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# MEI numerical Methods

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Introduction

Numerical methods coursework

Introduction:

Consider the problem x + ktanx = 1, the value of x lies between 0 and π/2 for different values of K, it is measured in radians. Normally a solution of a mathematical problem of this nature requires the root of an equation. If we attempt to find the root algebraically, we begin by rearranging the equation to give 0, hence the equation is x + ktanx – 1 = 0. Now lets assume the value of K is 1, we are left with x + tanx – 1 = 0. The only information we are provided is that the value of x is between 0 and π/2. Since x appears twice in the equation, the problem cannot be solved algebraically. In the sense that there are equations which cannot be integrated and require numerical methods to approximate the answer, eg the midpoint rule. In the same way there are equations which cannot be solved algebraically and require numerical methods to approximate the root.

To solve this solution I shall attempt to use numerical methods. I will be using the method of bisection, fixed point iteration, secant method and false position method. All these methods require the use of an excel spreadsheet since all methods require iterations which will get closer and closer to the real answer (converge). This is why the Newton raphson method isn’t being used as it requires differentiation which can be hard to use on computers. The other methods can be converted into formulas and inputted into excel which can easily perform multiple iterations, simply by dragging down the cells.

Strategy:

Method of bisection:

An approximation of the root can be obtained by this method; in order to use the method of bisection the first thing which must be done is to find an interval estimate of the root.

Middle

Formulas used: Fixed point iteration:

To use fixed point iteration we need to rearrange the equation so that it equals x. There are two possible rearranged forms are:

• X= 1 – ktanx
• X= arctan(1-x)/k

Fixed point iteration uses recurrence relation so by finding the answer of one iteration we produce X(1), we then use X(1) as x in the formula to produce X(2), this is how we get closer to the root by using this method.

For example the first iteration for x = 1 – ktanx assuming that k = 1 is:

X = 1 – tan(π/4)

The value of tan(π/4) = 1

1-1 = 0, hence the value of the first iteration is 0, for the second iteration we would do:

X = 1 – tan(0)

The value of tan(0) = 0

1-0 = 1 hence the value of the second iteration is 1 and we would carry on. This task can be tedious to carry on doing it again and again especially for different values of K and if it has a lot of iterations, hence we use a spreadsheet. Secant method:

The secant method requires two approximations of the root in order to work. Fortunately we’ve been provided with two roots, 0 and π/2. We can now call x(0) as π/2 and x(1) as 0, we can then substitute these values into the following equation.  False position method:

Similar to the secant method, the false position method also requires two

approximations of the root. I will use one root a as 0 and the other root b as π/4. Bearing this in mind we can substitute the values into the formula: The formula is the same as the secant method however remember the false position method roots approximation must support the sign change argument. Proof of root: Conclusion 0.4797310065 < x < 0.4797310075, to test the bounds are correct we can do the f of both of these values and it would have to support the sign change argument, as proved above.

This means the root is 0.479731007 ± 0.0000000005. We can however lower the boundaries and thereby create a better approximation of the root. In order to lower the boundaries we merely continue reducing/increasing them until we hit the boundary of sign changes.

f(0.4797310073) = 0.0000000000444833

f(0.4797310072= -0.0000000002

Conclusion

The root to the equation is, 0.4797310065 < x < 0.4797310075 hence 0.479731007 to 9 D.P. This is my approximation of the root compared to the actual root, bear in mind this is zoomed in by a lot. However how valid is this approximation of the root?

If we look back we found out there were problems with the formulas themselves however that was merely for the purpose of finding the best method, this cannot be used as a limitation because it doesn’t apply to this equation.

However earlier I did mention that excel is accurate to 15 D.P, my answer is only valid to 9 D.P, this is definitely a limitation, if I had more time I would change my figures and calculate it to 15 D.P.

Another way to test its validity we merely insert it back into the original equation,

x + tanx – 1 =0. Notice the =0 part, if this approximation is valid then by using our approximation (x) in this equation we should end up with an answer close to 0. The result of the equation is -6.37 x 10^-10. Notice the degree of the polynomial is a high minus number meaning this number is tiny and extremely close to 0. This means my approximation of the root is valid. In order to improve on this approximation we could find the root to more D.P as explained earlier.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

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