algebric solutions by newton raphson method and secant method
- 1. Solution of algebraic & transdental equation with Newton Raphson method & Secant method 1S.N.P.I.T. & R.C.
- 2. Presented by: 130490106065 – Modi Nagma 130490106085 – Patel Pinal 140493106015 – Padhiyar Sagar 140493106025 – Taylor Kishan 2S.N.P.I.T. & R.C.
- 3. Two Fundamental Approaches 1. Bracketing or Closed Methods - Bisection Method - False-position Method (Regula falsi). 2. Open Methods - Newton-Raphson Method - Secant Method - Fixed point Methods Roots of Equations 3S.N.P.I.T. & R.C.
- 4. Newton Raphson’s Method The equation for Newton’s Method can be determined graphically! 4S.N.P.I.T. & R.C.
- 5. Continue..... The equation for Newton’s Method can be determined graphically! From the diagram tan Ө = ƒ'(x0) = ƒ(x0)/(x0 – x1) 5S.N.P.I.T. & R.C.
- 6. Continue.... The equation for Newton’s Method can be determined graphically! From the diagram tan Ө = ƒ'(x0) = ƒ(x0)/(x0 – x1) Thus, x1=x0 -ƒ(x0)/ƒ'(x0). 6S.N.P.I.T. & R.C.
- 7. In general, if the nth approximation is xn and f’(xn) ≠ 0, then the next approximation is given by: 3 1 ( ) 12 '( ) n n n n f x x x f x Equation/Formula 2 Continue… 7S.N.P.I.T. & R.C.
- 8. Starting with x1 = 2, find the third approximation x3 to the root of the equation x3 – 2x – 5 = 0 Example Of Newton Raphson’s method Example 1 8S.N.P.I.T. & R.C.
- 9. We apply Newton’s method with f(x) = x3 – 2x – 5 and f’(x) = 3x2 – 2 Newton himself used this equation to illustrate his method. He chose x1 = 2 after some experimentation because f(1) = -6, f(2) = -1 and f(3) = 16 Continue… Example 1 9S.N.P.I.T. & R.C.
- 10. Equation 2 becomes: Continue… 10S.N.P.I.T. & R.C. 3 2 2 5 1 3 2 n n n n n x x x x x
- 11. With n = 1, we have: Continue….. 11S.N.P.I.T. & R.C. 3 1 1 2 1 2 1 3 2 2 5 3 2 2 2(2) 5 2 3(2) 2 2.1 x x x x x
- 12. With n = 2, we obtain: It turns out that this third approximation x3 ≈ 2.0946 is accurate to four decimal places. Continue…. 12S.N.P.I.T. & R.C. 3 2 2 3 2 2 2 3 2 2 5 3 2 2.1 2(2.1) 5 2.1 3(2.1) 2 2.0946 x x x x x
- 13. Use Newton’s method to find correct to eight decimal places. First, we observe that finding is equivalent to finding the positive root of the equation x6 – 2 = 0 So, we take f(x) = x6 – 2 Then, f’(x) = 6x5 Example 2 Newton Raphson’s method 6 2 6 2 13S.N.P.I.T. & R.C.
- 14. So, Formula 2 (Newton’s method) becomes: Continue… 14S.N.P.I.T. & R.C. 6 1 5 2 6 n n n n x x x x
- 15. Choosing x1 = 1 as the initial approximation, we obtain: As x5 and x6 agree to eight decimal places, we conclude that to eight decimal places. Continue… 2 3 4 5 6 1.16666667 1.12644368 1.12249707 1.12246205 1.12246205 x x x x x 6 2 1.12246205 15S.N.P.I.T. & R.C.
- 16. Newton Raphson’s method Use Newton’s method to find correct to four decimal places. Where, N=12 q=3 = 1 S.N.P.I.T. & R.C. 16 1 1 1 1n n q n N x x q q x 0x 3 12
- 17. continue…. So, Formula 2 (Newton’s method) for find out qth root becomes: S.N.P.I.T. & R.C. 17 1 3 1 1 12 3 1 1 3 n n n x x x 1 4.6667x
- 18. Continue…. Choosing x1 = 1 as the initial approximation, we obtain: As x5 and x6 agree to four decimal places, we conclude that to four decimal places. S.N.P.I.T. & R.C. 18 3 12 2.2894 2 3 4 5 6 7 3.2948 2.5650 2.3180 2.2898 2.2894 2.2894 x x x x x x
- 19. Secant Method 19S.N.P.I.T. & R.C.
- 20. Continue.... First we find two points(x0,x1), which are hopefully near the root (we may use the bisection method). A line is then drawn through the two points and we find where the line intercepts the x-axis, x2. 20S.N.P.I.T. & R.C.
- 21. Continue.... If f(x) were truly linear, the straight line would intercept the x-axis at the root. However since it is not linear, the intercept is not at the root but it should be close to it. 21S.N.P.I.T. & R.C.
- 22. Continue.... From similar triangles we can write that, 10 10 1 21 xfxf xx xf xx 1x 0x2x 1xf 0xf 22S.N.P.I.T. & R.C.
- 23. Continue.... From similar triangles we can write that, Solving for x2 we get: 10 10 1 21 xfxf xx xf xx 10 10 112 xfxf xx xfxx 23S.N.P.I.T. & R.C.
- 24. Continue.... Iteratively this is written as: nn nn nnn xfxf xx xfxx 1 1 1 24S.N.P.I.T. & R.C.
- 25. S.N.P.I.T. & R.C. 25 Approx. f '(x) with backward FDD: Substitute this into the N-R equation: to obtain the iterative expression: i 1 i i 1 i f (x ) f (x ) f '(x) x x i i 1 i i f (x ) x x f '(x ) i i 1 i i 1 i i 1 i f(x )(x x ) x x f(x ) f(x ) Continue….
- 26. x f(x) 1 2 new est. x f(x) 1 new est. 2 FALSE POSITION SECANT METHOD The new estimate is selected from the intersection with the x-axis 26S.N.P.I.T. & R.C.
- 27. Example of Secant Method The floating ball has a specific gravity of 0.6 and has a radius of 5.5cm. You are asked to find the depth to which the ball is submerged when floating in water. The equation that gives the depth x to which the ball is submerged under water is given by 010993.3165.0 423 xx 27S.N.P.I.T. & R.C.
- 28. Use the secant method of finding roots of equations to find the depth to which the ball is submerged under water. Conduct three iterations to estimate the root of the above equation. Find the absolute relative approximate error at the end of each iteration, and the number of significant digits at least correct at the converged iteration. 28S.N.P.I.T. & R.C.
- 29. S.N.P.I.T. & R.C. 29 Secant Method From the physics of the problem 11.00 )055.0(20 20 x x Rx x water Figure 2 : Floating ball problem
- 30. S.N.P.I.T. & R.C. 30 Let us assume 11.0,0 UL xx 4423 4423 10662.210993.311.0165.011.011.0 10993.310993.30165.000 fxf fxf U L Hence, 010662.210993.311.00 44 ffxfxf UL
- 31. 31 Iteration 1 0660.0 10662.210993.3 10662.2010993.311.0 44 44 UL ULLU m xfxf xfxxfx x 5 423 101944.3 10993.30660.0165.00660.00660.0 fxf m 00660.00 ffxfxf mL 0660.0,0 UL xx S.N.P.I.T. & R.C.
- 32. 32 Iteration 1 0.0000 0.1100 0.0660 N/A -3.1944x10-5 2 0.0000 0.0660 0.0611 8.00 1.1320x10-5 3 0.0611 0.0660 0.0624 2.05 -1.1313x10-7 4 0.0611 0.0624 0.0632377619 0.02 -3.3471x10-10 Lx Ux mx %a mxf 010993.3165.0 423 xxxfTable 1: Root of for secant Method. S.N.P.I.T. & R.C.