Mathematical equations can be solved in many ways; however some equations cannot be solved algebraically. I am going to show the three methods of solving these types of equations numerically

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Maths Core 3: Coursework

Introduction

Mathematical equations can be solved in many ways; however some equations cannot be solved algebraically. I am going to show the three methods of solving these types of equations numerically, also I will show each method working, and each method failing.

The three methods are:

Change Of Sign Method
Rearranging f(x)= 0 Into The Form x=g(x)
Newton-Raphson Method

Change of Sign Method

Change of Sign Method Working

I am going to solve the equation f(x) = 0, where f(x) =. The graph of y=f(x) is shown here.

In order to calculate the value of the root in [0.3, 0.4] to 3 decimal places I must check whether the value is closer to 0.309 than 0.310 meaning that the value 0.3095 needs to be calculated. If this value is negative then the root is 0.309 to 3 decimal places however if the value is positive then the root would have to be given as 0.310.

The value of f(x) when x=0.3095 is 0.00614708 meaning that I can deduce that the value of the root in [0, 1] to 3 decimal places is 0.310.

The error bounds are 0.3095 and 0.310.

Change of Sign Method Failing

Consider the equation f(x) =0, where f(x) = I shall try to locate the roots by making a table of values that correspond to this function of x.

Rearranging f(x) =0 into x=g(x)

Rearranging f(x) =0 into x=g(x) failing

I am going to solve the equation f(x) =0 where f(x) =. The graph of y=f(x) is shown below.

The sequence formed is converging; however it converges to the incorrect root meaning that it is considered as a failure. , therefore . As we know that the root <-3, g’(x)>1 meaning that it fails as the range which enables the method to succeed -1<g’(x) <1 is not fulfilled by this equation.

Rearranging f(x) =0 into x=g(x) working

I am going to solve the equation f(x) =0 where f(x) =. The graph of y=f(x) is shown below.

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Rearranging f(x) =0 into x=g(x) working

I am going to solve the equation f(x) =0 where f(x) =. The graph of y=f(x) is shown below.

By starting with the value of -3 we are now able to start finding the root in [-3,-2]. The iterations are shown below:

=-3.301927

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.

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 ±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.

I shall now solve it using this rearrangement:

The tables shows that the sequence is converging towards the root and finds the root at x=0.310.

I shall now solve it using the Newton-Raphson method to find the root in [0, 1] to the same degree of accuracy.

I shall start at 0 as it is a reasonable starting point in order to find the root.

Speed of Convergence

To calculate the root to 3 decimal places using decimal search it requires 4 tables of values to calculate the root whereas with x=g(x) iteration and also Newton Raphson only 1 table of values is required. We can thereby conclude that the Decimal search method is the slowest. Newton Raphson only takes 3 rows of calculations to find 2 results which match to 3 decimal places which is the specified degree of accuracy for the initial decimal search calculation. This is faster than the x=g(x) iteration as this method takes 4 rows of calculations to find 2 results that match to the specified degree of accuracy.