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Numberical method

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In this coursework I am going to use knowledge of numerical methods to produce an approximation root which cannot be solved using analytical methods.

Some particular equations Iwill be using during the coursework because they’re cannot be integrated by any analytical method therefore approximation method need to be use, because of this numerical method is a suitable method to find out the solution of these function.

1 Change of sign method

Consider this equation image02.pngimage02.png


                From the graph we can see point A lies between -2 and -3

When x= -2 given   image10.pngimage10.png

When x= -3 given image22.pngimage22.png

Y= -28.1

There is a change in sign between – 2 and -3 means that there is a root between the two points.

Decimal research

For function f(x) = image26.pngimage26.png

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We can see that the function of f(x) is change sign from -2.2 to -2.1 therefore the root must lie between these two points


  • I now repeat the process by change the increment from 0.1 to 0.01 between -2.2 and -2.1 given that

To work out x I’ll use the formula image04.pngimage04.pngr+1= image04.pngimage04.pngr+0.01 to give the function of x

 f(x)+1= image08.pngimage08.png

























From the table there is a sign change from -2.19 to -2.18 which means this become the new interval in which the root is between.image09.png

  • The interval had narrow down to -2.18 and -2.19, I’ll then now change the increment size to 0.001 which will zoom in the interval even more
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In other cases where this method would fails to work is when there’re roots are very much close together because they are likely to be in the same interval result a missing of change in sign.
And also when a function given a discontinuous curve the whole fraction has an asymptote between some specific intervals that make the change in size think there is a root present but there is no real root present.

Fixed point Iteration using  image19.pngimage19.png method image20.pngimage20.png

Given the equation image21.pngimage21.png= 0

         This can be rearrange into the form of x=g(x)

image23.pngimage23.png= 0



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