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

Maths change of sign coursework

Extracts from this document...

Introduction

                Osaamah Mohammed

Maths A2 Coursework

Change of Sign Method:-

3x3 – 6x + 1 = 0

Sketch: -

image15.pngimage25.pngimage33.pngimage31.pngimage25.pngimage25.pngimage09.pngimage14.pngimage00.pngimage01.png

Looking at the graph we find that the roots lie between these intervals:
[-2,-1]  [0,1]  [1,2]

Finding root interval between [0,1] using a Decimal Search

image03.pngimage32.pngimage04.pngimage02.pngimage34.pngimage06.pngimage05.pngimage06.png

Root [0 , 1]

X1 = 0

...read more.

Middle

image07.png

0.17

-0.005261

Root [0.16 , 0.17]

X

f(X)

0.16

0.052298

0.161

0.046520

0.162

0.040755

0.163

0.034992

0.164

0.029233

0.165

0.023476

0.166

0.017723

0.167

0.011972

0.168

0.0062249

0.169

0.00048043image08.pngimage07.png

0.170

-0.005261

Root [0.169 , 0.170]

X

f(X)

0.1690

0.00048043image07.pngimage08.png

0.1691

-0.0000923852

Root [0.1690 , 0.1691]

X

f(X)

0.1690

0.00048043

0.16901

0.00042299

0.16902

0.00036557

0.16903

0.00030814

0.16904

0.00025071

0.16905

0.00019328

0.16906

0.00013586

0.16907

0.00078427

0.16908

0.00021image08.pngimage07.png

0.16909

-0.000036426

From this method we find that the root lies between[0.16908 , 0.16909]

Root = 0.169085 ± 0.000005

testf (0.16908) = 0.00021

        f (0.16909) = -0.000036426

There is a change

...read more.

Conclusion

Start X1 = 0        Use Xn+1 = 2xn3 + 1
                                      
5

X2 = 2(0)3 + 1 = 0.2      

                  5

X3 = 2(0.2)3 + 1 = 0.216      

                  5

X4 = 2(0.216)3 + 1 = 0.22016      

                  5

X5 = 0.22134

X6= 0.22169

X7 = 0.22179

X8 = 0.22182

X9 = 0.22183                        root to 5 s.f (took 9 iterations)image30.png

X10 = 0.22183

Failure: -

Solve 2x3 – 5x + 1 = 0

Re-arrange this so that “X=….”

  • 2x3 – 5x + 1 = 0

      2x3 +1 = 5x

      2x3 = 5x – 1

      x3 = 5x – 1

                 2

      x = (5x – 1)

                 2

      x = g(x)

Sketch : -

...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. Numerical Method (Maths Investigation)

    A tangent to the curve at point (0.5, f(0.5)) is drawn and observes where it cut the x-axis. It cut at a point on the x-axis that is out of the domain of the equation and hence Newton-Raphson Method cannot be continue on and it come to a halt, i.e.

  2. The method I am going to use to solve x−3x-1=0 is the Change ...

    So my starting value for x(x0) would be 0. x0 = 0 ( I substitute x0 = 0 into iterative formula to work out x1 ) x1 = V[(1-3(x0)^5)/5] = 0.4472136 x2 = V[(1-3(x1)^5)/5] = 0.4350481 x3 = V[(1-3(x2)^5)/5] = 0.4366342 x4 = V[(1-3(x3)^5)/5] = 0.4364376 x5 = V[(1-3(x4)^5)/5] =

  1. Mathematics Coursework - OCR A Level

    One then looks for where this new formula (y=g(x)) crosses the line of y=x as it will cross at the same points as where the original formula crosses the x-axis thus giving me the roots of the original equation. Original equation I found an equation, the graph of which is shown below, and rearranged into from the form f(x)=0 to x=g(x).

  2. Methods of Advanced Mathematics (C3) Coursework.

    I began by using the change of sign method. This gave me an Idea of where my solutions would be. I then found the respective figures to substitute into the Newton-Raphson iterative formula: Xn+1=xn- (f (xn)/(f' (xn) For this my equation will be f (x)=2x^5-5x^3+1 This was the beginning stages of this method and shows the change of signs I had to investigate.

  1. Change of Sign Method

    However the integer search does not show a change of sign between x=2 and x=3 and therefore misses the 2 roots shown in the graph: x -6 -5 -4 -3 -2 -1 0 1 2 3 4 y -71 1 43 61 61 49 31 13 1 1 19 This is therefore a failure of the change of sign method.

  2. 'Change of Sign Method'

    However, as the two roots are so close together, the decimal search method does not detect this and so in this instance, this method cannot be used to help solve the equation y=0.6918x3 - 0.2159x2 - 3.019x + 2.77 Newton Raphson Method Using this method, a first approximation, x1 is made.

  1. Change of Sign Method.

    = -0.00218772512 When x = 0.8395, f(0.8395) = 0.00280856463 The error bounds of the root 0.839 are 0.839 � 0.0005. However, I am able to say that I have a more accurate solution, as I know that the root lies in the interval [0.8389,0.8390]. Failure of the Change of Sign Method There are a number of situations that can cause problems for change of sign methods.

  2. Change of sign method.

    If I look at the numbers that are present in the spreadsheet once the equation has been entered, we can find the root to a give number of decimal places. Change of sign method y=5x^3-7x+1 a b (a+b)/2 f(a) f(b)

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