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# C3 Mei - Numerical Methods to solve equations

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Introduction

C3 Coursework

In this coursework, I will use numerical methods to solve the following equation,  as I cannot solve it algebraically. I can only obtain an approximation of the solution as it is impossible or hard to find the exact value of the function.

Decimal Search

The graph below is the function  I will use decimal search in order to find an approximation of one of the roots.   The table below shows decimal search. Each boundary is tested for sign change which indicates that a root exists between them. The x where the sign change occurs in now the new boundaries and tested for sign change again.

This method is repeated until an approximation of the root is found to a suitable number of decimal places.

 x f(x) 0 2 0.1 1.9501 0.2 1.8016 0.3 1.5581 0.4 1.2256 0.5 0.8125 0.6 0.3296 0.7 -0.2099 0.8 -0.7904 0.9 -1.3939 1 -2 x f(x) 0.66 0.011747 0.661 0.006295 0.662 0.000838 0.663 -0.00462 0.664 -0.01009 0.665 -0.01556 0.666 -0.02104 0.667 -0.02652 0.668 -0.032 0.669 -0.03749 0.67 -0.04299 x f(x) 0.662 0.000838 0.6621 0.000292 0.6622 -0.00025 0.6623 -0.0008 0.6624 -0.00135 0.6625 -0.00189 0.6626 -0.00244 0.6627 -0.00299 0.6628 -0.00353 0.6629 -0.00408 0.663 -0.00462
 x f(x) 0.6 0.3296 0.61 0.277958 0.62 0.225763 0.63 0.17303 0.64 0.119772 0.65 0.066006 0.66 0.011747 0.67 -0.04299 0.68 -0.09819 0.69 -0.15383 0.7 -0.2099

[0, 1]                   [0.6, 0.7]                      [0.66, 0.67]                 [0.662, 0.663]

Root intervals [0.6621, 0.6622]
I know that the root is 0.662 to 3 decimal places

Middle and  are the same to 6 decimal places, indicating that an approximation of the root has been found.

The other two roots were calculated in the same manner.   x 2.8464 2.8283 2.8281 2.8281     x - 0.62766 - 0.46882 - 0.4343 - 0.43266 - 0.43265 - 0.43265

For the last approximation, the error bounds are -0.43265 ± 0.000005

In order to confirm the approximation as the root, I will check for sign change;  As there is a sign change, when the boundaries of the approximation are put into the equation, a root exists.

Failure

In order to show the failure of this method, I will use following equation:

y = 3.5x+2.8x³+5.4x+3    x -9.499 Overflow

The above graph and the table shows that the Newton-Raphson method has failed even thought the starting root was close to the actual root.

Rearrangement equation

In the rearrangement method, an equation is rearranged to find .

The graph below shows the equation  which I will rearrange to find one root. In order to find the root, I rearranged my equation to      The graphs above are and . The two functions intersect at the roots of .

The graph below shows the success of the rearrangement method.

Conclusion

The Decimal Search takes more iteration; but, this method is the easiest and easily understood. However, this method is best done on a spreadsheet, where you would be able to spot the sign change easily.

The Rearrangement Method takes slightly more iteration but it provides the root to any degree of accuracy. Also, the formula is iterative, therefore, it is not very time consuming. However, finding  can be tricky.

Software

In terms of the software used, Decimal Search was the easiest as it only required spreadsheet which is not difficult to use. Although making the tables can be repetitive, any faults can easily be rectified.

Both the Newton-Raphson Method and the Rearrangement Method used a calculator to work out the iterative steps. This was often very time-consuming and frustrating as simple mistakes could let to the wrong route.

Autograph was used to draw all the graphs and show the methods at work. It was not hard to use but tricky, due to the different options available.

Page |

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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