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Integral Equations, Eigenvalue, Function Interpolation

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Integral Equations, Eigenvalue, Function Interpolation Integral equations, eigenvalue, function interpolation Marcin Chrząszcz [email protected] Monte Carlo methods, 26 May, 2016 1 /19 Marcin Chrząszcz (Universität Zürich) Integral equations, eigenvalue, function interpolation 1/19 . Integral equations, introduction V Fredholm integral equation of the second order: ∫ b ϕ(x) = f(x) + K(x, y)ϕ(y)dy a V The f and K are known functions. K is called kernel. V The CHALLENGE: find the ϕ that obeys the above equations. V There are NO numerical that can solve this type of equations! V Different methods have to be used depending on the f and K func- tions. V The MC algorithm: construct a probabilistic algorithm which has an expected value the solution of the above equations. There are many ways to build this! V We assume that the Neumann series converges! 2 /19 Marcin Chrząszcz (Universität Zürich) Integral equations, eigenvalue, function interpolation 2/19 . Integral equations, approximations V The following steps approximate the Fredholm equation: ∫ b ϕ0(x) = 0, ϕ1(x) = f(x) + K(x, y)ϕ0(y)dy = f(x) a ∫ ∫ b b ϕ2(x) = f(x) + K(x, y)ϕ1(y)dy = f(x) + K(x, y)f(y)dy a a ∫ ∫ ∫ ∫ ϕ (x) = f(x) + b K(x, y)dy = f(x) + b K(x, y)f(y)dy + b b K(x, y)K(y, z)f(z)dydz 3 a a a a V Now we put the following notations: ∫ b K(1)(x, y) = K(x, y) K(2)(x, y) = K(x, t)K(t, y)dt a V One gets: ∫ ∫ b b (1) (2) ϕ3(x) = f(x) + K (x, y)f(y)dy + K (x, y)f(y)dy a a 3 /19 Marcin Chrząszcz (Universität Zürich) Integral equations, eigenvalue, function interpolation 3/19 . Integral equations, approximations V Continuing this process: ∫ b − K(n)(x, y) = K(x, t)K(n 1)(t, y)dt a and the n-th approximation: ∫ ∫ b b (1) (2) ϕn(x) = f(x) + K (x, y)f(y)dy + K (x, y)f(y)dy+ a a∫ b ... + K(n)(x, y)f(y)dy a V Now going with the Neumann series: n → ∞: ∫ ∑n b (n) ϕ(x) = limn→∞ϕn(x) = f(x) + K (x, y)f(y)dy i=1 a V The above series converges only inside the square: a ¬ x, y ¬ b for: ∫ ∫ b b |K(x, y)|2dxdy < 1 a a 4 /19 Marcin Chrząszcz (Universität Zürich) Integral equations, eigenvalue, function interpolation 4/19 . Integral equations, algorithm V The random walk of particle happens on the interval (a, b): • In the t = 0 the particle is in the position x0 = x. • If the particle at time t = n − 1 is in the xn−1 position then in time t = n the position is: xn = xn−1 + ξn. The numbers ξ1, ξ2, ... are independent random numbers generated from ρ p.d.f.. • The particle stops the walk once it reaches the position a or b. • The particle life time is n when xn ¬ a and xn ­ b. • The expected life time is given by the equation: ∫ b τ(x) = ρ1(x) + [1 + τ(y)] ρ(y − x)dy a where: ∫ ∫ a−x ∞ ρq(x) = ρ(y)dy + ρ(y)dy −∞ b−x is the probability of particle annihilation in the time t = 1. 5 /19 Marcin Chrząszcz (Universität Zürich) Integral equations, eigenvalue, function interpolation 5/19 . Integral equations, algorithm 2 V The above can be transformed: ∫ b τ(x) = 1 + τ(y)ρ(x − y)dy (1) a V Now if p(x) is the probability that the particle in time t = 0 was in position x gets annihilated because it crosses the border a. V The probability obeys the analogous equation: ∫ b p(x) = ρ(x) + p(y)ρ(y − x)dy (2) a where ∫ a−x ρ(x) = ρ(y)dy −∞ is the probability of annihilating the particle in the first walk. V For the functions τ and ρ we got the integral Fredholm equation. V So the above random walk can be be used to solve the Equations 1 and 2. 6 /19 Marcin Chrząszcz (Universität Zürich) Integral equations, eigenvalue, function interpolation 6/19 . Integral equations, algorithm 3 V The ρ(x) is the p.d.f. of random variables ξn. V We observe the random walk of the particle. The trajectory: γ = (x0, x1, x2, ..., xn). This means for t = 0, 1, 2..., n − 1 and xn ¬ a or xn ­ b. Additionally we mark: γr = (x0, x1, ..., xr), r ¬ n. V We defined a random variable: ∑n S(x) = V (γr)f(xr−1) r=1 where V (γ0) = 1, K(xr−1, xr) V (γr) = V (γr−1) ρ(xr − xr−1) V One can prove that E [S(x)] treated as a function of x variable is the solution to the integral equation. 7 /19 Marcin Chrząszcz (Universität Zürich) Integral equations, eigenvalue, function interpolation 7/19 . Integral equations, algorithm 4 V We define a new random variable: { V (γn−r )f(xn−r ) , r ¬ n, ρ(xn−r ) cr(x) = 0, r > n where ρr(x) is defined as: ∫ ∫ a−x +∞ ρ1(x) = ρ(y)dy + ρ(y)dy, ∫−∞ ∫ b−x b b ρr(x) = ... ρ(x1 − x)ρ(x2 − x1)...ρ(xr−1 − xr−2)ρ1(xr−1)dx1...dxr−1 a a is the probability that the particle that is at given time in the x coordinate will survive r moments. V One can prove that E [cr(x)] treated as a function of x variable is the solution to the integral equation. 8 /19 Marcin Chrząszcz (Universität Zürich) Integral equations, eigenvalue, function interpolation 8/19 . Integral equations, general remark . There is a general trick: . Any integral equation can be transformed to linear equation using quadratic form. If .done so one can use the algorithms form lecture 8 to solve it. Bullet prove solution! 9 /19 Marcin Chrząszcz (Universität Zürich) Integral equations, eigenvalue, function interpolation 9/19 . Eigenvalue problem V The Eigenvalue problem is to find λ that obeys the equation: −→ −→ H x = λ x V For simplicity we assume there the biggest Eigenvalue is singular and it’s real. V The numerical method is basically an iterative procedure to find the biggest Eigen- value: • −→ We choose randomly a vector x 0. • The m vector we choose accordingly to formula: −→ −→ x m = H x m−1/λm where λm is choose such that ∑n −→ |( x m)j | = 1 j=1 −→ −→ the ( x )j is the j coordinate of the x vector, j = 1, 2, 3, ..., n V The set λm is converging to the largest Eigenvalue of the H matrix. 10 /19 Marcin Chrząszcz (Universität Zürich) Integral equations, eigenvalue, function interpolation 10/19 . Eigenvalue problem V From the above we get: ∑n −→ m−→ m−→ λ1λ2...λm( x j ) = (H x 0)j ; λ1λ2...λm = (H x 0)j j=1 V For big k and m > k one gets: ∑ n m−→ (H x 0)j − ∑j=1 = λ λ ...λ ≈ λm k n k−→ k+1 k+2 m j=1(H x 0)j from which: [∑ −→ ] 1 n m m−k (H x 0)j λ ≈ ∑j=1 n k−→ j=1(H x 0)j m−→ V This is the Eigenvalue estimation corresponding to H x 0 for sufficient large m. 11 /19 Marcin Chrząszcz (Universität Zürich) Integral equations, eigenvalue, function interpolation 11/19 . Eigenvalue problem, probabilistic model V Let Q = (qij ), i, j = 1, 2, ..., n is the probability matrix: ∑n qij ­ 0, qij = 1 j=1 V We construct a random walk on the set: {1, 2, ....n} accordingly to the below rules: • In the t = 0 the particle is in a randomly chosen state i0 accordingly to binned p.d.f.: pj . • If in the moment t = n − 1 the particle is in in−1 state then in the next moment it goes to the state in with the probability qin−1j . • For γ = (i0, i1, ...) trajectory we define a random variable: −→ ( x )i0 hi1i0 hi2i1 hi3i2 ...hir ir−1 Wr(γ) = pi0 qi1i0 qi2i1 qi3i2 ...qir ir−1 V Now we do: E [W (γ)] − m ≈ λm k E [Wk(γ)] V So to estimate the largest Eigenvalue: 12 /19 Marcin Chrząszcz (Universität Zürich) Integral equations, eigenvalue, function interpolation 12/19 . [ ] 1 W (γ) m−k λˆ = m Wk(γ) . Function interpolation V Lets put f(x1) = f1, f(x2) = f2, which we know the functions. V The problem: calculate the f(p) for x1 < p < x2. V From the interpolation method we get: p − x1 x2 − p f(p) = f2 + f1 x2 − x1 x2 − x1 V I am jet-lagged writing this so let me put: x1 = 0 and x2 = 1: f(p) = (1 − p)f1 + pf2 V For 2-dim: ∑ f(p1, p2) = r1r2f(δ1, δ2) δ where: { 1 − pi, δi = 0 ri = pi, δi = 1 V the sum is over all pairs (in this case 4). 13 /19 Marcin Chrząszcz (Universität Zürich) Integral equations, eigenvalue, function interpolation 13/19 . Function interpolation V For n-dim we get a monstrous: ∑ f(p1, p2, ..., pn) = r1r2...rnf(δ1, ..., δn) δ the sum is over all combinations (δ1, ..., δn), where δi = 0, 1. V The above sum is over 2n terms and each of it has (n + 1) temrs. It’s easy to imagine that for large n this is hard... Example n = 50 then we have 1014 ingredients. V There has to be a better way to do this! V From construction: ∑ 0 ¬ r1r2...rn ¬ 1, r1r2...rn = 1 δ V We can treat the ri as probabilities! We define a random varaible: ξ = (ξ1, ..., ξn) such that: P(ξi = 0) = 1 − pi, P(ξi = 1) = pi The extrapolation value is then equal: f(p1, p2, ..., pn) = E [f(ξ1, ..., ξn)] 14 /19 Marcin Chrząszcz (Universität Zürich) Integral equations, eigenvalue, function interpolation 14/19 .
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