- Level: AS and A Level
- Subject: Maths
- Word count: 1687
Newton Raphson Method for Solving 6x3+7x2-9x-7=0
Extracts from this document...
Introduction
Page of
Newton-Raphson method
The equation I am going to Solve is 6x3+7x2-9x-7=0
Newton-Raphson is a fixed point iteration method. The first approximation has to be near the root so the method will work. A tangent has to be drawn at the first point y=f(x) and when it touches the x-axis this gives another root. The tangent cuts the second point on the curve. This process is then repeated. The Newton Raphson method uses an algebraically formula to calculate the value of roots. .
The formula is first worked out by the gradient of the tangent there are steps of algebra to get to the final equation. Autograph will do these calculations automatically.
f `(xn) =
(x1-x2)f `(x1) = f(x1)
x1-x2 =
x2 =x1 -
f(x) is negative The root lies between 1.1155 and 1.11555
1.11545 1.1155 1.11555
f(1.11545) = -0.002199645 f(1.1155) = 0.00020309 f(1.11555)= 0.000701867.
X = 1.1155
Error Bounds = ±0.00005
Solution bounds = (1.11545≤ x ≤1.11555).
n | xn | f(xn) | f `(xn) | xn+1 |
0 | x0 = 1 | f(x0)= -3 | f '(x0)= 23 | x1 = 1.1304 |
1 | x1 = 1.1304 | f(x1) = 0.4376 | f '(x1) = 29.826 | x2 = 1.1157 |
2 | x2 = 1.1157 | f(x2) = -0.000748 | f '(x2) = 29.015 | x3 = 1.1155 |
Middle
X= -0.63488 (5 sig fig)
Error bounds = ±0.000005
Solution bounds = (-0.634875≤ x ≤-0.634885).
n | xn | f(xn) | f `(xn) | xn+1 |
0 | x0 = -1 | f(x0)= 3 | f '(x0) = -5 | x1 = -0.4 |
1 | x1 = -0.4 | f(x1) = -2.664 | f '(x1) = -11.72 | x2 = -0.6273 |
2 | x2 = -0.6273 | f(x2) = -0.0808 | f '(x2) = -10.6991 | x3 = -0.63486 |
3 | x3 = -0.63486 | f(x3) = -0.000200809 | f '(x3) = -10.63319005 | X4 = -0.63488 |
Solving 6x3+7x2-9x-7=0 for third root
f(x) is negativeThe root lies between -1.6473 and -1.64725
-1.64735 -1.6473 -1.64725
f(-1.64735)= -0.000611049 f(1.6473)=0.00013765 f(-1.64725)=0.001067206
Conclusion
The equation of y=1.7ln(x-3)+3 shows the limitations of Newton Raphson where failure can be experienced because the function is discontinuous. My first guess was x=4. The tangent at that point on the curve hits the x-axis at 2.2353 the graph does not meet at that point so therefore the method is stopped. It has asymptotes at three so the second value cannot touch the graph this shows that any values lower than three is not defined for that equation. Overflow is shown after the second value this indicates that no solution cannot be found after x=2.2353.
Newton Raphson method
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