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# Numerical Solutions of Equations.

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Introduction

P2 Maths Coursework Numerical Solutions of Equations HANNAN SHAH Introduction In this coursework I intend to use numerical methods to find the solutions of equations that cannot be solved algebraically. Many simple equations can be solved using methods such as factorising or the quadratic formula but such methods cannot be applied to more complicated equations. My intention is to use three methods of numerical analysis to solve some of these more complex equations, show problems associated with these methods and then to compare their relative merits. The methods I will be using are decimal search, fixed point iteration using Newton-Raphson and fixed point iteration after rearranging f(x) = 0 into the form x = g(x). Decimal Search Assuming that you are looking for the roots of the equation f(x) = 0 you want to find the values of x for which the graph of y = f(x) crosses the x-axis. As y = f(x) crosses the x-axis, f(x) changes sign (from + to -), provided that f(x) is continuous. Using decimal search I will attempt to find one root of the equation y = x5+4x�-1. The graph for this function is shown below. The graph clearly shows that there are three roots in the intervals [-2, -1]. [-1,0] and [0,1]. I am going to attempt to find the value of the root in the interval [0,1]. I will start at one decimal place by taking values of x between 0 and 1 in increments of 0.1. The table on the left shows that there is a change of sign between x = 0.4 and x = 0.5. ...read more.

Middle

-1.618 as this is nearest to the point where the tangent of x = 0.8 meets the x axis and the following tangents do not have such a low gradient that may cause them to diverge away from x = -1.618 and back towards the root in the interval [0,1]. Fixed Point Iteration (Rearranging f(x) = 0 in the form x = g(x) This method uses an estimate for the value of x using an iterative process. By rearranging f(x) = 0 into the form x = g(x) provides an iterative formula which will converge to a required root by picking an appropriate starting value. On a graph the roots will be equal to the values of x where the curve y = f(x) intersects the line y = x. The graph for y = x5-3x+1 shows the roots lie in the intervals [-2,-1], [0,1] and [1,2]. x5-3x+1 = 0, therefore x = x5+1 3 This provides the iterative formula: xn+1 = xn5+1 3 The curves above show y = x and y = g(x) = x5+1. 3 With the equation in the form x = g(x) means that you are now finding the points of intersection between the curves rather than when the curve y = x5-3x+1 crosses the x-axis. I will first attempt to find the root in the interval [0,1]. I will take the end point of this interval (x = 1) as the starting point. xn xn+1 = x5+1 3 x0 = 1 x1 = (15 + 1) � 3 = 0.66667 x1 = 0.66667 x2 = (0.666675 + 1) ...read more.

Conclusion

Yet decimal search also had a problem by not showing a change of sign in the first set of calculations when the root had two decimal places. Therefore, ignoring any potential problems I think that fixed point iteration and Newton-Raphson are quicker than decimal search at converging to the required root. I have already made some mention of the use of technology to help speed up the process of finding the roots. Microsoft Excel was of great benefit when making calculations, as it only required the input of the initial values and the formulae. A few mouse clicks gave the results for the entire set of values within a few seconds. Autograph was another very useful piece of software as it could be used to instantly visualise curves and gradient functions. This meant that within seconds I could produce an accurate curve instead of having to make some calculations and draw the curve myself. Autograph also allowed me to see the intervals of the roots (or points of intersection in fixed point iteration). This saved a lot of time as I could easily pick appropriate starting values without having to waste time using trial and error to pick my starting value. Autograph was very useful for both Newton-Raphson and fixed point iteration as it had an inbuilt function to make the various calculations in with a few mouse clicks and these could be used to confirm the values obtained from Microsoft Excel. It was also excellent for showing the iterative processes for both methods by zooming in on the roots as I have shown many times when using these numerical methods. HANNAN SHAH P2 Maths Coursework Hannan Shah Numerical Solutions of Equations Page 2 of 14 ...read more.

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=2*B4^3-3*B4^2-8*B4+23 =2*C4^3-3*C4^2-8*C4+23 =(B4+C4)/18 =2*F4^3-3*F4^2-8*F4+23 =IF(G4>0,B4,F20) =IF(G4>0,F4,C20) =2*B4^3-3*B4^2-8*B4+24 =2*C4^3-3*C4^2-8*C4+24 =(B4+C4)/19 =2*F4^3-3*F4^2-8*F4+24 =IF(G4>0,B4,F21) =IF(G4>0,F4,C21) =2*B4^3-3*B4^2-8*B4+25 =2*C4^3-3*C4^2-8*C4+25 =(B4+C4)/20 =2*F4^3-3*F4^2-8*F4+25 =IF(G4>0,B4,F22) =IF(G4>0,F4,C22) =2*B4^3-3*B4^2-8*B4+26 =2*C4^3-3*C4^2-8*C4+26 =(B4+C4)/21 =2*F4^3-3*F4^2-8*F4+26 =IF(G4>0,B4,F23) =IF(G4>0,F4,C23) =2*B4^3-3*B4^2-8*B4+27 =2*C4^3-3*C4^2-8*C4+27 =(B4+C4)/22 =2*F4^3-3*F4^2-8*F4+27 Bisection failure I am going to show the failure of bisection method. • Over 160,000 pieces
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