Cauchy distribution with p(z) = 1 1 * 1+2² is a solution for the case f(z) = ln(1+z²) and c = 2ln 2, and obtain the Lagrange multipliers in this case. You may assume that the methods of the calculus of variations apply on the infinite interval (-00,00). The values of the following integrals may be useful: dz 1 1+2² dz In(1+1²) 1+22 2x In 2. eigenvalue problem x²+2xy +(3x²+ + ( 3a λ (z². 2+1) 3 y= 0, y(-1)=y(1) = 0, with cigenvalue A, can be written as a constrained variational problem with functional dz S[y] = dr (z²² - 3x²²) and constraint 32 C[y]= dz 1, (1+z²)³ For a probability density function p(x) > 0 on the interval (-00,00) the entropy functional S[p] is given by Sp - dz p(z) log p(x). with boundary conditions y(-1)=y(1) = 0. (ii) Assume now that the system in part (b)(i) has eigenvalues > with <+1, and cigenfunctions (±), k = 1, 2, ………. Using the trial function (z; A) = A(1 - 2²) show that 32 Suppose that the function p(x) is subject to the constraints L dz p(z) = 1 and L dz f(x)p(x) = c, -00 briefly outlining the ideas of any theory you use: detailed proofs are not required. You may use the integral Li (1-2)² T dr -1 (1+x²)3 4 ?? 0 0 where f(x) is a fixed function and c is a constant. stationary path for S[p] is given by p(x) = exp(-1-λ – µƒ(±)), where A and are Lagrange multipliers.