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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
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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
· 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.
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: 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.
This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.
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