Unit-5

Numerical Methods

This method is consists of finding the root of the equation which lie between a and b (say).

The function is a continuous function between a and b and f (a) and f (b) are of opposite signs then there is a least one root between a and b.

Suppose f (a) is negative and f (b) is positive, then the first approximate value of the root is

If, then the correct root is .But if, then the root either lie between a and or and b according to is positive or negative, we again bisect the interval as above and the process is repeated the root is found to the desired accuracy.

Note: Remember that root of an equation lie between its positive and negative values and we take an average of them to come closer to its accurate root.

Example1 Find a real root of using bisection method correct to five decimal places.

Let then by hit and trial we have

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

Now,

I.e. positive so the root of the given equation must lie between

Now,

I.e. negative so the root of the given equation lie between

Now,

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

Now,

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

Now,

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

Now,

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

Now,

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

Now,

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

Hence the approximate root of the given equation is 1.32421

Example2 Find the root of the equation, using the bisection method.

Let then by hit and trial we have

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

Now,

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

Now,

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

Now,

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

Now,

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

Now,

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

Now,

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

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

Example3 Find the root of the equation between 2 and 3, using bisection method correct to two decimal places.

Let

Where

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

Now,

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

Now,

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

Now,

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

Now,

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

Now,

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

Now,

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

Now,

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

Now,

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

Hence the root of the given equation correct to two decimal places is 2.1269

This method is an improvement over the Regula Falsi Method as it does not require the condition

The general formula for successive approximation is,

Where

We keep on calculating until we get the desired root to the correct decimal places.

Example1 Using the Secant Method find the root of the equation correct to three decimal places

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.

Example2 Using the Secant Method to find the root of the equation correct to four decimal places

Let

By Secant Method

Let the initial approximation be

For n=1, the first approximation

Now,

For n=2, the second approximation

Now,

For n=3, the third approximation

Hence the root of the given equation correct to four decimal places is 3.25636

Example3 Using the Secant Method find the root of the equation correct to four decimal place

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 lie between 2 and 1.96590

For n=3, the third approximation

Now,

So the root of the equation lie between 2 and 1.96600

For n=4, the fourth approximation

Now,

So the root of the equation lie between 2 and 1.96687

For n=5, the fifth approximation

Now,

So the root of the equation lie 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.

Key takeaways-

The general formula for successive approximation is,

Where

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 lie between and and so,

The Regula Falsi formula

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

Proceed in this manner until the desired accurate root is found.

Example1Find a real root of the equation near, correct to three decimal places by the Regula Falsi method.

Let

Now,

And also

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

By Regula Falsi Method

Now,

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

By Regula Falsi Method

Now,

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

By Regula Falsi Method

Now,

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

By Regula Falsi Method

Now,

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

By Regula Falsi Method

Now,

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

By Regula Falsi Method

Now,

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

Example2 Find the real root of the equation

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 lie between 2 and 3 and also

By Regula Falsi Method

Now,

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

By Regula Falsi Method

Now,

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

By Regula Falsi Method

Now,

So, the 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

Example3 Apply Regula Falsi Method to solve the equation

Let

By hit and trail

And

So the root of the equation lie between and also

By Regula Falsi Method

Now,

So, the root of the equation lies between 0.60709 and 0.61 and also

By Regula Falsi Method

Now,

So, the root of the equation lies between 0.60710 and 0.61 and also

By Regula Falsi Method

Hence the root of the given equation correct to five decimal places is 0.60710.

Key takeaways-

The Regula Falsi formula

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 the desired root to the correct decimal places.

Example1 Using the Newton-Raphson method, find a root of the following equation correct to 3 decimal places:.

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 places 2.79838.

Example2 Using the Newton-Raphson method, find a root of the following equation correct to 3 decimal places: near to 4.5

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

Example3 Using the Newton-Raphson method, find a root of the following equation correct to 4 decimal places:

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

Successive Approximation Method:

The successive approximation is also known as the iteration method. To start the solution using this method we need one or more approximate values which is not necessarily the root of the given equation.

We are finding the root of the given equation

..(1)

We rewrite the given equation in the form

..(2)

We know that the root of the equation lie between its positive and the negative values

Let

So, the interval of the root of the equation be .

Now, let be an approximate root of the given equation (1).

Putting it in equation (2) we get

Successive substitution give the approximations

…………..

If the above values converge to a definite number then that number will be the root of the given equation.

Example1: Find the real root of the equation

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.

Example2: Find the real root of the polynomial correct to three decimal places?

Given equation ….(1)

Here

Also

Therefore the 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.

Example3: By iteration method, find the value of , correct to three decimal places.

Let

Let .

Also

Therefore the root of the equation lie between 3 and 4.

Given equation can rewrite .

Or …(2)

Let , in the interval .

The successive approximation we have

Hence the root of the equation correct to three decimal places is 3.634.

Convergence and stability-

Convergence-

A sequence of iterates is said to converge with order if

For some constant c > 0.

Note-

Stability-

Let there are n steps to evaluate a function. Then the total process of evaluating this function is said to have instability if at least one step is ill-conditioned. If all the steps are well-conditioned, then the process is said to be stable.

In this method, we eliminate successively the unknown so that the equation (1) remain with the single unknown and reduce to the upper triangular system. At last with help of back substitution, we calculate the values of the remaining unknowns.

Consider a system of n linear equation in n unknown

… …… ….. … (1)

… …… …. …

To convert the above system into an upper triangular matrix we eliminate from the second, third, fourth …., n equations above by multiplying the first equation by added them to the corresponding equations second, third, fourth,…., n equation. We get

… …… ….. …

… …… …. …

Repeating the above method for we get finally the upper triangular form.

Upper Triangular form of above

……………………….. (i)

Thus .

Then we calculate the values of .

Example1 Apply Gauss Elimination method to solve the equations:

Given Check Sum (sum of coefficient and constant)

-1 …. (I)

-16 …. (ii)

5 …. (iii)

(I)We eliminate x from (ii) and (iii)

Apply eq(ii)-eq(i) and eq(iii)-3eq(i) we get

-1 ….(i)

-15 ….(iv)

8 ….(v)

(II) We eliminate y from eq(v)

Apply

-1 ….(i)

-15 ….(iv)

73 ….(vi)

(III) Back Substitution we get

From (vi) we get

From (iv) we get

From (i) we get

Hence the solution of the given equation is

Example2 Solve the equation by Gauss Elimination Method:

Given

Rewrite the given equation as

… (i)

….(ii)

….(iii)

…(iv)

(I) We eliminate x from (ii),(iii) and (iv) we get

Apply eq(ii) + 6eq(i), eq(iii) -3eq(i), eq(iv)-5eq(i) we get

…(i)

….(v)

….(vi)

…(vii)

(II) We eliminate y from (vi) and (vii) we get

Apply 3.8 eq(vi)-3.1eq(v) and 3.8eq(vii)+5.5eq(v) we get

…(i)

….(v)

…(viii)

…(ix)

(III) We eliminate z from eq (ix) we get

Apply 9.3eq (ix) + 8.3eq (viii), we get

… (i)

….(v)

…(viii)

350.74u=350.74

Or u = 1

(IV) Back Substitution

From eq(viii)

Form eq(v), we get

From eq(i) ,

Hence the solution of the given equation is x=5, y=4, z=-7 and u=1.

Example3

Apply Gauss Elimination Method to solve the following system of the equation:

Given … (i)

… (ii)

… (iii)

(I) We eliminate x from (ii) and (iii)

Apply we get

… (i)

… (iv)

… (v)

(II) We eliminate y from (v)

Apply we get

… (i)

… (vi)

… (vii)

(III) Back substitution

From (vii)

From (vi)

From (i)

Hence the solution of the equation is

The method is based on the fact that every matrix A can be expressed as the product of a lower triangular matrix and an upper triangular matrix, provided all the principal minors of A are non-singular.

Which means-

If , then-

Now consider the equations-

We can write it as-

Where-

Let

Where-

Equation (1) becomes-

Writing-

Equation (3) becomes-

which is equivalent to the equations-

Solving these for we know V. Then equation (4) becomes-

From which can be found by back substitution.

We write (2) as to find the matrix L and U-

Multiplying the matrix on the left and equating corresponding elements from both sides, we get-

3.

4.

5.

We compute the elements of L and U in the following manner-

Example: Solve the equations-

Sol.

Let

So that-

3.

4.

5.

So

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-

Jacobi’s Iteration method:

Let us consider the system of simultaneous linear equation

(1)

The coefficients of the diagonal elements are larger than all other coefficients and are non-zero. Rewrite the above equation we get

(2)

Take the initial approximation we get the values of the first approximation of .

By the successive iteration, we will get the desired result.

Example1Use Jacobi’s method to solve the system of equations:

Since

So, we express the unknown with a large coefficient in terms of other coefficients.

(1)

Let the initial approximation be

2.35606

0.91666

1.932936

0.831912

3.016873

1.969654

3.010217

1.986010

1.988631

0.915055

1.986532

0.911609

1.985792

0.911547

1.98576

0.911698

Since the approximation in the ninth and tenth iteration is the same up to three decimal places, hence the solution of the given equations is

Example2 Solve by Jacobi’s Method, the equations

Given equation can be rewritten in the form

… (i)

..(ii)

..(iii)

Let the initial approximation be

Putting these values on the right of the equation (i), (ii), and (iii) and so we get

Putting these values on the right of the equation (i), (ii), and (iii) and so we get

Putting these values on the right of the equation (i), (ii), and (iii) and so we get

0.90025

Putting these values on the right of the equation (i), (ii), and (iii) and so we get

Putting these values on the right of the equation (i), (ii), and (iii) and so we get

Hence solution approximately is

Example3Use Jacobi’s method to solve the system of the equations

Rewrite the given equations

(1)

Let the initial approximation be

1.2

1.3

0.9

1.03

0.9946

0.9934

1.0015

Hence the solution of the above equation correct to two decimal places is

Gauss-Seidel method:

This is the modification of Jacobi’s Iteration. As above in Jacobi’s Iteration, we take first approximation as and put in the right-hand side of the first equation of (2) and let the result be . Now we put right-hand side of the second equation of (2) and suppose the result is now put in the RHS of the third equation of (2) and suppose the result be the above method is repeated till the values of all the unknown are found up to the desired accuracy.

Example1 Use the Gauss-Seidel Iteration method to solve the system of equations

Since

So, we express the unknown of larger coefficients in terms of the unknowns with smaller coefficients.

Rewrite the above system of equations

(1)

Let the initial approximation be

3.14814

2.43217

2.42571

2.4260

Hence the solution correct to three decimal places is

Example2 Solve the following system of equations

By Gauss-Seidel method.

Rewrite the given system of equations as

(1)

Let the initial approximation be

Thus the required solution is

Example3 Solve the following equations by Gauss-Seidel Method

Rewrite the above system of equations

(1)

Let the initial approximation be

Hence the required solution is

References

1. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.

2. N.P. Bali and Manish Goyal, A textbook of Engineering Mathematics, Laxmi Publications.

3. Higher engineering mathematic, Dr. B.S. Grewal, Khanna publishers