• 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

Numerical Solutions of Equations

Extracts from this document...

Introduction

P2 Maths Coursework                

Numerical Solutions of Equations                Page  of

P2 Maths Coursework

Numerical Solutions of Equations

Introduction

The aim of this coursework is to compare three different numerical methods of solving equations. This will allow us to determine which one is the most efficient, quickest and easiest method to use.

The three methods I will use are:

  • Decimal Search method
  • Newton-Raphson method
  • Rearranging method

Decimal Search

The decimal search is named as it employs the tactic of splitting the current interval of x values into 10 equal intervals of equal size and looking for a change of sign.

This process is then repeated, again splitting the current interval into 10 equal intervals of equal size and this can be continued until the root has been found to the required degree of accuracy.

The equation that I have chosen to solve is y = x5 – 2.7x + 1.8

It is illustrated by the graph below.

image10.pngimage00.pngimage01.png

This is the same equation but has been zoomed in.

y = x5 – 2.7x + 1.8

image11.pngimage02.pngimage03.png

y = x5 – 2.7x + 1.8 crosses the x-axis between  x = -2 and x = -1. To find the smallest root, I will take x = -4 as my starting value, and see the number of iterations required to find the root.

x

Y=x^5-2.7x+1.8

-1

3.5

-1.1

3.15949

-1.2

2.55168

-1.3

1.59707

-1.4

0.20176

-1.5

-1.74375

-1.6

-4.36576

-1.7

-7.80857

-1.8

-12.23568

-1.9

-17.83099

-2

-24.8

x

Y=x^5-2.7x+1.8

 -1.4

0.20176

-1.41

0.03391633

-1.42

-0.139533923

-1.43

-0.318710894

-1.44

-0.503736422

-1.45

-0.694734062

-1.46

-0.891829098

-1.47

-1.095148551

-1.48

-1.304821197

-1.49

-1.520977575

-1.5

-1.74375

     

        


x

Y=x^5-2.7x+1.8

-1.411

0.01682557

-1.4111

0.015113406

-1.4112

0.013400679

-1.4113

0.011687391

-1.4114

0.009973541

-1.4115

0.008259128

-1.4116

0.006544153

-1.4117

0.004828615

-1.4118

0.003112515

-1.4119

0.001395852

-1.412

-0.000321374

x

Y=x^5-2.7x+1.8

-1.41

0.03391633

-1.411

0.01682557

-1.412

-0.000321374

-1.413

-0.017524621

-1.414

-0.034784292

-1.415

-0.052100505

-1.416

-0.069473381

-1.417

-0.086903041

-1.418

-0.104389604

-1.419

-0.121933191

-1.42

-0.139533923

                                     
image04.png

x

Y=x^5-2.7x+1.8

-1.4119

0.001395852

-1.41191

0.001224155

-1.41192

0.001052452

-1.41193

0.000880743

-1.41194

0.000709029

-1.41195

0.000537309

-1.41196

0.000365584

-1.41197

0.000193853

-1.41198

2.21161E-05

-1.41199

-0.000149626

-1.412

-0.000321374

...read more.

Middle

image14.png

In this case the function touches the x-axis but doesn't cross it, so no change of sign exists; hence we can’t use this method to calculate an estimate of the root

x

y = x3 + 5x2 + x + 0.051

0.1

0.202

0.11

0.222831

0.12

0.244728

0.13

0.267697

0.14

0.291744

0.15

0.316875

0.16

0.343096

0.17

0.370413

0.18

0.398832

0.19

0.428359

0.2

0.459


Newton Raphson Method

This method involves fixed-point estimation, whereby a tangent to the curve from an initial value of x is drawn then it is calculated where the tangent intercepts the x-axis. This gives the next approximation to the root.  Repeating the process gives more and more accurate values for the root.

Newton-Raphson formula is:  

f(x) = x³-13x+14

image15.png

 Using the Newton-Raphson method, I will choose a starting point (x1) of positive 3 and will draw a tangent to the line at this point on the x-axis – where x = 3.  Where the tangent cuts the x-axis, I will have as my x2, my second point.  I will draw a new tangent here and continue the process.  I will illustrate this root graphically, but the other two purely numerically.  I need to differentiate this equation, in order to use this method as the Newton-Raphson formula is as follows.

Differentiate the equation    f’(x) = 3x²-13.

x2= 3 – 3³- (13*3)+14

        (3*3²)-13

x2= 3 -  2

        14

x2= 2.85714

By using this same method, but shortened significantly by using the ANS button on my calculator, I will find x3,, x4 etc

x3 = 2.84423

x4 = 2.84181

x5 = 2.84134

x6 = 2.84125

x7 = 2.84123

x8 = 2.84122

x9 = 2.84122

...read more.

Conclusion

Newton-Raphson was the quickest of the three numerical methods.  Newton-Raphson took only four steps and fixed point iteration took six steps to converge to the required root.  The use of Microsoft Excel did not make a huge difference because a scientific calculator can easily be programmed with the required iterative formulae.  A few presses of the ‘equals’ button would converge to the required root.

The most likely problem with Newton-Raphson was finding the wrong root if the tangent diverges away from the required root. Decimal search also had a problem by not showing a change of sign in the first set of calculations when the root had two decimal places.  

The Rearrangement method takes a lot of time to find the right arrangement of an equation, because not all g(x) graphs will allow convergence to the required root. Graphamatica was a very useful piece of software as it could be used to instantly visualise curves and gradient functions.  This meant that within seconds I could produce an accurate curve instead of having to make some calculations and draw the curve myself. This saved a lot of time as I could easily pick appropriate starting values without having to waste time using trial and error to pick my starting value.  

...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)

    From my graph, I can see that my X0 will be -2. To work out X1 I will use the formula above to work out X1 in steps and will repeat it to find a root to 5d.p and summarise in a table: Here are the results from the Iterations:

  2. Marked by a teacher

    The Gradient Function

    5 star(s)

    5 625 1250 1000 My theory was correct. I shall now try and binomially prove this... 2(x+h)4 - 2x4 = 2(x4 + h4 +4hx� + 6x�h� + 4xh�) - 2x4 = x + h - x h 2x4 + 2h4 +8hx� + 12x�h� + 8xh� - 2x4 = 2h4 +8hx�

  1. C3 Coursework: Numerical Methods

    If this method was to be carried out manually it would probably take more time than the other two methods in terms of using the calculations as the formulas for the Newton Raphson method is the most complex. However overall, as it would take the smallest amount of time to converge, the Newton Raphson method would still be the fastest.

  2. MEI numerical Methods

    Formula used for spreadsheet: Proof of root: Another way to check the root is to check the upper and lower bounds of the interval in which my root lies in. The interval in which the root has to lie in is, 0.4797310065 < x < 0.4797310075 (this is due to the error being � 0.0000000005, as explained later)

  1. Solutions of equations

    Error: -1.01203 + 0.000005 Error bounds: [-1.012025, -1.012035] F(x) = ex-x3-1.4 F(-1.012025) = -0.0000073 F(-1.012035) = 0.0000198 A change of sign occurs and so confirms there is a root. Finding Root B X0 = 0.8 X1 = -0.22618 X2 = 0.69116 X3 = 0.21887 X6 = 0.37282 ( 5.d.p.)

  2. Mathematical equations can be solved in many ways; however some equations cannot be solved ...

    It is reasonable to start at x=2. When x = 2, the graph shows that the Newton Raphson method does not work as the starting point on the graph is a stationary point. As a stationary point, this means that the tangent never touches the x-axis hence the Newton Raphson method does not work.

  1. 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.

  2. Numerical integration can be described as set of algorithms for calculating the numerical value ...

    Notice that f1, f2, f3 ect. to fn-1 do double duty as right-hand and left-hand body. Note that is the dependent on the number of trapezia. Thus T16 means that area under the curve is split in 16 rectangles whereas T32 means its split up in 32.

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