Roots of polynomials


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Roots of polynomials

  1. 1. Maria Fernanda Vergara Mendoza Petroleum Engineering UIS-COLOMBIA
  2. 2. <ul><li>In this chapter, you will learn some methods to find the roots of polynomial equations of the general form: </li></ul><ul><li>Where n= the order of the polynomial; a= constant coefficients. </li></ul><ul><li>RULES: </li></ul><ul><li>For an n th-order equation, there are n real or complex roots. </li></ul><ul><li>If n is odd, there is at least one real root </li></ul><ul><li>The complex roots exsist in conjugate pairs (a+bi and a-bi), i=√(-1) </li></ul>
  3. 4. <ul><li>The Muller’s method, is like the secant method, just that this one projects a parabola through three points unlike secant method, who projects a straight line. </li></ul><ul><li>This method consists of deriving the coefficients of the parabola that goes through the three points. </li></ul>
  4. 5. <ul><li>Write the parabolic equation in this form: </li></ul><ul><li>The coefficients a, b, and c can be evaluated by substituting each of the three points to give: </li></ul>
  5. 6. <ul><li>Two of the terms of are zero, it can be solved for c=f(x i+1 ). </li></ul><ul><li>Using algebraic manipulations, we solve the remaining coefficients: </li></ul>
  6. 7. <ul><li>These can be substituted to give: </li></ul><ul><li>The results can be summarized as </li></ul><ul><li>Once you know the approximate coefficients you have to find the approximated root using the quadratic equation : </li></ul>
  7. 8. <ul><li>The error can be calculated as: </li></ul><ul><li>There is a problem with equation, this equation yields two roots, in this method the sign is chosen with this strategies: </li></ul><ul><li>1. If only real roots are being located, we choose the two original points that are nearest the new root estimate, x i+2 . </li></ul><ul><li>If both real and complex roots are being evaluated, a sequential approach is employed. That means: x i , x i+1 , x i+2 take the place of x i-1 , x i , x i+1 </li></ul>
  8. 9. If you have as initial values respectively, find the root of the equation: FIRST: Evalue the equation in its initial values
  9. 10. SECOND: This values are used to calculate: THIRD: Find the a, b, c coefficients:
  10. 11. The error is: This is a huge error, so its necesary to do other iterations: Repeat the calculations and get a low percent of error: Iteration Xr Ea% 0 5 -- 1 3.976487 25.74 2 4.00105 0.6139 3 4 0.0262 4 4 0.0000119
  11. 12. <ul><li>Is an iterative approach related loosely to both the Muller and Newton Raphson methods. </li></ul><ul><li>It is based on the idea of synthetic division of the given polynomial by a quadratic function and can be used to find all the roots of a polynomial. </li></ul><ul><li>The idea is to do a synthetic division of the polynomial P n (x) by the quadratic factor (x 2 - rx - s). </li></ul>
  12. 13. <ul><li>The synthetic division can be extended to quadratic factors: </li></ul><ul><li>When you multiply and match factors have: </li></ul>
  13. 14. <ul><li>The idea is to find values of r and s, making b 1 and b 0 zero. </li></ul><ul><li>The method works taking an initial approach (r 0, s 0 ) and getting better approaches (r k , s k ), this is an iterative procedure, the process ends when the residue of dividing the polynomial by (x 2 - r k x - s k ) its zero. </li></ul><ul><li>B 1 =f(s, r) </li></ul><ul><li>B 0 =g(s, r) </li></ul>
  14. 15. <ul><li>Because both b o and b 1 are functions of both r and s, they can be expanded using a Taylor series: </li></ul><ul><li>The changes, Δr and Δs, can be estimated by setting the expansion equal to zero: </li></ul>
  15. 16. <ul><li>“ If the partial derivatives of the b’s can be determined, these are a system of two equations that can be solved simultaneously for the two unknowns, Δr and Δs.” </li></ul><ul><li>According to Bairstow, the partial derivatives can be obtained by a synthetic division of the b’s. </li></ul>
  16. 17. <ul><li>Then the system of equations can be written as: </li></ul>
  17. 18. <ul><li>APPROXIMATED ERROR </li></ul><ul><li>When both of these error estimates fall below a stopping criterion, the values of the roots can be determined by: </li></ul>
  18. 19. <ul><li>Employ Bairstow’s method to determine the roots of the polynomial </li></ul><ul><li>Use initial guesses of r=s=-1 and iterate to a level of tolerance of 1% </li></ul><ul><li>SOLUTION: </li></ul><ul><li>b 5 =1 b 4 =-4.5 b 3 =6.25 b 2 =0.375 b 1 =-10.5 b 0 =11.375 </li></ul><ul><li>c 5 =1 c 4 =-5.5 c 3 =10.75 c 2 =-4.875 c 1 =-16.375 </li></ul><ul><li>Thus, the simultaneous equations to solve Δr and Δs are : </li></ul>
  19. 20. <ul><li>Which can be solved for Δr=0.3558 and Δs=1.1381. </li></ul><ul><li>r=-0.6442 </li></ul><ul><li>S=0.1381 </li></ul><ul><li>And the approximate errors are: </li></ul><ul><li>The computation can be continued with the result that after four iterations the metod converges on velues of r=-0.5 and s=0.5 </li></ul>
  20. 21. <ul><li>CHAPRA, Steven C. “Numerical methods for engineers”, Fifth edition. Mc Graw Hill. </li></ul><ul><li>CARRILLO, Eduardo. “Raices de polinomios”. PPT. </li></ul>