An equation of the form a$_{0}$x$^{n}$ + a$_{1}$x$^{n-1}$  + a$_{2}$x$^{n-2}$ + .........+ a$_{n}$ = 0 where a$_{0}$ $\neq$ 0 and a$_{1}$, a$_{2}$, ........, a$_{n}$ are constants and n is a positive integer called an algebraic equation of degree n.

If f(x) = a$_{0}$x$^{n}$ + a$_{1}$x$^{n-1}$ + a$_{2}$x$^{n-2}$ + .........+ a$_{n}$ then above equation becomes f(x) = 0
If f(x$_{0}$) = 0 for x = x$_{0}$ then x$_{0}$ is called a root of f(x) = 0

There are various numerical methods to solve algebraic equations like Iteration method, newton raphson method, finite differences, algebraic equations using gauss eliminations and gauss seidel method, rate of convergence, lagrange's method, numerical differentiation and integration etc.
Numerical analysis is the study of algorithms that use numerical approximation based on mathematical analysis. It does not seek answers as it is impossible to obtain in practice. It's applications can be found in all fields of engineering and physical sciences. Numerical weather prediction is feasible.

## Theoretical Numerical Analysis

Different numerical methods are explained below.

Iteration Method
To find the roots of equation f(x) = 0 by successive approximations,
we write the equation,  f(x) = a$_{0}$x$^{n}$ + a$_{1}$x$^{n-1}$ + a$_{2}$x$^{n-2}$ + .........+ a$_{n}$, in the form
x = $\phi$(x)

Let x = x$_{0}$ be an initial approximation of the desired root $\alpha$. Then the first approximation x$_{1}$ is given by x$_{1}$ = $\phi$(x$_{0}$)

Second approximation x$_{2}$ is given by x$_{2}$ = $\phi$(x$_{1}$)

Proceeding in this way, the nth approximation is given by x$_{n}$ = $\phi$(x$_{n-1}$)

The roots of f(x) = 0 are same as the point of intersection of the straight line y = x and y = $\phi$(x)
The form x = $\phi$(x) can be chosen in many ways. We choose $\phi$(x) in such a way that  |$\phi$(x)| < 1, $\forall$ x = x$_{0}$, x$_{1}$, x$_{2}$,......
Sequence {x$_{n}$} converge to the root $\alpha$
The rate of convergence is more, if the value of $\phi$'(x) is smaller.
This method is useful for finding the real roots of an equation given in the form of an infinite series.

Newton raphson method

Let the given equation be f(x) = 0.

Let x$_{0}$ be an appropriate value of the desired root and h is the correction factor

Then x$_{1}$ = x$_{0}$ + h      ........... ( 1 )
is the root of the equation f(x) = 0

f(x$_{1}$) = 0

f(x$_{0}$ + h) = 0

Using Taylor's theorem expand f(x$_{0}$ + h) we get

f (x$_{0}$ + h) = f (x$_{0}$) + hf ' (x$_{0}$) + ................. + hf ' (x$_{n}$) = 0

To get  f(x$_{0}$) + hf '(x$_{0}$) = 0, neglect h$_{2}$ and higher order derivatives since h is very small

h = − [f(x$_{0}$)f ′(x$_{0}$)]       ........... ( 2 )

Plug in equation (2) in  equation (1), we get the first approximation to the required root as

x$_{1}$ = x$_{0}$ −[f(x$_{0}$)f′ (x$_{0}$)]     ................. ( 3 )

Successive approximations are written as,

x$_{2}$ = x$_{1}$ − [f(x$_{1}$)f′ (x$_{1}$)]             ............. ( 4 )
x$_{(n+1)}$ = x$_{n}$ − [f(x$_{n}$)f′ (x$_{n}$)]        ............ ( 6 )

The sequence {x$_{n}$ } if it converges gives the root. The Newton - Raphson method is also known as the method of tangent.

## Numerical Analysis Solutions

### Solved Examples

Question 1: Find the real root of the equation 2x$^{3}$ + x$^{2}$ = 2 by iteration method.
Solution:

The given equation can be written as
2x$^{3}$ + x$^{2}$ - 2 = 0

Let f(x) = 2x$^{3}$ + x$^{2}$ - 2
f(0) = -2

f(2) = 16 + 4 - 2 = 0
f(2) = 18

Hence the root value lie between -2 to 18.

2x$^{3}$ + x$^{2}$ - 2 = 0
x$^{2}$ (2x + 1) = 2
x$^{2}$ = $\frac{2}{2x+1}$

x = $\frac{1}{\sqrt{x + 0.5}}$

$\phi$(x) = $\frac{1}{\sqrt{0.5 + 0.5}}$

Initial approximation x$_{0}$ = 0.5

x$_{1}$ = $\phi$(x$_{0}$) = $\frac{1}{\sqrt{0.5 + 0.5}}$

= 1

x$_{2}$ = $\phi$(x$_{1}$) = $\frac{1}{\sqrt{1 + 0.5}}$

= 0.82

x$_{3}$ = $\phi$(x$_{2}$) = $\frac{1}{\sqrt{0.82+ 0.5}}$
= 0.87

x$_{4}$ = $\phi$(x$_{3}$) = $\frac{1}{\sqrt{0.87 + 0.5}}$
= 0.85

x$_{5}$ = $\phi$(x$_{4}$) = $\frac{1}{\sqrt{0.85 + 0.5}}$
= 0.86

x$_{6}$ = $\phi$(x$_{5}$) = $\frac{1}{\sqrt{0.86 + 0.5}}$
= 0.86

The root is 0.86 since x$_{5}$ and x$_{6}$ are same.

Question 2: Find x$_{1}$ and x$_{2}$ for equation x$^{3}$ - 5x + 3 = 0 by using Newton's Raphson method.
Solution:

Here, x$_{0}$ is not given
f(x) = x$^{3}$ - 5x + 3
f'(x) = 3x$^{2}$ - 5
f(0) = 3
f(1) = -1
$\Rightarrow$ f(0) > 0 and f(1) < 0
Then root lies between (0, 1).
Here, f(1) = -1 is nearer to zero,
then f(0) = 3
x$_{0}$ = 1

Now, f(x$_{0}$) = f(1) = 1$^{3}$ - 5 * 1 + 3 = 1 - 5 + 3 = -1
f'(x$_{0}$) = f'(1) = 3 * 1$^{2}$ - 5 = 3 - 5 = -2

x$_{1}$ = x$_{0}$ - $\frac{f(x_{0})}{f'(x_{0})}$

= 1 - $\frac{-1}{-2}$

= 0.5
now f(x$_{1}$) = f(0.5) = (0.5)$^{3}$ - 5(0.5) + 3
= 0.125 - 2.5 + 3
= 0.625
f'(0.5) = 3(0.5)$^{2}$ - 5 = -4.25

x$_{2}$ = x$_{1}$ - $\frac{f(x_{1})}{f'(x_{1})}$

= 0.5 - $\frac{0.625}{-4.25}$

= 0.5 + 0.1471
= 0.6471

Question 3: Find the equations of the tangent to the curve y = x$^{2}$ - x + 13 at the point (1, 2).
Solution:

Equation of the given curve is

y = x$^{2}$ - x + 13

$\frac{dy}{dx}$ = 2x - 1

$\frac{dy}{dx}$ at point (1,2) is

= 2 -1 = 1

Therefore, the required equation of the tangent is

y - y$_{1}$ = m  (x - x$_{1}$)

y - 2 = x - 1

y - x = 1