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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

806.0. "Even terms of a polynomial (from USENET)" by CLT::GILBERT (Builder) Wed Dec 23 1987 15:10

I don't know the answers to these questions...
 
If  p  is a multivariate polynomial (or formal power series), let even(p)
be the polynomial (or f.p.s.) consisting of just those terms of  p  which have
even degree in each of the variables.
Example:  even(13*x^2*y^6 + z - 7*z^4 - 2*x^2*y^3) = 13*x^2*y^6 - 7*z^4
 
Q1 (vague): What nice things can be said about the function even()?
 
Q2:  Let  n  be an arbitrary positive integer.
Let  mu(k) = x[1]^k + x[2]^k + ... + x[n]^k  for nonnegative integer k.
 
What is  even(mu(1)^r)?
 
[Obviously zero if r is odd.  If r is even, it can be written in terms of
 even mu's:
        even(mu(1)^2) = mu(2)
        even(mu(1)^4) = 3*mu(2)^2 - 2*mu(4)
        even(mu(1)^6) = 15*mu(2)^3 - 30*mu(2)*mu(4) + 16*mu(6)
 What is the general expression?]
 
Q3:  What about even(mu(1)^r*mu(3)^s), and so on?  It seems that all such
things can be written in terms of even mu's, but how?
 
Q4:  What about even( exp(mu(1)^r) )?  [Interpret exp() as its formal power
series.]  Believe it or not, I have an application for this one.
 
- - - - - -
Please e-mail any suggestions (even if you post them, as news often gets
lost on the way here).   Thanks.
 
Brendan McKay
 
Paper:     Computer Science Department, Australian National University,
           GPO Box 4, Canberra, ACT 2601, Australia.
Telephone: + 61 62 49 3845                Telex: AA 62760 NATUNI
ACSnet:    bdm@anucsd.oz                  ARPA:  bdm%anucsd.oz@uunet.uu.net
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UUCP: {uunet,ubc-vision,ukc,mcvax,prlb2,hplabs,enea,mulga}!munnari!anucsd!bdm
BITNET:  anucsd.oz!bdm@uunet.uu.net (or similar)
T.RTitleUserPersonal
Name		DateLines
806.1Incremantal helpAKQJ10::YARBROUGHWhy is computing so labor intensive?Thu Apr 07 1988 15:3114
Q2:  Let  n  be an arbitrary positive integer.
Let  mu(k) = x[1]^k + x[2]^k + ... + x[n]^k  for nonnegative integer k.
 
What is  even(mu(1)^r)?
 
[Obviously zero if r is odd.  If r is even, it can be written in terms of
 even mu's:
        even(mu(1)^2) = mu(2)
        even(mu(1)^4) = 3*mu(2)^2 - 2*mu(4)
        even(mu(1)^6) = 15*mu(2)^3 - 30*mu(2)*mu(4) + 16*mu(6)
 What is the general expression?]

Messy.
        even(mu(1)^8) = 35*mu(2)^4 - 70*mu(4)^2 - 112*mu(2)*mu(6) + 148*mu(8)
806.2call me crazy, but ...ZFC::DERAMOTrust me. I know what I'm doing.Thu Apr 07 1988 16:2014
    Did you notice in .-1 that all of the right hand side coefficients
    except the last, and the last coefficient minus one, are all divisible
    by one less than "r"?
    
        even(mu(1)^2) = mu(2)
    by 1                0
        even(mu(1)^4) = 3*mu(2)^2 - 2*mu(4)
    by 3                3          -3
        even(mu(1)^6) = 15*mu(2)^3 - 30*mu(2)*mu(4) + 16*mu(6)
    by 5                15          -30               15
        even(mu(1)^8) = 35*mu(2)^4 - 70*mu(4)^2 - 112*mu(2)*mu(6) + 148*mu(8)
    by 7                35          -70          -112               147
    
    Dan
806.3Correction and partial general solutionSSDEVO::LARYFri Apr 08 1988 02:1157
>      even(mu(1)^8) = 35*mu(2)^4 - 70*mu(4)^2 - 112*mu(2)*mu(6) + 148*mu(8)

I believe this is wrong, although it does work for binomials and trinomials.

The even terms in mu(1)^8 fall into five classes:

	 8	
1)	a , where a is one of the variables in mu(1)

	   6 2
2)	28a b , where a and b are two variables in mu(1)

	   4 4
3)	70a b ,       "

	    4 2 2
4)	420a b c , where a, b, and c are variables in mu(1)

	     2 2 2 2
5)	2520a b c d , where a, b, c, and d are variables in mu(1)

If even(mu(1)^8) is expressed as sums of weighted products of mu(2*k), you
wind up with five simultaneous equations for the coefficients of each product.
in the above case, if you use the four products from reply .1 the equations
come out overconstrained and no solution is possible. If you introduce the
additional product mu(2)^2*mu(4), you can solve the system to get:

even(mu(1)^8) = 105*mu(2)^4 - 420*mu(2)^2*mu(4) + 448*mu(2)*mu(6)
		+ 140*mu(4)^2 - 274*mu(8)

In general, the product terms you use in the right-hand side of the equation
correspond to the possible product terms in the even(mu(1)^r) expansion; i.e.
		      8				 2 2 2 2
mu(8) corresponds to a , mu(2)^4 corresponds to a b c d , etc.
What you wind up with is P(r/2) simultaneous equations in an equal number of
unknowns, where P(n) is the number of ways n can be partitioned into distinct
sets of summands (these summands being one-half the exponents in the terms of
even(mu(1)^r)). The equations are sparse and should be easy to solve, but I
couldn't come up with a general solution. I did find that the coefficient
of mu(2)^(r/2) is always positive and given by:

			     r!
			-----------
				r/2
			(r/2)!*2


and, for r > 2, the coefficient of mu(2)^(r/2-2)*mu(4) is always negative
and given by:

			    - r!
			------------
			   r/2
			3*2   *(r/2-2)!

but after that it gets complicated....