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

Numerical solution of equations

Extracts from this document...

Introduction

Pure Mathematic 2 Coursework Numerical solution of equations By Michael Pang I am going to show 3 of the numerical methods for solving the equation which cannot be solved algebraically. They are interval estimation, fixed point estimation and Newton-Raphson method. Those of these numerical methods are used when algebraic ones are not available. When you found any equations which cannot be solved algebraically, probably you will draw the graph and see where the roots are. However, the other problem is found that you cannot get all roots accuracy. Actually, the numerical methods cannot find the exactly root but the answers are more accuracy than sketching the graphs. Therefore, we usually provide the answer to 5 or 6 decimal places depending on what the questioner needs. Finally, we check the answer by setting the lower bound and upper bound to see whether it has sign change or not. Even if the three numerical can solve the equation non-algebraically, they have its advantages and disadvantages. And now I am going to show how these methods works and their problems by using the equation F(x) = x�-9x+3. Interval estimation Assume that the roots of the equation F(x) = x�-9x+3, and I am looking for the roots which F(x) = 0. The roots of the equation are the values of x which the graph of y = x�-9x+3 crosses the x axis. By using the computer, I recognise that there are 3 roots on the graph and the value of these 3 roots. However, if we cannot use the computer, we have to use one of the numerical methods - interval estimation. ...read more.

Middle

They are - The curve touches the x-axis - The roots are close together - There is a discontinuity The curve touches the x-axis In the above graph, the curve touches the x-axis. Therefore, the sign change could not be found and none of the methods of interval estimation can be used. The roots are close together In the graph, there 3 roots close to together between the interval (0,1). In decimal search, F(0.1) = 0, so that 1 of the root is 0.1 and so further interval cannot be found. In addition, interval bisection and linear interpolation would be unaware of the existence of other roots since the 3 roots are too close together. There is a discontinuity in F(x) This equation y = 1/(x-1.1) has no root, but all change of sign methods will converge on a false root at x = 1.1 Avoid problem To avoid those problems occurring, it is important to start by sketching the graph and therefore the problem will be found. Fixed point estimation It is much different from the method of interval bisection. It involves an iterative process rather than finding interval within the roots must lie. Generally, iterative process means that a sequence of numbers by continued repetition of the same procedure. But this method has some limited which will be introduced when the process are showing. How does it method work? Before starting to use this method, the equation F(x) = 0 has to be rearranged into the form of x = g(x). ...read more.

Conclusion

The other methods to show they are right, I should check their upper bound and lower bound where I set it in Fixed point estimation. And now I show it again to make it clear. - The largest root : F(2.816915) = 0.0000142, F(2.816905) = -0.00013403 - The medium root : F(0.337615) = -0.00005233 F(0.337605) = 0.00003425 - The smallest root : F(-3.154525) = -0.00004153 F(-3.154515) = 0.00016701 Advantage Newton-Raphson method is fast and convenient, the processes are short and that why I didn't use a lot of presentation to show my understanding. It also gives an extremely rapid rate of convergence. I only use 4 to 5 steps for each of the root, so I find that it was much easier to finish than the first 2 methods. Disadvantage Even this method is the fastest and the most convenient, it still has several problems. Firstly, in my case, if I didn't choose the right initial value, I couldn't find the root by only 4 to 5 steps or even faster in other case. However, if the first value I h chose is not close to the root, or near the turning point, a turning point of y = F(x), the iteration may diverge, or converge to another root. And if the initial point is at a stationary point, F'(X) = 0 so the method cannot proceed. Secondly, all numerical methods for solving equations cannot be proceeded when the function is discontinuous function as it doesn't have any real roots. Thirdly, the function is not defined over the whole of. The tangent at the graph may meet the axis at a point outside the domain of the function. ...read more.

The above preview is unformatted text

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)

    -2.0 -2 -1.9 2.79901 -1.8 6.34432 -1.7 8.84143 -1.6 10.47424 -1.5 11.40625 -1.4 11.78176 -1.3 11.72707 -1.2 11.35168 -1.1 10.74949 -1.0 10 X f(x) -1.970 -0.41553 -1.969 -0.36493 -1.968 -0.31448 -1.967 -0.26416 -1.966 -0.21399 -1.965 -0.16395 -1.964 -0.11405 -1.963 -0.0643 -1.962 -0.01468 -1.961 0.034796 -1.960 0.084135 x f(x)

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

    Relative Merits of the Three Methods Decimal Search Rearrangement Newton Raphson Overall � Easy to use and also easy to comprehend. � Gives an estimate of the point faster than decimal search. � Gives an estimate of the point faster than the other two methods.

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

    Graph 2.1--- In this case, I choose [0, 1] as the root presented graphically as follows. First of all, I am going to show the way to find the root in the interval [0, 1] Graph 2.2---Newton-Raphson for By using the help of autograph, in interval [0, 1], I get a root of 3 decimal places 0.908.

  2. The open box problem

    To prove this again I will construct two graphs. Here it looks like the maximum volume of the square 15x15 is 250 but I am unsure of x; so I will construct another graph to show it closer up. We can clearly see on this graph that x is 2.5 and that the maximum volume is 250.

  1. This coursework is about finding the roots of equations by numerical methods.

    = [1/3(-x�+x+2)]1/3 g'(x) =1/9*((-x^2+x+2)^(-2/3))*(-2*x+1) The root was x= 0.8878 Hence g'(root) = -0.0.03618 Since -1< g'(x) <0 then we would expect the iteration to converge as a Cobweb y=-3x�-x�+x+2 Rearrange to x=3x�+x�-2 leads to failure. x1=1 x y 1 2 2 14 14 782 x'=9x^2+2x g'(0.8878)= 8.86929956 Since g'(root)

  2. C3 Coursework: Numerical Methods

    The x=g(x) method took fewer calculations to find he root but not as few as the Newton Raphson method. To find an estimate of the root to 5 decimal places it took 15 iterations using the x=g(x) method. As with the Change of Sign method, the software made it easy to replicate the formula used.

  1. MEI numerical Methods

    Furthermore on excel all you need is the formula, once you've created this that's all you need. For example we can change the value of K and the formula would take this into account and would calculate the new root accordingly.

  2. Solving Equations. Three numerical methods are discussed in this investigation. There are advantages and ...

    Microsoft excel can be set up in order to reduce human error and increase efficient of calculation. A spread sheet is set up for n a f(a) B f(b) 1 0 =B2^3-5*B2^2+4*B2+2 2 =D2^3-5*D2^2+4*D2+2 2 =IF(G2>0,F2,B2) =B3^3-5*B3^2+4*B3+2 =IF(G2>0,D2,F2) =D3^3-5*D3^2+4*D3+2 3 =IF(G3>0,F3,B3)

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