## Introduction To Numerical Analysis.

Introduction to Numerical Analysis by using numerical methods, which are techniques by which mathematical problems are formulated to be solved with arithmetic operations again and again.

You can click on any picture to enlarge, then press the small arrow at the right to review all the other images as a slide show.

The numerical analysis involved the study of methods of computing numerical data. In many problems, this implies producing a sequence of approximations by repeating the procedure

## Brief description of the videos.

The subject is the difference between the analytical and numerical methods. one example is given for which it is required to find the roots analytically, then the same example will be solved by different numerical methods as an introduction to numerical analysis.

The second video I used in the illustration.

You can click on any picture to enlarge, then press the small arrow at the right to review all the other images as a slide show.

## Difference between Numerical and analytical methods.

It is important as an introduction to numerical analysis to find the difference between Numerical and analytical methods and to understand what is a polynomial.

Numerical methods provide an approximation to the problem in question, while the analytical method gives exact solutions with more time-consuming.

Numerical methods only give the solution at a certain point, like finite difference and finite element; these methods are considered Numerical since they give results at certain points.

### Definition of the Quadratic function.

A quadratic function is a function that can be expressed in the form of **f(x)=ax^2+bx+c.** There are three cases to get real solutions. Depending on the relation between b^2 and 4ac. The complete illustration is shown in the next slide image.

Refer to this link for more information. What is discrimination? Therefore, the discriminant formula for the general quadratic equation is Discriminant, D = b2 – 4ac, Where

a is the coefficient of x2, b is the coefficient of x, and c is a constant term.

### Definition of the polynomial function.

The polynomial function with degree n has a leading coefficient a_{n} and has a constant a_{0} and can be written as **f(x)=a _{n}x^n+a_{n-1}*x^n-1+a_{n-2}*x^n-2+……..+a_{0}.**

## Roots findings.

As an introduction to numerical analysis, we will review the concept of root-finding of a function. The roots of a function, or zeros, are the points where the function crosses the x-axis, where the value of f(x)=0; for analytical methods, we use synthetic division, and for quadratic functions, we use the following equation for the solution.

### The remainder theorem.

The remainder theorem is a formula that is used to find the remainder when a polynomial is divided by a linear polynomial.

When a certain number of things are divided into groups with an equal number of things in each group, the number of leftover things is known as the remainder. It is something that “remains” after division. Let us learn the concept of the remainder theorem.

### Roots for y=x^2-4.

If we have a graph. Y = x^2 – 4, we draw the x and Y coordinate and while the intersection of the x-axis.

With the graph, will give us the value of f(x) = 0. Suppose to cross at +2 & -2. These two points are points of zero. How do we get these two points? we let f(x),or y=0 . We make bracketing (x^2-4 ), which will give us multiplication of (x+2) *(x-2).We equate to zero, so( x= -2) .

The first solution or there is another solution that x =+2. We have two solutions x = – 2 or x =+ 2 ,they are the Zeros? The second or last step is that we have to check that our solution is correct. x= +2, this equation y =x^2-4 ,we put x=+2 so y=2 to power of 2=4-4 ,gives us zero .

And for the other solution x=-2, we will write -2 ^ 2 – 4, it will give us zero. Once we have satisfied the f (x ), =0

We have these two zero, + 2, and -2. In synthetic division, to get the zero of the function sometimes, we are guessing the value of the X. with -the so Called Rational roots. In the next slide image, a representation of a function y=x^2-4 is required to determine the roots of that function.

[latexpage]

\begin{equation}

x=-\frac{b \pm \sqrt{b^{2}-4 a c}}{2 a}

\end{equation}

This is a graphical representation for the solution of the function, it means it has a root or more, if one root is shown in the graph one point is shown for the curve of function touching the x-axis with y=f(x)=0.

On the other hand if no solution then there will be no intersection of the graph of the function to the x-axis.

This is a graphical representation of the solution of the function; the solution could be multiple points or only one point based on the number of intersections with the x-axis.

## What is synthetic division?

As an introduction to numerical analysis, we will review synthetic division.

Synthetic division is a shorthand, or shortcut, method of polynomial division in the special case of dividing by a linear factor and it *only* works in this case.

Synthetic division is generally used not for dividing out factors but for finding zeroes (or roots) of polynomials.

In synthetic division, to get the zero of the function sometimes, we are guessing the value of the X. with -the so Called Rational roots. The rational Root is a handy way of obtaining a list of first guesses when you are trying to find the Zeros of the polynomial. So in our example. Y = x ^2 – 5 x + 6.

We have a leading coefficient of x^2 = +1. This one can be analyzed at the multiplication (+1*+1)=1, and( -1*-1 )will give us 1, while the 6, the constant, can be analyzed. as either (1*6) or(-1*-6) or (+2*+3) or (-2*-3).

So we have 1, and 6 + and – will put it here.

And for the leading coefficient, we have 1 and 1 + and -(1*1),(-1*-1).

We put it down here in the denominator and comma; plus or minus, we have 3 either plus or minus divided by the leading coefficient. Now we will Proceed to make the synthetic division. We are writing the coefficients only without the x^2 or x, for x^2. The coefficient of x^2 is 1.

We put (+1*+6),(-1*-6),(+2*3),(-2*-3).The first guessing is (x+3)=0 for the first time.

So x will be = – 3, will go there. We Put this – 3 here at the left side portion. The coefficient of x ^2 or the first coefficient. It will go down as 1.

Then we multiply one by -3; we put it here we add -5 to -3, we get -8, then we multiply -3 by -8. -8 multiplied -3, we get + 24, which we put here, and we add 24+6. We get + 30; as long as we do not get the zero, this point is not a root point, then we proceed with guessing another value. So these are the steps.

1-Our next guess is (x-2)=0; the coefficient of x ^2 will be carried down, multiplied by (+2), then add -+5 to add to +2, and we will get -3.

2-(-3) to be carried down, again multiply by(+2), we will have (-6), add to +6, we get a zero, this means that (x-2) is a zero point. Our solution x- 2 is one of the solutions

3- we utilize (x-3), the next guess is (x-3), we use +3, 1 will go down, multiply (+1*3)=3, add to (-3), we get zero value.

4- the two roots are (x-2) and (x -3 ). Now multiply by each other, and you will have (x^2-5*x+6) as the same function y=(x^2-5*x+6).

The next slide image shows the graphical solution of the previous function by tabulating different values of x and estimating the corresponding y value.

### A solved problem for a synthetic division with a remainder.

The next slide image shows division of f(x)=x^3-2x^2+x-3 by(x-2) using synthetic division .There will a remainder of (-1).

## A Solved problem for dividing a given function by a value.

Our new function is

1- It is required to divide the function by(x+2).

2- We have the following coefficients: for x^5 is 4, for x^4=0, for x^3=+6, for x^2=-9, for the constant is +6

3- We start with the solution of (x+2)=0; we divide by (-2).

4- number 5 will continue down, multiplied by -2, the product which is -10 will be placed as a second item and added to +6, the value will be -4.

5- (-4) to be multiplied by -2, the product which is=+8, to be placed as a third item.

6- Add +8to(-9), and the value which is -1 to be placed down.

We proceed to step 8, as shown in the next slide image.

Since we have no zero value, there will be a remainder which is expressed as shown in the next slide image.

## A Solved problem for dividing a given function by a value-long division.

The same problem will be solved by using the long division procedure; instead of putting (-2), we will use(x+2).

The values of 5x^4+6x^3-9x^2-7x+6 will be written on the right side of the (x+2). Divide 5x^4/x=5x^3.

Multiply(5x^3)*(x+2)=5x^4+10x^3. Subtract this value from (5x^4+6x^3). The result will be -4x^3.

Divide -4x^3/x=-4x^2.

Multiply(-4x^2)*(x+2)=-4x^3-8x^2.+10x^3. Subtract this value from (-4x^3-9x^2. The result will be -x^2.

Divide -x^2/x=-x.

Multiply(-x)*(x+2)=-x^2-2x. Subtract this value from (-x^2-7x+6. The result will be -5x+6.

Divide -5x /x=-5. Multiply( -5 )*(x+2)=-5x-10.

Subtract (-5x-10) from (-5x+6). The result will be 16. There will be a remainder of 16/x+2. The full details are shown in the next slide image.

This is a link to download the pdf data used in the illustration.

The next post is about the Bisecting method, a numerical method for rooting findings.

A very intersection source is Holistic Numerical methods.