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

# Am going to use numerical methods to solve equations that can't be solved algebraically

Extracts from this document...

Introduction

Greig Boyd

C3 Coursework

Solution of equations by numerical methods

Introduction

In this coursework I am going to use numerical methods to solve equations that can’t be solved algebraically, for example if the largest power of the equation was 3 or over , we would have to use the following methods: the change in sign method, Newton-Raphson method and the method of rearranging f(x) =0 in the form x = g(x).

I am going to use an equation which cannot be factorised or solved using the formula x = -b .

2a

Change of sign method

My equation is -6x3 +9x2 +5x-6. My equation has four roots I am going to use change of sign to find one of them. For the change in sign method I am going to do a decimal search, i.e. I have taken the negative and positive y values that are in between zero. I did my first decimal search to two decimal places and carried on the same procedure all the way up to four decimal places.

 x Y 0 -6 0.1 -5.416 0.2 -4.688 0.3 -3.852 0.4 -2.944 0.5 -2 0.6 -1.056 0.7 -0.148 0.8 0.688 0.9 1.416 1 2

The first thing I did was find out were the formula crosses the x-axis, which was 0.7 and 0.8.

Middle

Thus proves that the root lies no greater nor less than 0.00005 away from 0.717. It also shows it’s correct by the change in sign. Failure of “change of sign”

Failure will occur if:

• The curve touches the x-axis.
• There are several roots close together
• There is a discontinuity in f(x)

I am going to use decimal search to find the roots of another equation

-2x2-3x-1.12

I cannot use decimal search for this equation as it doesn’t cross the x-axis.

There are some cases where the roots cannot always be found. Reasons for this could be if the roots are too close together Newton-Raphson method

X5-6x +2

The Newton raphson method is a fixed point estimation method. It is important to use an estimation of the root as a starting point.

I will give an estimation (x0) for a root of f(x) =0 I will then draw a tangent to the curve y=f(x) at the point (x0, f(x)). The point where the tangent cuts the x-axis gives a closer approximation for the root, and then the process is repeated From this it is clear that D is a better estimation than B. We can continue this to get closer to A.

Conclusion

3 of decimal search I got 1.4675+ 0.0005. For Newton Raphson I got at x3 1.474406026 and for the rearrangement I got x3 1.496651121. Although Newton Raphson and the rearrangement were more accurate as it is to more decimal places. Even though the decimal search method was the most efficient it took longer to do.

Comparing methods in terms of ease of use with hardware and software

With all three methods I used a graphical calculator as my hardware. With the decimal search method I only had to be aware of where the roots lie in the intervals to be able to use the method. With Newton-Raphson method I just had to be able to check on the graphics calculator what the roots were to know whether it was converging towards the root or diverging from the root. This made me use the arrow buttons to move the curser to where the roots lie. On the contrary, with rearranging I had to plot the y=x and the function on the same graph and for this I had to use Microsoft excel. In conclusion in terms of ease of use with involved hardware and software; the decimal search was the most efficient method to use, then Newton- Raphson and the most involved one was the rearranging method.

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.  ## C3 Coursework - different methods of solving equations.

5 star(s)

Out of the three, some are faster at converging to the root that others and some are easier to use than others, especially with the recent technology made available to everyone. Decimal Search method This method is by far the easiest and requires least amount of mathematical skill as it

2. ## Best shape for gutter and further alegbra - using Excel to solve some mathematical ...

change according to the intentions of the designer, and so give high value feedback or other information, to the user" Using a spreadsheet in question 2 allows interactivity. Linked values are a defining feature of the spreadsheet. This capability to link and respond enables the spreadsheet to offer interactivity (Mathematical Association, 2002, p39).

1. ## Mathematical equations can be solved in many ways; however some equations cannot be solved ...

This resulting staircase approaches the root, which lies at the intersection of the line and the curve. At the root, the curve has a positive gradient as it is sloping upwards. Also, it is shallower than y=x. Therefore, at the root, 0<g'(x)

2. ## Three ways of reading The Bloody Chamber.

That is to say the third order signified will be of the form: (third order understaning) of second order signified for example: (third order understanding) of a 'freudianised' key In order to have a third order understanding one must have a second order understanding, a symbolic or mythic interpretation, and

1. ## Numerical solutions of equations

However, this was a very time-consuming method to do on a calculator. Although I only used a calculator for this coursework, it would have been quicker and more efficient if I had used Excel for the Decimal Search method.

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

into the iterative formula of Rearrangement A, which is y= [(1- 5x�)/3]^(1/5) and the starting value for x(x0) is 0, the same as before. Zx0 = 0 x1 = V[(1-3(x0)^5)/5] = 0.8027416 x2 = V[(1-3(x1)^5)/5] = -0.9417235 x3 = V[(1-3(x2)^5)/5] = -1.0274040 x4 = V[(1-3(x3)^5)/5] = -1.0735437 Immediately, I can

1. ## OCR MEI C3 Coursework - Numerical Methods

is therefore a root of f(x)=0. Speed of convergence This shows that Newton-Raphson converges the quickest. In this case fixed point iteration using x=g(x) was the next fastest method, though in some cases its rate of convergence can be very slow: e.g., using the rearrangement x=(4x-3)1/5, it takes 19 iterations to find the root 1.002 of f(x)=0 to three decimal places.

2. ## C3 COURSEWORK - comparing methods of solving functions

= 0.879385241 and f(0.879395) = 0.879385241 The root is trapped between 0.879385 and 0.879395 We can see that the root is 0.879385± 0.000005 or 0.87939 to 5 decimal places. Comparisons among all three methods Types of method Ranking of speed Number of calculations Change of sign 2nd 36 Newton Raphson 1st 5 x=g(x) • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to
improve your own work 