A quick and clever method for solving nonlinear systems of equations

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SUMMARY

The discussion focuses on solving the polynomial equation s^6 + as^5 + bs^4 + cs^3 + ds^2 + es + k by leveraging the roots of the polynomial expressed as (s + alpha)(s + beta)(s + gamma)(s + lambda)(s^2 + 2w*zeta*s + w^2). To efficiently find the constant term "k," participants emphasize that substituting s = 0 into the expanded product simplifies the process, allowing for the identification of k without fully multiplying out the polynomial. This method significantly reduces the complexity and time required for solving the nonlinear system of equations.

PREREQUISITES
  • Understanding of polynomial equations and their coefficients
  • Familiarity with nonlinear systems of equations
  • Knowledge of polynomial multiplication and expansion
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study polynomial root-finding techniques
  • Learn about the properties of polynomial coefficients
  • Explore methods for solving nonlinear equations efficiently
  • Investigate the use of symbolic computation tools for polynomial manipulation
USEFUL FOR

Mathematicians, engineers, and students involved in algebra, particularly those working with polynomial equations and nonlinear systems. This discussion is beneficial for anyone looking to enhance their problem-solving skills in polynomial analysis.

royzizzle
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if we are given a polynomial s^6+as^5+bs^4+cs^3+ds^2+es+k

(if a,b,c,d are known) what is a clever method to solving for the value k if we are given the following:

the above polynomial is equal to the following(zeta is given as some constant, say 1 for simplicity):
(s+alpha)(s+beta)(s+gamma)(s+lamda)(s^2+2w*zeta*s+w^2)

we would have to multiply out this expression that separate out coefficients for s and equation to a,b,c, and d

we then have a system of 6 nonlinear equations to solve for 6 unknowns

what is the fastest way to do this by hand within a 20 minute timeframe?
 
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royzizzle said:
we would have to multiply out this expression that separate out coefficients for s and equation to a,b,c, and d

If you only want to know the value of the constant term "k", you'd only have to think about all the terms in the product that can be formed without have any variable "s" in them. You wouldn't have to multiply out the whole product.

If you only want to know the value of the constant term "k", you could substitute s = 0 in the second function.
 

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