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

# In my coursework I will be using three equations to investigate their solutions using three numerical methods which are: change of sign using Newton-Raphson by finding the fixed point iteration fixed point iteration after rearranging the equa

Extracts from this document...

Introduction

Numerical Methods

In my coursework I will be using three equations to investigate there solutions using three numerical methods which are:

1. change of sign using
2. Newton-Raphson by finding the fixed point iteration
3. fixed point iteration after rearranging the equation f(x) = 0 into the form x = g(x)

Change of sign

This method involves finding an interval in which f(x) changes sign.

I am using the function f(x) = 2x³ – 5x² – 4x + 1 for this method I have used autograph to show the graph of y= 2x³ – 5x² – 4x + 1

Here is the table of values

 -2 -27 -1 -2 0 1 1 -6 2 -11 3 -2 4 33

This shows that there are three intervals containing roots:

[–1,0], [0,1] and [3,4].  It can be seen that f(-1) > 0 and f(0) < 0 so there is a root in the interval [-1,0].

Bisection

In this method the interval is successively halved by looking at the value of f(x) at its

mid-point. For the root in the interval [-1,0], you next try an x-value of-0.5.

f(-0.5) = 1.5

 -0.75 0.34375 -0.875 -0.667969 -0.8125 -0.123535 -0.78125 0.119568 -0.79688 0.000365464

Since this is positive the root lies in the interval [-1,-0.5]. The next value I’ll try is the midpoint of the new interval, -0.75

f(-0.75) = 0.34375

f(-0.875) =

Middle f(x)= x³+6x²+4x

f’(x)= 3x²+12x+4

Using excel I have found the roots

By using the Newton-Raphson equation

 x0 -2 x1 -1 x2 -0.8 x3 -0.765217391 x4 -0.76393381 x5 -0.763932023 x6 -0.763932023 x7 -0.763932023

This is fixed point iteration   .

I have used autograph to both draw my graphs and display

a table of results to show where the root is

By using Newton Raphson method  So for my example Xo= -2

So (-2)³+6*(-2)²+4*(-2) = 8

3*(-2)²+12*(-2)+4 = -8

Then (-2) – 8/-8 = -1

And then carrying on the principle x would be replaced with -1

by looking for a change of sign I can identify my error bounds and the sign changes between -1 and -0.5 so -1= 1 and -0.5= -0.625

so by using excel I have imputed the x values into the equation y = x³+6x²+4x  and here are the results:

x        y

 -0.5 -0.625 -0.6 -0.456 -0.7 -0.203 -0.8 0.128

I can see there is a sign change

between -0.7 and -0.8
so I use 2 d.p

x        y

 -0.75 -0.046875 -0.76 -0.013376 -0.77 0.020867 -0.78 0.055848 -0.79 0.091561

I can see a sign change

Between -0.76 and -0.77

 -0.761 -0.00998508 -0.762 -0.00658673 -0.763 -0.00318095 -0.764 0.000232256 -0.765 0.00365287

Conclusion

G’(x)= √12x+2 From the graph above you can see where both the rearrangement is successful and where it fails this is because of the gradient of y=³√6x²+2x-1 at where the graph y=x interact near (-0.5,-0.5) is more than 1. the reason for this fail is because the gradient of y=³√6x²+2x-1 is much larger than one meaning that section neither converges or diverges correctly so it was impossible to use both staircase diagram and cobweb diagram to determine the root.

x        y

 -0.05 -1.02757 -1.02757 1.48583 1.48583 2.47809 2.47809 3.44265 3.44265 4.25425 4.25425 4.87841 4.87841 5.33153 5.33153 5.64846 5.64846 5.86482 5.86482 6.01023 6.01023 6.10695 6.10695 6.17086

The

f(x)= 2x³–5x²–4x+1

f’(x)= 6x²–10x–4

 -1 -0.833333 -0.833333 -0.798475 -0.798475 -0.79693 -0.79693 -0.796927 -0.796927 -0.796927 -0.796927 -0.796927

x0-f(x)/f’(x)= 1-

I found that each method had its own advantages and disadvantages for me I found Newton raphson method easy to use on the computer because autograph has a special tool that works everything out for you.

One of the equations used above is selected and the

Other two methods are applied successfully to find the Same root. There is a sensible comparison of the relative merits of The three methods in terms of speed of convergence. There is a sensible comparison of the relative merits of the three methods in terms of ease of use with available hardware and software.

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.  ## C3 Coursework - different methods of solving equations.

5 star(s)

We can generalise this into the formula: So, now using the formula, we can work out the next X value: X1+1 =-0.58824 - f(X-0.5882)/f'(X -0.5882) X2 = 0.58824 - X2 = 0.44402 This summarised and finalised in a table from the Software Autograph: After the fifth iteration, we started repeating

2. ## MEI numerical Methods

tanx - 1, I could then compare my approximation of the root which the one autograph provided. Formula application: For my spreadsheets to work I had to create a formula for; * The method of bisection * Secant method * False position method * Fixed point iteration * I used

1. ## Numerical solutions of equations

My negative root is x = -1.220744 (6 decimal places) Now I will find the positive root of this function. The starting value (x1) is approximately 1.5 when f(x)=0. x1 = 1.5 x2 = 1.116379 x3 = 0.862068 x4 = 0.745761 x5 = 0.725049 x6 = 0.724492 x7 = 0.724491959 x8 = 0.724491959 I can see some convergence from x6.

2. ## This coursework is about finding the roots of equations by numerical methods.

is negative and f(3) is positive tells us the graph must cross the x axis between 2 and 3. Hence there is a root in the range (2, 3) X y 2.56150 -0.00034 2.56151 -0.00028 2.56152 -0.00021 2.56153 -0.00015 2.56154 -8.2E-05 2.56155 -1.8E-05 2.56156 4.63E-05 This graph shows us how the root is finally trapped in the range (2.56155, 2.56156)

1. ## C3 Numerical Solutions to Equations

= 1.16*10^-6. f(-1.5340701) = -1.08*10^-6. Therefore there is a root as the function is continuous. These two diagrams show the convergence of the iterations at different magnifications. It is shown that the root lies between -1.53407035 and -1.53407025. Taking 1 as the first guess gives the following results Therefore x=0.48269595 � 0.000000005 f( 0.48269594)

2. ## Fixed point iteration: Rearrangement method explained

By looking at the magnitude of g?(x) at the point of intersection with y = x I can tell whether or not method will work. At this point I see that g(x) line has a gradient of less than 1.

1. ## Solving Equations Using Numerical Methods

Method 2: The Newton-Raphson Method Root ? Root ? For the second method, the Newton-Raphson Method, I will be using a different function. The equation I will be using is x3-7x2+x+3=0. This means that I will be drawing the function of f(x)=x3-7x2+x+3.Root ? Firstly, I started by entering the equation into Autograph to get a sketch of the graph.

2. ## Evaluating Three Methods of Solving Equations.

can be difficult if it happens to be along & complicated one; but after that, one has to apply an iterative formula which the computer can do instantaneously, leaving you with the root. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 