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The method I am going to use to solve x&amp;#8722;3x-1=0 is the Change Of Sign Method involving the Decimal Search method

Extracts from this document...

Introduction

Mathematics C3 Coursework

Numerical solutions of equations

Decimal Search Method --------------------------------------------p.1

Rearrangement Method ------------------------------------------- p.5

Newton-Raphson Method ----------------------------------------- p.9

Comparison ----------------------------------------------------------- p.12

Solving  x³−3x-1=0 using the “Change Of Sign” Method:

The method I am going to use to solve  x³−3x-1=0 is the Change Of Sign Method involving the Decimal Search method, which is that you are looking for the roots of the equations f(x)=0. This means that I want the value of x for which the graph of y=f(x) crosses the x axis. As the curve crosses the x axis, f(x)changes sign, so provided that f(x) is continuous function, once I have located an interval in which f(x) changes sign, I know that that interval must contain a root. Now, I have drawn the graph of x³−3x-1=0by using the Autograph software, and the graph is shown below: The point that the arrow pointing is the root I need to find.

From my graph above, I can see that the root of the equation is between x=0 and x=-1. The table of x values and f(x) values is shown below. I can work out the f(x) values by substituting the x-values into the equation.

x                      f(x)

-0.1                    -1.299

-0.2                    -0.408

-0.3                    -0.127

-0.4                     0.136 Middle

x3 = √[(1-3(x2)^5)/5] = 0.4366342

x4 = √[(1-3(x3)^5)/5] = 0.4364376

x5 = √[(1-3(x4)^5)/5] = 0.4364621

x6 = √[(1-3(x5)^5)/5] = 0.4364590

x7 = √[(1-3(x6)^5)/5] = 0.4364594

x8 = √[(1-3(x7)^5)/5] = 0.4364594

I can see convergence from x4 , and there is no change between x7 andx8 for this number of decimal places. I know that this method has worked successfully to find a root. This is shown graphically below:

Therefore, my root is x=0.43646( 5d.p.) I know that using Rearrangement B can help me find a root of   3x^5+5x²-1=0. However, Rearrangement A of this function would not find the same root as I have found. This is explained with a diagram with of Rearrangement A and its iterative formula below. This failure is shown graphically below: From the graph above, I can see that there is no convergence to that particular root I am looking for, but I want to test it again by substitution into the iterative formula of Rearrangement A, which is

y= [(1- 5x²)/3]^(1/5) and the starting value for x(x0) is 0, the same as before.

Zx0  = 0

x1  = √[(1-3(x0)^5)/5] = 0.8027416

x2  = √[(1-3(x1)^5)/5] = -0.9417235

x3  = √[(1-3(x2)^5)/5] = -1.0274040

x4  = √[(1-3(x3)^5)/5] = -1.0735437

Immediately, I can see that  I can see that there is really no convergence to that particular root I am looking for(the one in Rearrangement B and my observation from the graph is correct.

Conclusion

x3 = x2 - [(x2^3-3x2+1)/ (3x2^2-3)] =  -0.347296

x4 = x3 - [(x3^3-3x3+1)/ (3x3^2-3)] =  -0.347296

For the Newton-Raphson method, I find the same root as the other methods had found. Therefore, the root for all methods is -0.34730 (6d.p.) and the error is 0.000005.

Now, I can compare the efficiency of each methods.

After a lot of examples, I found that the Decimal Search method was the simplest out of the three, as it does not involve any iteration, so mistakes with iterative formulas and rearrangements are not applicable as with the other two methods used. However, the efficiency is not that good, though it is

simple, it has to work many steps if an accurate root is needed (e.g. 5 decimal place), it really consume time to have an accurate root.

Both the Newton-Raphson method and the Rearrangement method were fixed point estimates and involved an iterative process. Therefore, these methods were very similar. However, these two methods differ because there is a specific formula for the Newton-Raphson method. Although the Newton-Raphson method gave the much more rapid rate of convergence. In fact, the Newton-Raphson method gave the most rapid rate of convergence to 6d.p. , whereas, for the Decimal Search method, it took me very many calculations to converge to 5d.p. . So the Decimal Search method is not as efficient as the iterative methods( Rearrangement method and Newton-Raphson method). However, it was really easy to make mistakes on the calculator due the order of the terms in the iterative formulae.

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