• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13
14. 14
14
15. 15
15

# C3 Mei - Numerical Methods to solve equations

Extracts from this document...

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.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related AS and A Level Core & Pure Mathematics essays

1. ## C3 Coursework - different methods of solving equations.

5 star(s)

requires no complex calculations whatsoever except subbing in X values into the original equation. But with Computers being able to do even that, it makes it even easier to use it. So the only human interaction needed is to look at the results and manually looking at between which two points (limits)

2. ## The Gradient Function

5 star(s)

2 4 2 4 8 8 3 9 18 12 4 16 32 16 The gradient here is equal to 4x - I can see this by investigating the values of x, as compared to the gradient. I predict that the gradient for 5 will be 20.

1. ## Estimate a consumption function for the UK economy explaining the economic theory and statistical ...

3 star(s)

0.000 0.9962 sigma 8845.94 RSS 4.14728156e+009 R^2 0.99624 F(1,53) = 1.404e+004 [0.000]** log-likelihood -576.847 DW 0.498 no.of observations 55 no.of parameters 2 mean(C) 339918 var(C) 2.00545e+010 C = + 1.37e+004 + 0.9207*Y (SE) (3e+003) (0.00777) Figure1 Table 2 EQ (2) Modelling LC by OLS (using project12min.xls) The estimation sample is: 1948 to 2002 Coefficient Std.Error t-value t-prob Part.R^2 Constant 0.750412 0.09542 7.86 0.000 0.5385 LY 0.938329 0.007520 125.

2. ## In this coursework, I am going to solve equations by using the Numerical Methods. ...

know the root lies between -1.6 and -1.5, but as I said I will try to find the root to 4 decimal place so I need to continue my calculation. The table below shows x values for 2 decimal places.

1. ## Numerical solutions of equations

0.68071 -3.816012x10-4 0.68072 -3.164083x10-4 0.68073 -2.512140x10-4 0.68074 -1.860182x10-4 0.68075 -1.208210x10-4 0.68076 -5.562232x10-5 0.68077 9.577756x10-6 I can see that the change of sign is between x = 0.68076 and x = 0.68077 As there is a change of sign, I can finally say that the root of this function is in the interval [0.68076, 0.68077].

2. ## I am going to solve equations by using three different numerical methods in this ...

(a+b)/2 is the mid-point. By using the Excel, we can easily find the x value in the many terms satisfy my required degree of accuracy which is answer to 4 decimal places. Root is -1.8007 to 4d.p Error bounds is -1.8007�0.00005 Root bounds is -1.80075<x<-1.80065 Check X Y -1.80075 -0.000688(negative)

1. ## Solving Equations. Three numerical methods are discussed in this investigation. There are advantages and ...

0.025129 0.007813 9 1.523438 0.025129 1.53125 -0.00827 1.527344 0.008436 0.003906 10 1.527344 0.008436 1.53125 -0.00827 1.529297 8.43E-05 0.001953 11 1.529297 8.43E-05 1.53125 -0.00827 1.530273 -0.00409 0.000977 12 1.529297 8.43E-05 1.530273 -0.00409 1.529785 -0.002 0.000488 13 1.529297 8.43E-05 1.529785 -0.002 1.529541 -0.00096 0.000244 14 1.529297 8.43E-05 1.529541 -0.00096 1.529419 -0.00044 0.000122

2. ## Numerical solution of equations, Interval bisection---change of sign methods, Fixed point iteration ---the Newton-Raphson ...

r Xr f(Xr) f'(Xr) f(Xr)/f'(Xr) 1 2 29 47 0.617021277 2 1.382978723 7.671007388 23.2775011 0.329546 3 1.053432723 1.678766068 13.4149463 0.125141468 4 0.928291256 0.205236732 10.1818519 0.020157112 5 0.908134144 0.004995226 9.68744176 0.000515639 6 0.907618504 3.23625E-06 9.67489018 3.345E-07 7 0.90761817 1.36247E-12 9.67488204 1.40825E-13 Spreadsheet 2.3 From spreadsheet 2.3 above, rounding up the solution to 5 decimal places, I get the solution of 0.90762.

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to
improve your own work