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 ) = e x 2 3 sin( y + z ) + 1 . 1 + x 2 + z 2 √· , sin, +, / . Composition of: exp, squaring, × , 3 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 ) = e x 2 3 sin( y + z ) + 1 . 1 + x 2 + z 2 √· , sin, +, / . Superposition of: exp, squaring, × , 3 Ziqin Feng Around Hilbert’s 13th Problem

Solving Functions for Polynomials Quadratic az 2 + bz + c → z 2 + pz + q Solution map: � p 2 − 4 q ), z ( a , b , c ) → z ( p , q ) = 1 2 ( − p + p = b / a , q = c / a . Quintic General Quintic → z 5 + 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 → z 7 + pz 3 + qz 2 + 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 az 2 + bz + c → z 2 + pz + q Solution map: � p 2 − 4 q ), z ( a , b , c ) → z ( p , q ) = 1 2 ( − p + p = b / a , q = c / a . Quintic General Quintic → z 5 + 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 → z 7 + pz 3 + qz 2 + 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 az 2 + bz + c → z 2 + pz + q Solution map: � p 2 − 4 q ), z ( a , b , c ) → z ( p , q ) = 1 2 ( − p + p = b / a , q = c / a . Quintic General Quintic → z 5 + 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 → z 7 + pz 3 + qz 2 + 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 of 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 of 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 z 7 + pz 3 + qz 2 + 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 φ 1 . . . , φ 2 n +1 in C ( I n ) such that 2 n +1 � ∀ f ∈ C ( I n ) f = g i ◦ φ i , for some g i ∈ C ( R ) . i =1 Step 2 Further, can choose the φ 1 , . . . , φ 2 n +1 such that: n � φ i ( x 1 , . . . x n ) = ψ ij ( x j ) j =1 Ziqin Feng Around Hilbert’s 13th Problem

Kolmogorov– Idea of Proof n = 2 Theorem (KST: n=2) Step 1 There exist φ 1 . . . , φ 5 in C ( I 2 ) such that 5 � ∀ f ∈ C ( I 2 ) f = g i ◦ φ i , for some g i ∈ C ( R ) . i =1 Step 2 Further, can choose the φ 1 , . . . , φ 5 such that: φ i ( x 1 , x 2 ) = ψ i 1 ( x 1 ) + ψ i 2 ( x 2 ) Ziqin Feng Around Hilbert’s 13th Problem

Ideas in Proof of KST: The ψ pq and φ q 0.10 0.08 0.06 0.04 0.02 0.4 0.3 0.4 0.2 0.3 0.1 0.2 0.1 Ziqin Feng Around Hilbert’s 13th Problem

Proof of KST: Contour Map of φ φ = φ ( x 1 , x 2 ) 0.5 0.4 0.3 0.2 0.1 0.0 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 0.10 0.08 0.06 0.04 0.02 0.00 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 0.10 0.08 0.06 0.04 0.02 0.00 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 0.056 0.054 0.052 0.050 0.048 0.046 0.044 0.042 0.040 0.20 0.22 0.24 0.26 0.28 0.30 Ziqin Feng Around Hilbert’s 13th Problem

Proof of KST: Looking Deeper at ψ 1e 8+3.999999e 2 2.5 2.0 1.5 1.0 0.20 0.22 0.24 0.26 0.28 Ziqin Feng Around Hilbert’s 13th Problem

Proof of KST: Looking Deeper at ψ 1e 8+3.999999e 2 2.5 2.0 1.5 1.0 0.20 0.22 0.24 0.26 0.28 Ziqin Feng Around Hilbert’s 13th Problem

Proof of KST: Approximate f φ f 0.5 0.5 0.4 0.4 0.3 0.3 0.2 0.2 0.1 0.1 0.0 0.0 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 g ◦ φ , ( � g ◦ φ � ≤ � f � ) 0.5 0.4 f 0.5 0.3 0.4 0.2 0.3 0.1 0.2 0.0 0.0 0.1 0.2 0.3 0.4 0.5 g 0.1 3 2 1 0.0 0.0 0.1 0.2 0.3 0.4 0.5 0 1 2 Ziqin Feng Around Hilbert’s 13th Problem 3 0.00 0.02 0.04 0.06 0.08 0.10 0.12

Proof of KST: Approximate f g ◦ φ , ( � g ◦ φ � ≤ � f � ) 0.5 0.4 f 0.5 0.3 0.4 0.2 0.3 0.1 0.2 0.0 0.0 0.1 0.2 0.3 0.4 0.5 g 0.1 3 2 1 0.0 0.0 0.1 0.2 0.3 0.4 0.5 0 1 2 Ziqin Feng Around Hilbert’s 13th Problem 3 0.00 0.02 0.04 0.06 0.08 0.10 0.12

Proof of KST: Replicate, Shift . . . φ 1 φ 2 φ 3 φ 4 φ 5 0.5 0.5 0.5 0.5 0.5 0.4 0.4 0.4 0.4 0.4 0.3 0.3 0.3 0.3 0.3 0.2 0.2 0.2 0.2 0.2 0.1 0.1 0.1 0.1 0.1 0.0 0.0 0.0 0.0 0.0 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 g 1 ◦ φ 1 g 2 ◦ φ 2 g 3 ◦ φ 3 g 4 ◦ φ 4 g 5 ◦ φ 5 0.5 0.5 0.5 0.5 0.5 0.4 0.4 0.4 0.4 0.4 0.3 0.3 0.3 0.3 0.3 0.2 0.2 0.2 0.2 0.2 0.1 0.1 0.1 0.1 0.1 0.0 0.0 0.0 0.0 0.0 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 � 5 1 f 3 g q ◦ φ q q =1 0.5 0.5 0.4 0.4 0.3 0.3 0.2 0.2 0.1 0.1 0.0 0.0 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