• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14

Change of sign.

Extracts from this document...

Introduction

Change of sign

Bisection

Success

The success of the bisection method can be proved by using the following equation:

f(x) = 0.64x4+0.28x2-2x+0.28

For this method, I have chosen to find the root between 0 and 1.  Below is the process taken to find this root, each table represents zooming in once (halving the window e.g. first window is 1 unit, second window is 0.5 units).image12.jpg

Lower Bound

Middle Value

Upper Bound

Lower Bound

Middle Value

Upper Bound

0.000000

0.500000

1.000000

x

0.000000

0.250000

0.500000

x

0.280000

-0.610000

-0.800000

f(x)

0.280000

-0.200000

-0.610000

f(x)

Lower Bound

Middle Value

Upper Bound

Lower Bound

Middle Value

Upper Bound

0.000000

0.125000

0.250000

x

0.125000

0.187500

0.250000

x

0.280000

0.034531

-0.200000

f(x)

0.034531

-0.084365

-0.200000

f(x)

Lower Bound

Middle Value

Upper Bound

Lower Bound

Middle Value

Upper Bound

0.125000

0.156250

0.187500

x

0.125000

0.140625

0.156250

x

0.034531

-0.025283

-0.084365

f(x)

0.034531

0.004537

-0.025283

f(x)

As to achieve the desired accuracy (of 3 decimal places) this process had to be repeated many times, the last of which is shown below:

Lower Bound

Middle Value

Upper Bound

Lower Bound

Middle Value

Upper Bound

0.142975

0.14299

0.143005

x

0.14299

0.142998

0.143005

x

4.14E-05

1.23E-05

-1.7E-05

f(x)

1.23E-05

-2.3E-06

-1.7E-05

f(x)

Shown in the lightly shaded area; are the X values for which the first 3 decimal places are the same.

image13.jpg

From the image above it can be seen that the process has succeeded in finding the root, although on the diagram it appears that the line is cutting at 0.143, it does indeed cut just before, however due to the limitations of the software, it cannot zoom in anymore to show properly labelled numbers.

Failure

The theory behind bisection is by splitting up two boundaries in two (by halves) so to defeat this method, there will need to be two roots within 1 unit (e.g. between 0 and 1)

...read more.

Middle

f(x) = -1.11x5+3.2x4+0.15x3-4x2- 2x

image16.jpg

Looking at the gradients at the integer values of X it can be seen that zooming into the root close to x = -1 will be possible, however looking at the gradient at X = 0, it does not seem steep enough for it to find the root closer to 0.  We can see below that it does not.

Xn

Xn+1

Xn

Xn+1

-1.00000

-0.94769

0.00000

-3.33333

-0.94769

-0.93838

-3.33333

-2.62212

-0.93838

-0.93810

-2.62212

-2.07028

-0.93810

-0.93810

-2.07028

-1.65018

-0.93810

-0.93810

-1.65018

-1.34104

-0.93810

-0.93810

-1.34104

-1.12809

-0.93810

-0.93810

-1.12809

-1.00120

-0.93810

-0.93810

-1.00120

-0.94803

-0.93810

-0.93810

-0.94803

-0.93840

-0.93810

-0.93810

-0.93840

-0.93810

-0.93810

-0.93810

-0.93810

-0.93810

-0.93810

-0.93810

-0.93810

-0.93810

-0.93810

-0.93810


Xn

Xn+1

Xn

Xn+1

Xn

Xn+1

1.00000

-6.40000

2.00000

5.73333

3.00000

2.66440

-6.40000

-5.04384

5.73333

4.75957

2.66440

2.44056

-5.04384

-3.96843

4.75957

3.99593

2.44056

2.31850

-3.96843

-3.11991

3.99593

3.40488

2.31850

2.27792

-3.11991

-2.45573

3.40488

2.95804

2.27792

2.27355

-2.45573

-1.94257

2.95804

2.63506

2.27355

2.27350

-1.94257

-1.55480

2.63506

2.42267

2.27350

2.27350

-1.55480

-1.27336

2.42267

2.31080

2.27350

2.27350

-1.27336

-1.08498

2.31080

2.27661

-1.08498

-0.98005

2.27661

2.27352

-0.98005

-0.94283

2.27352

2.27350

-0.94283

-0.93817

2.27350

2.27350

-0.93817

-0.93810

2.27350

2.27350

-0.93810

-0.93810

-0.93810

-0.93810

image05.png

image08.pngimage09.pngimage06.pngimage07.pngimage16.jpg

As seen in the diagram above, the iterative process used to solve the root failed as instead of finding the desired root, it zoomed in on another, thus resulting in a failure.

Error bound

The error bound for the demonstrated point of success is ±0.000002.  This was calculated by subtracting the mean of the last two different values of Xn+1 from those values.


Rearrangement equations

From previous knowledge, we know that if two functions were rearranged into one, the points of intersection will be the roots to the singular (rearranged) equation.

However by rearranging f(x) into g(x) and h(x), where h(x) =x.  We will have a straight like with a newly formed g(x)

...read more.

Conclusion

Excel was used to solve the function as this software makes iterative process very easy by using its auto-fill function (which looks at the pattern within the selected area, then repeats it throughout the dragged region).  This function was used repeatedly throughout all methods, however, was easier to implement in some methods (details of which are mentioned below).

The bisection method was very easily achieved in Omnigraph as it only need meet one requirement, that the successful root intersects the X axis between 2 integers where there isn’t another intersect.  However, using Excel to solve the root was more difficult, as each step involved its own table, which had to be copied and pasted for each step.  Then the figures of the previous step had to be entered also, resulting in a long and repetitive process, with a relatively inaccurate result.

The Newton Raphson method was completed with much more ease, as after the differentiation of the function, the rest becomes easy as the equation is to be inserted into the top cell, then auto-filled until the sequence converges (or diverges).  Also this method would be much easier to achieve much higher levels of accuracy, as each line of the table is one step and a subsequent increase in accuracy.

The rearrangement method may be the easiest of the methods to implement as all that is needed is for the g(x) to be placed into the cell, and the cells dragged down until the sequence converges (or diverges).  Again like in the Newton Raphson method, each line of the table represents one step.  So the more that is auto-filled, the more accurate the root is.

...read more.

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

See related essaysSee related essays

Related AS and A Level Core & Pure Mathematics essays

  1. Marked by a teacher

    The Gradient Function

    5 star(s)

    Here I have taken the point x = 2. I first get the co-ordinates for both points, which are 2 and 22, and Q, 3 and 32. After this I used the formula: y - x 2 2 = 3^2 - 2^2 = 9-4 = 5 y - x 3

  2. Marked by a teacher

    Estimate a consumption function for the UK economy explaining the economic theory and statistical ...

    3 star(s)

    became bigger and the scale in Figure1 (b) followed the same. However, a fundamental assumption behind OLS is that the behavior of the residuals is random. The Figure 1(b) shows that the residuals are related to one another, therefore the absolute consumption equation not fits data well as the graph is pattern.

  1. Change of Sign Method.

    The above illustration shows that for x0=2, the gradient of y=g(x)= 1/4( x3+2x2-4.58) is greater than that of y=x, i.e. greater than 1. Without further calculation, this can be seen by the fact that the line y=g(x) is steeper than the line of y=x. The gradient of y=g(x) 1/4( x3+2x2-4.58)

  2. Mathematical Investigation

    value of "b" is the horizontal stretch factor of the graph along the x-axis. The stretch factor is used as 1/b. The new period can be calculated from 2pi/b. * The phase shift represented by "c". Varying values of "c" tell how much each cycle shifts along the x-axis.

  1. Triminoes Investigation

    I'm doing this because to find what "c" is worth when substituting a = 2, b = 3 into equation 6. 19a + 5b + c = 10 19 x 1/6 + 5 x 1 + c = 10 19/6 + 5 + c = 10 3 + 1/6 +

  2. Although everyone who gambles at all probably tries to make a quick mental marginal ...

    Entry is $100 a ticket, and there is a set limit of 222 000 tickets sold. For the purposes of the calculations, all of the prizes from $1 000 000 to $22 207 shall be ignored. This is because the Payoff and the Odds are quite incongruous throughout these prizes.

  1. The method I am going to use to solve x−3x-1=0 is the Change ...

    �)] = -1.380277569 x8 = x7 - [(x7 ^4+x7 �-1=0)/( 4x7 �+3x7 �)] = -1.380277569 I can see some convergence from x6. There has been no change in the x-values between x7 and x8 for this number of decimal place.

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

    In order to create the SierpiÅksi Triangle, we need to set variables. Variable A and B both mean to go forward, variable + means to turn left by an angle of sixty degrees, while variable – means to turn right by sixty degrees.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work