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# Decimal Search.

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Introduction

Pure 2 Coursework

Decimal Search

This method works by drawing a graph, seeing where it crosses the axis, then re-plotting the graph, with a smaller scale.  This is repeated until required accuracy is achieved.  I require an accuracy of four decimal places.

The equation I shall investigate will be:

y = x4+2x3−5x2+1 I shall find the root by working out values of x and looking for a change of sign.  I can see by looking at the graph that the change of sign occurs between 3 and 4.  I shall plot a graph, to find a more accurate estimate.

 x f(x) -4 49 -3 -17 -2 -19 -1 -5 0 1 1 -1 2 13 3 91

I can now see that the change of sign occurs between 3.4 and 3.5.  Again, I shall draw another graph to find a more accurate answer. x f(x) -3 -17 -3.1 -14.2799 -3.2 -10.8784 -3.3 -6.7319 -3.4 -1.7744 -3.5 4.0625 -3.6 10.8496 -3.7 18.6601 x f(x) -3.4 -1.7744 -3.41 -1.23143 -3.42 -0.6796 -3.43 -0.11884 -3.44 0.450921 -3.45 1.029756 -3.46 1.617735 -3.47 2.214927 x f(x) -3.43 -0.11884 -3.431 -0.06227 -3.432 -0.00561 -3.433 0.051138 -3.434 0.107979 -3.435 0.16491 -3.436 0.221931 -3.437 0.279043

Solution = -3.4325 ± 0.0005

f (−3.432) = -0.00561

f (−3.433) =  0.051138

Middle

1).  This process is repeated a few times until the value of x appears to have stopped changing.

The Newton-Raphson iterative formula is:

xn+1 = xn – The equation I am going to solve is:

f(x) = 15x4 + √2 – x -  (x) = 60x3 - x0 = 1

x1 = x1 = 1 – x1 = 1 – 0.2114

x1 = 0.7886

First Root:  x0 = 1

 x |Δx| 0.7886 0.2114 0.6559 0.1327 0.5921 0.06378 0.5762 0.01584 0.5753 0.000933 0.5753 3.137E-06

x = 0.5753

Second Root:

x0 = 0

 x |Δx| 0.05227 0.05227 0.05231 3.574E-05 0.05231 9.95E-11

x = 0.05231 ± 0.000005

f (0.052305) =   4.82872E-06

f (0.052315) = −2.65013E-05

This method will fail if x0 is on a turning point, or if the graph has a discontinuity:

E.g. A graph with a discontinuity:

y = ⅓ Ln (x – 2) +1 The tangent to the line crosses the axis after the root.  The process cannot continue from here.

Rearrangement

y = 2x5 – This equation shall be rearranged to find two separate equations in the form x=g(x).  Instead of then finding the root of 0 = 2x5 – , we find the point of intersection between x=g(x) and the line y=x.  To do this, the

Conclusion

It is clear to see that decimal search is the simplest way to solve equations.  It is however the slowest, as it a table of values has to be made each time, to improve accuracy and the equation has to be written into Excel, which can take time.  Doing it by hand would also take a long time, as you would have to put numbers through the equation up to ten times before the change of sign is found.  This will be very monotonous and time consuming.

The Newton-Raphson method is the fastest, especially if you have a computer with Autograph to draw the graphs and to the maths behind it for you.  A few more steps are necessary if you use the formula each time to improve your answer.

The rearrangement method required the largest number of steps, and on top of that, the equation had to be rearranged twice.

Overall, I believe the Newton-Raphson method is the best method to use when solving equations, because it is the shortest by hand and without a doubt the shortest and easiest using a computer.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

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