• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# 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

x ≤-1.64735). So the final answer to 5 significant figure is x=-1.6473. The error bounds shows the possible root is between x= -1.64725 and -1.6473 because the f(x) is positive. The table values indicate that the rate of convergence is high because the level of accuracy has increased from -1.7429 to -1.6473.

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

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related AS and A Level Core & Pure Mathematics essays

1.  5 star(s)

Using these data, I predict that the next table containing the gradient of will display 1500. 5 625 1875 1500 Binomial proof shall verify this for us, though realistically enough, the table provides enough evidence. 3 (x+h) 4 - 3x4 = 3(x4 + h4 +4hx� + 6x�h� + 4xh�)

2. ## Numerical Method of Algebra.

Look up for the change of sign here and it is between x = -0.6 and x = -0.7. Hence, the roots is now between x = -0.6 and x = -0.7. Gph DS-02 Now I move on to step 3, step 4 and so on until the fifth decimal place.

1. ## Change of Sign Method.

The chosen equation is y=f(x)=x3+2x2-4x-4.58. I have already applied the fixed-point iteration method to this equation in order to determine the root that lies in the interval [-1,0]. This value is -0.917 correct to three decimal places. Change of Sign Method The decimal search method will be applied to the above equation in order to find the root.

2. ## The method I am going to use to solve x&amp;amp;#8722;3x-1=0 is the Change ...

The starting value (x0) is 1 as it lies between x=0 and x=1 when f(x)=0. x0 = 1 x1 = x0 - [(x0 ^4+x0 �-1=0)/( 4x0 �+3x0 �)] = 0.857142857 x2 = x1 - [(x1^4+ x1 �-1=0)/( 4x1 �+3x1 �)] = 0.821252204 x3 = x2 - [(x2^4+x2 �-1=0)/( 4x2 �+3x2

1. ## Investigation into combined transoformations of 6 trigonometric functions

I'll be using the same values as I did for a, b for a and c. Graph 1 Graph 2 Graph 3 Graph 4 Graph 5 a 1 2 2 -2 -2 c 1 2 -2 2 -2 Predictions Graph 1 - Will have a maximum at (-?,1), and a minimum at (0?,-1)

2. ## newton raphson

Change of sign method has big advantage because they provide bounds (the two ends of the interval) within root lies. Knowing that the root lies in the interval [0.680760, 0.680770] means that you can take the root as 0.680765 with a maximum error of +/- 0.000005.

1. ## Investigate the relationships between the lengths of the 3 sides of the right angled ...

5a + b = 12 -eqn4 - 3a + b = 8 -eqn5 2a = 4 A = 4/2 A = 2 Substitute A =2 into equation 5 I am doing this to find what B is worth. 3a + b = 8 3 x 2 + b = 8

2. ## Change of Sign Method

x -5 -4 -3 -2 -1 0 1 2 3 4 5 y -155 -79 -31 -5 5 5 1 -1 5 25 65 These show that there are three roots. One between x=-2 and x=-1, one between x=1 and x=2 and another one between x=2 and x=3. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 