• 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

Prital Upadhayay

C3 Coursework

In this coursework, I will use numerical methods to solve the following equation, image47.pngimage48.pngas 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 image56.pngimage56.pngI will use decimal search in order to find an approximation of one of the roots.

image69.pngimage00.pngimage01.png

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

...read more.

Middle

image63.png and image64.pngimage64.png 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.

image65.png

image02.png

image03.png

x

2.8464

2.8283

2.8281

2.8281

image66.png

image04.pngimage04.png

image05.png

image06.png

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;

image67.png

image68.png

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

image70.png

image71.png

image04.png

image07.png

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 findimage72.png.

The graph below shows the equation image73.pngimage73.png which I will rearrange to find one root.

image74.png

In order to find the root, I rearranged my equation to image75.png

image76.png

image08.png

image09.png

image11.png

image12.png

The graphs above are image77.pngandimage78.png. The two functions intersect at the roots ofimage80.png.

The graph below shows the success of the rearrangement method.

...read more.

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 image110.pngimage110.png 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 |

...read more.

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

See related essaysSee related essays

Related AS and A Level Core & Pure Mathematics essays

  1. Marked by a teacher

    C3 Coursework - different methods of solving equations.

    5 star(s)

    Because this only works out the value to 3 decimal places, I will use the Newton Raphson Formula to work it out to 5 decimal places. I will summarise it in a table like previously So I have to work out F' of my equation.

  2. MEI numerical Methods

    the upper bounds and lower are two opposite signs the root must lie within the range of values of the interval. As K varies As we proved above, if K = 1 the root is 0.479731007. However what if K, was 2, 3 etc, could a pattern be distinguished, the

  1. This coursework is about finding the roots of equations by numerical methods.

    To find the root in the interval (0, 1) Decimal search X y 1 -1 0.9 -0.097 0.8 0.624 0.89 -0.01701 0.88 0.061184 0.889 -0.00911 0.888 -0.00123 0.887 0.006639

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

    Newton Raphson method This is a typical method of fix point iteration. The concept of fix point iteration is very similar to interval estimation but instead of bisect two points to find an approximation of roots we will use a 'fix point' and try to find a better result

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

    To conclude the results I have found, Newton-Raphson Method is the one has the fastest speed of convergence, which can be explained as in this method, finding a root by the intercepts of tangents to the curve with x-axis seems to be more efficient, by comparing to the Interval Bisection Method and the method of Rearranging the equation f(x)

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

    =2*B4^3-3*B4^2-8*B4+23 =2*C4^3-3*C4^2-8*C4+23 =(B4+C4)/18 =2*F4^3-3*F4^2-8*F4+23 =IF(G4>0,B4,F20) =IF(G4>0,F4,C20) =2*B4^3-3*B4^2-8*B4+24 =2*C4^3-3*C4^2-8*C4+24 =(B4+C4)/19 =2*F4^3-3*F4^2-8*F4+24 =IF(G4>0,B4,F21) =IF(G4>0,F4,C21) =2*B4^3-3*B4^2-8*B4+25 =2*C4^3-3*C4^2-8*C4+25 =(B4+C4)/20 =2*F4^3-3*F4^2-8*F4+25 =IF(G4>0,B4,F22) =IF(G4>0,F4,C22) =2*B4^3-3*B4^2-8*B4+26 =2*C4^3-3*C4^2-8*C4+26 =(B4+C4)/21 =2*F4^3-3*F4^2-8*F4+26 =IF(G4>0,B4,F23) =IF(G4>0,F4,C23) =2*B4^3-3*B4^2-8*B4+27 =2*C4^3-3*C4^2-8*C4+27 =(B4+C4)/22 =2*F4^3-3*F4^2-8*F4+27 Bisection failure I am going to show the failure of bisection method.

  1. Best shape for gutter and further alegbra - using Excel to solve some mathematical ...

    The benefit of the replication of formulae is also a benefit to using a spreadsheet in question 2. The first two rows were calculated first using the formula given in the table on the previous page. By filling down with this formula, the other rows contain the same calculation for

  2. Different methods of solving equations compared. From the Excel tables of each method, we ...

    Let Use Autograph to draw the graph of and. Suppose we want to find the root that lies between x=-1 and x=0. With a starting point at x=-1, we can see that it finally diverges away from the required root and enters a recursive cycle.

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