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Solving equations by numerical methods - The Interval Bisection method

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Introduction

Maxim Protsenko

Pure Mathematics 2 Coursework

Solving equations by numerical methods

The Interval Bisection method

The coursework on Numerical Methods shows the techniques that can be applied to find the solutions of equations, which have no algebraic solution. The first method is the Interval Bisection Method. It involves finding an interval of x in which f(x) changes sigh. If f(x) is a continuous function, it follows that it has a root within that interval. The equation for this method that I took is x³+2x²-x-3=0.

image00.png

It is clearly seen on the graph that one of the roots of the equation is between 1 and 2. Within this interval f(x) changes the sigh from negative to positive. To find the root I built a table where a is the lower limit of the interval (in this case 1) and b is the upper limit of the interval (in this equation it is 2). The table looks like this.

The Interval Bisection Table

x³+2x²-x-3=0

f(x)<0

f(b)>0

A

b

(a+b)/2

f((a+b)/2)

possible error

1

2

1.5

3.375

0.5

1

1.5

1.25

0.828125

0.25

1

1.25

1.125

-0.169921875

0.125

1.125

1.25

1.1875

0.307373047

0.0625

1.125

1.1875

1.15625

0.06338501

0.03125

1.125

1.15625

1.140625

-0.054592133

0.015625

1.140625

1.15625

1.1484375

0.004064083

0.0078125

1.140625

1.1484375

1.14453125

-0.025346935

0.00390625

1.14453125

1.1484375

1.146484375

-0.0106621757

0.001953125

1.146484375

1.1484375

1.1474609375

-0.00330423657

0.0009765625

1.1474609375

1.1484375

1.14794921875

0.000378625351

0.00048828125

So the answer is 1.14794921 to (6 d. p.)

The possible error value represents the distance between the estimated root and the interval limits. The error bunds are a and b, so that x is between the two values. The error bounds for this root of the equation x³+2x²-x-3=0 are

1.1474609375 <x< 1.1484375

...read more.

Middle

, for a root of f(x)=0. Then drawing the tangent to the curve y=f(x) at the point (image02.png, f(image02.png)). The point at which the tangent cuts the x axis then gives the next approximation for the root and the process is repeated.

The equation I took for this method is x³-7x+1=0.

It looks like this:

image24.png

The gradient of the tangent at (image02.png, f(image02.png)) is f ‘(image02.png). The equation of the straight line can be written as

y-image03.png=m(x-image02.png),

the equation of the tangent is:

y-f(image02.png)=f ‘(image02.png)[x-image02.png].

This tangent cuts the x axis at (image04.png, 0), so:

0 -f(image02.png)=f ‘(image02.png)[image04.png-image02.png].

Rearranging this gives:

image04.png=image02.png-image06.png.

=> the Newton-Raphson iterative formula:

image07.png

As my equation is x³-7x+1=0, with the root in the interval [2, 3], I can rewrite it as:

f(x)= x³-7x+1=0 and f ‘(x)=image08.png.

The iterative formula is therefore:

image07.png

=image09.png

Starting with the point image02.png=3 gives

x(n)

f(x)

f '(x)

x(n+1)

3

7

20

2.65

2.65

1.059625

14.0675

2.574676

2.574676

0.044678977

12.886864

2.571209

2.571209

0.00009

12.833342

2.571201

2.571201

0.0000000004

12.83323

2.571201

The Newton-Raphson method gives an extremely rapid rate of convergence.

The first root of the equation I found is 2.571201 with the error bounds of 2.571200<x<2.571202.

...read more.

Conclusion

        But despite of this it takes lots of time to obtain one root and would take ages to do it without any computer or graphic calculator aid. Also the initial search may miss one or more root, for example when the x-axis is a tangent to the urve or when several roots are very close together.

        Newton Raphson method is the easiest way to estimate the root and takes much less time than to estimate the root using the first method. It doesn’t take that much time but it can go wrong very quickly and jump to another root.

        And the last method rearranging is the hardest one. It takes plenty of time to rearrange the equation and then to estimate the root. And it’s very likely to do something wrong using it.

        In conclusion I can say that the most efficient and easiest way for me to estimate the root, was the first one, because it takes not so much time if you get used to it. Very useful were PC programs such as Excel and Autograph. And at last the fact that in excel it is easier to change one equation on another and to get other roots, I can definitely say that it is the best method for me.

...read more.

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