C3 Coursework

USE AS EXAMPLE ONLY. DO NOT COPY, EVEN IN PART.

HideShow resource information
  • Created by: Daniel
  • Created on: 19-03-12 10:25
Preview of C3 Coursework

First 77 words of the document:

C3 Coursework
I have chosen the equation 2x3+2x2-3.6x+1=0. The function is f(x) =2x3+2x2-3.6x+1, illustrated as a
graph above. I shall attempt to solve this equation and thus find the roots of the function.
Change of sign
To find a root, I shall use the Decimal Search Method to find a change of sign. This will find roughly
where the root is.
1 | Page

Other pages in this set

Page 2

Preview of page 2

Here's a taster:

There is a change of sign between x=-3 and x=-2. This is noted as [-3,-2] and suggests a root being
present between these values. Remember that a root will always have the value y=0, which is
between a positive value and a negative value.
I shall now continue using the Decimal Search Method to find the root to 3 decimal places, using
smaller steps each time to "home in" on the root.…read more

Page 3

Preview of page 3

Here's a taster:

The solution to this is to use smaller steps. This finds a change of sign in the region [0, 0.5]
However, on closer inspection you can see that there are not one, but two roots in this region.
Without the graph for analysis you would fail to see one of the roots.
Newton-Raphson MethodTM
I will now carry out the Newton-Raphson MethodTM on a new equation.…read more

Page 4

Preview of page 4

Here's a taster:

The Newton-Raphson MethodTM relies on the general iterative formula:
For this function, it becomes:
Starting at 2, the method quickly converges to the root
4 | Page…read more

Page 5

Preview of page 5

Here's a taster:

The root is found to be between 2.0557 and 2.0556. This x value can be written as:
2.0556<x<2.0557, or as a value with error bounds: -2.055695±0.0000005
The Newton-RaphsonTM method will not work if we start at x=0 as it encounters a division by 0 in its
process, when xn is 0:
1
x1 = 0 - 0
(Green line represents method working)
The Method will draw a tangent at 1, and then try to find the x axis.…read more

Page 6

Preview of page 6

Here's a taster:

Rearrangement of f(x) =0 to give x=g(x)
I will now carry out the Rearrangement of f(x)=0 to give x=g(x) on a new function and equation.
New Function:
F(x) =0.2x3+1.4x2+2.5x+2.5
New Equation:
0.2x3+1.4x2+2.5x+2.…read more

Page 7

Preview of page 7

Here's a taster:

Rearranging f(x) gives x = -2.5-0.2x -1.4x
2.5
This now provides our iterative formula:
-2.5-0.2(xn)3-1.4(xn)2
xn+1 = 2.5
This is called g(x) and is displayed as the blue curve:
The function y=x is also added to the graph. Starting at x=0, a line is drawn vertically to meet the g(x)
line. A horizontal line is drawn from this meeting point to the line y=x. This is repeated until the
method converges to a root.…read more

Page 8

Preview of page 8

Here's a taster:

Page 9

Preview of page 9

Here's a taster:

After 17 iterations the method converges to -5.00005±0.000005.
However, if we choose a different rearrangement of f(x), we can get:
3
f(x)= -2.5-0.2x2-1.4x2
0.…read more

Page 10

Preview of page 10

Here's a taster:

Clearly, the method will continue to infinity, not providing a solution, even though the lines
y=x and y=g(x) cross.
The magnitude of g'(x) (the gradient of g(x)) must be less than 1 (the gradient of y=x), otherwise the
series will be divergent and never reach a solution.
Comparison of Methods
Now I will carry out all three methods on one equation.
Function:
F(x) =0.2x3+0.…read more

Comments

NumNub


What grade/mark did you get for this? Thanks :)

Similar Mathematics resources:

See all Mathematics resources »See all resources »