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

Solve the equation: X3- 5X-5 = 0. This equation cannot be factorised nor solved algebraically therefore it must be solved through interval bisection.

Extracts from this document...

Introduction

Interval Bisection.

Ben Ward

Solve the equation:  X3– 5X-5 = 0. This equation cannot be factorised nor solved algebraically therefore it must be solved through interval bisection.

To calculate the approximate position of the roots and the number of roots I used Omnigraph to sketch the graph (below shows where the graph crosses the x-axis).

This is the root I want to find and shows how the solution lies between 2.6 and 2.7image00.pngimage01.png

[f (X) = X3 – 5X - 5]

X0 = 2.6   f (X0) =  -0.424

X1 = 2.7   f (X1) = + 1.183

As f(X0)< 0 and f(X1) > 0 it shows that the solution must lie between X0 and X1, thus creating an error bound  2.6 < X < 2.7.

In order to reduce the error bound and ultimately find the solution I set-up a table of values for  X and f(X) using Excel.

...read more.

Middle

0.001199

1

2.62734375

11

2.6273926

0.0004319

1

2.62734375

12

2.6273682

4.837E-05

1

2.62734375

13

2.627356

-0.000143

-1

2.62736816

14

2.6273621

-4.75E-05

-1

2.62736816

15

2.6273651

4.335E-07

1

2.62736206

16

2.6273636

-2.35E-05

-1

2.62736511

After 16 literations it gives the answer of 2.627 to 3 d.p or 2.627 +/- 0.0005. Therefore it can be said that that 2.627<X<2.628.

  • As the sign of (X) when its equal to 2.627 is negative

            i.e. 2.627 3 – 5*2.627  - 5 =  -0.00573

  • And the sign of (X) when it equals 2.628 is positive

            i.e. 2.628 3 – 5*2.628  - 5 =  -0.00998

To give an example of an equation that cannot be solved using this method is an equation that follows two main points:

  1. It must be an equation that just touches the x-axis but doesn’t cross the axis i.e. X4 (shown on omnigraph below).image02.png
  1. Must not touch the x-axis at the origin- therefore the graph of X4 needs to be ‘shifted’ either left or right i.e. (X-7)4   [shown below on omnigraph].image03.png

...read more.

Conclusion

An example of this is: X3-3X2+X+1 = 0. image06.png

The diagram below shows that the gradient of the curve X3-3X2+X+1 = 0 although is not too steep to allow the Newton Raphson method to calculate the value of the root but highlights how the solution lies close to a maximum point on the graph.

image07.png

It is also clear that by using excel to create a spread sheet, as used to calculate the solution for: 2X4- X3 + 3X2-2 = 0, shows how the tangent and therefore the answer moves further away and doesn’t give a correct solution.

Newton raphson

n

X(n)

0

0.5

1

1.2

2

0.991489

3

1.000001

4

1

5

1

6

1

7

1

8

1

9

1

10

1

11

1

...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)

    1 2 1.001 2.006006002 6.012006 x x3 2x3 gradient 1 1 2 6 2 8 16 24 3 27 54 54 4 64 128 96 After looking through each value, I have observed a pattern. I have noticed that the gradient is always equal to 6x�.

  2. Best shape for gutter and further alegbra - using Excel to solve some mathematical ...

    31 3.33333 3.33333 14.42935 The formulae to calculate the area of this cross section are given in Appendix 1. Triangular cross section(V shape) - Area = x () x () sin? A =w2 sin? By varying ? by 1�, where 0� is at the point of the two sides touching each other to make a straight line.

  1. Finding the root of an equation

    Because of the nature of the answer when given to five significant figures, the amount of error is +0.000005.

  2. The Gradient Fraction

    The graph follows on the next page Results x Gradient 1 3 2 12 3 27 4 48 x=1: Gradient = 1.030 - 1 = 0.03 1.01 - 1 = 0.01 = 3 x=2: Gradient = 8.120 - 8 = 0.12 2.01 - 2 = 0.01 = 12 x=3: Gradient

  1. In this coursework I will be looking at equations that cannot be solved algebraically

    between the integer bounds [2, 3], and it seems there may be roots between [5, 6] but it is not clear. Using decimal search, I have calculated the integer values for x from [2, 6] x y 2 -6.24 3 1.98 4 2.101 5 0.1214 6 2.042 This table shows

  2. Investigate the relationships between the lengths of the 3 sides of the right angled ...

    = an� + bn + c F (n) = 2n� + 2n Try n = 1 F (1) = 2 x 1� + 2 x 1 = 2 + 2 = 4 Try n = 2 F (2) = 2 x 2� + 2 x 2 = 8 + 4 = 12 Try n = 100 F (100)

  1. Examining, analysing and comparing three different ways in which to find the roots to ...

    I then took this a stage further, and did the decimal search to 2 decimal places. The results were: 4.1 4.11 4.12 4.13 4.14 4.15 4.16 4.17 4.18 4.19 4.2 -3.58 -3.22 -2.87 -2.51 -2.14 -1.78 -1.41 -1.04 -0.67 -0.29 0.09 The result here was an interval between [4.19, 4.2], and so I then went further, to 3 decimal places.

  2. Numerical Methods used to solve those equations which cannot be solved analytically.

    1 0.227 1.1 0.1088 1.2 0.03188 1.3 0.0004885 1.4 0.01874 1.5 0.0908 1.6 0.2208 1.7 0.4129 1.8 0.6712 1.9 1 The graph does not display any change of sign, implying that there are no roots. However, it is evident from the graph that there are two roots (see magnified version of graph).

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