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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

1808.0. "Math Magazine 1430" by RUSURE::EDP (Always mount a scratch monkey.) Wed Oct 13 1993 13:26

    Proposed by David Doster, Choate Rosemary Hall, Wallingford,
    Connecticut.
    
    Solve the recurrence
    
                 x[n]^4 + 18x[n]^2 + 9
    	x[n+1] = ---------------------,		n >=0, x[0] = 2.
    	           4x[n]^3 + 12x[n]
T.RTitleUserPersonal
Name
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1808.1AUSSIE::GARSONHotel Garson: No VacanciesFri Oct 15 1993 01:583
    re .0
    
    Well it converges to sqrt(3) but that doesn't answer the question.
1808.2Newton-RaphsonHERON::BLOMBERGTrapped inside the universeFri Oct 15 1993 08:029
1808.3is in Hackmem...GAUSS::ROTHGeometry is the real life!Sat Oct 16 1993 11:5012
    One of the Hackmem items gives an analytic flow for Newton's method,
    which would produce a solution to this problem.  My copy isn't
    handy or I'd go look it up, I don't remember how the flow was
    derived exactly.

    An analytic flow is a continuous function of the "iteration number"
    for an iterative process - simple examples would be the exponential
    expression for the Fibonacci numbers, the Gamma function for the
    factorials, etc. but flows have been derived for many other
    processes as well,  A related subject is Schroeder functions.

    - Jim
1808.4RUSURE::EDPAlways mount a scratch monkey.Tue Oct 25 1994 17:0431
    Solution by Stephen C. Lock, Florida Atlantic University, Boca Raton,
    Florida.

    We will show that x[n] = p[n]/q[n], where p[n] and q[n] are integers
    such that p[n]+q[n]sqrt(3) = (2+sqrt(3)^4^n.

    Let x[n] = p[n]/q[n], for integers p[n] and q[n].  Then

		p[n+1]   p[n]^2+18*p[n]^2q[n]^2+9q[n]^4
		------ = ------------------------------.
		q[n+1]     4p[n]^3*q[n]+12p[n]q[n]^3

    It is natural to set p[n+1]=p[n]^4+18p[n]^2*q[n]^2+9q[n]^4, and
    q[n+1]=4p[n]^3*q[n]+12p[n]q[n]^3.

    Then

    p[n+1]+/-q[n+1]sqrt(3) =
    	p[n]^4+6p[n]^2*(q[n]sqrt(3))^2+(q[n]sqrt(3)^4 +/-
    		 4p[n]^3*(q[n]sqrt(3) +/- 4 p[n](q[n]sqrt(3)^3 =
    	(p[n]+/-q[n]sqrt(3)^4.

    Since we may choose p[0]=2 and q[0]=1, this establishes

    p[n]+q[n]sqrt(3)=(2+sqrt(3))^4^n, p[n]-q[n]sqrt(3)=(2-sqrt(3))^4^n.

    Solving for p[n] and q[n] yields

		       p[n]           (2+sqrt(3))^4^n + (2-sqrt(3))^4^n
		x[n] = ---- = sqrt(3) ---------------------------------.
		       q[n]           (2+sqrt(3))^4^n - (2-sqrt(3))^4^n
1808.5RTL::GILBERTSat Nov 12 1994 18:225
There's a minor typo in .-1, indicated below below the v.
			      v
		p[n+1]   p[n]^4+18*p[n]^2q[n]^2+9q[n]^4
		------ = ------------------------------.
		q[n+1]     4p[n]^3*q[n]+12p[n]q[n]^3