2(y),

u4 = g′′(ψ1(x) + ψ2(y)), u5 = ψ′′

1(x),

u6 = ψ′′

2(y), and

u7 = g(3)(ψ1(x) + ψ2(y)). Differentiate f relentlessly! Simplify using the ui’s. P(21) = ∂f ∂y∂x = ψ′

1(x) · ψ′ 2(y) · g′′(ψ1(x) + ψ2(y)) = u2u3u4.

Ziqin Feng Around Hilbert’s 13th Problem

A Simplified Example

Consider functions of the form f (x, y) = g(ψ1(x) + ψ2(y)) where g, ψ1, ψ2 are analytic. We show that any such f satisfies a (fixed) algebraic PDE. Set u1 = g′(ψ1(x) + ψ2(y)), u2 = ψ′

1(x),

u3 = ψ′

2(y),

u4 = g′′(ψ1(x) + ψ2(y)), u5 = ψ′′

1(x),

u6 = ψ′′

2(y), and

u7 = g(3)(ψ1(x) + ψ2(y)). Differentiate f relentlessly! Simplify using the ui’s. P(11) = ∂f ∂x∂x = u2

2u4 + u5u1.

Ziqin Feng Around Hilbert’s 13th Problem

A Simplified Example

Consider functions of the form f (x, y) = g(ψ1(x) + ψ2(y)) where g, ψ1, ψ2 are analytic. We show that any such f satisfies a (fixed) algebraic PDE. Set u1 = g′(ψ1(x) + ψ2(y)), u2 = ψ′

1(x),

u3 = ψ′

2(y),

u4 = g′′(ψ1(x) + ψ2(y)), u5 = ψ′′

1(x),

u6 = ψ′′

2(y), and

u7 = g(3)(ψ1(x) + ψ2(y)). Differentiate f relentlessly! Simplify using the ui’s. P(22) = ∂f ∂y∂y = u2

3u4 + u6u1.

Ziqin Feng Around Hilbert’s 13th Problem

A Simplified Example

Consider functions of the form f (x, y) = g(ψ1(x) + ψ2(y)) where g, ψ1, ψ2 are analytic. We show that any such f satisfies a (fixed) algebraic PDE. Set u1 = g′(ψ1(x) + ψ2(y)), u2 = ψ′

1(x),

u3 = ψ′

2(y),

u4 = g′′(ψ1(x) + ψ2(y)), u5 = ψ′′

1(x),

u6 = ψ′′

2(y), and

u7 = g(3)(ψ1(x) + ψ2(y)). Differentiate f relentlessly! Simplify using the ui’s. P(121) = ∂f ∂x∂y∂x = u3(u5u4 + u2

2u7).

Ziqin Feng Around Hilbert’s 13th Problem

A Simplified Example

Consider functions of the form f (x, y) = g(ψ1(x) + ψ2(y)) where g, ψ1, ψ2 are analytic. We show that any such f satisfies a (fixed) algebraic PDE. Set u1 = g′(ψ1(x) + ψ2(y)), u2 = ψ′

1(x),

u3 = ψ′

2(y),

u4 = g′′(ψ1(x) + ψ2(y)), u5 = ψ′′

1(x),

u6 = ψ′′

2(y), and

u7 = g(3)(ψ1(x) + ψ2(y)). Differentiate f relentlessly! Simplify using the ui’s. P(212) = ∂f ∂y∂x∂y = u2(u6u4 + u2

3u7).

Ziqin Feng Around Hilbert’s 13th Problem

A Simplified Example

Consider functions of the form f (x, y) = g(ψ1(x) + ψ2(y)) where g, ψ1, ψ2 are analytic. We show that any such f satisfies a (fixed) algebraic PDE. Set u1 = g′(ψ1(x) + ψ2(y)), u2 = ψ′

1(x),

u3 = ψ′

2(y),

u4 = g′′(ψ1(x) + ψ2(y)), u5 = ψ′′

1(x),

u6 = ψ′′

2(y), and

u7 = g(3)(ψ1(x) + ψ2(y)). Differentiate f relentlessly! Simplify using the ui’s. P(122) = ∂f ∂x∂y∂x = u2

3u2u7 + u6u4.

Ziqin Feng Around Hilbert’s 13th Problem

P(1) = u2u1 P(2) = u3u1 P(21) = u2u3u4 P(11) = u2

2u4 + u5u1

P(22) = u2

3u4 + u6u1

P(121) = u3(u5u4 + u2

2u7)

P(212) = u2(u6u4 + u2

3u7)

P(122) = u2

3u2u7 + u6u4

8 equations, but only 7 variables (u1, . . . , u7). Eliminate them!

Ziqin Feng Around Hilbert’s 13th Problem

To get a (rational) polynomial Q such that Q (. . . , Pα, . . .) = 0. In fact: P2

(1)P(2)P(122)−P(1)P2 (2)P(212)−P2 (1)P(12)P(22)+P2 (2)P(12)P(11) = 0.

And remember: Pα = ∂

∂x

α f . So f of the given form satisfies the PDE: Q(. . . , ∂ ∂x α f , . . .) = 0.

Ziqin Feng Around Hilbert’s 13th Problem

Counting Partial Derivatives

Let f (x, y) be a superposition of analytic functions of 1–variable, along with + and ×, in the fixed form of the previous theorem. Fix k. Differentiate f repeatedly wrt x, y up to order k. This gives N equations of the form: ∂ ∂x α f = Pα(u1, . . . , um), the ui (i = 1, . . . , m) are the functions arising by differentiating the given representation of f , and Pα are polynomials. N = O(k2), m = O(k) = ⇒ ∃k : N > m. Then there is a polynomial Q so that Q (. . . , Pα, . . .) = 0. And f satisfies the algebraic PDE: Q

• . . . ,

∂

∂x

α f , . . .

• = 0.

Ziqin Feng Around Hilbert’s 13th Problem