Unit 1 | unit 1 numerical analysis

MFAI

Unit - 1

Numerical analysis

Q1) Define algebraic and transcedental equations.

A1)

Algebraic Equation: If f(x) is a pure polynomial, then the equation is called an algebraic equation in x.

Ex:

Transcendental Equation: If f(x) is an expression contain function as trigonometric, exponential and logarithmic etc. Then is called transcendental equation.

Q2) Using Newton-Raphson method, find a root of the following equation correct to 3 decimal places:.

A2)

Given

By Newton Raphson Method

The initial approximation is in radian.

For n =0, the first approximation

For n =1, the second approximation

For n =2, the third approximation

For n =3, the fourth approximation

Hence the root of the given equation correct to five decimal place 2.79838.

Q3) Using Newton-Raphson method, find a root of the following equation correct to 4 decimal places:

A3)

Let

By Newton Raphson Method

Let the initial approximation be

For n=0, the first approximation

For n=1, the second approximation

For n=2, the third approximation

Since therefore the root of the given equation correct to four decimal places is -2.9537

Q4) What is the method of false position?

A4)

This is the oldest method of finding the approximate numerical value of a real root of an equation.

In this method we suppose that and are two points where and are of opposite sign .Let

Hence the root of the equation lies between and and so,

The Regula Falsi formula

Find is positive or negative. If then root lies between and or if then root lies between and similarly we calculate

Q5) Find a real root of the equation near, correct to three decimal place by the Regula Falsi method.

A5)

Let

Now,

And also

Hence the root of the equation lies between and and so,

By Regula Falsi Mehtod

Now,

So the root of the equation lies between 1 and 0.5 and so

By Regula Fasli Method

Now,

So the root of the equation lies between 1 and 0.63637 and so

By Regula Fasli Method

Now,

So the root of the equation lies between 1 and 0.67112 and so

By Regula Fasli Method

Now,

So the root of the equation lies between 1 and 0.63636 and so

By Regula Fasli Method

Now,

So the root of the equation lies between 1 and 0.68168 and so

By Regula Fasli Method

Now,

Hence the approximate root of the given equation near to 1 is 0.68217

Q6) Find the real root of the equation

By the method of false position correct to four decimal places

A6)

Let

By hit and trail method

0.23136 > 0

So, the root of the equation lies between 2 and 3 and also

By Regula Falsi Mehtod

Now,

So, root of the equation lies between 2.72101 and 3 and also

By Regula Falsi Mehtod

Now,

So, root of the equation lies between 2.74020 and 3 and also

By Regula Falsi Mehtod

Now,

So, root of the equation lies between 2.74063 and 3 and also

By Regula Falsi Mehtod

Hence the root of the given equation correct to four decimal places is 2.7406

Q7) Solve the equations-

A7)

Let

So that-

Thus-

Writing UX = V,

The system of given equations become-

By solving this-

We get-

Therefore the given system becomes-

Which means-

By back substitution, we have-

Q8) Solve, using Taylor’s series method and compute .

A8)

Here This implies that .

Differentiating, we get

The Taylor’s series at ,

(1)

At in equation (1) we get

Q9) Solve numerically, start from and carry to using Taylor’s series method.

A9)

Here .

We have

Differentiating, we get

implies that or

implies that or .

implies that

The Taylor’s series at ,

Here

The Taylor’s series

Q10) Using Euler’s method solve the differential equation for y at x=1 in five steps

A10)

Given equation

Here

No. Of steps n=5 and so that

So that

Also

By Euler’s formula

, n=0,1,2,3,4 ……(i)

For n=0 in equation (i) we get

For n=1 in equation (i) we get

For n=2 in equation (i) we get

For n=3 in equation (i) we get

For n=4 in equation (i) we get

Hence

Q11) Use modified Euler’s method to compute y for x=0.05. Given that

Result correct to three decimal places.

A11)

Given equation

Here

Take h = = 0.05

By modified Euler’s formula the initial iteration is

)

The iteration formula by modified Euler’s method is

-----(i)

For n=0 in equation (i) we get

Where and as above

For n=1 in equation (i) we get

For n=3 in equation (i) we get

Since third and fourth approximation are equal .

Hence y=1.0526 at x = 0.05 correct to three decimal places.

Q12) Use Runge Kutta method to find y when x=1.2 in step of h=0.1 given that

A12)

Given equation

Here

Also

By Runge Kutta formula for first interval

Again

A fourth order Runge Kutta formula:

To find y at

A fourth order Runge Kutta formula:

Q13) Using Runge Kutta method of fourth order, solve

A13)

Given equation

Here

Also

By Runge Kutta formula for first interval

)

A fourth order Runge Kutta formula:

Hence at x = 0.2 then y = 1.196

To find the value of y at x=0.4. In this case

A fourth order Runge Kutta formula:

Hence at x = 0.4 then y=1.37527

Q14) Using Runge Kutta method of order four, solve to find

A14)

Given second order differential equation is

Let then above equation reduces to

(say)

Or .

By Runge Kutta Method we have

A fourth order Runge Kutta formula:

Q15) Solve the differential equations

for

A15)

Using four order Runge Kutta method with initial conditions

Given differential equation are

Let

And

Also

By Runge Kutta Method we have

A fourth order Runge Kutta formula:

And

Q16) Find the solution of the differential equation in the range for the boundary conditions y = 0 and x = 0 by using Milne’s method.

A16)

By using Picards method-

Where

To get the first approximation-

We put y = 0 in f(x, y),

Giving-

In order to find the second approximation, we put y = in f(x,y)

Giving-

And the third approximation-

Now determine the starting values of the Milne’s method from equation (1), by choosing h = 0.2

Now using the predictor-

X = 0.8

And the corrector-

, ................(2)

Now again using corrector-

Using predictor-

X = 1.0,

And the corrector-

Again using corrector-

, which is same as before

Hence

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