• 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

# Maths change of sign coursework

Extracts from this document...

Introduction

Osaamah Mohammed

## Maths A2 Coursework

Change of Sign Method:-

3x3 – 6x + 1 = 0

##### [-2,-1]  [0,1]  [1,2]

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

X1 = 0

Middle

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 0.00048043 0.170 -0.005261

#### Root [0.169 , 0.170]

 X f(X) 0.1690 0.00048043 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 0.00021 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

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)

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 : -

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

# Related AS and A Level Core & Pure Mathematics essays

1. ## Numerical Method (Maths Investigation)

= 2. Making X the subject: 5X = 7X4 - 1 X = , g(X) = Now the 2 possibilities of f(X) are written down on the top of this and then from Gph RM-01, we take note of the roots of the equation f(X).

2. ## The method I am going to use to solve x&amp;amp;#8722;3x-1=0 is the Change ...

I can stop it now as only want a root of 5 decimal places and the root is -0.347295. The maximum error is 0.000005. Now, I use the Rearrangement method trying to find the same root. Rearrangement 1: x= (x^3 -1)/3 I set my starting value for x is 0

1. ## Change of Sign Method.

The iterative formula calculates these values as they converge. xr+1 = 1/4( x3r+2x2r-4.58) x1 = 1/4(( -1)3+2(-1)2-4.58) =1/4(-1+2-4.58) =1/4(-3.58) x1 = -0.895 The following table was obtained using a Microsoft Excel spreadsheet, with the formulae shown. The sequence works using this particular iterative formula because when the gradient of y=g(x)

2. ## Change of Sign Method

bounds are likely to be between x=-1.65115 and x=-1.65105 but this must be justified by showing that there is a change of sign between the two. When x=-1.65115 y=-0.00067 and when x=-1.65105 y=0.00051. As there is a change of sign between the two there must be a root there.

1. ## Change of sign method.

1.1054687500 1.1035156250 -0.0275378227 0.0164708495 -0.0055966303 1.1035156250 1.1054687500 1.1044921875 -0.0055966303 0.0164708495 0.0054213097 1.1035156250 1.1044921875 1.1040039063 -0.0055966303 0.0054213097 -0.0000916085 1.1040039063 1.1044921875 1.1042480469 -0.0000916085 0.0054213097 0.0026638633 1.1040039063 1.1042480469 1.1041259766 -0.0000916085 0.0026638633 0.0012858806 1.1040039063 1.1041259766 1.1040649414 -0.0000916085 0.0012858806 0.0005970744 1.1040039063 1.1040649414 1.1040344238 -0.0000916085 0.0005970744 0.0002527175 1.1040039063 1.1040344238 1.1040191650 -0.0000916085 0.0002527175 0.0000805507 1.1040039063 1.1040191650 1.1040115356

2. ## Functions Coursework - A2 Maths

We have therefore found the root to 0 decimal places, which equals 2. One can not determine the root at this stage to 1 decimal place because either: x<1.85, in which case the root would be x=1.8 to 1 decimal place or: x?1.85, in which case the root would be

1. ## 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.

2. ## 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).

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