Solving Equations Using Numerical Methods

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Solving Equations Using Numerical Methods

Numerical methods are used for solving equations which can’t be solved with algebra. Quadratics would not need to use numerical methods as an accurate answer can be found using factorisation or by using the quadratic formula. I will be using numerical methods to solve cubic equations that have decimal answers. I will be using three methods in this coursework; ‘Decimal Search’, ‘Newton-Raphson’ and ‘Graphical Convergence’.

Decimal Search

This method solves the equation by looking for a change of sign. We know that the root of the equation that we are looking for is in between the 2 numbers that the change of sign occurs in. We then repeat the same process as before but this time to a greater level of accuracy. When I have done this process to the level of accuracy required, I will be able to tell where the root of the angle is.

I will now solve an equation using the decimal search method. The equation I will be using is y=x³-2x+0.5

I will be trying to find this root using decimal search. I will start with the integer immediately below the root (-2) and this will be the lower bound. I will find the values of f(x) for this and consecutive increments of 0.1 until a change of sign is found. Here is my table of values.

From this table, I can see that the change in sign appears in between -1.6 and -1.5. To be certain that the root is between -1.6 and -1.5, I will calculate f(x) in each case to make sure that a change of sign occurs.

F(x)        = y=x³-2x+0.5

F(-1. 6) = -0.39

F(-1. 5) = 0.125

As a change of sign has been found and confirmed here, the root does lie between these points. I will now use the same method and use increments of 0.01 to find the root between -1.6 and -1.5. Here is the table of results.

From this table, I can see that the change in sign appears in between -1.53 and      -1.52. To be certain that the root is between -1.53 and -1.52, I will calculate f(x) in each case to make sure that a change of sign occurs.

F(x)        = y=x³-2x+0.5

F(-1. 53) = -0.0216

F(-1. 52) =  0.0282

As a change of sign has been found and confirmed here, the root does lie between these points. I will now use the same method and use increment of 0.001 to find the root between -1.53 and -1.52. Here is the table of results.

I can now see that the root lies between -1.526 and -1.525. As I only intend to find the root to 3 decimal places I don’t need to continue the method. To be certain that the root is between -1.526 and -1.525, I will calculate f(x) in each case to make sure that a change of sign occurs.

F(x)        = y=x³-2x+0.5

F(-1.526) = -0.00156

F(-1.525) =  0.00342

        As a change of sign has been found and confirmed here, this is where the root lies.

        I have successfully found one of the roots for this equation to my required 3 decimal places. Although I have been successful in finding this root for this equation, sometimes the decimal search method fails to find the root. I will now show an example of the decimal search method failing and find out why it fails. I will use the equation
y=1.676x³ + 3.018x² -1.851x + 0.2427. Here is the equation.

I will try to find the middle root of the equation using the decimal search method.

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I will start with the integer immediately below the root (0) and this will be the lower bound. I will find the values of f(x) for this and consecutive increments of 0.1 until a change of sign is found. Here is my table of values.

When looking at this table of values, no change of sign occurs. This means that I can’t continue with this method as I do not know which two numbers look between for the next stage. I cannot use this method to find this root with this equation. The reason that the decimal search method ...

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