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# 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

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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

-

-

-

-

-

-

-

-

-

+ 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 - - - - - - - - + + 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 - - - + + + + + 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  +/-0.00005 • The root can be -1.50 to (2 decimal place)

Error bounds:

...read more.

Conclusion

When decimal search method works: 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.       Figure 4 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)} 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  Figure 6

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

figure 7 shows zoom-in version of figure 6.

...read more.

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