Unit – 1

Solution of Algebraic and Transcendental Equations

Q1) What are algebraic and transcendental 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.

Ex

Q2) What do you understand by iteration?

A2)

Numerical methods are often repetitive in nature. They consist of repetitive calculation of the same process where in each step the result of preceding values are used (substitute).  This is known as iteration process and is repeated till the result is obtained to desired accuracy.

Q3) What is Newton-Raphson method?

A3)

Newton-Raphson Method:

Let be the approximate root of the equation.

By Newton Raphson formula

In general,  

Where n=1, 2, 3…… we keep on calculating until we get desired root to the correct decimal places.         

Q4) Using Newton-Raphson method, find a root of the following equation correct to 3 decimal places: .

A4)

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.

Q5) Using Newton-Raphson method, find a root of the following equation correct to 3 decimal places: near to 4.5

A5)

Let

  The initial approximation

By Newton Raphson Method

     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 equation correct to three decimal places is 4.5579

Q6) Using Newton-Raphson method, find a root of the following equation correct to 4 decimal places:

A6)

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

Q7) What is Newton’s iteration method?

A7)

To find the roots of equation f(x) = 0 by successive approximations,

We will write it as-

Now let be an initial approx. Of the root(desired) , then the first approx. Is given as-

The second approximation is-

Same as the n’th approximation is given as-

Q8) Find the real root of the equation cos x = 3x – 1 correct to three decimal points by using iteration method.

A8)

Here we have-

Now,

A root lies between 0 and .

We can rewrite the equation as-

We have-

And

Here we can apply iteration method, starting with

Then the successive approximation are-

x1 = (x0) = 1/3 (cos 0 + 1) = 0.6667

x2 = (x1) = 1/3 (cos 0.6667 + 1) = 0.5953

x3 = (x2) = 1/3 (cos (0.5953) + 1) = 0.6093

x4 = (x3) = 0.6067

x5 = (x4) = 0.6072

x­6 = (x5) = 0.6071

Here last two approximations are almost same, the root is 0.607 correct to 3 decimal places.

Q9) Starting with x = 0.12, solve x = 0.21 sin (0.5 + x) by using the iteration method.

A9)

Here

First approximation of x is gives as-

x(1) = 0.21 sin (0.5 + 0.12) = 0.122

x(2) = 0.21 sin (0.5 + 0.122) = 0.1224

x(3) = 0.12242 ,  x(4) = 0.12242

Here last two approx are same, hence required root is 0.12242.

Q10) What is the method of Newton-Rapshon for non linear equations?

A10)

Suppose the equations be given by

f(x, y) = 0,  g(x, y) = 0 …… (1)

Whose real roots are required within a specified accuracy.

Let ( be an initial approximation to the root of the system (1).

If is the root of the system, the we must have,

Assuming that f and g are sufficiently differentiable, we expand above equations by using Taylor’s series to obtain,

Where

Neglecting the second and higher order terms, we get the following system of linear equations:

If the Jacobian

Does not vanish, then the linear equations (3) possesses a unique solution given by

The new approximations are then given by

We repeat the process till we get the desired roots with accuracy.

Q11) Find the roots of the equations

A11)

Before proceeding towards the solution, we take y = x as our first approximation. This gives

And therefore

Where

Further

Hence

And therefore the convergence criterion is satisfied.

We then have

These equations gives,

Hence the first approximation to the root is given by

For the second approximation, we have

Then

Clearly the condition of convergence is satisfied and we have the simultaneous equations,

Solving these equations, we get

And the second approximation therefore given by

And

The values above may be compared with the true values, which are given by

Unit – 1

Solution of Algebraic and Transcendental Equations

Q1) What are algebraic and transcendental 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.

Ex

Q2) What do you understand by iteration?

A2)

Numerical methods are often repetitive in nature. They consist of repetitive calculation of the same process where in each step the result of preceding values are used (substitute).  This is known as iteration process and is repeated till the result is obtained to desired accuracy.

Q3) What is Newton-Raphson method?

A3)

Newton-Raphson Method:

Let be the approximate root of the equation.

By Newton Raphson formula

In general,  

Where n=1, 2, 3…… we keep on calculating until we get desired root to the correct decimal places.         

Q4) Using Newton-Raphson method, find a root of the following equation correct to 3 decimal places: .

A4)

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.

Q5) Using Newton-Raphson method, find a root of the following equation correct to 3 decimal places: near to 4.5

A5)

Let

  The initial approximation

By Newton Raphson Method

     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 equation correct to three decimal places is 4.5579

Q6) Using Newton-Raphson method, find a root of the following equation correct to 4 decimal places:

A6)

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

Q7) What is Newton’s iteration method?

A7)

To find the roots of equation f(x) = 0 by successive approximations,

We will write it as-

Now let be an initial approx. Of the root(desired) , then the first approx. Is given as-

The second approximation is-

Same as the n’th approximation is given as-

Q8) Find the real root of the equation cos x = 3x – 1 correct to three decimal points by using iteration method.

A8)

Here we have-

Now,

A root lies between 0 and .

We can rewrite the equation as-

We have-

And

Here we can apply iteration method, starting with

Then the successive approximation are-

x1 = (x0) = 1/3 (cos 0 + 1) = 0.6667

x2 = (x1) = 1/3 (cos 0.6667 + 1) = 0.5953

x3 = (x2) = 1/3 (cos (0.5953) + 1) = 0.6093

x4 = (x3) = 0.6067

x5 = (x4) = 0.6072

x­6 = (x5) = 0.6071

Here last two approximations are almost same, the root is 0.607 correct to 3 decimal places.

Q9) Starting with x = 0.12, solve x = 0.21 sin (0.5 + x) by using the iteration method.

A9)

Here

First approximation of x is gives as-

x(1) = 0.21 sin (0.5 + 0.12) = 0.122

x(2) = 0.21 sin (0.5 + 0.122) = 0.1224

x(3) = 0.12242 ,  x(4) = 0.12242

Here last two approx are same, hence required root is 0.12242.

Q10) What is the method of Newton-Rapshon for non linear equations?

A10)

Suppose the equations be given by

f(x, y) = 0,  g(x, y) = 0 …… (1)

Whose real roots are required within a specified accuracy.

Let ( be an initial approximation to the root of the system (1).

If is the root of the system, the we must have,

Assuming that f and g are sufficiently differentiable, we expand above equations by using Taylor’s series to obtain,

Where

Neglecting the second and higher order terms, we get the following system of linear equations:

If the Jacobian

Does not vanish, then the linear equations (3) possesses a unique solution given by

The new approximations are then given by

We repeat the process till we get the desired roots with accuracy.

Q11) Find the roots of the equations

A11)

Before proceeding towards the solution, we take y = x as our first approximation. This gives

And therefore

Where

Further

Hence

And therefore the convergence criterion is satisfied.

We then have

These equations gives,

Hence the first approximation to the root is given by

For the second approximation, we have

Then

Clearly the condition of convergence is satisfied and we have the simultaneous equations,

Solving these equations, we get

And the second approximation therefore given by

And

The values above may be compared with the true values, which are given by

Unit – 1

Unit – 1

Unit – 1

Unit – 1

Unit – 1

Solution of Algebraic and Transcendental Equations

Q1) What are algebraic and transcendental 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.

Ex

Q2) What do you understand by iteration?

A2)

Numerical methods are often repetitive in nature. They consist of repetitive calculation of the same process where in each step the result of preceding values are used (substitute).  This is known as iteration process and is repeated till the result is obtained to desired accuracy.

Q3) What is Newton-Raphson method?

A3)

Newton-Raphson Method:

Let be the approximate root of the equation.

By Newton Raphson formula

In general,  

Where n=1, 2, 3…… we keep on calculating until we get desired root to the correct decimal places.         

Q4) Using Newton-Raphson method, find a root of the following equation correct to 3 decimal places: .

A4)

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.

Q5) Using Newton-Raphson method, find a root of the following equation correct to 3 decimal places: near to 4.5

A5)

Let

  The initial approximation

By Newton Raphson Method

     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 equation correct to three decimal places is 4.5579

Q6) Using Newton-Raphson method, find a root of the following equation correct to 4 decimal places:

A6)

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

Q7) What is Newton’s iteration method?

A7)

To find the roots of equation f(x) = 0 by successive approximations,

We will write it as-

Now let be an initial approx. Of the root(desired) , then the first approx. Is given as-

The second approximation is-

Same as the n’th approximation is given as-

Q8) Find the real root of the equation cos x = 3x – 1 correct to three decimal points by using iteration method.

A8)

Here we have-

Now,

A root lies between 0 and .

We can rewrite the equation as-

We have-

And

Here we can apply iteration method, starting with

Then the successive approximation are-

x1 = (x0) = 1/3 (cos 0 + 1) = 0.6667

x2 = (x1) = 1/3 (cos 0.6667 + 1) = 0.5953

x3 = (x2) = 1/3 (cos (0.5953) + 1) = 0.6093

x4 = (x3) = 0.6067

x5 = (x4) = 0.6072

x­6 = (x5) = 0.6071

Here last two approximations are almost same, the root is 0.607 correct to 3 decimal places.

Q9) Starting with x = 0.12, solve x = 0.21 sin (0.5 + x) by using the iteration method.

A9)

Here

First approximation of x is gives as-

x(1) = 0.21 sin (0.5 + 0.12) = 0.122

x(2) = 0.21 sin (0.5 + 0.122) = 0.1224

x(3) = 0.12242 ,  x(4) = 0.12242

Here last two approx are same, hence required root is 0.12242.

Q10) What is the method of Newton-Rapshon for non linear equations?

A10)

Suppose the equations be given by

f(x, y) = 0,  g(x, y) = 0 …… (1)

Whose real roots are required within a specified accuracy.

Let ( be an initial approximation to the root of the system (1).

If is the root of the system, the we must have,

Assuming that f and g are sufficiently differentiable, we expand above equations by using Taylor’s series to obtain,

Where

Neglecting the second and higher order terms, we get the following system of linear equations:

If the Jacobian

Does not vanish, then the linear equations (3) possesses a unique solution given by

The new approximations are then given by

We repeat the process till we get the desired roots with accuracy.

Q11) Find the roots of the equations

A11)

Before proceeding towards the solution, we take y = x as our first approximation. This gives

And therefore

Where

Further

Hence

And therefore the convergence criterion is satisfied.

We then have

These equations gives,

Hence the first approximation to the root is given by

For the second approximation, we have

Then

Clearly the condition of convergence is satisfied and we have the simultaneous equations,

Solving these equations, we get

And the second approximation therefore given by

And

The values above may be compared with the true values, which are given by