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Introduction to Numerical Analysis: Introduction

The bulk of the material on numerical analysis was adopted from the notes by Dr. Karen Dow. I wish to acknowledge her and thank her for providing her notes to me.

Numerical analysis is the science of finding numerical approximations to well defined mathematical problems. As an example, consider the problem that seeks to find a real number x\in\mathbb{R} that satisfies:

    \[ x^2=2 \]

In other words, the problem states: “Find a real number such that when it is multiplied by itself, the result is equal to 2”. This problem has led to the development of the modern definition of the set of real numbers consisting of the rational and irrational numbers. The analytical solution for this problem is

    \[ x=\pm \sqrt{2} \]

\sqrt{2} is an irrational number. In order to use it for practical or engineering applications, an approximation in the form of a rational number that is “close enough” is used instead. There are various methods to find this rational number approximation of \sqrt{2}. For example, we will employ the Babylonian method which starts by rearranging the equation as follows:

    \[ x=\frac{2}{x}\Rightarrow 2x=x+x=x+\frac{2}{x}\Rightarrow x=\frac{x+\frac{2}{x}}{2} \]

The approximation is then found using an iterative method. We first assume an initial value for x and designate it x_0. For example, we set x_0=1. Then, for the iteration number n, we calculate a new approximation x_n by using the formula:

    \[ x_n=\frac{x_{n-1}+\frac{2}{x_{n-1}}}{2} \]

So, in the first iteration we have:

    \[ x_1=\frac{x_0+\frac{2}{x_0}}{2}=\frac{1+\frac{2}{1}}{2}=1.5 \]

Now, x_1 is not a very good approximation because x_1^2=1.5^2=2.25. For the next iteration, x_2 can be calculated as:

    \[ x_2=\frac{x_1+\frac{2}{x_1}}{2}=\frac{1.5+\frac{2}{1.5}}{2}\approx 1.4167 \]

x_2 is a better approximation because x_2^2\approx=1.4167^2\approx 2.007. For the next iteration:

    \[ x_3=\frac{x_2+\frac{2}{x_2}}{2}=\frac{1.4167+\frac{2}{1.4167}}{2}\approx 1.41422 \]

x_3 is practically equal to \sqrt{2} because when we square it we get x_3^2=1.41422^2\approx 2.00002! For the sake of completion, let’s try the next iteration:

    \[ x_4=\frac{x_3+\frac{2}{x_3}}{2}=\frac{1.41422+\frac{2}{1.41422}}{2}\approx 1.414214 \]

For all practical purposes, x_3 and x_4 are very close to each other (The first four decimal places). The closeness of the two values will be clearly quantified in terms of error measurements in the next section. Once the two values are close up to an acceptable level, we say that convergence has been achieved.

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