# Numerical solution of equations

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) = x³-9x+3.

Interval estimation

Assume that the roots of the equation F(x) = x³-9x+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 = x³-9x+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. A root must be lying on the interval a and b where an interval have been marked in which F(x) change signs. We thus know that there must be a root in that interval. The equation F(x) = x³-9x+3 has 3 roots which are in the interval of (-3,-4), (0,1) and (2,3). We can find the roots through decimal search, interval bisection and linear interpolation.

Decimal search

It is one of the methods of interval estimation through dividing each interval into 10 parts and looking the sign change. Here are three steps of this method:

- Take increment in x of size 0.1 within the interval of the root. Stop when there is a sign change.

- The interval has been narrowed down and now keeps on taking increment in x of size 0.01 within the narrowed interval. Stop again if there is a sign change.

- Repeat the step by taking increment in x of size smaller up to 6 decimal places (the final answers are recorded to 5 decimal places.

In the diagram (Appendix A), I have followed all the steps and finally found all the roots. They are 2.81691, 0.33761 and -3.15452 which are the same as the result from the graph.

The only advantage of this method is that the upper bound and lower are found automatically. However, the procedure is long and bored. Therefore, mistakes are easily made if you didn't check it. In addition, without the help of computer, it wastes much time to record every number you have found.

Interval bisection

It is very similar to the decimal search. It has to look at the sign change and divides the interval. However, the interval is bisected and narrowed down the interval depends on how the sign change (negative or positive). Here are the following steps:

- Find the interval which the root is between it and start by taking the mid-point of the interval. For example, the interval between the biggest roots is (1,0), I have chosen it since there is a sign change. And the mid-point of the interval 0.5.

- Put the ...