Unit - 1

Rate of convergence

Q1) Define order of convergence.

A1)

Let   be the given equation.  Let α be its exact root.

Then

Then the positive number p is said to be order of convergence if,

If p=1, then convergence is linear and if p=2, then it is quadratic.

An iterative method is said to be kth order convergent if k is the largest positive real

Number such that

Here A is a non-zero finite number called asymptotic error constant and it depends on derivative of f(x) at an approximate root x.

and are the errors in successive approximation.

Q2) What is an absolute and relative error?

A2)

Absolute Error:

The Approximate error is the numerical difference between the quantity of exact solution and the approximate value of the solution.

Let X   be the exact solution and the approximate value of the solution is denoted by , then the absolute error is defined as

Relative Error:

The error obtained by dividing the approximate solution to the exact solution is known as relative error and is denoted by

The percentage error is the 100 times of the relative error i.e.

The absolute accuracy is magnitude absolute error.

The relative accuracy is the magnitude of the relative error.

Q3) An approximate value of   is given by and its exact value is Find the absolute and relative errors.

A3)

The absolute error is

.

The relative error is

Q4) Find the difference   to three significant figure?

A4)

Correct to three decimal places.

Q5) Find the absolute error if the number is truncate to three decimal places?

A5)

The given number is

After truncate it to three decimal places the rounded value is

Therefore the absolute error is

=

.

Q6) Explain generalized error formula.

A6)

Let-

Be the function of and let the error in each be .

The error in u is given by-

By using Taylor’s theorem we expand RHS-

Lets assume that the errors in are small, so that the higher powers of can be neglected,

The above relation yields-

The formula for the relative error follows-

Q7) Find the real root of the equation

A7)

Correct to three decimal places in the interval  ]

The given equation is       ..(1)

Or

Or =      ..(2)

Or

Let , in the interval .

The successive approximation we have

Hence the root of the equation correct to three decimal places is 1.524.

Q8) Find the real root  of the polynomial correct to three decimal places?

A8)

Given equation    ….(1)

Here

Also

Therefore root of the equation lies between .

Again

    ….(2)

Let , in the interval .

The successive approximation we have

Hence the root of the equation correct to three decimal places is 0.755.

Q9) What do you understand by stable and unstable methods?

A9)

In an initial value problems, we need the solution for and usually up to a point x = b.

The length of step h for application of any numerical method for the initial value problem should be chosen properly.

There are only two types of errors in computation- truncation and round off error.

We can handle the truncation error but round-off error can be handled, i.e. this is not in the hands of user. They can increase and can destroy the solution too.

Then we say that the method is numerically unstable.

When the step length is chosen larger than the allowed limiting value, in this case we come across with this situation.

Many explicit methods have no restrictions on the step length can be used.

These methods are called unconditionally stable method.

Q10) Write a note on arithmetic mathematical software.

A10)

Mathematical software is used to mode, calculate, analyze or visualize the numeric or geometric dataset.

There are various categories of mathematical software for different task relate to mathematical data.

A calculator, which is also a software allows the user to perform simple mathematical calculations such as, addition, subtraction, trigonometry, geometry, exponential, etc. we get the output in text format after entering the data manually.

To solve the algebraic equations, there are various computer algebraic systems. They are especially designed to solve the algebraic equations, and give the accurate solutions.

A lot of computer tools are available to analyze, visualize the statistical data sets, such as R programming, SAS, MS- excel. SPSS, stata, statistica, Julia, dataplot etc.

MATLAB is widely used software for numerical computations. It has its own programming language. We can implement the numerical algorithms.

GNU octave is the high-level language, generally made for numerical computations.

We can solve linear and non-linear problems numerically by using it.

1. FFTW is a collection of fast C routines for computing the discrete Fourier transform in one or more than one dimensions. It has complex, real and parallel transforms and can handle arbitrary array sizes efficiently.
2. LAPACK is a standard library for numerical linear algebra.
3. SuperLU is a library which perform an LU decomposition for the direct solution of large and non-symmetric system of linear equations on high performance machines. It is written in C.

Q11) Find a real root of using bisection method correct to five decimal places.

A11) Let then by hit and trial we have

Thus .So the root of the given equation should lies between 1 and 2.

Now,

i.e. positive so the root of the given equation must lies between

Now,

i.e. negative so the root of the given equation lies between

Now,

i.e. positive so the root of the given equation lies between

Now,

i.e. negative  so that the root of the given equation lies between

Now,

i.e. positive so that the root of the given equation lies between

Now,

i.e. positive so that the root of the given equation lies between

Now,

I.e. negative so that the root of the given equation lies between

Now,

i.e. negative so that the root of the given equation lies between

Hence the approximate root of the given equation is 1.32421

Q12) Find the root of the equation, using the bisection method.

A12)

Let then by hit and trial we have

Thus .So the root of the given equation should lies between 2 and 3.

Now,

i.e. negative so the root of the given equation must lies between

Now,

i.e. positive so the root of the given equation must lies between

Now,

i.e. negative so the root of the given equation must lies between

Now,

i.e. negative so the root of the given equation must lies between

Now,

i.e. positive so the root of the given equation must lies between

Now,

i.e. negative so the root of the given equation must lies between

Hence the root of the given equation correct to two decimal place is 2.67965.

Q13) Find a real root of the equation near, correct to three decimal place by the Regula Falsi method.

A13)

Let

Now,

And also

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

By Regula   Falsi Method

Now,

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

By Regula Falsi Method

Now,

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

By Regula Falsi  Method

Now,

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

By Regula Falsi  Method

Now,

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

By Regula Falsi  Method

Now,

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

By Regula Falsi  Method

Now,

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

Q14) Find the real root of the equation

A14)

By the method of false position correct to four decimal places

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 Method

Now,

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

By Regula   Falsi Method

Now,

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

By Regula   Falsi Method

Now,

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

By Regula   Falsi Method

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

Q15) Using the Secant Method find the root of the equation correct to three decimal places

A15)

Let

By Secant Method

Let the initial approximation be

For n=1, the first approximation

Now,

For n=2, the second approximation

563839

Now,

For n=3, the third approximation

56717

Now,

For n=4, the fourth approximation

567143

Hence the root of the given equation correct to four decimal places is 0.5671.

Q16) Using Secant Method find the root of the equation correct to four decimal place

A16)

Let

By Secant Method

Let the initial approximation be

For n=1, the first approximation

Now,

So the root of the equation lies between 2 and 1.92857

For n=2, the second approximation   ,

Now,

So the root of the equation lies between 2 and 1.96590

For n=3, the third approximation

Now,

So the root of the equation lies between 2 and 1.96600

For n=4, the fourth approximation

Now,

So the root of the equation lies between 2 and 1.96687

For n=5, the fifth approximation

Now,

So the root of the equation lies between 2 and 1.96690

For n=6, the sixth approximation

Now,

Hence the root of the given equation correct to four decimal places is 1.9669.

Q17) Using Newton-Raphson method, find a root of the following equation correct to 3 decimal places: near to 4.5

A17)

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

Q18) Using Newton-Raphson method, find a root of the following equation correct to 4 decimal places:

A18)

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

Q19) Explain fixed point iteration formula.

A19)

Fixed point iteration method is also called general iteration method.

In the very first step in this method, we write the equation f(x) = 0 in an equivalent form such as-

Now we find a root of f(x) = 0 is same as finding a number such that that is a fixed point of .

Using equation (1), the iteration method can be written as-

The function is known as the iteration function.

We compute the next approximation as, (starting with the initial approximation)-

...........

...........

...............

................

Q20) Find the smallest positive root of the equation by using fixed point iteration method.

A20)

Here we have-

Since f(2) f(2) <  0, the smallest positive root lies in the interval (2, 3).

Now write-

We define the iteration method as-

We get-

We find | < 1 for all x in interval (2, 3), hence the iteration converges

Let

, we get the following results-

Since

We take the required root as  

x = 2.3089