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) = … 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.

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

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

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

Iteration No. 1: 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 =

Iteration No. 3: Iteration No. 4: p 3 = p 2 – b 3 /c 2 = – / = p 4 = p 3 – b 3 /c 2 = – / =

Iteration No. 5: Iteration No. 6: p 5 = p 4 – b 3 /c 2 = – / = p 6 = p 5 – b 3 /c 2 = – / =

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 = To verify: f ( ) = 2* – 5* = ≈ ≈ Performing further division by x = , we get the deflated polynomial as; Therefore f (x) = (2x x – ) * (x – ) If (2x x – ) = 0 then x = or x = To verify: f (0.2008) = ≈ 0 and f ( ) = ≈ 0 Thus f (x) has three roots as x = , x = and x =

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

Iterationxaxa xbxb XrXr Hence x = is root of the equation. f( ) = ≈ 0

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

Iterationxaxa xbxb xrxr Hence x = is root of the equation. f( ) = ******* ≈ 0