- Level: AS and A Level
- Subject: Maths
- Word count: 2565
Solutions of Equations by Numerical Methods - Change of sign Method and Decimal Search Method.
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Introduction
Solutions of Equations by Numerical Methods Change of sign Method Decimal Search Method f(x)=x�+2x�-3x+2 The solutions of f(x)=0 are the place where the curve cuts the x-axis. The graph shows that there is one real root lying in the interval [-4,-3] f(-4) = -64+32+12+2 = -18 f(-3 )= -27+18+9+2 = 2 The function f(x) is continuous therefore the change of sign form negative to positive in moving from x=-4 to x=-3 shows that there must be a root between the two values. The decimal search method will locate more accurate values of the root systematically by taking increments of size 0.1 in the interval [-4,-3] until a change of sign occurs. x -4 -3.9 -3.8 -3.7 -3.6 -3.5 -3.4 -3.3 -3.2 -3.1 -3 f(x) -18.000 -15.199 -12.592 -10.173 -7.936 -5.875 -3.984 -2.257 -0.688 0.729 2.000 There is a change of sign between -3.2 and -3.1, this show that the root is in the interval [-3.2,-3.1] Continuing the decimal search method will increase the accuracy by narrowing down the interval further. x -3.20 -3.19 -3.18 -3.17 -3.16 -3.15 -3.14 -3.13 -3.12 -3.11 -3.10 f(x) -0.688 -0.540 -0.393 -0.247 -0.103 0.039 0.180 0.320 0.457 0.594 0.729 x -3.16 -3.159 -3.158 -3.157 -3.156 -3.155 -3.154 -3.153 -3.152 -3.151 -3.15 f(x) ...read more.
Middle
The graph shows that there are three real roots lying in the intervals [-2,-1], [0,1] and [2,3]. By substituting y=g(x) into the iterative formula accurate values of the roots can be located. xn+1 = g(xn ) = 1/2 { (xn)�-2(xn)�+2) } I shall use method x=g(x) to obtain the root lying in the interval [0,1]. X 1 2 3 4 5 6 7 8 X 0 1.0000000000 0.5000000000 0.8125000000 0.6080322266 0.7426925379 0.6532395018 0.7126539366 X 67 68 X 0.6888921825 0.6888921825 This shows that the root lies within the interval [0.68889218245, 0.68889218255] The root therefore is 0.6888921825 with a maximum error of � 0.00000000005 For iterations to converge the derivative of g(x) must lie between -1 and 1. -1< g'(x) <1 g(x) = 1/2 (x�-2x�+2) g'(x)= 1/2 (3x�-4x) g'(x)= 1/2 { 3(0.6888921825)�-4(0.6888921825) } = -0.6659257063 This shows that g'(x) lies within the limits and therefore converges to the root. Failure of x = g(x) Method f(x) can be rearranged and written in the form of x=g(x) to locate more accurate values of the roots systematically. f(x) = x�-2x�-2x+2 g(x)= ?(2x�+2x-2) xn+1 = g(xn ) = ? { 2(xn)�+2(xn)-2 } I shall use method x=g(x) ...read more.
Conclusion
However, there is a lot of room for error and mistakes are hard to see until the very end. This method requires the most work and need for the use of different software and hardware. It has a slower speed of convergence than the Newton-Raphson method and can often diverge away towards another root. Despite the unreliability checks can be made to ensure correct results such as making sure that the derivative of g(x) must lie between -1 and 1. This is what confirms the high level of accuracy within the results. Overall each method has its own attributes which make it better than the other two, so it depends on what type of method you wish to apply; simple calculations which can be carried out with ease would be the decimal search method; fast rate of convergence would be the Newton-Raphson method; and high level of accuracy would be x=g(x) method. However, the Newton-Raphson method stands out as being the best method as it combines all the qualities of the other method by having fairly simple calculations; high speed of convergence; and accurate results. Also although each method is less time-consuming with the use of various software and hardware it is not vital for this particular method. 12.02.03 Rachel Withey ...read more.
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