Slide 1

C3 Coursework

Examples of the “Rearrangement Method”

Marks 1 and 2

Student A: Slides 2 to 4

Student B: Slides 5 to 7

Student C: Slide 8 & 9

Failure of Rearrangement Method – Marks 3 and 4

Student Z: Slides 10 &11

Student Y: Slides 12 & 13

Student X: Slide 14

Slide 2 Student A ( page 1 of 3 )

3. Rearranging f(x) = 0 in the form x = g(x)

For this method, you rearrange the equation f(x) = 0 into the form x = g(x). By doing this you are then able to simultaneously plot the graphs of y = g(x) and y = x. Where these graphs intersect will be the x values of the roots of the original equation. The equation I will use to test this method is:

f(x) = x 5 – 5.8x + 3 = 0

This function can be seen in the following graph. As you can see there are 3 roots in the following intervals: [­2, ­1], [0, 1] and [1, 2]:

I will now rearrange f(x) = x 5 – 5.8x + 3 into x = g(x).

(1) f(x)= x 5 –5.8x+3 (3) f(x)= x 5 –5.8x+3 g(x)= (5.8x – 3) 0.2 g(x)= (5.8x – 3)

x 4

(2) f(x)= x 5 –5.8x+3 (4) f(x)= x 5 –5.8x+3 g(x)= x 5 –4.8x+3 g(x)= x 5 +3

5.8

Slide 3 Student A ( page 2 of 3 )

By drawing a graphs of any of the functions y = g(x) against y = x then the roots of f(x) = 0 can be found where the two graphs intersect. I will use the first rearrangement to test this method: g(x)= (5.8x – 3) 0.2

This can be seen in the following graph:

As you can the graphs intersect in the intervals [­2,­1], [0,1] and [1,2] agreeing with the intervals in the original graph.

If I take x = xn in order to iterate then:

xn+1= (5.8xn – 3) 0.2

Now I take x1 as a starting point so that I can begin to iterate using the above formula to converge on the root in the interval [1, 2]. I will use x1 = 3 as a starting point for my iteration since 3 is close to the root:

From my iteration my estimate of the root is: 1.37993 Upper Bound = 1.379935 Lower Bound = 1.379925

Error Bounds: x= 1.37993 ± 0.000005

n xn xn+1 1 3 1.70480 2 1.70480 1.47101 3 1.47101 1.40791 4 1.40791 1.38876 5 1.38876 1.38274 6 1.38274 1.38083 7 1.38083 1.38021 8 1.38021 1.38002 9 1.38002 1.37996 10 1.37996 1.37994 11 1.37994 1.37993 12 1.37993 1.37993 13 1.37993 1.37993

Slide 4 Student A ( page 3 of 3 )

Convergence to the root can be seen in the following staircase graph:

Magnitude of g’(x)

In­order for the iteration to converge to the root, the gradient of g(x) at the root must be between –1 and 1. I will test this for my g(x) to demonstrate this fact.

g(x)= (5.8x – 3) 0.2

g’(x)= 0.2(5.8x ­ 3) ­0.8 x 5.8

g’(x)= 1.16(5.8x ­ 3) ­0.8

The estimate for this root is 1.37993, using g’(x) I can show the gradient of the line at this point:

g’(1.37993) = 0.31991

Therefore, the gradient is less than 1 and greater than –1 and so this confirms that the method successfully converges to the root for my chosen g(x).

y= x g(x)= (5.8x – 3) 0.2

Iterations

Slide 5 Student B ( page 1 of 3 )

Rearranging f(x) = 0 in the form x = g(x):

This is the equation that I am going to solve: 3.5x 5 – 6.1x² + x + 1.2 = 0 To solve this equation I will use the fixed point iteration method. The graph below is of: f(x) = 3.5x 5 – 6.1x² + x + 1.2

I can see that there are 3 roots between the intervals [­1, 0], [0, 1] and [1, 2]. I will find a root by rearranging 3.5x 5 – 6.1x² + x + 1.2 = 0 into the form x = g(x) and using fixed point iteration to find the point where the rearranged formula cuts the line y=x.

I will now rearrange 3.5x 5 – 6.1x² + x + 1.2 = 0 into x = g(x), the general formula for fixed point iteration is xn+1 = g(xn).

1.) f(x) = 3.5x 5 –6.1x²+x+1.2 g(x) = 6.1x 2 ­3.5x 5 ­1.2

2.) f(x) = 3.5x 5 –6.1x²+x+1.2 g(x) = 0.2

3.) f(x) = 3.5x 5 –6.1x²+x+1.2 g(x) = 0.5

4.) f(x) = 3.5x 5 –6.1x²+x+1.2 g(x) = 6.1x 2 ­x­1.2

3.5x 4

5.) f(x) = 3.5x 5 –6.1x²+x+1.2 g(x) =

I will use the rearrangement 5.) to find a root where it crosses the line y = x using fixed point iteration.

6.1x 2 ­x­1.2 3.5

3.5x 5 +x+1.2 6.1

­1.2 (3.5x 4 ­6.1x+1)

xn+1 = 6.1xn 2 ­3.5xn 5 ­1.2

xn+1 = 0.2

xn+1 = 0.5

xn+1 = 6.1xn 2 ­xn­1.2 3.5xn 4

xn+1 =

6.1xn 2 ­xn­1.2 3.5

3.5x 5 +xn+1.2 6.1

­1.2 (3.5xn 4 ­6.1xn+1)

Pg.10

Slide 6

­1.2 (3.5xn 4 ­6.1xn+1)

Student B ( page 2 of 3 )

Key for the graphs below:

As you can see the points where the rearranged curve y = g(x) crosses with y =x, shown that it’s the root of the original function f(x) = 3.5x 5 – 6.1x² + x + 1.2 = 0 I will use x = ­1 as my starting value, so x1= ­1.

Now I will use Excel to iterate this formula: xn+1=

f(x) = 3.5x 5 –6.1x²+x+1.2

g(x) =

f(x) = x

­1.2 (3.5x 4 ­6.1x+1)

n xn xn+1 1 ­1 ­0.113208 2 ­0.113208 ­0.709580 3 ­0.709580 ­0.193058 4 ­0.193058 ­0.549824 5 ­0.549824 ­0.256751 6 ­0.256751 ­0.464866 7 ­0.464866 ­0.300065 8 ­0.300065 ­0.419760 9 ­0.419760 ­0.327047 10 ­0.327047 ­0.395384 11 ­0.395384 ­0.343114 12 ­0.343114 ­0.381982 13 ­0.381982 ­0.352464 14 ­0.352464 ­0.374527 15 ­0.374527 ­0.357838 16 ­0.357838 ­0.370348 17 ­0.370348 ­0.360906 18 ­0.360906 ­0.367996 19 ­0.367996 ­0.362652

20 ­0.362652 ­0.366668 21 ­0.366668 ­0.363643 22 ­0.363643 ­0.365918 23 ­0.365918 ­0.364205 24 ­0.364205 ­0.365493 25 ­0.365493 ­0.364524 26 ­0.364524 ­0.365253 27 ­0.365253 ­0.364704 28 ­0.364704 ­0.365117 29 ­0.365117 ­0.364806 30 ­0.364806 ­0.365040 31 ­0.365040 ­0.364864 32 ­0.364864 ­0.364997 33 ­0.364997 ­0.364897 34 ­0.364897 ­0.364972 35 ­0.364972 ­0.364916 36 ­0.364916 ­0.364958 37 ­0.364958 ­0.364926 38 ­0.364926 ­0.364950 39 ­0.364950 ­0.364932

40 ­0.364932 ­0.364946 41 ­0.364946 ­0.364935 42 ­0.364935 ­0.364943 43 ­0.364943 ­0.364937 44 ­0.364937 ­0.364942 45 ­0.364942 ­0.364938 46 ­0.364938 ­0.364941 47 ­0.364941 ­0.364939 48 ­0.364939 ­0.364940 49 ­0.364940 ­0.364939 50 ­0.364939 ­0.364940 51 ­0.364940 ­0.364940 52 ­0.364940 ­0.364940

Pg.11

Slide 7 Student B ( page 3 of 3 )

Table above shows the results of the iterative formula and my estimate of the root is x = ­0.364940 (6.dp) which is the root between the interval [­1, 0].

The cobweb graph for my iterations is shown below. It shows convergence to the root in the interval [­1, 0].

Magnitude of g’(x) The gradient of g(x) at the points where it crosses y = x must be between ­1 and 1, inorder for the iteration to converge to the root.

g(x) = g’(x) =

The estimate root is x = ­0.364940: substituting this into g’(x) gives: g’ (x) = ­ 0.75252 As shown above, ­1 < ­0.75252 < 1, this demonstrates that this rearrangement converges to the root.

16.8x 3 ­7.32 (3.5x 4 ­6.1x+1) 2

f(x) = 3.5x 5 –6.1x²+x+1.2

g(x) =

f(x) = x

Converging line

­1.2 (3.5x 4 ­6.1x+1)

Zoom in

Required root

X50= ­0.364940

Required root

­1.2 (3.5x 4 ­6.1x+1)

Pg.12

Slide 8 Student C ( page 1 of 2 )

Fixed Point Iteration

Now I’ll consider a new equation: 2x 3 + 3.5x 2 ­ 8x – 6 = 0 Shown below is the graph y = 2x 3 + 3.5x 2 ­ 8x – 6 It shows that 3 roots exist.

To use the fixed point iteration method, I’ll rearrange the equation into the form: x = g(x) x = (2x 3 + 3.5x 2 – 6) /8

Hence the iterative formula: xn+1 = (2xn 3 + 3.5xn 2 – 6) /8

y = g(x) and y = x are now shown below:

-4 -2 2 4

-10

10

20

x

y

-6 -4 -2 2 4 6

-4

-2

2

x

y

Slide 9 Student C ( page 2 of 2 )

I’ve used fixed point iteration to find where they 2 lines cross. A starting estimation of x1 = ­0.5

The graph below shows convergence in a Cobweb Diagram:

There is a root at x = ­0.63708

To check this, I will use the derivative: g‘(x) = 3/4x 2 + 7/8x

g‘(­0.63708) = 3/4(­0.63708) 2 + 7/8(­0.63708) = ­0.25304

­1 < ­0.25304 < 1 and this shows that this root will converge

Xn X (2x 3 + 3.5x 2 – 6) /8 X1 ­0.5 ­0.67188 X2 ­0.67188 ­0.62833 X3 ­0.62833 ­0.63929 X4 ­0.63929 ­0.63652 X5 ­0.63652 ­0.63722 X6 ­0.63722 ­0.63704 X7 ­0.63704 ­0.63708 X8 ­0.63708 ­0.63707 X9 ­0.63707 ­0.63708 X10 ­0.63708 ­0.63708 X11 ­0.63708 ­0.63708

-0.8 -0.7 -0.6 -0.5

-0.7

-0.65

-0.6

-0.55

-0.5

x

y

Slide 10 Student Z ( page 1 of 2 )

Failure of this method

This method fails when the gradient of the graph at the root is greater than +1 or lower than –1. This can be illustrated through the second rearrangement of my original equation: f(x) = x 5 – 5.8x + 3 = 0 This rearrangement is: (2) g(x) = x 5 – 4.8x + 3

This can be seen in the following graph:

I have previously found the root x = 1.37993 using the Rearrangement Method.

(From looking at this graph it is clear that where the two lines intersect, between x = 1 and x = 2, the gradient of the function g(x) is greater than 1 so that the method will fail).

If I use a starting point of near to the root I previously found, (x = 1.37993), then I can illustrate how this second rearrangement fails. I will use a starting point of 1.4:

Clearly the method fails after just 3 iterations.

n xn xn+1 1 1.4 1.65824 2 1.65824 7.57870 3 7.57870 Overflow

Slide 11 Student Z ( page 2 of 2 )

These points can be seen in the following graph which staircase away from the root, (x = 1.37993), where the iterations diverge away.

This illustrates that when a starting value close to the root is taken, the method fails to converge to the root. The failure of this is due to the gradient of g(x) at the actual root:

Magnitude of g’(x)

g(x)= x 5 –4.8x+3 g’(x)= 5x 4 –4.8

The estimate for this root using the successful method was x= 1.37993. The gradient at this estimate for the unsuccessful method is: g’(1.37993)= 13.33

The gradient of the curve at the root is significantly greater than 1 and so this explains why the method fails to converge and why it diverges so quickly.

n= 3

n= 2 n= 1

Iterations stair ­ casing away from the root

Slide 12 Student Y ( page 1 of 2 )

Example where one rearrangement fails to converge to root

The equation that I am going to solve: 3.5x 5 – 6.1x² + x + 1.2 = 0

I will use rearrangement 1.) g(x) = 6.1x 2 ­ 3.5x 5 ­ 1.2 to demonstrate how this method fails to find the root x = ­0.364940 (6.dp) which I previously calculated using rearrangement 5.): g(x) =

The points where the rearranged curve y = g(x) intersects with y = x, correspond with the roots of the original function y = f(x) = 3.5x 5 – 6.1x² + x + 1.2. (Shown by the vertical dotted lines).

I will use x = ­0.5 as a starting value to show how the method fails to find the root in the interval [­1, 0], which contains the root that I found previously with rearrangement 5.)

f(x) = 3.5x 5 –6.1x²+x+1.2

g(x) = 6.1x 2 ­3.5x 5 ­1.2

f(x) = x

n xn xn+1 1 ­0.5 0.434375 2 0.434375 ­0.103166 3 ­0.103166 ­1.135035 4 ­1.135035 13.252135 5 13.252135 Overflow

­1.2 (3.5x 4 ­6.1x+1)

Slide 13 Student Y ( page 2 of 2 )

The table and graph above shows the iterations diverging away from the root, which shows how the method failed to find the root in the interval [­1, 0] with rearrangement 1.) This is because the gradient of g(x) is not between ­1 and 1 at the root.

Magnitude of g’(x)

g(x) = 6.1x 2 ­3.5x 5 ­1.2 g’(x) = 12.2x­17.5x 4

From the previous successful method the estimate root between interval [­1, 0] was x= ­0.364940, substitute this into g’(x).

g’(x) = ­4.7627

As you can see that the value of g’(x) does not lies between ­1 and 1, this explain why the method with rearrangement 1.) fails to find the required root, and this is how fixed point iteration method fails.

Slide 14 Student X ( page 1 of 1 )

Failure of this method

0 = 2x 3 + 3.5x 2 ­ 8x – 6 The root I discovered with the previous method was x = ­0.63708.

My new rearrangement is: x 3 = ­((3.5x 2 ­8x­6)/2) x = (­((3.5x 2 ­8x­6)/2)) 1/3

y = g(x) = (­((3.5x 2 ­8x­6)/2)) 1/3 intersects with the line y = x as follows:

Fixed point iteration with this rearrangement cannot find the root ­0.63708, even though it clearly exists where y = x intersects with y = (­((3.5x 2 ­8x­6)/2)) 1/3 .

The 2 rearrangements began close to the root starting from ­0.7 and ­0.5; but they went away from the root at ­0.67308 to the roots near 3 and ­2.

This is because g(x) is not between +1 and ­1. g(x) = (­((3.5x 2 ­8x­6)/2)) 1/3 g ‘(x) = ­ 7x­8 / 7(3.5x 2 ­8x­6) 2/3 g(­0.67308) = 2.7625 1 < 2.7625 so this rearrangement cannot find the root of ­0.67308. This is a disadvantage of the Fixed Point Iteration method.

-6 -4 -2 2 4 6

-4

-3

-2

-1

1

2

3

4

x

y

-6 -4 -2 2 4 6

-4

-3

-2

-1

1

2

3

4

x

y

Slide 15