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Tutorial-01, B.Tech. Semester-I (Leibnitz’s rule, Partial Differentiation)

Problem 1: Find the th derivative of the following functions: (a) tan (b) sin 4 cos 6 Ans: (a)(−1) (b) Problem 2: If

(

=

(

)

(10 +

log ), prove that

( )

10) −

=

( − 1)!

. sin

(2 + tan

− 1)! and hence show that

+(

= ! {log +1+ + + … + } Problem 3: If = ( − 1) , use Leibnitz’s theorem to show that (1 − ) −2 + ( + 1) = 0. ), show that Problem 4: If = sin( sin (1 − ) − (2 + 1) +( − ) = 0 and hence evaluate Problem 5: If cos

= log

+ (2 + 1)

, prove that

+2

Problem 6: If = [ − ( − 1)] , prove that (1 − ) − (2 + 1) +( − ) = 0. Problem 7:(a) If = establish the relation that = . / ( ) (b) If = + + , show that . + . + =− + = 0. (c) If =φ ( − ) + ( + ) , show that − = 0. (d) If = sin

, show that

+ tan

.

10)

+

.

( ) . = 0.

and

+

= 0.

(e) If =

, then show that + + = 0. 1 1 (f) If = , show that = (1 + 3 + ) . ), then prove that (g) If =log( + + −3 1

+ Problem 8:(a) If (b) If

+

= ( ) and r= =

(b) If (c) If

)

. , then

+

, show that at =

(c) If Problem 9:(a) If

= −(

+

+

(

) +

=

+ ,

= "( ) + = −(

=

, where

+

= ( , ), where

Then prove that

.

= 1 prove that +(

) = 2(

.

+

.

, find the value of n will make

=

)

′( ).

+ . (

) Ans:

)=

= sin find ? Ans:3( cos − = cos , = sin find the value of

=

=−

cos ,

(

) +

= , find the value of

=(

) +

(d) If

+

.

(e) If

= ( − , − , − ) , then prove that

(

sin ) .

) . Ans:

+

+

= 0. 1

Problem 10: State and prove Euler’s theorem for a homogeneous function ( , ) of degree in two variables x and y. Also deduce that (i) . + . = ( − 1) (ii) . + . = ( − 1) (iii) . + . +2 = ( − 1) Problem 11: Prove the following results:

tan sin

u ( / ) + + + +

Result . + . = 0. . + . = sin 2 . .

+ +

sin

+

+ +

log

+

.

.

+

.

x sin

.

= tan = − tan

.

=3

+

.

+

.

+2

=0

.

+2

=2

.

+ cot

+ cos

+

.

+

=0

√ +

Problem 12: If (1 −

= )

Problem 13: If

= sin

, Establish the relation − (2 + 1) −( + + cos

Problem 14: If = [ + ( evaluate ( ) .

prove that

) =m

+ 1)] ,prove that (1 +

= 0 and hence evaluate

( )

{1 + (−1) sin 2mx} )

+

−

= 0 and hence

2

Tutorial-02, B.Tech Semester-I (Expansions of functions of several variables and Curve Tracing) Problem 1: Find the equation of the tangent plane and the normal to the surface = 4(1 + + ) at (2,2,6) Ans: 4 + 4 − 3 = −2 ; =

=

Problem 2: Expand e cos near the point (1, ) by Taylor’s theorem. Ans:

√

1 + ( − 1) −

−

+

(

)

− ( − 1)

−

−

−

+ ⋯……

Problem 3: Obtain Taylor’s expansion of tan about (1,1) upto and including the second degree terms. Hence compute (1.1,0.9). Ans: tan = − ( − 1) + ( − 1) + ( − 1) − ( − 1) +… and 0.7862 Problem 4: Expand x in powers of ( − 1) and ( − 1)upto the third degree terms. Ans: x =1+( − 1)+( − 1)( − 1) + ( − 1) ( − 1)+……… Problem 5: Expand e sin in powers of and upto the third degree terms. Ans: Problem 6: Trace the following curves 1. a y = x ; Cubical parabola 3. (2 − ) = Cissoid 5. = ℎ Catenary 7. ( + 4 ) = 8

+

+

= 1 or

=

25. = ( − 1)( − 2)( − 3) 27. 30. 32.

= ; = ), = ( + ), = ( +

= 26.

28. = = (1 + = (1 −

(

) !

+ ⋯………

2. = 4 (2 − ); 4. + =5 6. ( − 1)=( + 1) 8. = (2 − )

9. x + y = a or = = Astroid ), = (1 − 10.. = ( − ) Cycloid 12. = (1 + ) Cardiod 14. = (1 − ) Cardiod ( ) 16. = + , = (1 + ) Cycloid 18. = + , < Limecon )=2 20. (1 − 22. 2 = Hyperbola 24.

+

=

11. = + , > 13. = 3 Three leaves rose 15. = 2 Four leaves rose 17. = (2 − ) 19. = 2 )=2 21.. (1 + 23.y = x Semi- cubical parabola

Hypocycloid cos

; = ) Cycloid ) Cycloid

+

,

=

sin t { Tractrix }

29. = , = − ), = (1 + 31. = ( −

) Cycloid

3

Tutorial-3 B.Tech. Semester-I (Double integrals and their applications) 1. Find the area of the loop of the curve

+

. Also find the area bounded between the

=

curve and its asymptote.

Ans:

and

.

2. Evaluate ∬ (x + y ) dxdy over the area A enclosed by the curves = 4 , + = 3, = 0 and

Ans:

= 2.

3. Evaluate the following double integrals (i)

(ii)

(iv)

( + )

Ans: (i)

−

(v)

.(ii)

4. Show that

(iii)

(

+

−

−

(√ + ) (iv)-1 (v) =

)

5. Show that

(

+

)

=

6. Show that

{

7. Show that Ans: . . =

(iii)

}{

(

)

=

}

≠

, . . =−

(

{ )

}{

}

, also find the values of two integrals.

Give a conclusion on the basis of the results in Q. 4 to 7.

8. Evaluate ∬ (

)

over the region x + y ≤ 1.

9. Evaluate ∬ (

)

over the region A bounded by the curves Ans:

= .

Ans:

10. Evaluate ∬ r dθdr over the area of the circle =

= , and

+

Ans:

11. Find by double integration the area lying inside the circle = parabola (1 +

= ,

and outside the

)=

Ans:

(

)

12. Change the order of the following double integrations: (i)

Ans:

(ii) (iii) (iv)

Ans:

( , ) ( , ) ( , )

Ans:

( , ) Ans:

( , ) +

( , )

( , ) +

( , ) 4

(v)

Ans:

13. Express as single integral and evaluate:

√

Ans:

√

;

+ + √

√

14. Convert into polar co-ordinates 15. Using transformation + = , =

Ans: show that

16. Using transformation − = , + =

=

show that∬

=

Where R is

the region bounded by = , = , + = . 17. Find the whole area of the curve = ( − ) by double integration. Ans: πa 18. Find the area enclosed by the curve = + by double integration. Ans: 11π 19. Find the volume of the torus generated by revolving the circle + = about the line = 3. Ans: 24π 20. Find the Center of gravity of the area bounded by the parabola = and the line + = . Ans: ( , − )

21. Find the Center of gravity of the loop of the curve = 22. Find the Center of gravity of an arc of the curve = ( + = ( −

) in the positive quadrant.

. Ans:(

√

, 0)

), Ans: [ ( − ,

]

5

6

Tutorial-05, B. Tech Sem-I Jacobians 1. If

=

+

+ + ,

=

+

− − ,

0 and hence find a relation between , ,

=

−

, =

and . Ans:

+

=

−

−

( , . , )

show that

( , , ,)

=

+2

2. Prove that the following functions are not independent. Find the relation between them =

(i)

+

+ ,

=

(ii)

=

+

+

+

+

,

=

, =

+

+ ,

+ =

+

−3

+

+

. Ans: u = 3uv + w

. Ans: v = u + 2w ( , )

, , are connected by a functional relation ( , , ) = 0, show that

3. If

4. If λ, μ, ν are the roots of the equation in , (

( , , ) ( , , )

=−(

)(

)(

)

)(

)(

)

= ( −

5. If that

( , . ) ( , , )

=( −

=

6. If

) ,

= ( −

,

,

= sin

( , , ) ( , . )

(1 − +

9. Prove that J =1 10. If = , =

+

11. If =f(

=

.

= , prove that

= ( −

) , where

=

+

+

show

+

,, … . . ,

), =

+

, show that =

being given =

…… =

( , , )

( −

find J( ,

……

, Ans:

being given

( , . )

cos ϕ,

=−

(

)(

)(

(

),

,.,

)(

=

) )

( −

.

) ,…..,

)

}. = sin √1 −

,

) where m + n = 1.

+

{Hint:

,

+ ⋯..+

8. Find the Jacobian

) ,

=

( −

…… + +

.

) .

7. Find the Jacobian of

{Hint:

+

=

.

, = + + ,

=

+

( , )

(

, Ans:

− ,

(0) = 0 and f (x) =

=

+

)

+ , Find

}.

( , , ) ( , . )

Ans:

(

)

, prove without using the method of integration, that

). {Hint: Let = ( ) +

( ) and

=

( )+

the ind J(u, v)}

7

( )

Tutorial-6, B. Tech. Sem-I, (Triple integrals and their application) Problem 1: Evaluate the triple integrals Problem 2: Evaluate∭( +

, ≥ , +

+

Ans:

+ )

over the region bounded by

≤ .

≥ ,

≥

Ans:

Problem 3: Evaluate the following triple integrals

(

(i)

+

)

+

(ii)

(iii) √

(iv)

(

(v) (

Ans: (i)

)

.(ii)

≤ , 0≤,

determined by 0≤

≤

0≤

( )−

(iii)

Problem 4: Evaluate the triple integral ∭ ( −

+ ) ≤

(iv)8π (v) , where R is the region

+ .

Problem 5: Find the volume of the region bounded by the surface planes

+ )

Ans: =

= , = .

,

=

and the

Ans:

Problem 6: Find the volume of the tetrahedron bounded by the co-ordinate planes and the plane

(

+ + = . { Hint:

)

(

)

}

Ans:

Problem 7: Find the volume common to the cylinder x +y = a and x +z = a . (

{ Hint:

) (

( )

) (

}

)

Ans:

Problem 8: Find the mass of the tetrahedron bounded by the co-ordinate planes and the plane

+ + = { Hint:

the variable density (

)

(

=

)

}

Ans:

( √ − )

Problem 9: Find the moment of inertia of the solid about its major axes generated by revolving the ellipse

=

about minor axes.

8

1 0 1. Find rank of 3 1 0 5

Tutorial-07, B.Tech. Sem-I (Matrices)

0 0 Ans: 3. 2

2. Find the rank of the matrices (i) A=

1 2 3 1

3 2

1 1 (ii) B= 1 2 0 −1

(ii) ( ) = 2. 3. Each entry of a matrix is unity. Show that its rank is one. 4 2 3 4 6 4. Determine the rank of the matrix (i) = 8 −2 −1 −3/2 1 1 1 1 5 4 5. If A= 0 3 2 , B= 2 2 2 , Find ( ), ( ), ( + 3 3 3 2 3 10 Ans: ( ) = 2, ( ) = 1, ( ) = 1and , ( ) = 1 ) ≤ ( Hint: ( ) 1 1 −1 1 6. Find the rank of the matrix = 1 −1 2 −1 Ans: 3 1 0 1 1 2 3 2 7. Find the rank of the matrix 2 3 5 1 Ans: = 1 3 4 5 3 −2 0 −1 −7 0 2 2 1 −5 8. Find the rank of the matrix Ans: 1 −2 −3 −2 1 0 1 2 1 −6 1 4 9. Determine the value of b if the rank of 3, where =

2 3 −1

Ans: (i)

( )=1

Ans: ), (

) and

so that

Also find ( ) . Ans: 2 12. Find the rank of the followings matrices:

).

= . −1 0 −3 1 2 2 3

{Hint: After using Row transformations ↔ 1 1 −1 0 ~ 0 0 1 1 R − 3R 0 0 +6 0 −2 0 0 −2 ∶ (1) If b=2, Then | | = 0 , ( ) = 3. (2) If = −6 Then, number of non-zero rows is 3. Therefore 1 2 −1 4 2 4 3 4 10. Reduce the matrix A to its Normal form when A= 1 2 3 4 −1 −2 6 −7 rank of the matrix? Ans: 3. ,

(

( )=

1 4 2 9 9

11. Find non-singular matrices

( ) = 2.

is a normal form where

( ) = 3. Hence, find the

1 1 = 1 −1 3 1

1 −1 1

9

1 1 2 3

2 3 4 7

1 2 3 4 2 4 1 1

2 3 0 2 2 2 (i) . 3 (ii) −6 42 24 54 .2 4 21 −21 0 15 6 −2 0 6 2 −2 0 6 2 0 2 4 2 0 2 (iii) .3 (iv) .3 −1 0 3 1 −1 0 3 −2 1 2 1 −2 1 2 0 4 −12 8 9 1 2 −5 (v) 0 2 −6 2 5 : (vi) −4 1 −6 . 0 1 −3 6 4 6 3 −4 0 −8 24 3 1 1 2 3 4 1 1 2 (vii) : (ix) .3 3 4 1 2 1 2 2 4 3 1 2 2 2 3 3 2 5 7 12 (x) 1 1 2 3 5 .2 3 3 6 9 15 −1 0 1 13. The rank of the diagonal matrix is (a)2 , (b)1 , (c)4, (d*)3 . 0 4 0 14. If A is a non-zero column vector ( × 1) then rank of (A)2, (B)n, (C)0, (d*)1 . −1 0 15.The rank of the matrix 0 −1 is 2, for is equal to (a) Arbitrary (b*) 1, (c) 2, −1 0 (d)3. 2 0 0 1 2 3 16. If A= 0 2 0 and B= 0 1 3 , the value of |AB| is (a) 4, (b)8,(c*)16 ,(d)32 0 0 0 0 0 2 x+a b c d a x+b c d 17. The determinate of is (a*) x (x + a + b + c + d), a b x+c d a b c x+d (b)(x+a)(x+b)(x+c)(x+d),(c)x + ax + bx + cx + d (d) None of the above. x x 1+x 18. If y y 1 + y =0, then =(a*) -1, (b)1, (c) 0, (d) 3 z z 1+z 19. If the system of equations + 4 + = 0, + 3 + = 0, + 2 + = 0 has a non-trivial solution, then , , , are in (a) A.P, (b)G.P, (c*) H.P., (d) None of these. 1 log y log z 1 log z is (a) 1, (b*)0,(c)-1 ,(D) 2 20. If , , ∈ ℝ , then log x log x log y 1 a −x a a a a a b b − x b b b b 21. If =0, for = −1,2,3 then the value of (a) -6, c c c −x c c c (b*)6, (c) -4, (d) 4 10

1 2 3

22. If A= 1 3 4 then |AdjA| is (a) -4, (b*)4,(c)-2 ,(d)2 1 4 3

23. If A be a square matrix such that A,A , A are non-zero matrices but A is a zero matrix.Then (I − A)

is (a) A+ A + A (b)I+A+ A (c) I+ A + A , (D*)I+A+ A + A 1 2 3 24. The rank of the matrix 2 4 7 is (a) 1, (b*)2, (c)3 ,(d) None of these. 3 6 10 25. If ρ(A × )=2 and ρ(B × )=3 then ρ(AB)= (a) 5, (b)3,(c*)2 ,(d) 1 26. If the system of equations 3 − + = 1, 2 + + = 2, + 2 − = −1 has a unique solution if λ= (a) any value, (b) , (c*) ≠ , (d) ≠

27. If A and B are square matrices of the same order, which of the following is true

(a)( + ) = +2 + ,( )( + )( − ) = − ,( ) ( − ) ( + ) = − ,( ∗) ( + )( − ) + ( − ) ( + ) = 2 − 2 28. Prove that the eigen values of a triangular matrix are just diagonal elements of the matrix. 29. Prove that all the eigen values of a Hermitian matrix are real. 30. Find all the eigen values and eigenvectors of the matrices 3 1 4 1 1 3 (i) A= 0 2 6 . Ans: 2,3,5 ; −1 , 0 , 2 (UPTU SE 2002,04) 0 0 5 0 0 1 c −c −2 2 −3 d −2c (ii) B= 2 Ans: -3,-3,5 ; , , (Utt.TU 2006) 1 −6 c −1 −2 0 1 2 2 (iii)C= 0 2 1 . Ans: (UPTU 2006) −1 2 2 31. Verify Cayley-Hamilton theorem for the following matrices and hence find A 4 3 1 5 −1 −7 (i) A= 2 1 −2 A = −4 3 10 (UPTU 2001) 1 2 1 3 −5 −2 2 −1 1 3 1 −1 (ii) A= −1 2 −1 A = 1 3 1 (UPTU 2003,04,05) 1 −1 2 −1 1 3 1 2 −2 32. A square matrix is defined by A= 1 2 1 , Is it diagonalizable ? If yes, reduce it −1 −1 0 into diagonal form. Ans: Diag(1, −1,3) cos α sin α cos n α sin nα 33. If A= , show that A = , where n∈ℤ − sin α cos α − sin nα cos nα 1 0 0 0 1 1 3 1 −1 0 0 34. Prove (i) 5 is an 2 6 is a nilpotent matrix of index 3; (ii). 1 −2 1 0 −2 −1 −3 1 −3 3 −1 cos sin 1 1+ involutory matrix (iii) is an orthogonal matrix.; (iv) is √ − sin cos 1− −1 a unitary matrix. 35. Prove that if s an eigen value of a matrix A, then k , ± , λ , , of kA, A± , , , Adj A respectively.

| |

are the eigen values

11

36. If A has all entries −1 then rank of A is (a) 7; (b*) 1; (c) 5; (d) 0. 37. The rank of a × matrix where < , can not be more than (a*) m; (b) n; (c) mn; (d) None. 1 0 −1 38. If R= 2 1 −1 then top row of R is (a) [5 6 4]; (b*) [5 −3 1]; (c) 2 3 2 [2 0 −1]; (d) [2 −1 1/2] 1 0 1 0 1 1 2 2 −2 5 0 2 39. If A= 0 3 0 then A is (a) 0 0 0 ; (b) −1 −1 1 ; (c) −2 2 −2 ; 0 1 1 1 0 1 10 2 2 2 0 1 1/2 1/2 −1/2 (d*) −1/2 1/2 −1/2 0 0 1 40. Check the consistency for the following linear systems. If system is consistence find the +

+2 + = 5 2 +3 − −2 =2 4x1+5 +3 = 7,

2 + 6 + 11 = 0 6 + 20 − 6 = −3 6 − 18 + 1 = 0

Ans: No solution

Ans: No solution

+ + =6 2 +3 −2 = 2 5 + +2 =0 Ans: = , = , =

2 +5 +3 = 1 − +2 + =2 + + =0 Ans: = − , = ; =

+2 − = 3 3 − +2 =1 2 −2 +3 = 2 − + = −1 Ans:

=− ,

= , =

7 2

+ 2 + 3 = 16 + 11 + 5 = 25 + 3 + 4 = 13 Ans: = , = ,

=

41. For what values of and so that the equations + + = 6; + 2 + 3 = 10; +2 + = have (i) No solution; (ii) a unique solution and (iii) an infinite number of solutions ANS: (i) λ = 3 no solution; (ii)λ ≠ 3, μ ≠ 10 , unique solution; (iii)λ = 3, μ = 10 , In inite many solutions. 42. Determine b such that the system 2 + + 2 = 0; + + 3 = 0; 4 + 3 + =0 has (i) trivial solution(ii) non- trivial solution. 43. Find the values of a , b and c for which the system has (i) No solution (ii) unique

solution (iii) Infinitely many solutions for −2 + + = ; −2 + = ; + −2 = Ans: ( ) + + ≠ 0 ( ) + + = 0,Infinite solutions

+ + =3 +2 +2 = +5 +3 =9 Ans: (i) = −1, (ii) ≠ −1, (iii) = −1, = 6 44. Solve the following system of homogeneous equations + 3 − 2 = 0; 2 − + 4 = 0; − 11 + 14 = 0 Ans: = , = = ( ).

,

≠6 ,

solutions.

4 + 2 + + 3 = 0; 6 + 3 + 4 + 7 = 0; 2 + + =0 Ans: = , = , = −2 − , = −

12

Tutorial-8, B.Tech Sem-I (Vector Calculus) Note: In this exercise bold face letters (say) F represents vector f and i,j,k represents unit vectors ı̂,ȷ̂,k respectively. Problem1: Evaluate the following (i) ∇∙(r ) (ii) ∇∙(r ∇( )) (iii) ∇ (∇∙(r ))(iv) grad Div( ) Ans: (i)6r , (ii)3r , (iii)2r , (iv) − Problem2:If A= 2yz i − x y j + xz k , B= x i + yz j − xy k and φ=2x yz , then find (i) ( ∙ ) ; Ans: (ii) ∙ ; (iii) ( ∙ ) ; (iv) ( × ∇) ; (v) ×( ) Problem3: If A and B are differentiable vector functions, ϕ and φ are differentiable scalar functions of position ( , , ), then prove the following results: (i) ∙ ( + ) = ∙ + ∙ (ii) ∙( )= ( ∙ ) + ∙ × (iii) × ( )= ( × ) + × (iv) ∙( × ) = ∙ ( × )− ∙( × ) ( ∙ ) =( ∙ ) +( ∙ ) + ×( × )+ ×( × ) (v) (vi) × ( ∇ ) =0 Problem4: Evaluate the grad. of log| | Ans: Problem5: Show that ∇ is a vector perpendicular to the surface ( , , )= constt. Problem6: Find the directional derivative of ( , , )=x yz + 4xz at (1, −2, −1) in the direction 2 − − 2 . Ans:37/3 Problem7:Prove that the vector A= 3y z i + 4x z j − 3x y k is solenoidal. Problem8: Prove following identities (i) Div(∇φ×∇ψ)=0 (ii) If A and B are irrotational then A×B is solenoidal. (iii) ∇×(φ ∇Φ)=0 (iv) Div ( f ∇g) = f ∇ g+∇f ∙ ∇g ( ∙ )( ∙ ) ∙ (v) b∙∇(a∙∇ ) = − where a and b are constt. Vectors. (vi) ∙( − ∇ )=U∇ − ∇ (vii) = Problem9: (a) Prove that = ( + ) + (2 − 4) + (3 + 2) is a conservative force field. (b)Find the scalar potential for F. Ans: ( + −4 +2 + ) (c)Find the work done in moving an object in this field from (0,1, −1) to ( , −1,2). Ans: 15+4π Problem10: Show that V=2xyz + ( +2 ) + is irrotational. Express V as gradient of a scalar function φ. ( , ) Problem11: Evaluate ( , ) (10 x − 2xy )dx − 3x y dy along the path x − 6xy = 4y Ans: 60. {Hint: Use Exact differential} Problem12: Use Green’s theorem to evaluate ∮ ( − )dx + sinx cosy dy ,where 13

C≡

+

= .

Ans: 0 .

Problem13: Verify Green’s theorem in the plane for ∮ (xy + y ) dx + x dy ,where C is the closed curve of the region bdd. by the line = and curve = x . Problem14: Find the work done in moving a particle once around a circle C in the xy-plane, if circle has center at the origin and radius 3, Force field is given by = (2 − + ) + ( + − ) + (3 − 2 + 4 ) Ans: 18π Problem15: State and prove Green’s theorem. Problem16: Prove that the area bounded by a simple closed curve C is given by ∮ xdy − ydx{Hint: Use Green’s theorem } Problem16: Find the constants a,b,c such that V= ( + 2 + ) + ( − 3 − ) + (4 + + 2 ) is irrotational. Express V as gradient of a scalar function φ. Problem18: State Green’s theorem and hence evaluate ∮ (cosy)dx + (x − xsiny) dy ,where C is the closed curve + = . Problem19: Use Green’s theorem to evaluate ∮ ( + )dx + ( x + y ) dy ,where C is the square formed by the lines = ±1, = ±1. Problem20: Use Stoke’s theorem to evaluate ∮ ( + 2 )dx + (x − z) dy + (y − z)dz ,where C is the boundary of the ∆ with the vertices (2,0,0),(0,3,0),(0,0,6) oriented in the anticlockwise direction. Problem21: Verify Stoke’s theorem for F= x y k − y j + xzi and S is the surface of the region bounded by = 0, = 0, = 0,2 + + 2 = 8, which is not included in the xzplane . Problem22: Evaluate ∬ (∇ × ) ∙ n ds ,where F= y i + (x − 2xz) j − xy k and S is the surface of the sphere + + = above the xy-plane. Ans: Zero. Problem23: Evaluate ∬ ∙ n ds ,where F= 4xz i − y j + yz k and S is the surface of the cube bounded by = 0, = 1, = 0, = 1, = 0, = 1. Ans: . Problem24: If F= (x + y − 4)i + 3xy j + (2xz + z ) k ,evaluate ∬ (∇ × ) ∙ n ds , and S is the surface of the sphere + + = above the xy-plane. Ans: −16 Problem25: Evaluate ∬(y z i + x z j + x y k) ∙ n ds ,where S is the part of the surface of the sphere + + = above xy-plane. Problem26: If V is the volume enclosed by the surface S, Find the value of ∙ n ds Ans: 3V Problem27: State and prove Gauss Divergence theorm. Problem28: Evaluate ∬(ax i + by j + cz k) ∙ n ds ,where S is the surface of the sphere + + = . [P.U. 2004] Problem29: Evaluate ∬ ∙ n ds ,where A= 18z i − 12 j + 3y k and S is the part of the plane 2 + 3 + 6 = 12 which is located in the first octant. Ans: 24

14

Problem30: Evaluate ∬ cylinder + Ans: 90

∙ n ds ,where A= z i + x j − 3y z k and S is the surface of the = included in the first octant between = 0 and = 5.

Problem 31: Prove that (i) ∬

∙

ds=∭

;

(ii) ∬

∙

ds=∭ 5r

dv

(iii) ∬ (∇ × ) ∙ n ds =0 for any closed surface S. Problem 32: Use divergence theorem to evaluate ∬ ∙ ds ,where A= 4x i − 2y j + z k and S is the surface of the cylinder + = bdd. between = 0 and = 3. Problem 33: Verify Stoke’s theorem for F= (x + y ) i − 2xy j taken around rectangle bdd. by the lines = ± ; = 0, .

15

Tutorial–1, B. Tech. Sem III, 24 July, 2016 (Root Findings and Linear System of Equations)

1. Find the root of the equation ex = 3x lying in [0,1] correct to three decimal places using Bisection, Regula-Falsi and Newton-Raphson methods. 2. Use the interval halving method to improve a root of the equation x4 + 2x3 − x − 1 = 0 lying in [0,1] correct to three decimal places.

ON

I

3. For smallest positive root of the equation: x3 − 5x + 1 = 0 correct to 3 D places. Use the methods Bisection, Regula-Falsi and Newton-Raphson methods. 4. Find positive root of the equation: tan x + tanh x = 0 by using Bisection and Regula-Falsi methods correct to 4 D places. 5. Solve the following system of equations up to 2D places by Gauss– Seidel i . 20x + y − 2z = 17; 3x + 20y − z = −18; 2x − 3y + 20z = 25 with Initial root (0,0,0); Ans: x = 1, y = −1, z = 1

R. K. S

ii . 10x + 2y + z = 9; x + 10y − z = −22; −2x + 3y + 10z = 2 with Initial root (0,0,0). Ans: x = 1, y = −2, z = 3 iii . 11x1 − 7x2 + x3 = 32; x1 + 5x2 − 2x3 = 18; −2x1 + 2x2 + 7x3 = 19; upto 1D places with Initial root (5,5,5) iv . x − 2y + z = 8; x + y + 2z = 9; 3x − y + z = 6 with Initial root (3,3,2); Ans: .......

6. Verify that each of the following equations has a root on the interval (0,1). Next, perform the bisection method to determine p3 , the third approximation to the location of the root (a) ln(1 + x) − cos x = 0

(b) x5 + 2x − 1 = 0 (c) e−x − x = 0

(d) cos x − x = 0

7. It was noted that the function f (x) = x3 + 2x2 − 3x − 1 has a zero on the interval (−3, −2) and another on the interval (-1,0). Approximate both of these zeroes to within an absolute tolerance of 5 × 10−5 . √ 8. Approximate 3 13 to three decimal places by applying the bisection method to the equation x3 − 13 = 0. 9. Approximate 1/37 to five decimal places by applying the bisection method to the equation 1/x − 37 = 0. 10. Consider the function g(x) = cos x. (a) Graphically verify that this function has a unique fixed point on the real line. (b) Can we prove that the fixed point is unique using the theorems of this section? Why or why not? 1

11. Consider the function g(x) = 1 + x − 18 x3 . (a) Analytically verify that this function has a unique fixed point on the real line. (b) Can we prove that the fixed point is unique using the theorems of this section? Why or why not?

I

12. Each of the following equations has a root on the interval (0,1). Perform Newton’s method to determine p4 , the fourth approximation to the location of the root.

(b) x5 + 2x − 1 = 0 (c) e−x − x = 0 (d) cos x − x = 0

ON

(a) ln(1 + x) − cos x = 0

13. The equation x3 + x2 − 3x − 3 = 0 has a root on the interval (1,2), namely x = √ 14. The equation x7 = 3 has a root on the interval (1,2), namely x = 7 3.

√

3.

R. K. S

15. The equation 1/x − 37 = 0 has a zero on the interval (0.01,0.1) namely x = 1/37. 16. Show that when Newton’s method is applied to the equation x2 − a = 0, the resulting iteration function is g(x) = 12 (x + xa ). 17. Show that when Newton’s method is applied to the equation 1/x − a = 0, the resulting iteration function is g(x) = x(2 − ax). 18. For each of the functions given below, use Newton’s method to approximate all real roots. Use an absolute tolerance of 10−6 as a stopping condition. (a) f (x) = ex + x2 − x − 4

(b) f (x) = x3 − x2 − 10x + 7

(c) f (x) = 1.05 − 1.04x + ln x

19. Each of the following equations has a root on the interval (0,1). Perform the secant method to determine p4 , the fourth approximation to the location of the root. (a) ln(1 + x) − cos x = 0

(b) x5 + 2x − 1 = 0 (c) e−x − x = 0

(d) cos x − x = 0 20. For the following i-iii (a) Using scaled partial pivoting during the factor step, find matrices L, U and P such that LU = P A. (b) Solve the system Ax = b for each of the given right-hand-side vectors.

2

    1 2 3 4 10 −4  −1 1   5   −5 2 3     (i) A =   1 −1 1 2  , b1 =  3  , b2 =  −3 −1 1 −1 5 4 −4      1 0 2 0 3 −1  −1 4 3 6   12   −6     (ii) A =   0 −2 5 −3  , b1 =  0  , b2 =  −4 3 1 1 0 5 3       2 7 5 14 −4 (iii) A =  6 20 10  , b1 =  36  , b2 =  −16  , 4 3 0 7 −7





 −2     , b3 =  −3    1  −8    3     , b3 =  −8    10  2   −3 b3 =  −12  6

ON

I



21. In Exercises 1-3, use the Gauss-Seidel method to solve the indicated linear system of equations. Take x(0) = 0, and terminate iteration when ||x(k+1) − x(0) ||∞ falls below 5 × 10−6 .

1.

2.

4x1 − x2 = 2 −x1 + 4x2 − x3 = 4 − x2 + 4x3 = 10

3.

7x1 + 3x2 + −3x1 + 9x2 + x3 x2 + 3x3 − x4 −x3 + 10x4 − 4x5 −4x4 + 6x5

R. K. S

+ x3 + x4 + 2x3 + 3x4 − 5x3 + 2x3 + 4x4

= −5 = 23 = 9 = 4

4x1 + x2 x1 + 8x2 x1 + 2x2 x1

= 4 = −6 = 3 = 7 = 2

22. Solve the following system of linear equations by Triangularization /Factorization or Crout’s method: (a)

2x + 3y + z = 9; 3x + y + 2z = 8; x + 2y + 3z = 6; Ans: ....

(b)

x + y + z = 3; 2x − y + 3z = 16; 3x + y − z = −3; Ans: x = 1, y = −2, z = 4

(c) x+2y+3z+4w = 20; 3x−2y+8z+4w = 26; 2x+y−4z+7w = 10, 4x+2y−8z−4w = 2; Ans: x = 4, y = 3, z = 2, w = 1 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗∗

3

1. Prove the following operator relations: 2 (i). E = 1 + ∆ = (1 − 5)−1 , (ii) ∆ = µδ + δ2 , (x) (iii). ∆ log f (x) = log(1 + ∆f ) f (x) (iv). ∆[f (x)g(x)] = f (x)∆g(x) + g(x + h)∆f (x), (v). 45 = 54 =q4 − 5 = δ 2 , 2

I

Tutorial–2, B. Tech. Sem III, 26 July, 2016 ( Difference Operators and Interpolation)

ON

(vi). ∆ = 21 δ 2 + δ 1 + δ4 , (vii). δ 3 y 1 = y2 − 3y1 + 3y0 − y−1 , 2 (viii). δ(fk gk ) = µfk δ(gk ) + µgk δ(fk ), = √2−5 , (x). ∆ = ehD − 1, (xi). µδ = sinh(hD), (ix). √2+∆ 1−5 1+∆ (xii). e−hD = 1 − 5, 1 (xiii). f [x, y, z] = x + y + z, f = x3 , (xiv). f [a, b, c] = abc , f = a1 , n (xv). f [a, b, c, ...(n symbols)...l] = 4n!hf (a) n

4n ( x1 ), (v).

R. K. S

2. Find for (h = 1) 2 3 (i). 42 (abex ), (ii). 4Exx3 , (iii). 4 tan−1 (ax), (iv). (vi). 4(x + cosx).  2   x E e 3. Prove that: ex = 4E ex . 4 and 2 ex f0 + xf1 +

x2 f 2! 2

+ ... = ex (f0 + x 4 f0 +

x2 2!



42 E



x3 ,

4 f0 + ...)

4. Let Pn (x) = (x − x0 )(x − x1 )(x − x2 ).......(x − xn−1 ), where xi = x0 + ih, i being integer; n! hr Pn−r , r = 1 (1) n − 1, and show that 4Pn = nhPn−1 , hence show that 4r Pn = (n−r)! n n 4 Pn = n!h . 5. Find the third divided difference with argument 2, 4, 9, 10 of the function f (x) = x3 − 2x. 6. Form a divided difference table for f (x) = x4 + 6x2 + x − 2 for values of x = −3 (1) 3, show that 5th order diferences are zero. 7. Given that f (0) = 8, f (1) = 68, and f (5) = 123 determine f (2); calculate the error also. 8. Tabulate sin x for x0 = 30 (2) 40 and interpolate sin 310 and sin 330 . Compare with exact values. 9. Tabulate ex for x = 1.7 (0.1) 2.2 and interpolate at x = 1.71, 2.15. 10. Find log10 1152.5 and log10 1161.3 using the following data: x 1150 1155 1160 1165 1170 1175 log10 (x) 3.06069 3.06258 3.06445 3.06632 3.06818 3.07003 and error in the result. 11. Use Lagrange’s formula to interpolate the values of f (5) from x 1 2 3 4 7 f (x) 2 4 8 16 128 How much it deviates from 25 .?

4

1180 3.07188

12. A third degree polynomial passes through the points (0, −1), (1, 1), (2, 1), (3, −2). Find the polynomial. Ans: − 16 x3 − 12 x2 + 83 x − 1

I

13. From the following data, find the number of students who obtained less than 45 marks: Marks 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 No. of students 31 42 51 35 31 Ans: approx. 48

ON

14. Find the form of f (x), given that f (0) = 8, f (1) = 11, f (4) = 68 and f (5) = 123 also determine f (2). Ans: x3 − x2 + 3x + 8, 18 15. Given that f (0) = −18, f (1) = 0 = f (3) = f (6), f (5) = −248 and f (9) = 13104 find the form of f (x) assuming it to be a polynomial of degree 5th. {Hint: f (x) = (x − 1)(x − 3)(x − 6)φ(x) , φ(x) is a polynomial of degree 2 with φ(0) = 1, φ(5) = 31, φ(9) = 91 } Ans: x5 − 9x4 + 18x3 − x2 + 9x − 18

R. K. S

16. Find log10 301 using Newton’s Divided difference interpolation formula from the data: x 300 304 305 307 Ans: 2.4786 log10 (x) 2.4771 2.4829 2.4843 2.4871 √ 17. The values of y = x are listed below: x 4 6 7 10 y 2 2.449 2.646 3.162 Compute x corresponding to y = 2.5. Ans: 6.25148 18. What should be the minimum number of tabular points required for the piecewise linear interpolation for f (x) = cos(x) on [0, π], such that error does not exceed by 1/2 × 10−6 ? Ans: The number of subdivisions required n = 1571. ∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗

5

Tutorial–3, B. Tech. Sem III, 22 August, 2016 (Numerical Differentiation, Integration and Differential Equations)

I

1. Find the gradient of the road at the initial point of the elevation above a datum line of seven points of road which are given below: x: 0 300 600 900 1200 1500 1800 y : 135 149 157 183 201 205 193

ON

2. Find the first three derivatives of the function at x = 1.5 from the data x: 1.5 2.0 2.5 3.0 3.5 4.0 y : 3.375 7.0 13.625 24.0 38.875 59.0 Ans: 4.75, 9.0, 6.0

3. The table given below reveals the velocity v of a body during the time ’t’ specified. Find its acceleration at t = 1.1 t : 1.0 1.1 1.2 1.3 1.4 v : 43.1 47.7 52.1 56.4 60.8 Ans: 44.917

R. K. S

4. Derive Newton Cote’s quadrature formula. Hence deduce (i) Trapezoidal rule, One third Simpson’s and 3/8 Simpson’s rule of Numerical integrations. (ii) Calculate Truncation error as well as Max. Global error in the rules. R1 1 5. Evaluate 0 1+x 2 dx, with h = 0.2 (up to 3D) by using Trapezoidal rule of Numerical integration. Hence find an approximate value of π. Give your decision about the statement ”Can we apply One third Simpson’s and 3/8 Simpson’s rule of Numerical integrations for this problem” Ans: 3.135, No R1 6. Evaluate 0 ex dx by One third Simpson’s correct to 5 D places with proper choice of h. Ans: 1.71828 R6 1 7. Evaluate 0 1+x 2 dx, up to 3D by using Trapezoidal rule and Simpson’s rules of Numerical integrations. Also check your results by actual integration. Ans: 1.411, 1.366, 1.357, Actual value is 1.406 8. Construct the divided difference table for the following data set, and then write out the Newton form of the interpolating polynomial. x -1 0 y 3 -1

1 -3

2 1

9. Construct the divided difference table for the following data set, and then write out the Newton form of the interpolating polynomial. x -7 y 10

-5 5

-4 2

-1 10

10. Write out the Newton form of the interpolating polynomial for f (x) = sin x that passes through the points (0, sin 0), (π/4, sin π/4), and (π/2, sin π/2).

6

11. Apply Picard’s Method with 4-iterations upto 4D places to find the values of y at x = dy 0.1(0.3)0.3, given that dx = y − x, y(0) = 2. Ans: 2.21, 2.42, 2.65 12. Apply Picard’s Method with 3-iterations to find the values of y at x = 0.1, 0.2, given that dy = y 2 + x2 , y(0) = 0. Ans: 0.00033, 0.00267 dx

I

13. By using Picard’s Method with 5-iterations, find y at x = 0.1, 0.2 upto 4D places from dy = x + x2 y, y(0) = 1. Ans: 1.0053, 1.0227 the differential equation dx

ON

14. Write out the Newton form of the interpolating polynomial for f (x) = ex that passes through the points (−1, e−1 ), (0, e0 ), and (1, e1 ). 15. Use Runge–Kutta method of fourth order to find a numerical solution at x = 0.5 for dy = 12 (x − y), y(0) = 1 taking h = 0.25. dx dy 16. Solve the differential Equation dx = x+y +xy, y(0) = 1 by using Taylor’s series expansion to get y at x = 0.1(0.1)0.5 (use terms up to x5 in the expansion ) by shifting the origin for each value.

R. K. S

17. Evaluate y upto 4D places for x = 0.1(0.1)0.5 by using Taylor’s series method for x + y 2 , y(0) = 0 .

dy dx

=

dy = x + y 2 , y(0) = 0 for x = 0.1(0.1)0.5 by using modified Euler’s method correct 18. Solve dx to 4D places. dy 19. Solve dx = x + y + xy, y(0) = 1 by using modified Euler’s method correct to 4D places to obtain y at x = 0.1(0.1)0.5

20. Using Runge–Kutta fourth order method compute y at x = 0.2(0.2)0.6 for y 2 , y(0) = 0.9 correct to 3D places.

dy dx

= 1+

dy 21. Using fourth order Runge–Kutta method solve dx = −xy, y(0) = 1 in the interval [0,0.6] by taking h = 0.2 and compare the result with the values obtained from the exact solution.

22. Using Taylor’s series method up to 5th terms to get y(1.1), y(1.2) for Ans 2.6384, 3.7080

dy dx

= x2 +y 2 , y(1) = 2.

dy = x−y 2 , y(0) = 23. Find y at x = 0.2 by using Euler’s method correct to 3D places to solve dx 1 taking h = 0.1. What is the value if modified Euler’s method is applied under same conditions. Ans 0.8, 0.858

∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗

7

24. Review: Numerical Integration Formulae: are based on Polynomial Interpolation x f (x)

x0 f (x0 )

x1 f (x1 )

x2 f (x3 )

... ...

xn f (xn )

I

Types of Newton-Cotes Formulae:

ON

1. Trapezoidal Rule (Two point formula) Z b h f (x)dx = [f (x0 ) + 2 ∗ (f (x1 ) + f (x2 ) · · · f (xn−1 )) + f (xn )] 2 a Error in composite Trapezoidal rule Error in the ith interval [xi−1 , xi ] is E CT (I) = −

h3 00 (xi − xi−1 )3 00 f (ξi ) = − f (ξi ) 12 12

R. K. S

where ξi ∈ (xi−1 , xi )). Hence, the Max. error in composite rule (Global Error) is n

E

CT

or

=

h3 X 00 = f (ξi ) 12 i=1

(b − a)h2 00 f (ξ), 12

where a < ξ < b, 2. Simpson’s 1/3 Rule (Three Point formula); Then Z b

Z

x2

f (x)dx =

Z

x4

x0

x2i

f (x)dx + · · · +

f (x)dx +

a

Z

x2

Z

x2n

f (x)dx + · · · + x2i−2

f (x)dx x2n−2

h {f (x0 ) + 4 × [f (x1 ) + f (x3 ) + f (x5 ) + · · · + f (x2n−1 )]+ 3 +2 × [f (x2 ) + f (x4 ) + f (x6 ) + · · · + f (x2n−2 )] + f (x2n )}

=

The Error in the Composite Simpson’s

1 3

Rule i.e. error in the ith interval (x2i−2 , x2i ) is

h5 (4) f (ξi ), ξi ∈ (x2i−2 , x2i ) 90 Max. Error in composite Simpson’s 1/3-rule is given by E cs (I) = −

E CS = (b − a) 8

h4 (4) f (ξ), 180

where ξ ∈ [a, b] 3. Composite Simpson’s

3 8

rule or (Four point formula);

The [a, b] is divided into 3n equal subintervals. (h = of the n intervals [x3i−3 , x3i ] for i = 1, 2, 3, · · · , n.

b−a .) 3n

and we apply

3 8

rule on each

Hence, b

Z

x3

Z

f (x)dx '

Z

x3n =b

f (x)dx + · · · +

f (x)dx + x3

x0 =a

a

x6

f (x)dx

I

Z

x3n−3

ON

3h [f0 + 3f1 + 3f2 + 2f3 + 3f4 + 3f5 + 2f6 + 3f7 + · · · + 3f3n−1 + f3n ] 8 Remember: =

f with suffices of multiple 3 are multiplied by 2. Others by 3, except the end points. T. Error Es = 3h4 (4) f (ξ) 80

where ξ ∈ [a, b]

ξi ∈ (x3i−3 , x3i )

R. K. S

G. Error E s =

−3h5 (4) f (ξi ), 80

9

Tutorial–4, B. Tech. Sem III, 5 Sep., 2016 (Introduction of Complex numbers & Analytic functions)

1. Find the locus of z in each of the following relations: (i)|z − 5| = 6, (ii)|z + 2i| ≥ 1, (iii)Re(z + 2) = −1, (iv)|z − i| = |z + i|, (v)|z + 3| + |z + 1| = 4, (vi)1 ≤ |z − 3| ≤ 2, (vii) |z + 3| − |z + 1| = 1.

ON

3. Define an analytic function at a point and in a domain.

I

2. For which complex number following are true, justify in each (i)z = −z, (ii)−z = z −1 , (iii) z = z −1 , (iv)z = z¯

4. Prove that an analytic function of constant modulus is always constant. 5. Prove that Real and imaginary parts of an analytic function are harmonic. 6. Prove that an analytic function is always continuous but converse need not be true. Give an example. 7. State and prove the necessary and sufficient condition for a function f (z) = u + iv to be analytic.

R. K. S

8. Define an analytic function at a point. Illustrate such a function. 9. If f (z) = origin.

(¯ z )2 ,z z

6= 0; f (0) = 0 then f (z) satisfies Cauchy–Riemann equations (CR) at

10. Using Milne–Thomson method construct an analytic function f (z) = u + iv for which 2u + 3v = 13(x2 − y 2 ) + 2x + 3y. 11. State and Prove Cauchy–Riemann equations in polar coordinate system. p 12. Let f (x, y) = (|xy|), then (a) fx , fy do not exist at (0, 0); (b) fx (0, 0) = 1; (c) fy (0, 0) = 0; (d) f is differentiable at (0, 0). Ans: c  2 13. If function f (z) = zz¯ , when z 6= 0, f (z) = 0 for z = 0, Then f (z) (a) satisfies C.R. equations at z = 0; (b) is not continuous at z = 0; (c) is differentiable at z = 0; (d) is analytic at z = 1 Ans: b 2

14. Show that f (z) = (¯zz) , when z 6= 0, f (z) = 0 for z = 0, satisfies C.R. equations at z = 0; but is not differentiable at z = 0. 15. The harmonic conjugate of u = x2 − y 2 + xy is (a) x2 − y 2 − xy; (b) x2 + y 2 − xy; (c) 1/2(−x2 + y 2 ) + 2xy. (d) 2(−x2 + y 2 ) + 1/2 Ans: c 16. f (z) = (|z|)2 is (a) continuous everywhere but nowhere differentiable; (b) continuous at z = 0 but differentiable everywhere; (c) continuous nowhere; (d) none of these. Ans: d 17. Examine the nature of the function f (z) = including the origin.

10

x2 y 5 (x+iy) x4 +y 10

; if z 6= 0 , otherwise 0, in a region

3

3

(1−i) 18. Prove that the function f (z) = u + iv, where f (z) = x (1+i)−y ; if z 6= 0 , otherwise x2 +y 2 0 0 is continuous and that Cauchy–Riemann equations are satisfied at the origin, yet f (z) does not exists there.

2

20. If f (z) = xyx2(x+iy) ; if z 6= 0 and f (0) = 0, prove that +y 4 radius vector but not as z → 0 in any manner.

f (z)−f (0) z

→ 0 as z → 0 along any

f (z)−f (0) z

→ 0 as z → 0 along any

I

3

19. If f (z) = x xy(y−ix) ; if z 6= 0 and f (0) = 0, show that 6 +y 2 radius vector but not as z → 0 in any manner.

ON

21. The function f (z) = z is (a) analytic at z = 0, (b) differentiable only at z = 0;(c) satisfies C.R. equations everywhere; (d) nowhere analytic. Ans: d 22. Derive the C.R. equations for an analytic function f (r, θ) = u(r, θ) + i v(r, θ) and deduce that urr + 1r ur + r12 uθθ = 0. 23. Find the point where the C.R. equations are satisfied for the function f (z) = xy 2 + ix2 y. 0 In which region f (z) exists?

R. K. S

24. f (z) = (|z|)2 is (a) continuous everywhere but nowhere differentiable except at 0; (b) continuous at z = 0 but differentiable everywhere (c) continuous nowhere; (d) none of these. Ans: a 25. Prove that the function f (z) = z|z| is nowhere analytic. 0

26. If f (z) is an analytic function such that Ref (z) = 3x2 − 4y − 3y 2 and f (1 + i) = 0 then f (z) is (a) z 3 + 6 − 2i, (b) z 3 + 2iz 2 + 6 − 2i, (c) z 3 + 2iz 2 − 2i, (d) z 3 + 2z 2 + 6 − 2i Ans: b Hint: ux = 3x2 − 4y − 3y 2 = φ1 (x, y)(say), integrating partially w.r.t. y we get u = x3 − 4xy − 3xy 2 + g(y) Therefore, uy = −4x − 6xy + g 0 (y) or −uy = vx = φ2 (x, y)(say) = 4x + 6xy − g 0 (y) Thus φ1 (z, 0) = 3z 2 , φ2 (z, 0) = 4z + g 0 (0) Now, applying Milne Thomson we get R 2 f (z) = (3z + i 4z)dz + constt. or f (z) = z 3 + 2i z 2 + constt. and applying f (1 + i) = 0 implies Constt = 6 − 2i 27. The orthogonal trajectory of u = ex (x cos y − y sin y) is (a) ex (cos y + x sin y) + c; (b) ex x sin y + c ; (d) ex (y cos y + x sin y) + c. Ans: c 28. Find the locus of points in the plane satisfying the relation |z + 5|2 + |z − 5|2 = 75. Ans: circle 29. The function f (z) = z¯ is (a) analytic at z = 0, (b) continuous at z = 0;(c) differentiable only at z = 0; (d) analytic anywhere. Ans: b 30. If f (z) = u + iv is an analytic function and u − v = (x − y)(x2 + 4xy + y 2 ) then f (z) is (a)−z 3 + c, (b) −iz 2 + ic, (c)−iz 3 + β, (d) z 3 − ic, Ans: c −y

x+sin x−e ; find f (z) if f (z) 31. If f (z) = u + iv, is an analytic function of z and u − v = cos 2 cos x−ey −e−y subject to the condition f (π/2) = 0. Ans: 1/2{1 − cot(z/2)}

11

32. If f (z) = u(r, θ) + iv(r, θ) is an analytic function and u = −r3 sin 3θ then construct the analytic function f (z). 33. If f (z) = u + iv, is analytic function of z and u + v = of z. Ans: 21 (1 + i) cot z + d

2 sin 2x ; −2 cos 2x+e2y −e−2y

find f (z) in terms

i ii iii (a) p r q (b) r p q (c) p q r

B (f (z) = u + iv is an analytic function) (i). sin z + ci (ii).z 3 + 3z + 1 + ci (iii).i(z 3 + c)

Ans: b

x2 cosh2 β

+

y2 sinh2 β

R. K. S

35. if sin(α + iβ) = x + iy prove that (a)

ON

A (u is given) p. x3 − 3xy 2 + 3x + 1 q. y 3 − 3x2 y r. sin x cosh y

I

34. Choose the correct code for matching list A and B.

36. Prove that (

= 1, (b)

x2 cos2 α

−

y2 sin2 α

= 1.

∂2 ∂2 + )|f (z)|2 = 4|f 0 (z)|2 for an analytic function f (z) = u + iv. ∂x2 ∂y 2

37. Find the harmonic conjugate of u = x3 − 3xy 2 + 3x2 − 3y 2 + 2x + 1 and corresponding analytic function f (z) = u + iv. Ans: v = 3x2 y − y 3 + 6xy + 2y + d, f (z) = z 3 + 3z 2 + 2z + 1 + id. 38. Find orthogonal trajectory of v = e2x (x cos 2y −y sin 2y) Ans: −e2x (x sin 2y +y cos 2y)+d ∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗

12

Tutorial–5, B. Tech. Sem III, 14 Sep., 2016 (Complex Integrations) 1. Define simply and multiply connected regions? State and prove Cauchys theorem for an analytic function? Is it true for multiply connected regions?

I

2. State and prove Cauchy integral formula for nth derivative? Is it true for multiply connected regions? If yes, give your explanation with necessary proofs.

ON

3. If f (z) is analytic in a simply connected region D and a, z are two points in D, then Rz f (z)dz is independent of the path in D joining a and z. a

R. K. S

R (2,4) 4. Evaluate (0,3) z¯ dz along (a) parabola x = 2t, y = t2 + 3, (b) a straight line joining (0, 3) and (2, 4). Find whether both values are different, if yes, justify reason why it is so? R 5. Evaluate C z¯ dz from z = 0 to z = 4 + 2i along the curve C given by (i)z = t2 + it; (ii) the line from z = 0 to z = 2i and then the line from z = 2i to z = 4 + 2i. Ans: (i) 10 − 8i3 ; (ii)10 − 8i. H 2 dz, where C is the ellipse (x/2)2 + (y/3)2 = 1. Find the value of 6. If F (a) = z +2z−5 z−a C

F (4.5). Ans: Zero H 1 7. Evaluate z−a dz, where C is any simple closed curve and z = a is (i) outside C; (ii) C

inside C. Ans: (i) 0 ; (ii) 2πi. H 1 + 8. Evaluate (z−a) n dz ; n 6= 1, n ∈ Z , where C is any simple closed curve and z = a is C

inside C. Ans: Zero

9. Suppose f (z) and on a simple closed curve C and z = a is inside C. H is1 analytic H inside 1 Prove that z−a dz = z−a dz where C1 is a circle center at a and totally contained in C

C1

simple closed curve C. H 2 10. Evaluate (¯ z ) dz ; where C is the circle (i)|z − 1| = 1, (ii)|z| = 1 Ans: 4πi, 0. C

11. Find

H

z¯dz around (a) the circle |z − 2| = 3, (b) the ellipse |z − 3| + |z + 3| = 10, (c) the

C

square with vertices 0, 2, 2i and 2 + 2i. Ans: 18πi, 40πi, 8i

12. Suppose f (z) is integrable along a curve C having finite length l and there exists a positive R real number M such that |f (z)| ≤ M on C. Prove that | C f (z)dz| ≤ M l. Remark: This result is helpful to evaluate the upper bound of an integral without evaluating it. 13. Work the following integrals around the contour prescribed against it H out πz 2 +sin πz 2 dz; C: |z| = 3. Ans: 4πi. (i) cos(z−1)(z−2) CH 1 (ii) z2 +16 dz; C: |z| = 6. Ans: 0 C

13

e5z dz; z−2i

(iii)

H

(iv)

C 1 2πi

(v)

H

C: |z − 2| + |z + 2| = 6. Ans: 2πie10i

ez dz; z−2

C: |z| = 3. Ans: e2

C sin πz dz; 2 z −1

H C

• C is the rectangle with vertices 2 + i, 2 − i, −2 + i, −2 − i;

ON

I

• C is the rectangle with vertices −i, 2 − i, 2 + i, i. Ans: 0, 0. H eπiz (vi) z2 −4z+5 dz; C: |z − 1 − 2i| = 2. Ans: πeiπ CH ez +cos z dz; where C is the boundary of a triangle with vertices: −1, 1, −7/2i. (vii) (z−5)(z+5i) C

Ans: 2πi(πi−5) 2 +25 (2 cos πi − sin πi) H πsin 6z (viii) z−π/6 dz; C: |z| = 2. Ans: πi2−5 C 2 R ez (ix) Show that C z2 (z−1−i) dz = πe2i ; where C consists of |z| = 2 anticlockwise and |z| = 1 clockwise.H 1 ez (x) 2πi dz; Ans: e−2 (z+2)2

R. K. S

C:|z|=4

14. Evaluate

H

C

1 dz z 2 −1

around z = −i + 5eit . Ans: 0

H 15. Evaluate (z 2 + 3 + 4/z)dz, C: |z| = 4. Ans: 8πi C

16. Evaluate the upper bound of the integral without evaluating it ? √ R (a) C (z 4 + 1)dz, C : Line segment from 0 to 1 + i. Ans: 5 2. R (b) C zdz, C : Line segment from 0 to i. Ans: 1. √ R (c) C 2zdz, C : Line segment from i to 2 + i. Ans: 4 5. R (d) C (x2 + iy 2 )dz, C : z = eit , −π/2 < t < π/2. Ans: π 17. Which of the following integrals are compatible to apply the Cauchy theorem? H sinz 1. z+2i dz, C : |z| = 1. C

2.

H

C

3.

H

sinz z+2i

dz, C : |z + 3i| = 1.

ez¯ dz, C : |z − 3i| = 6.

C

4.

H |z|=b

5.

H |z|=3

1 z 2 +bz+1 1 1−ez

dz, 0 < b < 1.

dz.

14

6.

1+i R

z 3 dz, along y = x.

0

H

1. Evaluate

|z|=3

2. Evaluate

H C

1 z 3 −z

z2 z−2

dz, Ans: 0

dz where C is the boundary of a triangle with vertices −1, 0 and 2i. Ans:

I

0

R. K. S

ON

3. Evaluate around the contour prescribed against it H z+ezthe following integrals it dz; C: z = 7e , 0 ≤ t ≤ 2π. Ans: −πi. (i) (z+iπ) 3 CH 1 −3πi (ii) z3 (z−2) . 2 dz; C: |z − 3| = 2, Ans: 8 C H z+ez it (iii) (z+iπ) 3 dz; C: z = 7e , 0 ≤ t ≤ 2π. Ans: −πi. CH 1 (iv) z3 (z−2) 3 dz; C: |z − 1| = 3, Ans: 0. C H 3z 4 1 dz. Ans: 3 × 64 (v) 2πi (z−6i) H|z|=10 1 (vi) dz. Ans: 0. z 4 −1 |z−1|=5 H 1 dz. Ans: −4πi. (vii) z 4 (z+i) |z−i|=3/2 H ez −z 2 (viii) dz. Ans: 2πi(2e4 − 1); (z−2)3 |z−1|=3 H sin z (ix) dz. Ans: 2πi cos 1; (z−1)2 |z|=2 H cos z (x) dz. Ans: 0; (z+3i)6 |z+i|=3/2 H 3 (xi) z2 (z+i) 2 dz; C: |z| = 5, Ans: 0. C H ez (xii) dz. Ans: 2πi sin a; z 2 +a2 a |z|=2a H ia ez (xiii) dz. Ans: e a π z 2 +a2 |z−ia|=a H z (xiv) z+1−e dz; C: |z − i| = 2, Ans: 0. z(z+3) C

H 4. Evaluate (x2 + iy 2 )ds; C: |z| = 2 where s is the arc length. C

∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗

15

Ans: 8π(1 + i)

Tutorial–6, B. Tech. Sem III, 5 October, 2016 (Laurent’s Expansions) 1. State and prove Laurents series expansion of a function f (z).

I

2. Find Laurents series expansion about the indicated singularity for each of the following functions. Name the singularity in each case and give the region of convergence of each series. e2z (i). (z−1) 3; z = 1

2

ON

2(z−1) 2 2 4 1 + ........; z = 1 is a pole of order 3 and Series Ans: e2 [ (z−1) 3 + (z−1)2 + (z−1) + 3 + 3 converges for all z 6= 1. 1 ; z = −2; (ii). (z − 3) sin (z+2) 5 1 5 1 Ans: 1 − z+2 − 6(z+2) 2 + 6(z+2)3 + 120(z+2)4 − ........; z = −2 is an essential singularity and Series converges for all z 6= −2. ; z = 0; (iii). z−sinz Z3 4

R. K. S

Ans: 3!1 − z5! − z7! ..; z = 0 is a removable singularity. Series converges for all z. z (iv). (z+1)(z+2) ;z = −2; 2 Ans: z+2 + 1 + (z + 2) + (z + 2)2 + ..; z = −2 a simple pole and Series converges for all z in 0 < |z + 2| < 1. 1 (v). z2 (z−3) 2; z = 3 (z−3) 1 2 1 Ans: 9(z−3) 2 − 27(z−3) + 27 − 243 + ........; z = 3 is a pole of order 2 and Series converges for all z in 0 < |z − 3| < 3.

1 3. Expand f (z) = (z+1)(z+3) in a Laurent’ s series valid for expansion (a). 1 < |z| < 3, (b). |z| > 3, (c) 0 < |z + 1| < 2, (d) |z| < 1, 1 z z2 z3 Ans: (a). · · · · · · − 2z14 + 2z13 − 2z12 + 2z − 16 + 18 − 54 + 162 − ······; 4 13 40 1 (b). z2 − z3 + z4 − z5 + · · · · · · ; 1 1 (c). 2(z+1) − 14 + (z+1) − 16 (z + 1)2 + · · · · · · ; 8 (d). 31 − 4z + 13 z2 + · · · · · · · · · · · · . 9 27 z 4. Expand f (z) = (z−1)(2−z) in a Laurents series valid for (a). 1 < |z| < 2; (b). |z| > 2; (c). 0 < |z − 2| < 1; (d). |z| < 1; (e). |z − 1| > 1. 2 3 Ans: (a). 1 + z2 + z4 + z8 + · · · · · · z1 + z12 + · · · · · · ; (b). − 12 − z32 − z73 − 15 ······; z4 2 (c). 1 − z−2 − (z − 2) + (z − 2)2 − (z − 2)3 + (z − 2)4 · · · · · · ; 15 4 z − ······. (d). − 21 z − 43 z 2 − 87 z 3 − 19 1 2 2 (e). − (z−1) − (z−1)2 − (z−1)3 · · · · · · ; 1 5. Expand f (z) = z−3 in a Laurents series valid for (a). |z| < 3; (b). |z| > 3 Ans: (a). (i)−1/3 − 1/9z − 1/27z 2 − 1/81z 3 − · · · · · · ; (ii) z −1 + 3z −2 + 9z −3 + 27z −4 + · · · · · · .

6. Expand f (z) =

1 z(z−2) ∞ P

1 Ans: (a). − 2z −

n=0

in a Laurent’s series valid for (a). 0 < |z| < 2; (b). |z| > 2 ∞ n−1 P zn 2 ; (b). ; 2n+2 z n+1 n=1

16

R. K. S

ON

∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗

I

7. Find Taylor’s series expansion about the indicated points for each of the following functions. Give the region of convergence of each series. (i). ezz+1 ; z = 0; z (ii). zsin 2 +4 ; z = 0 Ans: (i) |z| < π; (ii) |z| < 2

17

Tutorial–7, B. Tech. Sem IV, 24 Feb, 2016 (Singularities and Residues)

(iπ+2kπ) 4

, ∀k = 0, 1, 2, 3, 4 each is a simple pole. z = ±ai are simple poles. z = 0, a pole of order 2. z = iπ is a simple pole. z = 0 is a simple pole. z = 0 is a pole of order 2. z = 0 is a pole of order 3. z = 0 is a simple pole. z = 1 is a pole of order 3 and z = i is a removable singularity.

e

R. K. S

(xi). z41+1 ; iz (xii). zze 2 +a2 ; 2z ; (xiii). 1−e z3 e2z (xiv). 1+ez ; e2z (xv). 1−e z; z−sinz (xvi). z5 ; (xvii). z2 (e1z −1) ; z ; (xviii). sinh z2 2 (z +1)ez (xix). (z−1)3 (z−i) ;

ON

I

1. Discuss the types of each singularity for the following functions: 1 (i). cos z−sin ; z = π/4 is a simple pole. z cot πz (ii) . (z−a)2 ; z = a is a pole of order 2. ez z = ±i is a simple pole and z = ∞ is an essential singularity. (iii). z2 +1 ; z z = −1, −2 are simple poles. (iv). (z+1)(z+2) ; ez (v). sin z ; z = kπ, k ∈ I is a simple pole. 1 (vi). arcsin z ; z = 1/nπ, (n ∈ I+ ) simple pole and z = 0 an isolated essential singularity. (vii). arctan πz; z = 1/n, (n ∈ I+ ) is simple pole and z = 0 isolated essential singularity. ; z = 0 is a removable singularity. (viii). z−sinz z3 (ix). zz+1 ; z = 0, 2 are simple poles. 2 −2z 2 z = −2 is a simple pole and z = 1 is a pole of order 2. (x). (z−1)z2 (z+2) ;

2. Examine the nature of singularity for the following functions and find its residue: 1 Hint: residue is defined as coefficient of (z−z in Laurent’s series expansion of a function 0) f (z) about a point z0 . Thus, 1 dm−1 m residue for a pole of order m is (m−1)! lim dz m−1 ((z − z0 ) f (z)) z→z0

residue for a removable singularity=0. residue for an essential singularity z0 can be find by finding coefficient of Laurent’s series expansion about a point z0 . z z (i). e +sin ; z = 0 is a removable singularity, Residue= 0 z4 1 (ii). (z+i)3 ; z = i is a a pole of order 3. Residue = 6i3 . (iii). (iv).

zeiz ; z 2 +a2 1 ; z 2 sinh z

z = ±ai are simple poles. Residues are = z = 0 a pole of order 3. Residue= − 61

3. Using theorem evaluate the following integrals: R 2πResidue 1 2π (i). 0 2+cos θ dθ, Ans: √ . 3 R 2π 1 (ii). 0 1−2a cos θ+a2 dθ, a > 1 Ans: .... R π 1+2 cos θ π (iii). 0 5+3 cos θ dθ, Ans: 24 . R 2π cos 2θ Ans: π6 (iv). 0 5+4 cos θ dθ, R 2π cos2 3θ (v). 0 5−4 cos 2θ dθ, Ans: 3π 8 18

e−a . 2

and=

1 z−z0

ea . 2

in its

Tutorial–9, B. Tech. Sem III, 22 October, 2016 (Bays’ rule) 1. Two third of the students in a class are girls and rest boys. If it is known that probability of a boy getting first class is 0.25, that of a girl is 0.28. Find the probability that a student chosen at random will get first class. Ans: ...

ON

I

2. The chances that doctor will diagnose a disease X correctly is 60%. The chances that a patient will die by his treatment after correct diagnosis is 40% and the chance of death by wrong diagnosis is 70%. A patient of the doctor, who had disease X, died. What is 6 the probability that his disease was diagnose correctly. Ans: 13 3. A manufacturing firm produces steel pipes in three plants with daily production volumes of 500, 1000 & 2000 units respectively. According to past experience it is known that the fraction of defective output produced by three plants are respectively 0.005, 0.008, 0.01. What is the probability that a pipe is selected at random from a day’s output and found to be defective? Ans: ...

R. K. S

4. In a university 4% of male students and 1% of female students are taller than 6 feet. Further 60% of the students are female. Now, if a randomly selected student is taller than 6 feet, what is the probability that the student is a female? Ans: ... 5. Two shipment of parts are received. The first shipment contained 1000 parts with 10% defectives and the second shipment contains 1500 parts with 5% defectives. One shipment is selected at random, two parts were tested at random and found to be good. Find the probability that tested parts were selected from first shipment. Ans: ... 6. A coin is tossed. If it turn up H, two balls will be drown from urn A, otherwise 2 balls will be drawn from urn B. Urn A contains 3 Red and 5 Blue balls, Urn B contains 7 Red and 5 Blue balls. What is the probability that urn A is used, given that both balls are blue. (Find in both cases, when balls were chosen with replacement and without replacement). Ans: ... 7. There are 10 urns of which each of 3 contains 1W, 9B balls, each others 3 contains 9W, 1B balls and remaining 4 each contains 5W, 5B balls. One of the urn was selected at random and a ball was chosen from it. (i) what is the probability that it is white ball? (ii) If ball is black, what is the chance that it comes out from the urn consisting 9W and 1B balls. Ans: ... 8. There are two bags A and B. A contains n white and 2 black balls and B contains 2 white and n black balls. One of the two bags is selected at random and two balls are chosen from it without replacement. If both the balls drawn are white and the probability that Ans: n = 4. the bag A was used to draw the ball is 67 , find out the value of n.

19

Numerical Integration If f : [a, b] −→ R is differentiable then, we obtain a new function f 0 : [a, b] −→ R, called the derivative of f . Likewise, if f : [a, b] −→ R is integrable, then we obtain a new function F : [a, b] −→ R defined by

Z

x

f (t)dt

F (x) =

∀x ∈ [a, b].

a

Z Observation: If f is nonnegative function, then

b

f (x)dx is represent a

the area under the curve f (x).

Dr. Raj Kumar, VBSPU Jaunpur.

1

Antiderivative Antiderivative: Let F : [a, b] −→ R be such that f = F 0, then F is called an antiderivative of f .

Recall

Fundamental Theorem of Calculus: Let f : [a, b] −→ R is integrable and has an antiderivative F , then Z

b

f (x)dx = F (b) − F (a). a

Dr. Raj Kumar, VBSPU Jaunpur.

2

Basic Problems

• Difficult to find an antiderivative of the function (for example f (x) = 2

e−x )

• Function is given in the tabular form.

Dr. Raj Kumar, VBSPU Jaunpur.

3

Newton-Cotes Methods/Formulae The derivation of Newton-Cotes formula is based on Polynomial Interpolation.

x

x0

x1

x2

...

xn

f (x)

f (x0)

f (x1)

f (x3)

...

f (xn)

Dr. Raj Kumar, VBSPU Jaunpur.

4

The idea is: Replace f by pn(x) and evaluate

Rb a

pn(x)dx

That is, b

Z

Z f (x)dx '

b

pn(x)dx = a

a

= =

Z bX n a i=0 n X

li(x)f (xi)dx Z

b

li(x)dx

f (xi)

i=0 n X

a

Aif (xi)

i=0

Where Ai =

Rb

l (x)dx a i

called weights.

Dr. Raj Kumar, VBSPU Jaunpur.

5

Types of Newton-Cotes Formulae

• Trapezoidal Rule (Two pint formula) • Simpson’s 1/3 Rule (Three Point formula) • Simpson’s 3/8 Rule (Four point formula)

Dr. Raj Kumar, VBSPU Jaunpur.

6

Trapezoidal Rule Since it is two point formula, it uses the first order interpolation polynomial P1(x).

Z

b

Z

x1

f (x) ≈ a

P1(x)dx x0

P1(x) = f (x0) + s∆f (x0) x − x0 s= h Now, dx = h ds at x = x0, s = 0 and at x = x1, s = 1. Dr. Raj Kumar, VBSPU Jaunpur.

7

Hence, Z

b

Z

1

f (x)dx ≈

(f (x0) + s∆f (x0))hds = 0

a

h [f (x0) + f (x1)] 2

OR Z

b

f (x)dx ≈ a

b−a [f (a) + f (b)] 2

Error (b − a)3 00 E =− f (ξ), 12 T

where a < ξ < b Remark: x0 = a and x1 = b.

Dr. Raj Kumar, VBSPU Jaunpur.

8

Basic Simpson’s To evaluate

Rb

1 3

Rule

f (x)dx.

a

• f will be replaced by a polynomial of degree 2 which interpolates f at a,

a+b 2

and b. . Here, x0 = a, x1 = Z a

b

a+b 2 , x2

=b

b−a b+a f (x) dx = [f (a) + 4f ( ) + f (b).] 6 2

Error −h5f (4)(ξ) E = 90 s

for some ξ ∈ (a, b). Dr. Raj Kumar, VBSPU Jaunpur.

9

Basic Simpson’s

3 8

Rule

• f is replaced by p3(x) which interpolates f at x0 = a, h,

x3 = a + 3h = b. where h =

x2 = a + 2h, Z

a

b

b−a 3 .

x1 = a +

Thus we get:

3h f (x)dx ' [f0 + 3f1 + 3f2 + f3] 8

−3h5 (4) f (ξ), where a < ξ < b. Error: E = 80 s

Dr. Raj Kumar, VBSPU Jaunpur.

10

Example Using Trapezoidal and Simpson

1 3

rules find

R2 0

4

x dx and

R2 0

sinxdx and

find the upper bound for the error.

Dr. Raj Kumar, VBSPU Jaunpur.

11

Composite Rules Note that if the integral [a, b] is large, then the error in the Trapezoidal rule will be large. Idea Error can be reduced by dividing the interval [a, b] into equal subinterval and apply quadrature rules in each subinterval.

Composite Trapezoidal Rule h=

b−a , xi = x0 + ih n

Dr. Raj Kumar, VBSPU Jaunpur.

12

Composite Rule Z

b

Z

xn

f (x)dx = a

f (x)dx = x0

n Z X i=1

xi

f (x)dx

xi−1

Now apply Trapezoidal rule on each [xi−1, xi], we have

Z a

b

h f (x)dx = [f (x0) + 2 ∗ (f (x1) + f (x2) · · · f (xn−1)) + f (xn)] 2

Dr. Raj Kumar, VBSPU Jaunpur.

13

Error in composite Trapezoidal rule

E

CT

h2 00 = −(b − a) f (ξ), ξ ∈ [a, b] 12

Dr. Raj Kumar, VBSPU Jaunpur.

14

The Composite Simpson’s

1 3

Rule

• [a, b] will be will be divided into 2n equal subintervals and we apply basic Simpson’s

1 3

rule on each of the n intervals [x2i−2, x2i] for i =

1, 2, 3, · · · , n. Thus here h =

b−a 2n .

Then Z

b

b=x2n

Z f (x)dx =

a

Z

x2

=

Z

x4

a=x0

Z

x2i

f (x)dx + · · · +

f (x)dx + x0

f (x)dx

x2 Dr. Raj Kumar, VBSPU Jaunpur.

Z

x2n

f (x)dx + · · · + x2i−2

f (x)dx x2n−2 15

h h = [f (x0) + 4f (x1) + f (x2)] + [f (x2) + 4f (x3) + f (x4)]+ 3 3 h + · · · + [f (x2n−2) + 4f (x2n−1) + f (x2n)] 3 h = {f (x0) + 4 × [f (x1) + f (x3) + f (x5) + · · · + f (x2n−1)]+ 3 +2 × [f (x2) + f (x4) + f (x6) + · · · + f (x2n−2)] + f (x2n)}

E

CS

h4 (4) f (ξ), = −(b − a) 180

where ξ ∈ [a, b]

Dr. Raj Kumar, VBSPU Jaunpur.

16

Example Evaluate the integral

R1 −1

x2 exp(−x)dx by composite Simpson’s

1 3

rule

with spacing h = 0.25 Solution: According to composite Simpson’s Z

1 3

rule:

1

h x2 exp(−x)dx = [f (x0) + 4f (x1) + 2f (x2) + 4f (x3) + 2f (x4)+ 3 −1 +4f (x5) + 2f (x6) + 4f (x7) + f (x8)]

Here f (x0) = f (−1) = 2.7183 f (x1) = f (−0.75) = 1.1908 Dr. Raj Kumar, VBSPU Jaunpur.

17

f (x2) = f (−0.5) = 0.4122

f (x3) = f (−0.25) = 0.0803

f (x4) = f (0) = 0

f (x5) = f (0.25) = 0.0487

f (x6) = f (0.50) = 0.1516

f (x7) = f (0.75) = 0.2657

f (x8) = f (1) = 0.3679 Dr. Raj Kumar, VBSPU Jaunpur.

18

Substituting these values in the above formula we get: Z

1

x2 exp(−x)dx ' 0.87965

−1

Dr. Raj Kumar, VBSPU Jaunpur.

19

Example

Find the minimum no. of subintervals, used in composite Trapezoidal Z 1 −x4 e dx such that and Simpson’s 1/3 rule in order to find the integral 0

the error can not exceed by .00001.

Sol. For the composite Trapezoidal rule, we have

00

13 max0