- Level: AS and A Level
- Subject: Maths
- Word count: 1816
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
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Introduction
Numerical Methods
In my coursework I will be using three equations to investigate there 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 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 | -0.66796875 |
-0.8125 | -0.123535156 |
-0.78125 | 0.119567871 |
-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 | -0.009985081 |
-0.762 | -0.006586728 |
-0.763 | -0.003180947 |
-0.764 | 0.000232256 |
-0.765 | 0.003652875 |
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 | -1.027566442 |
-1.027566442 | 1.485825571 |
1.485825571 | 2.47808672 |
2.47808672 | 3.442647822 |
3.442647822 | 4.254251613 |
4.254251613 | 4.878406213 |
4.878406213 | 5.331530284 |
5.331530284 | 5.648456535 |
5.648456535 | 5.864822571 |
5.864822571 | 6.010227998 |
6.010227998 | 6.106948094 |
6.106948094 | 6.170855114 |
The
f(x)= 2x³–5x²–4x+1
f’(x)= 6x²–10x–4
-1 | -0.833333333 |
-0.833333333 | -0.798474946 |
-0.798474946 | -0.796929982 |
-0.796929982 | -0.796926979 |
-0.796926979 | -0.796926979 |
-0.796926979 | -0.796926979 |
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.
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