• 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)

    After the f(X) equation has been rearrange to have one or more possibilities, draw the graph(s) of g(X) and the line y = x. The intersection between a graph of a g(X) function of f(X) and line y = x will be the root(s)

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

    �)] = 0.819179147 x4 = x3 - [(x3 ^4+x �-1=0)/( 4x3 �+3x3 �)] = 0.819172513 x5 = x4 - [(x4 ^4+x4 �-1=0)/( 4x4 �+3x4 �)] = 0.819172513 I can see some convergence from x3. There has been no change in the x-values between x4 and x5 for this number of decimal place.

  1. Change of Sign Method.

    = 0.75 - 1 = -0.25 The fact that this value is less than 1, tells us that the chosen value of x0 will lead to a root of the equation. It is necessary to substitute x0=-1 into the iterative formula to find the value of x.

  2. Change of Sign Method

    root of the equation at a point you were not looking for. For example if I was looking for the root of the equation between and . The iterative formula for would be: A sensibleto take would behowever the result would look like this.

  1. Change of sign method.

    so when: f(1.104005) = -7.9268 < 0 f(1.104015) = 3.3557 > 0 Therefore because my answer lies between the positive and negative, it must be accurate. However, there are some instances where this method does not succeed and we are unable to find the root of the equation using the change of sign method.

  2. Functions Coursework - A2 Maths

    x=1.9 to 1 decimal place This process can be repeated as many times to get the root to a desired number of decimal places. To illustrate the fact that the root lies in the interval [1.8,1.9], part of the graph of y=f(x)

  1. Methods of Advanced Mathematics (C3) Coursework.

    There were three routes as shown by the highlighted cells. -5 -5624 -4 -1727 -3 -350 -2 -23 -1 4 0 1 1 -2 2 25 3 352 4 1729 5 5626 After taking an estimate of the roots to be -1.5, 0.5 and 1.5 I started to use the iterative formula.

  2. Mathematics Coursework - OCR A Level

    x-value x-y 0.2509 0.0008579 0.2492 0.001614 0.2523 0.00303 0.2466 0.005714 0.2572 0.01068 0.2369 0.02031 0.2744 0.03749 0.2006 0.07381 0.3322 0.1316 Overflow Overflow - it reached overflow because there are certain x values where there is no point of the graph present.

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