Mathematical Analysis of Variable Function Compositions

x2 x 0 x 1 0 1 0 5 n.w
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Explore the step-by-step calculations involving various mathematical functions with neglecting and rejecting certain variables. Discover the outcomes of function compositions and their values at specific inputs. Follow the progression through a series of mathematical operations to understand the results obtained.

  • Mathematics
  • Function Compositions
  • Variable Analysis
  • Mathematical Operations
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  1. x2 = (x0+ x1) = (0 + 1) = 0.5

  2. F(0.5) = -0.25 is ve

  3. We neglect x1and we have

  4. X3 = (x0+ x2) = (0 + 0.5) = 0.25

  5. F(0.25) = 0.3125 ve

  6. We neglect x0and we have

  7. x4= (x2+ x3) = (0.5 + 0.25) = 0.375

  8. F (0.375) is +ve

  9. We reject x3and we have

  10. x5= (x2+ x4) = (0.5 + 0.375) = 0.4375

  11. F(0.4375) is ve

  12. We reject x2, and have

  13. x6= (x4+ x5) = (0.375 + 0.4375)

  14. F(x0) = +1

  15. F(x1) = -1

  16. X0= 9nx1= bn (x0+ x1) F (cn) 0

  17. 0

  18. 0.5

  19. 0.5

  20. 0.375 1

  21. 0.5

  22. 0.25

  23. 0.375

  24. 0.4375 0.5

  25. 0.25

  26. 0.375

  27. 0.4375

  28. 0.40625-0.25 ve

  29. 0.3125 +ve

  30. 0.0156 +ve

  31. -0.12109 -ve

  32. This is linear function x(y) with x(y0) = x0and x(y1) = x1

  33. The line intersects the x axis at the point obtained by putting y = 0 in (3.4)

  34. -y1 -y0

  35. y0 y1 y1 y0

  36. = x0y1 x1y0

  37. y1 y0

  38. This may be re-written on a linear interpolation form as x0F(x1) x1 F(x0)

  39. F(x1) F(x0)

  40. This can also be written as

  41. X2= x1 (x1 x0) F (x1)

  42. F (xi) F(x0)

  43. iterative scheme results. As in the case of binary search (bisection rule) the two estimates slid give opposite signs in F(x).

  44. The convergence in this method is faster than that in bisection rule.

  45. Example F(x) = x2 3x + 1 = 0

  46. Choose x0= 0

  47. x1= 1

  48. Since F(x0) is +ve = 1

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