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How to solve multiple power equations?

Discussion in 'Electronic Design' started by Don Lancaster, Mar 15, 2007.

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  1. What is the approach to solving, say, x^7 - y^7 = k (x^5 - y^5) ?

    In general totally independent polynomial simultaneous equations are not
    directly solvable above order 4. But special cases abound if there are
    interrelationships between the variables.

    What I really need is a deterministic (non iterative) solution to an
    equation set of form....

    w^1 - x^1 + y^1 - z^1 = k0
    w^3 - x^3 + y^3 - z^3 = k0*k1
    w^5 - x^5 + y^5 - z^5 = k0*k1*k2
    w^7 - x^7 + y^7 - z^7 = k0*k1*k2*k3

    preferably sanely extensible to w^27 and 14x14.

    Note that all leftside values have +- unity coefficients.
    k1 through k3 are ratios of small integers.

    the range of w through z is always 0 through 1.
    k0 also ranges from 0 to 1.

    Other approaches to the problem strongly suggest that a deterministic
    solution does in fact exist.


    --
    Many thanks,

    Don Lancaster voice phone: (928)428-4073
    Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
    rss: http://www.tinaja.com/whtnu.xml email:

    Please visit my GURU's LAIR web site at http://www.tinaja.com
     
  2. Robert Baer

    Robert Baer Guest

    It has been over 40 years since i was any good at seeing equation
    patterns / similarities by sight.
    So i may eaily be totally off the path with my first impression of
    Fourier Transform...or a hyperbolic "equivalent".
     
  3. The initial problem is indeed Fourier. Which we are trying to transform
    OUT of, not into.

    Curiously, an eighth order polynomial approximation to the true answer
    does show characteristic Fourier "ears". Instead of the Chebycheff
    equalripple I was sort of expecting.

    An approximate calculator appears as http://www.tinaja.com/demo28a.asp
    While fast and accurate, it is both iterative and presupposes one of
    what I hope are a class of solutions. It is also hard to change n.

    I am currently approaching finding a deterministic solution in four
    different ways. (a) making the iteration enough better that only one
    pass is needed. (b) improving the initial guess (bad because classes are
    excluded), (c) fitting eighth and higher order polynomials to the exact
    solutions and trying to find some underlying secrets behind them,

    And (d) our above technique of taking our basic equations, letting
    Chebycheff piss with them, and transforming them to the above power
    equations.

    I suspect (d) will be the hardest and the most useful.

    Another hint: the trig identity of cos(a+b) = cos(a)cos(b) -
    sin(a)sin(b) appears to play a crucial and pivotal role.





    --
    Many thanks,

    Don Lancaster voice phone: (928)428-4073
    Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
    rss: http://www.tinaja.com/whtnu.xml email:

    Please visit my GURU's LAIR web site at http://www.tinaja.com
     
  4. Robert Baer

    Robert Baer Guest

    It has been ages, but there is an electronic-based 11-term polynomial
    used as a Fourier approximation; the idea was all one needed to do was
    take 11 samples and turn the crank (ie: quick and dirty).
    Would that be a place to start?
     
  5. Gerry

    Gerry Guest

    I am not sure but
    x^7-y^7=(x-y)(x^6+x^5y+x^4y^2+x^3y^3+x^2y^4+xy^5+y^6)
    and
    x^5-y^5=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)
    so
    (x^6+x^5y+x^4y^2+x^3y^3+x^2y^4+xy^5+y^6)=k(x^4+x^3y
    +x^2y^2+xy^3+y^4)

    gives the sectic

    x^5(x+y)+(y^2-k)(x^4+yx^3+y^2x^2+y^3x+y^4)=0

    If we represent the roots for this sectic by
    (a1+ib1),(a1-ib1),(a2+ib2),(a2-ib2),(a3+ib3),(a3-ib3)
    (3 pairs complex conjugate roots)
    so
    -y=2(a1+a2+a3)
    2k-y^2=2(a1^2+a2^2+a3^2-b1^2-b2^2-b3^2)
    then
    k=3(a1^2+a2^2+a3^2)+4(a1a2+a1a3+a2a3)-(b1^2+b2^2+b3^2)

    If we represent the roots for this sextic by
    (a1+ib1),(a1-ib1),(a2+ib2),(a2-ib2),(a3+b3),(a3-b3)
    (2 pairs complex conjugate roots and 2 real roots)
    so
    -y=2(a1+a2+a3)
    2k-y^2=2(a1^2+a2^2)+(a3+b3)^2+(a3-b3)^2-b1^2-b2^2
    and
    k=3(a1^2+a2^2+a3^2)+4(a1a2+a1a3+a2a3)-(b1^2+b2^2)+b3^2

    I believe you can continue this for only real solutions of the sectic,
    but i am afraid this will not help.

    Gerry
     

  6. That is sort of what I am doing.

    Do you have a reference to this third party approximation?

    --
    Many thanks,

    Don Lancaster voice phone: (928)428-4073
    Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
    rss: http://www.tinaja.com/whtnu.xml email:

    Please visit my GURU's LAIR web site at http://www.tinaja.com
     
  7. Robert Baer

    Robert Baer Guest

    Did some snooping inmy "library"; i was wrong; it is 13 terms with
    another using 11 terms.
    Refer to Radiotron Designer's Handbook 1953, page 304 (near end of
    section 8: Fourier series and harmonics.
    They refer to Mouromtseff, I. E. and H. N. Kozanowski "A short-cut
    method for calculation of harmonic distortion in wave modulation", Proc.
    IRE 22.9 (Sept, 1934) 1090.
    Later, they mention the 11 points in place of the 13, and refer to
    chapter 13 sect 3(iv)D and Fig. 13.24 as well as RCA application note
    "Use of the plate family in vacuum tube power output calculations" No.
    78 (July, 1937).
    Chapter 13 sect 3(iv)C and D both are interesting (7 point and 11 point).
    If needed i might be able to scan and OCR but due to equations, etc
    probqbly better to print each scan to a PDF.
     
  8. Virgil

    Virgil Guest

    Note that when all the exponents are odd, as in the above, the
    subtractions can all be replaced by the corresponding additions with no
    loss of generality.

    I do not know how much work is involved in producing a Groebner basis
    for your set of equations, but you might consider it.
     
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