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Pure Mathematics 2: Solution of equation by Numerical Methods

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Introduction

Pure Mathematics 2: Solution of equation by Numerical Methods Introduction: In this coursework, I am going to solve equations by using the Numerical Methods. Numerical methods are used to solve equation that cannot be solved algebraically e.g. quadratic equations ax�+bx+c=0 can be solved using this formula: x= -b� V b� - 4ac 2a Therefore numerical methods would not be used for quadratic equations. I will be working with cubic equation because there is no formula to solve it. There are three methods, which I will be using: * Change of sign method * Newton-Raphson method * Rearranging f(x) = 0 in the form x = g(x) Change of sign method: This method is concerned with when a function crosses the x-axis, and by definition changes sign (+ and -). If we are looking the root of equation f(x) = 0. The point at which the curve crosses x-axis is the root. Once an interval where f(x) changes sign then the root must be in the interval. f(a) > 0 f(b) < 0 Therefore root must be between [a,b] f(a) < 0 f(b) > 0 Root is between the interval [a,b] To find the interval of each root for the equation, I'll be doing a decimal search first. Lets take the equation y = x� - 12x + 5 x -4 -3 -2 -1 0 1 2 3 4 f(x) -11 14 21 16 5 -6 -11 -4 21 There are 3 roots in this equation and they are in these intervals: [-4, -3] [ 0 , 1 ] [ 3 , 4 ] Now, I am going to use Interval Bisection to find out one of the root. This method is similar to decimal search but instead of dividing each interval into 10 parts, using interval bisection, only need to divide into 2, which is more simpler. f(x) = x� - 12x + 5 I am going to use the root in the interval [0,1] f(0) ...read more.

Middle

f(x(n)) f'(x(n)) x(n+1) 1 2 3 9 1.666667 2 1.666667 0.62963 5.333333 1.548611 3 1.548611 0.06804 4.194589 1.53239 4 1.53239 0.001218 4.044659 1.532089 5 1.532089 4.17E-07 4.04189 1.532089 6 1.532089 4.88E-14 4.041889 1.532089 7 1.532089 0 4.041889 1.532089 This shows the root is close to 1.532089 (6 d.p.) Failure of Newton-Raphson method: To test the failure of the Newton-Raphson method, I am going to choose another equation; y = x� -4x + 2 f(x) = x� - 4x + 2 xn+1 = xn - xn� - 4xn + 2 3xn� - 4 Let x1 = 1.1 This is close to the root in interval [0,1] and [1, 2], therefore using the Newton-Raphson method, it should converge to either one of these roots. x1 = 1.1 x2 = 1.1 - (1.1� - 4(1.1) + 2) 3(1.1�) - 4 = 1.1 - (1.331 - 4.4 + 2) 3.63 - 4 = 1.1 - (-1.069) -0.37 = 1.1 - 2.889189189 = -1.789189 x3 = -1.789189 - (-1.789189� - 4(-1.789189) + 2) 3(-1.789189�) - 4 = -1.789189 - (-5.7275456956 - (-7.156756) + 2) 9.603591833 - 4 = -1.789189 - 3.429210305 5.603591833 = -1.789189 - 0.611966468 = -2.401155 x4 = -2.401155 - (-2.401155� - 4(-2.401155) + 2) 3(-2.401155�) - 4 = -2.401155 - (-13.84396801 - (-9.60462) + 2) 17.296636 - 4 = -2.401155 - (-2.23934801) 13.296636 = -2.401155 - (-0.168414628) = -2.232740 n x(n) f(x(n)) f'(x(and)) x(n+1) 1 1.1 -1.069 -0.37 -1.789189 2 -1.789189 3.429208 5.603594 -2.401155 3 -2.401155 -2.239348 13.29664 -2.23274 4 -2.23274 -0.199539 10.95539 -2.214527 5 -2.214527 -0.002216 10.71238 -2.21432 6 -2.21432 -2.84E-07 10.70964 -2.21432 7 -2.21432 -5.33E-15 10.70964 -2.21432 8 -2.21432 0 10.70964 -2.21432 As you can see, the root is diverging towards the interval [-2, -3] instead of converging towards [0, 1]. This diagram shows clearly that when x1 is 1.1, the tangent diverts to -2, the reason that this happens is because the value that I chosen to draw the tangent is too close to the turning point of the curve, therefore when the tangent is drawn, it diverts to -2. ...read more.

Conclusion

However, the root that is found using this method is just as accurate as the other two methods, but using this method, the number of steps to work the root out is 20, and is the most steps taken in all three methods. * Newton-Raphson The Newton-Raphson method gave the fastest speed of convergence, although when calculating the root using the Newton-Raphson iteration formula takes a lot of time because xn, follows one after another. When the iterative formula is used, it can fail to find the root if the calculations show divergence (when x1 is too close to the turning point). Using the iterative formula, only 5 steps are needed to work out the root. * Rearranging method Rearranging the function of f(x) = 0 to g(x), using the iteration formula, and the staircase diagram, this converges to the root in [0, 1] but have to rearrange the formula again in order to work out the other two. Therefore, using this method may take longer to work out all three roots in the equation. Only 6 steps are needed to work out the root. All three methods are at 6 d.p. Uses of hardware and software: In this investigation, I had used quite a range of software packaging, which enables me to do complicated equations or graph instantly. A graph-drawing package (Autograph) benefited all the methods. I was able to create the graphs for the functions and equations. This package also enables me to zoom into a root to draw tangents. A spreadsheet software package (Excel) allowed me to customize the tables and implement formulae that helped to calculate the approximations at each value of x. On the graphics application, I could easily revise and correct the formulae to make changes where applicable. I was then able to import the formulae into the word processor and present them alongside the graphs and explanations. For hardware, I used a calculator to help me with complicated calculations to check for answers for Newton-Raphson and Rearranging formular. . ?? ?? ?? ?? Celene Leong 13.7 1 ...read more.

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