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
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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
· NewtonRaphson 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 xaxis. If we are looking for the root of the equation f(x) = 0. The point at which the curve crosses xaxis is the root. Once an interval where f(x) changes sign then the root must be in the interval.
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:
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: yy1 = m(xx1). Therefore yf(x1) = f’(x1) [xx1]. 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 x71 [5*3]x31 + [5*2]x21 5*0
= x615x2+10x
So now, using the NewtonRaphson 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 zoomin version of figure 6.
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