• 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

...read more.

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

...read more.

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

...read more.

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

See related essaysSee related essays

Related AS and A Level Core & Pure Mathematics essays

  1. Marked by a teacher

    The Gradient Function

    5 star(s)

    First I shall investigate fractional powers - I will try to investigate 2 or 3 of these to find a general rule. y = x1/2 x y second point x second point y gradient 3 1.73205080756888 3.1 1.760681686 0.286309 3 1.73205080756888 3.01 1.734935157 0.288435 3 1.73205080756888 3.001 1.732339459 0.288651 4

  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. newton raphson

    The Decimal Searching method disappoints, it is inexact. When the curve touches the x-axis there is no change of sign, so change of sign methods are doomed to failure. 2. Equation: x�+0.7x�-2.1775x+0.845 = 0 We are looking for the change in signs between -4 and 4.

  1. The method I am going to use to solve x−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

  2. The Gradient Fraction

    x=3: 12.41 (round down to 12) the 3 has been multiplied by 4. x=4: 16.8 (round down to 16) the 4 has also been multiplied by 4. x=-2.5: -9.23 (round to -10) the -2.5 has been multiplied by 4. x=-4: -17.77 (round down to -16) the -4 has been multiplied by +4.

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

    4a + 2b + c = 12 -eqn2 - a + b + c = 4 -eqn1 3a + b = 8 -eqn5 Equation 4 - Equation 5 I am doing this to eliminate B and finally work out what A is worth.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work