From: mckay@cs.concordia.ca (MCKAY john) Subject: Re: elementary symmetric functions Date: 19 Aug 2000 13:09:36 GMT Newsgroups: sci.math Summary: [missing] In article <399E7B19.BDB00A30@student.utwente.nl>, Wilbert Dijkhof wrote: >I've a question about the elementary symmetric >functions. > >Suppose we have n variables x_1 ... x_n and >define: > >s_1 = sum( x_j, j) >s_2 = sum( x_j*x_k, js_3 = sum( x_i*x_j*x_k, i... >s_n = x_1*x_2* ... * x_n > >Define the following functions: > >f_r = sum( (x_j)^r, j=1..n) > >It's possible to express f_j as a polynom in >s_1,...s_j with integer coefficients and j!*s_j >as a polynom in f_1,...,f_j with integers. > >For example: > >f_1 = s_1 >f_2 = s_1^2 - 2*s_2 >f_3 = s_1^3 - 3*s_1*s_2 + 3*s_3 >f_4 = s_1^4 - 4*s_1^2*s_2 + 2*s_2^2 > + 4*s_1*s_3 - 4*s_4 > >and > >s_1 = f_1 >2*s_2 = f_1^2 - f_2 >6*s_3 = f_1^3 - 3*f_1*f_2 + 2*f_3 >24*s_4 = f_1^4 - 6*f_1^2*f_2 + 8*f_1*f_3 > + 3*f_2^2 - 6*f_4 > > >I think the following is true (although I have no >clue how to prove it): > >Suppose we set f_t = p (an integer > 0) for all t and >divide both sides by j! (in j!*s_j = f_1^j - ...) >then we get >s_j = p!/(j! * (p-j)!) (an integer). > >Is this true, and how do you prove this? > >Wilbert The f_t are called power sums. The s_t are the elementary symmetric functions. They are related by a matrix (I think it is called the Kostka matrix) It is: a1 1 2a2 a1 1 3a3 a2 a1 1 4a4 a3 a2 a1 1 ... tat a[t-1] ..a1 [The first column is special. the others are constant on antidiagonals, the top right = 0.] The ak are the coefficients of q^k of a monic polynomial whose roots are of interest. They are the elementary symm fns. up to sign. The determinant of the above matrix give the power sum, f_t. The characteristic function of the matrix is known as a Faber polynomial. These are surveyed in an article by Curtis (?Curtiss) in Amer Math Monthly in ? 1978. They were discovered by Fber in 1903. They give the action of a Hecke operator on modular functions. Want more?? There is a simple recursive way to generate the Faber polynomial since they satisfy a Newton relation. See also Macdonald's standard reference book on Symmetric Polynomials. -- But leave the wise to wrangle, and with me the quarrel of the universe let be; and, in some corner of the hubbub couched, make game of that which makes as much of thee. ============================================================================== From: Wilbert Dijkhof Subject: Re: elementary symmetric functions Date: Sat, 19 Aug 2000 21:04:00 +0200 Newsgroups: sci.math MCKAY john wrote: > > In article <399E7B19.BDB00A30@student.utwente.nl>, > Wilbert Dijkhof wrote: > > > >Suppose we set f_t = p (an integer > 0) for all t and > >divide both sides by j! (in j!*s_j = f_1^j - ...) > >then we get > >s_j = p!/(j! * (p-j)!) (an integer). > > > >Is this true, and how do you prove this? > > > >Wilbert > > The f_t are called power sums. The s_t are the elementary > symmetric functions. > > They are related by a matrix (I think it is called the Kostka matrix) > > It is: > > a1 1 > 2a2 a1 1 > 3a3 a2 a1 1 > 4a4 a3 a2 a1 1 > ... > tat a[t-1] ..a1 > > [The first column is special. the others are constant on antidiagonals, > the top right = 0.] Accordingly to Eric the determinant of the matrix above equals f_t if a_j is replaced by s_j. "http://mathworld.wolfram.com/ElementarySymmetricFunction.html" > The ak are the coefficients of q^k of a monic polynomial whose roots are of > interest. They are the elementary symm fns. up to sign. So I don't understand this remark if you must substitute s_j for a_j. Are you sure you mean a monic polynomial and not an infinite polynomial? > The determinant of the above matrix give the power sum, f_t. > > The characteristic function of the matrix is known as a Faber polynomial. What's a Faber polynomial ? About the program you sent me, with t=2 (2x2-matrix): > e(2); 2 t - 2 a[1] > det(matrix([[a1-k,1],[a2,a1-k]])); 2 2 a1 - 2 a1 k + k - a2 Maybe a typo in your code? > These are surveyed in an article by Curtis (?Curtiss) in Amer Math Monthly > in ? 1978. They were discovered by Fber in 1903. > They give the action of a Hecke operator on modular functions. I've tried do hide it. But indeed these are questions arising from modular function-things I don't understand yet. I've to read about these operators yet :) > Want more?? > > There is a simple recursive way to generate the Faber polynomial since they > satisfy a Newton relation. > > See also Macdonald's standard reference book on Symmetric Polynomials. I will search this book on monday. Anyway this stuff about the Faber polynomial, how is this helpful solving the problem I have ? Wilbert