Pt1420 Unit 7 Homework
A2 Mathematics Coursework C3
Year 12
Numerical solutions of equations
Solving 0 = x5+x-5 using the “Change Of Sign” Method
The method I will use to solve 0 = x5+x-5 is the Change of Sign Method involving the Decimal Search method. I have drawn this graph using the Autograph Software, and the print screen of this is below:
From my graph above, I can see that the root of this equation is between x =1 and x = 1.5. The table of x values and f(x) values is shown below. I can work out the f(x) values by substituting the x-values into the equation.
x
1
1.1
1.2
1.3
1.4
1.5 f(x) -3
-2.28949
-1.31168
0.01293
1.77824
4.09375
From my table of values above, it is clear that the change of sign from negative …show more content…
Therefore, using this method, I can take the root of this equation to be x = 0.680765, which is similar to the answer I got when finding the same root using the “Rearrangement method”.
I will now use the Newton-Raphson method to find the same root for this function.
0=x5+4x2-2 f’(x) = 5x4+8x
The iterative formula for the Newton-Raphson method is: xn+1=xn- f(xn) f’(xn)
So my iterative formula is: xn+1=xn- xn5+4xn2-2 5xn4+8xn
I think that a good starting value for x1 would be 1.2 x1 = 1.2 x2 = 0.887083 x3 = 0.720596 x4 = 0.682450 x5 = 0.680772 x6 = 0.680769 x7 = 0.680769
I can see a definite convergence from x6. So the root is x = 0.680769 which is similar to what I obtained when finding the same root using the Rearrangement method and the Decimal Search
Year 12
Numerical solutions of equations
Solving 0 = x5+x-5 using the “Change Of Sign” Method
The method I will use to solve 0 = x5+x-5 is the Change of Sign Method involving the Decimal Search method. I have drawn this graph using the Autograph Software, and the print screen of this is below:
From my graph above, I can see that the root of this equation is between x =1 and x = 1.5. The table of x values and f(x) values is shown below. I can work out the f(x) values by substituting the x-values into the equation.
x
1
1.1
1.2
1.3
1.4
1.5 f(x) -3
-2.28949
-1.31168
0.01293
1.77824
4.09375
From my table of values above, it is clear that the change of sign from negative …show more content…
Therefore, using this method, I can take the root of this equation to be x = 0.680765, which is similar to the answer I got when finding the same root using the “Rearrangement method”.
I will now use the Newton-Raphson method to find the same root for this function.
0=x5+4x2-2 f’(x) = 5x4+8x
The iterative formula for the Newton-Raphson method is: xn+1=xn- f(xn) f’(xn)
So my iterative formula is: xn+1=xn- xn5+4xn2-2 5xn4+8xn
I think that a good starting value for x1 would be 1.2 x1 = 1.2 x2 = 0.887083 x3 = 0.720596 x4 = 0.682450 x5 = 0.680772 x6 = 0.680769 x7 = 0.680769
I can see a definite convergence from x6. So the root is x = 0.680769 which is similar to what I obtained when finding the same root using the Rearrangement method and the Decimal Search