• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

In this coursework, I am going to solve equations by using the Numerical Methods. Numerical methods are used to solve equation that cannot be solved algebraically

Extracts from this document...

Introduction

ﻔﻩﺩ

Pure mathematic 2 coursework

Introduction:

In this coursework, I am going to solve equations by using the Numerical Methods. Numerical methods are used to solve equation that cannot be solved algebraically e.g. quadratic equations ax²+bx+c=0 can be solved using this formula:

x= -b± √ b² - 4ac

2a

Therefore numerical methods would not be used for quadratic equations. I will be working with equation which don’t have a formula to solve it. There are three methods, which I will be using:

· Change of sign method

· Newton-Raphson method

· Rearranging f(x) = 0 in the form x = g(x)

Objective:

Our object is to investigate the solution of equations using the three different methods.  To solve an equation we must find all its roots; a method which misses one or more roots will fail to solve the equation.

Change of sign method:

There are three ways in which we can do this method which are:

  • Decimal searcher
  • Interval bisection
  • Linear interpolation

Description:

This method works when a function crosses the x-axis. If we are looking for the root of the equation f(x) = 0. The point at which the curve crosses x-axis is the root. Once an interval where f(x) changes sign then the root must be in the interval.                                      

...read more.

Middle

-1.009

-0.738

-0.472

-0.211

0.0437

sign

-

-

-

-

-

-

-

-

-

+

image20.png

From this we know that the root lie between -1.51 and -1.50, so now we will zoom between -1.51 and -1.50 to see where the root lies. The table below shows calculation for 3 decimal place.

F(x)

-1.509

-1.508

-1.507

-1.506

-1.505

-1.504

-1.503

-1.502

-1.501

-1.500

y

-0.1861

-0.1603

-0.1347

-0.109

-0.0834

-0.0579

-0.0324

-0.007

0.01841

0.04375

sign

-

-

-

-

-

-

-

-

+

+

image20.png

Now we can state that the root lies between -1.502 and -1.501, the calculation now would be from

-1.502 to -1.501. The table below shows value for x to 4 decimal place.  

F(x)

-1.5020

-1.5019

-1.5018

-1.5017

-1.5016

-1.5015

-1.5014

-1.5013

y

-0.007

-0.0044

-0.0019

0.00065

0.00319

0.00573

0.00826

0.0108

sign

-

-

-

+

+

+

+

+

image20.png

From the table above we can state that the root lie between {-1.5018 <x< -1.5017}, now we can do more calculation and get even more precise answer but we don’t know what is the precise accurate answer, the answer may be to 5 or even 10 decimal places. So it can take along time to find an accurate answer.

Root lies {-1.5018,-1.5017}

We can express this information as:

  • The root can be taken as -1.50175 with a maximum error of  +/-0.00005image13.png
  • The root can be -1.50 to (2 decimal place)

Error bounds:

...read more.

Conclusion

When decimal search method works:

image04.png

I will try solving: F(x) =

The reason why I choose this equation is because I need to prove that this method work and not only it has only few roots but it is very easy to differentiate this function. Figure 4 shows the graph of the function above. Figure 5 shows zoom in version of the graph.

image25.pngimage11.pngimage06.pngimage07.pngimage05.pngimage09.pngimage08.png

Figure 4

image21.png

Figure 5

As you can see there are three roots in this graph, they are in the interval, [-3,-2],

[-1, 0], [2, 3]

The gradient for the tangent to the curve at (x1, f(x1)) is f’(x1) (meaning dy/dx for x). The equation of the tangent is: y-y1 = m(x-x1). Therefore y-f(x1) = f’(x1) [x-x1]. This tangent passes through the point (x2, 0). Carrying on with this process, this will get closer and closer to the tangent. But there is a general formula for this process:

xn+1 = xn – {f(Xn)/f ’(Xn)}

image04.png

Differentiating the function:

dy/dx = f’(x) = [1/7]*7 x7-1- [5*3]x3-1 + [5*2]x2-1- 5*0

               = x6-15x2+10x

So now, using the Newton-Raphson iteration equation we can find the roots

Root A= interval [-1, 0]

Root B= interval [-3, -2]

Root C= interval [2, 3]

Root A

image22.pngimage12.png

Figure 6

Figure 6 shows the curve with a tangent, so xn=-1

figure 7 shows zoom-in version of figure 6.  

...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)

    After 19 iterations, I found the root of the equation as from then on, it repeats. So the root of the equation x=G(x) against y = x graph. Failure of Rearrangement method There are situations where this method doesn't work.

  2. OCR MEI C3 Coursework - Numerical Methods

    -1.56004 1.75E-05 Change of sign indicates root exists in interval [-1.56005,-1.56004] x=-1.560045�0.000005 x=-1.560 (3d.p.) x f(x) -1.560050 -0.00024 -1.560049 -0.00021 -1.560048 -0.00019 -1.560047 -0.00016 -1.560046 -0.00014 -1.560045 -0.00011 -1.560044 -8.5E-05 -1.560043 -5.9E-05 -1.560042 -3.4E-05 -1.560041 -8.1E-06 -1.560040 1.75E-05 Change of sign indicates root exists in interval [-1.560041,-1.560040] x=-1.5600405�0.0000005 x=-1.56004 (5d.p.)

  1. Methods of Advanced Mathematics (C3) Coursework.

    0.0000000005 as shown in red above. Rearranging method Iteration 1 - Working to find root [-2, -3] y = (5x - 2)^0(1/3) x y 5 2.802039 4 2.571282 3 2.289428 2 1.912931 1 1.259921 0 -1.44225 -1 -2 -2 -2.35133 -3 -2.62074 -4 -2.84387 -5 -3.03659 -1 -2 -2 -2.35133

  2. Numerical Methods Coursework

    Since the absolute error is proportional to h�. It is a second order method. Also trapezium rule is similar to the mid- point rule. Error is also proportional to h�. Therefore this also means halving h, or equivalently doubling n will reduce the error by a factor of = 0.25.

  1. Numerical solutions of equations

    I have now found the same root using the Decimal Search method, the Rearrangement method and the Newton-Raphson method. It is now possible to compare these three methods in terms of ease and the speed of convergence. I found that the Decimal Search method was the simplest out of the

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

    =0 where f(x) =. The graph of y=f(x) is shown below. ==> ==> ==> ==> By starting with the value of -3 we are now able to start finding the root in [-3,-2]. The iterations are shown below: ==> =-3.301927 The initial estimate for the root was -3.

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

    B. Here the b) equation is chosen to solve the equation. Graph 3.2 From the Graph 3.2, y = x and intersect at the 3 roots which are around 3.5, 0.5 and -4. Graph 3.3 x=(x�+8)/13 N x 1 -2 2 0 3 0.61538462 4 0.63331116 5 0.63492387 6 0.63507352 7 0.63508745 8 0.63508874

  2. C3 COURSEWORK - comparing methods of solving functions

    Failure Case Example: y= (0.5x³+1.5x²–x–0.25) 1/3 Reason cause to failure is that the graph crosses the x axis with a very steep gradient Graph is y=f(x) The steep crossing points were caused by the power 1/3 On the graph, the Newton Raphson method will fail to reach a root y= (0.5x³+1.5x²–x–0.25)

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