Unit-5

Numerical Solution of Algebraic and transcendental equation

Question-1: 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 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

Question-2: 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

Question-3: 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.

Question-4: 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

Question-5: Find 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 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 1 is 0.68217

Question-6: Apply Regula Falsi Method to solve the equation

Let 

By hit and trail

  And

So the root of the equation lies 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.

Question-7: 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

Question-8: 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.

Question-9: By iteration method, find the value of , correct to three decimal places.

Let 

Let .

Also

Therefore the root of the equation lies 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.

Question-10: 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.

Question-11: 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-

Question-12: Use 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

Question-13: Use the Gauss-Seidel Iteration method to solve the system of equations

Since

So, we express the unknown of a larger coefficient 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

Question-14: 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