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 e x 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: 1.Birge – Vieta Method 2.Bairstow Method 3.Bisection Method 4.False Position Method 5.Simple Fixed Point Iteration Method 6.Newton Raphson (Tangent) Method 7.Secant Method

5. b 1 =a 1 +p 0 b 0 1.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 p 0 = 1. Synthetic Division will be performed as below: Let f (x) = a 0 x 3 + a 1 x 2 + a 2 x + a 3 p0p0 a0a0 a1a1 a2a2 a3a3 b0b0 b1b1 b2b2 b3b3 c0c0 c1c1 c2c2 c3c3 p0b0p0b0 p1b1p1b1 p2b2p2b2 p0p0 s i m i l a r l y p 1 = p 0 – b 3 /c 2 Repeat synthetic division using p 1

6. 120-51 22-3-2 241 1 22-3 241 Iteration No. 1: 320-51 261340 21249187 3 61839 636147 Iteration No. 2: p 1 = p 0 – b 3 /c 2 = 1 – (-2)/1 = 3 Not required p 2 = p 1 – b 3 /c 2 = 3 – 40/49 = 2.1837

7. 2.183720-51 24.36744.537110.9076 28.734823.6113 2.1837 4.36749.53719.9076 4.367419.0742 Iteration No. 3: 1.721720-51 23.44340.92852.5986 26.886812.785 1.7217 3.44345.92851.5986 3.443411.857 Iteration No. 4: p 3 = p 2 – b 3 /c 2 = 2.1837 – 9.9076/23.6113 = 1.7217 p 4 = p 3 – b 3 /c 2 = 1.7217 – 2.5986/12.785 = 1.5185

8. 1.518520-51 23.037-0.38830.4104 26.0748.8351 1.5185 3.0374.6117-0.5896 3.0379.2234 Iteration No. 5: 1.472120-51 22.9442-0.66580.01986 25.88848.0025 1.4721 2.94424.3342-0.9801 2.94428.6683 Iteration No. 6: p 5 = p 4 – b 3 /c 2 = 1.5185 – 0.4104/8.8351 = 1.4721 p 6 = p 5 – b 3 /c 2 = 1.4721 – 0.01986/8.0025 = 1.469624

9. Click here to download Excel sheet for performing synthetic division and find p n.here Thus one of the roots of given f (x) is x = 1.47114 To verify: f (1.47114) = 2* 1.47114 3 – 5* 1.47114 + 1 = 0.012138021435088 ≈ 0 1.4711420-51 22.94228-0.67150.01213 ≈ 0 2.942284.3285-0.9879 Performing further division by x = 1.47114, we get the deflated polynomial as; Therefore f (x) = (2x 2 + 2.94228x – 0.6715) * (x –1.47114) If (2x 2 + 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 x a = 2 and x b = 4 so that f(x a ) * f(x b ) is negative. Hence the required root lies between the initial two guesses. Now x r = (x a + x b ) / 2If f(x r )*f(x a ) is negative then x b =x r else x a =x r R e p e a t Graphically: xaxa xbxb xrxr xbxb xrxr

11. Iterationxaxa xbxb XrXr 1233 2242.5 32 2.25 4 2.52.375 5 2.52.4375 62.3752.43752.40625 7 2.43752.421875 8 2.43752.429688 92.4218752.429688 102.4257822.4296882.427735 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 x a = 3 and x b = 6 so that f(x a ) * f(x b ) is negative. Hence the required root lies between the initial two guesses. If f(x r )*f(x a ) is negative then x b =x r else x a =x r R e p e a t Graphically: xaxa xbxb xrxr xbxb xrxr

13. Iterationxaxa xbxb xrxr 1363.320755 2 63.704273 3 64.075649 4 64.363149 5 64.547058 6 64.650068 7 64.703378 8 64.729822 9 64.742659 104.74265964.748827 Hence x = 4.748827 is root of the equation. f(4.748827) = ******* ≈ 0