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

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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.

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