• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13

Solution of equations by numerical methods.

Extracts from this document...

Introduction

Solution of equations by numerical methods

This investigation is to find equation solutions using three methods:

  1. Change of sign using bisection, decimal search or linear interpolation.
  2. Newton – Raphson.
  3. Rearrangement of the equation f(x) = 0 into the form x = g(x).

The change of sign methods are systematic searches, which use the positive and negative signs of the f(x) solutions to find the location of the root within the intervals found using the graph of the function curve. The change of sign method I have chosen to use within my investigation is the decimal search method.

Fixed point iteration requires finding a single value or point as an estimation for the value of x, rather than establishing an interval as in the change of sign methods. The Newton – Raphson method and the rearrangement of equation f(x) = 0 into the form x = g(x) will be used to investigate this form of numerical equation solution.

Once each of these methods have been investigated I will compare each of them, in order to find the easiest method for equation solution. This comparison will also include the negative and positive points of each method, such as problems that result in an inability to find the correct root and the speed of convergence to the correct root.

Fixed Point Iteration – Newton – Raphson Method.

To start off using the Newton – Raphson method, we must first take an estimate of the root as a starting point. For a root f(x) we shall start with estimation x1 –we then draw a tangent to the curve y = f(x) at the point (x1, f(x1)).

...read more.

Middle

From this we can see that there are three roots, within the intervals [-2,-1],[-1,0] and [1,2].

  • one rearrangement: x = g(x) = x^5 - 2 which gives us the graph:

                                                        4

Taking x = 2.0 as the starting point to find the root within the interval [-1,0] – we gain the following set of results:

image00.png

This rearrangement provides the basis for the iterative formula:

                        Xn+1 = x^5n - 2

                                   4

This gives the set of results:

x1

-0.8

x5

-0.50856

x2

-0.58192

x6

-0.50850

x3

-0.51668

x7

-0.50850

x4

-0.50921

x8

-0.50850

This shows the root to be –0.50850 within the interval [-1,0], which can be written:

  • -0.509 with a maximum error of ±0.0005, or
  • -0.5 (to 1 d.p.)

Displayed graphically:

image00.png

image00.png

From which we gain the root as –0.5085 (to 4 s.f.) from the results box, which again gives the root in both the ways written above.

As         f(x) = x^5 – 4x – 2

And            g(x) = x = (x^5 – 2)/4

  • g’(x) = (5x^4)/16

The magnitude of g(x) for the results gained above can be shown to be:

x1

-0.8

-0.128

x5

-0.50856

-0.021

x2

-0.58192

-0.036

x6

-0.50850

-0.021

x3

-0.51668

-0.022

x7

-0.50850

-0.021

x4

-0.50921

-0.021

x8

-0.58050

-0.021

This shows that the magnitude is always less than one, and as the method will only succeed if        -1 < g’(x) < 1 , and as this is the case for all the values of x found, the method works.

Method Failure

We can see that this rearrangement of f(x) works when trying to find the root within the interval [-1,0], but I shall now try to find the root within the interval [1,2] using the same method.

...read more.

Conclusion

The hardware and software that have been used in this investigation are a scientific calculator and a graph drawing program – Autograph. This allows us to find the roots of each graph without having to draw each of them out along with the tangents, and gives as accurate results for the roots with our own specification on the number of significant figures. This program proves little use for the decimal search method, other than establishing the intervals the roots lie in – the main body of work must then be done through calculations and recording of results, and each significant figure must be found one at a time.

The fixed-point iteration methods benefit greatly from the graph program, as they allow us to gain all the tangents and converging ‘staircases’ and ‘cobwebs’ simply by typing in the starting value of x. This gives us much faster convergence than calculating, and a graphical representation of what is happening, so that we can more fully understand the way each method works. Both are simple to use as long as the equation and the starting point for each interval are known.

...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

    C3 Coursework - different methods of solving equations.

    5 star(s)

    With the use of Autograph, plotting y = g(x) against y = x graph and selecting them both, clicking on x = g(x) iteration and then selecting a starting point to start the iterations.

  2. Sequences and series investigation

    - 2(102) + 2Y(10) - 1 = 13335 - 200 + 26Y - 1 = 1159 Sequence 15: N = 15 _(15�) - 2(152) + 2Y(15) - 1 = 4500 - 450 + 40 - 1 = 4089 This equation has correctly given me the number of squares in each sequence which again proves

  1. Numerical solutions of equations

    = 0. The graph and calculations of this is below: The function is: (x+21/2)1/2 ln (x+21/2) = 0 f'(x) = ((x+21/2)1/2 X 1 ) + (ln (x+21/2) X 1/2 (x+21/2) -1/2) x + 2 1/2 (x+21/2)1/2 + ln (x+21/2) (x+21/2)

  2. Numerical solution of equations, Interval bisection---change of sign methods, Fixed point iteration ---the Newton-Raphson ...

    -9.585E-05 6.10352E-05 15 1.212891 0.000166729 1.212952 -9.5853E-05 1.213 3.5436E-05 3.05176E-05 16 1.212921 3.54359E-05 1.212952 -9.5853E-05 1.213 -3.021E-05 1.52588E-05 17 1.212921 3.54359E-05 1.212936 -3.0209E-05 1.213 2.6133E-06 7.62939E-06 18 1.212929 2.61329E-06 1.212936 -3.0209E-05 1.213 -1.38E-05 3.8147E-06 19 1.212929 2.61329E-06 1.212933 -1.3797E-05 1.213 -5.592E-06 1.90735E-06 20 1.212929 2.61329E-06 1.212931 -5.5923E-06 1.213 -1.49E-06 9.53674E-07

  1. Numerical Solutions of Equations

    = -0.000171 Negative f(-4.056055) = 0.000193 Positive Failure of Newton Raphson method The method fails if the initial value is not close to the root, or is near a turning point of y = f(x), the iteration may diverge, or converge to another root.

  2. Maths - Investigate how many people can be carried in each type of vessel.

    multiply by (AE-DB) => y (AE-DB)(HA-BG) + z (AE-DB)(IA-CG) = (AE-DB)(AL-JG) We once again subtract the two equations from each other eliminating y and creating: z(AE-DB)(IA-CG) - z(AF-DC)(HA-BG) = (AE-DB)(AL-JG)-(AK-DJ)(HA-BG) z = [(AE-DB)(AL-JG) - (AK-DK)(HA-BG)] / [(AE-DB)(IA-GC) -(AF-DC)(HA-BG)] Now that the formula for 'z' has been determined, we now need to find 'x' and 'y'.

  1. C3 Coursework: Numerical Methods

    His is because as with like the Change of Sign method it takes many calculations, in the case of x=g(x), iterations, to find the roots. The fast numerical method appears to be the Newton Raphson method; this is because it takes the fewest number of iterations to find a root.

  2. Fractals. In order to create a fractal, you will need to be acquainted ...

    outputs : B ? A ? B + A + B + A + B ? A ? B + A + B + A + B ? A ?B + A + B + A + B ? A ? B + A + B + A + B ? A ? B.

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