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# Numerical solution of equations

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Introduction

Pure Mathematic 2 Coursework Numerical solution of equations By Michael Pang I am going to show 3 of the numerical methods for solving the equation which cannot be solved algebraically. They are interval estimation, fixed point estimation and Newton-Raphson method. Those of these numerical methods are used when algebraic ones are not available. When you found any equations which cannot be solved algebraically, probably you will draw the graph and see where the roots are. However, the other problem is found that you cannot get all roots accuracy. Actually, the numerical methods cannot find the exactly root but the answers are more accuracy than sketching the graphs. Therefore, we usually provide the answer to 5 or 6 decimal places depending on what the questioner needs. Finally, we check the answer by setting the lower bound and upper bound to see whether it has sign change or not. Even if the three numerical can solve the equation non-algebraically, they have its advantages and disadvantages. And now I am going to show how these methods works and their problems by using the equation F(x) = x�-9x+3. Interval estimation Assume that the roots of the equation F(x) = x�-9x+3, and I am looking for the roots which F(x) = 0. The roots of the equation are the values of x which the graph of y = x�-9x+3 crosses the x axis. By using the computer, I recognise that there are 3 roots on the graph and the value of these 3 roots. However, if we cannot use the computer, we have to use one of the numerical methods - interval estimation. ...read more.

Middle

They are - The curve touches the x-axis - The roots are close together - There is a discontinuity The curve touches the x-axis In the above graph, the curve touches the x-axis. Therefore, the sign change could not be found and none of the methods of interval estimation can be used. The roots are close together In the graph, there 3 roots close to together between the interval (0,1). In decimal search, F(0.1) = 0, so that 1 of the root is 0.1 and so further interval cannot be found. In addition, interval bisection and linear interpolation would be unaware of the existence of other roots since the 3 roots are too close together. There is a discontinuity in F(x) This equation y = 1/(x-1.1) has no root, but all change of sign methods will converge on a false root at x = 1.1 Avoid problem To avoid those problems occurring, it is important to start by sketching the graph and therefore the problem will be found. Fixed point estimation It is much different from the method of interval bisection. It involves an iterative process rather than finding interval within the roots must lie. Generally, iterative process means that a sequence of numbers by continued repetition of the same procedure. But this method has some limited which will be introduced when the process are showing. How does it method work? Before starting to use this method, the equation F(x) = 0 has to be rearranged into the form of x = g(x). ...read more.

Conclusion

The other methods to show they are right, I should check their upper bound and lower bound where I set it in Fixed point estimation. And now I show it again to make it clear. - The largest root : F(2.816915) = 0.0000142, F(2.816905) = -0.00013403 - The medium root : F(0.337615) = -0.00005233 F(0.337605) = 0.00003425 - The smallest root : F(-3.154525) = -0.00004153 F(-3.154515) = 0.00016701 Advantage Newton-Raphson method is fast and convenient, the processes are short and that why I didn't use a lot of presentation to show my understanding. It also gives an extremely rapid rate of convergence. I only use 4 to 5 steps for each of the root, so I find that it was much easier to finish than the first 2 methods. Disadvantage Even this method is the fastest and the most convenient, it still has several problems. Firstly, in my case, if I didn't choose the right initial value, I couldn't find the root by only 4 to 5 steps or even faster in other case. However, if the first value I h chose is not close to the root, or near the turning point, a turning point of y = F(x), the iteration may diverge, or converge to another root. And if the initial point is at a stationary point, F'(X) = 0 so the method cannot proceed. Secondly, all numerical methods for solving equations cannot be proceeded when the function is discontinuous function as it doesn't have any real roots. Thirdly, the function is not defined over the whole of. The tangent at the graph may meet the axis at a point outside the domain of the function. ...read more.

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