How to solve multiple power equations?

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".

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

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.

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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: don@tinaja.com

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

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?

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

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: don@tinaja.com

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

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.

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.