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Numerical Methods and Computational Techniques

Numerical Methods and Computational Techniques. Solution of Transcendental and Polynomial Equations. Dictionary meaning of Transcendent: More than ordinary, Supernatural, Superlative Transcendental Number: A number which is not ordinary. For example : Pi ( π ) = 3.14159… approximately.

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Numerical Methods and Computational Techniques

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  1. Numerical Methods and Computational Techniques Solution of Transcendental and Polynomial Equations

  2. Dictionary meaning of Transcendent: More than ordinary, Supernatural, Superlative Transcendental Number: A number which is not ordinary. For example : Pi (π) = 3.14159… approximately Or e Hence f (x) involving ex or ln (x) are transcendental functions. Further Sin (x) = Cos (x) = Hence f (x) involving sin (x), cos (x) etc are also transcendental functions.

  3. Root of transcendental or polynomial function is the value of x for which f (x) = 0 In graphical form it is point of intersection of graph of f (x) with X – axis. f (x) roots

  4. Numerical Methods to determine the roots: • Birge – Vieta Method • Bairstow Method • Bisection Method • False Position Method • Simple Fixed Point Iteration Method • Newton Raphson (Tangent) Method • Secant Method

  5. Birge – Vieta Method: • Used for finding roots of polynomial functions. • Uses “synthetic division” of polynomial to extract factor of the given polynomial in the form of (x – p). Problem: Find roots of f (x) = 2x³ – 5x + 1 using Birge – Vieta Method. Solution: Assume that x = 1 is root of the equation. Hence initial approximation of the solution is p0 = 1. Synthetic Division will be performed as below: Let f (x) = a0x3 + a1x2 + a2x + a3 p0 a0 a1 a2 a3 p0b0 p1b1 p2b2 p0 b0 b1=a1+p0b0 b1 b2 b3 p1 = p0 – b3/c2 s i m i l a r l y Repeat synthetic division using p1 c0 c1 c2 c3

  6. Iteration No. 1: 1 2 0 -5 1 2 2 -3 1 2 2 -3 -2 2 4 1 2 4 1 -1 p1 = p0 – b3/c2 = 1 – (-2)/1 = 3 Iteration No. 2: 3 2 0 -5 1 6 18 39 Not required 3 2 6 13 40 6 36 147 2 12 49 187 p2 = p1 – b3/c2 = 3 – 40/49 = 2.1837

  7. Iteration No. 3: 2.1837 2 0 -5 1 4.3674 9.5371 9.9076 2.1837 2 4.3674 4.5371 10.9076 4.3674 19.0742 2 8.7348 23.6113 p3 = p2 – b3/c2 = 2.1837 – 9.9076/23.6113 = 1.7217 Iteration No. 4: 1.7217 2 0 -5 1 3.4434 5.9285 1.5986 1.7217 2 3.4434 0.9285 2.5986 3.4434 11.857 2 6.8868 12.785 p4 = p3 – b3/c2 = 1.7217 – 2.5986/12.785 = 1.5185

  8. Iteration No. 5: 1.5185 2 0 -5 1 3.037 4.6117 -0.5896 1.5185 2 3.037 -0.3883 0.4104 3.037 9.2234 2 6.074 8.8351 p5 = p4 – b3/c2 = 1.5185 – 0.4104/8.8351 = 1.4721 Iteration No. 6: 1.4721 2 0 -5 1 2.9442 4.3342 -0.9801 1.4721 2 2.9442 -0.6658 0.01986 2.9442 8.6683 2 5.8884 8.0025 p6 = p5 – b3/c2 = 1.4721 – 0.01986/8.0025 = 1.469624

  9. Click here to download Excel sheet for performing synthetic division and find pn. Thus one of the roots of given f (x) is x = 1.47114 To verify: f (1.47114) = 2* 1.471143 – 5* 1.47114 + 1 = 0.012138021435088 ≈ 0 Performing further division by x = 1.47114, we get the deflated polynomial as; 1.47114 2 0 -5 1 2.94228 4.3285 -0.9879 2 2.94228 -0.6715 0.01213 ≈ 0 Therefore f (x) = (2x2 + 2.94228x – 0.6715) * (x –1.47114) If (2x2 + 2.94228x – 0.6715) = 0 then x = 0.2008 or x = -1.67195 To verify: f (0.2008) = 0.012192769 ≈ 0 and f (-1.67195) = 0.012155754 ≈ 0 Thus f (x) has three roots as x = 1.47114, x = 0.2008 and x = -1.67195

  10. 3. Bisection Method: Used for finding roots of polynomial functions or transcendental functions. Uses two guess points which are on either sides of the root. Problem: Find roots of f (x) = 2.5x³ – 17x² + 22x + 11 using Bisection Method. Solution: Let xa = 2 and xb = 4 so that f(xa) * f(xb) is negative. Hence the required root lies between the initial two guesses. Now xr = (xa + xb) / 2 If f(xr)*f(xa) is negative then xb=xr else xa=xr R e p e a t Graphically: xr xb xb xa xr

  11. Hence x = 2.427735 is root of the equation. f(2.427735) = -0.0140314332742740625 ≈ 0

  12. 4. False Position Method: Used for finding roots of polynomial functions or transcendental functions. Uses two guess points which are on either sides of the root. Normally, Faster than Bisection Method. Problem: Find roots of f (x) = 3x³ – 20x² + 25x + 15 using False Position Method. Solution: Let xa = 3 and xb = 6 so that f(xa) * f(xb) is negative. Hence the required root lies between the initial two guesses. If f(xr)*f(xa) is negative then xb=xr else xa=xr R e p e a t Graphically: xr xb xb xa xr

  13. Hence x = 4.748827 is root of the equation. f(4.748827) = ******* ≈ 0

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