• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Solution of equations by numerical methods.

Extracts from this document...

Introduction

## Solution of equations by numerical methods

This investigation is to find equation solutions using three methods:

1. Change of sign using bisection, decimal search or linear interpolation.
2. Newton – Raphson.
3. Rearrangement of the equation f(x) = 0 into the form x = g(x).

The change of sign methods are systematic searches, which use the positive and negative signs of the f(x) solutions to find the location of the root within the intervals found using the graph of the function curve. The change of sign method I have chosen to use within my investigation is the decimal search method.

Fixed point iteration requires finding a single value or point as an estimation for the value of x, rather than establishing an interval as in the change of sign methods. The Newton – Raphson method and the rearrangement of equation f(x) = 0 into the form x = g(x) will be used to investigate this form of numerical equation solution.

Once each of these methods have been investigated I will compare each of them, in order to find the easiest method for equation solution. This comparison will also include the negative and positive points of each method, such as problems that result in an inability to find the correct root and the speed of convergence to the correct root.

Fixed Point Iteration – Newton – Raphson Method.

To start off using the Newton – Raphson method, we must first take an estimate of the root as a starting point. For a root f(x) we shall start with estimation x1 –we then draw a tangent to the curve y = f(x) at the point (x1, f(x1)).

Middle

From this we can see that there are three roots, within the intervals [-2,-1],[-1,0] and [1,2].

• one rearrangement: x = g(x) = x^5 - 2 which gives us the graph:

4

Taking x = 2.0 as the starting point to find the root within the interval [-1,0] – we gain the following set of results: This rearrangement provides the basis for the iterative formula:

Xn+1 = x^5n - 2

4

This gives the set of results:

 x1 -0.8 x5 -0.50856 x2 -0.58192 x6 -0.5085 x3 -0.51668 x7 -0.5085 x4 -0.50921 x8 -0.5085

This shows the root to be –0.50850 within the interval [-1,0], which can be written:

• -0.509 with a maximum error of ±0.0005, or
• -0.5 (to 1 d.p.)

Displayed graphically:  From which we gain the root as –0.5085 (to 4 s.f.) from the results box, which again gives the root in both the ways written above.

As         f(x) = x^5 – 4x – 2

And            g(x) = x = (x^5 – 2)/4

• g’(x) = (5x^4)/16

The magnitude of g(x) for the results gained above can be shown to be:

 x1 -0.8 -0.128 x5 -0.50856 -0.021 x2 -0.58192 -0.036 x6 -0.5085 -0.021 x3 -0.51668 -0.022 x7 -0.5085 -0.021 x4 -0.50921 -0.021 x8 -0.5805 -0.021

This shows that the magnitude is always less than one, and as the method will only succeed if        -1 < g’(x) < 1 , and as this is the case for all the values of x found, the method works.

## Method Failure

We can see that this rearrangement of f(x) works when trying to find the root within the interval [-1,0], but I shall now try to find the root within the interval [1,2] using the same method.

Conclusion

The hardware and software that have been used in this investigation are a scientific calculator and a graph drawing program – Autograph. This allows us to find the roots of each graph without having to draw each of them out along with the tangents, and gives as accurate results for the roots with our own specification on the number of significant figures. This program proves little use for the decimal search method, other than establishing the intervals the roots lie in – the main body of work must then be done through calculations and recording of results, and each significant figure must be found one at a time.

The fixed-point iteration methods benefit greatly from the graph program, as they allow us to gain all the tangents and converging ‘staircases’ and ‘cobwebs’ simply by typing in the starting value of x. This gives us much faster convergence than calculating, and a graphical representation of what is happening, so that we can more fully understand the way each method works. Both are simple to use as long as the equation and the starting point for each interval are known.

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

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related AS and A Level Core & Pure Mathematics essays

1.  5 star(s)

This can be shown here: 5 25 50 20 My hypothesis is correct. It will be also be useful to prove this binomially, however: 2(x+h) � - 2x� = 2(x�+h�+2hx) - 2x� = 2x�+2h�+4hx -2x� = 2h�+4hx x + h - x h h h 2h + 4x = ((2)(0))

2. ## MEI numerical Methods

the formula described above we would have an approximation of the root closer to the real root X(3), then I would use x(2) and x(3). The secant method always uses the two most recent approximations of the root. I will produce an answer to 9 significant figures by the use

1. ## Sequences and series investigation

- 2(102) + 2Y(10) - 1 = 13335 - 200 + 26Y - 1 = 1159 Sequence 15: N = 15 _(15�) - 2(152) + 2Y(15) - 1 = 4500 - 450 + 40 - 1 = 4089 This equation has correctly given me the number of squares in each sequence which again proves

2. ## Numerical solutions of equations

However, my aim was to find the negative root, and I could not find it. Therefore, the Newton-Raphson method has failed to find that particular root, even though the starting value was close to it. Another example where the Newton-Raphson would not work is the function (x+21/2)1/2 ln (x+21/2)

1. ## In my coursework I will be using three equations to investigate their solutions using ...

Rearrangement method Rearranging equations- any equation f(x) = 0 can be rearranged in the form x = g(x) in any number of ways, and can be used as a basis for the iteration xn+1 = g(xn). So f(x) = x� - 6x� - 2x + 1 = 0 Can be written as x = 1/2( x� - 6x� + 1)

2. ## C3 Numerical Solutions to Equations

This is the root of the equation f(x)=0 At the root x = 1.75650874, g'(x) = 0.479 (3sf). �g'(x)�<1 for the iterations to converge. If �g'(x)�>1 the iterations will not converge on the correct root. For example sing the same rearrangement of the equation to find the root between

1. ## Solving Equations. Three numerical methods are discussed in this investigation. There are advantages and ...

Microsoft excel can be set up in order to reduce human error and increase efficient of calculation. A spread sheet is set up for n a f(a) B f(b) 1 0 =B2^3-5*B2^2+4*B2+2 2 =D2^3-5*D2^2+4*D2+2 2 =IF(G2>0,F2,B2) =B3^3-5*B3^2+4*B3+2 =IF(G2>0,D2,F2) =D3^3-5*D3^2+4*D3+2 3 =IF(G3>0,F3,B3)

2. ## Fractals. In order to create a fractal, you will need to be acquainted ...

In general, we can break a line segment into N1 self-similar pieces, each with magnification factor N. The two-dimensional square can be broken down into N2 self-similar pieces, with a magnification factor of N. However, we can break the cube into N3 self-similar pieces, each of which has magnification factor of N. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 