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The initial estimate for the root was -3. To find U2, this value must be substituted into the equation of the curve, g(x). This produces the value -3. 301927 which can be seen on the y-axis. To find U3, this new value must be substituted into the equation of the curve. So we go across to the line y=x, which reflects this value onto the x-axis, and then down to the curve. This produces the value -3. 382999. I then repeated the process, going across to the line and down to the curve each time.

This resulting staircase approaches the root, which lies at the intersection of the line and the curve. At the root, the curve has a positive gradient as it is sloping upwards. Also, it is shallower than y=x. Therefore, at the root, 0<g'(x) ;1 and so the method is successful. From the table we can see that two values match to 5 significant figures. U7 and U8 agree to 5 significant figures giving a value of -3. 4115. There is a change of sign, which confirms the root is -3. 4115 to 5 significant figures as we can denote that the graph is crossing the x-axis.

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Newton-Raphson Method I am going to solve the equation f(x) =0, where f(x) =, the graph of y=f(x) is shown below. The graph shows that the equation has root sin [-4,-3],[0,1] and [3,4]. I shall find the root in [0,1]. X=0 is a good first estimate for the root which can be obtained from the graph. If Un is a good estimate of the root, Un+1is a better one, where Un+1 is the x-co-ordinate of the point where the tangent at x= Un cuts the x-axis. To work it out, the following sequence was derived and the following data was calculated.

The root = 0. 46360 to 5 significant figures. The change of sign shows that the root is correct to 5 significant figures as 0. 46360 where the error bounds are i?? 0. 000005. Other Roots I can now calculate the other roots using this method relatively easily. The tables of calculations are shown below: Newton-Raphson Method Failing I will try to solve the equation f(x) =0, where f(x) =. The equation has roots in [1, 2], [2, 3] and [4, 5]. I am going to find the root in [1, 2]. It is reasonable to start at x=2.

When x = 2, the graph shows that the Newton Raphson method does not work as the starting point on the graph is a stationary point. As a stationary point, this means that the tangent never touches the x-axis hence the Newton Raphson method does not work. Comparison of Methods I am going to solve the equation I did in decimal search and use all of the other methods in order to compare them with each other in order to find the root to the same degree of accuracy. This was the equation f(x) =0 where f(x) =. Using decimal search I obtained the root x=0. 310.

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