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Newton Raphson Method for Solving 6x3+7x2-9x-7=0

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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.

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