Tutorial, B. Tech. Sem III, 24 July, 206 (Root Findings and Linear System of Equations). Find the root of the equation e x = 3x lying in [0,] 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 x 4 + 2x 3 x = 0 lying in [0,] correct to three decimal places. 3. For smallest positive root of the equation: x 3 5x + = 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 2 = 7; 3x + 20y = 8; 2x 3y + 20 = 25 with Initial root (0,0,0); Ans: x =, y =, = ii. 0x + 2y + = 9; x + 0y = 22; 2x + 3y + 0 = 2 with Initial root (0,0,0). Ans: x =, y = 2, = 3 iii. x 7x 2 + x 3 = 32; x + 5x 2 2x 3 = 8; 2x + 2x 2 + 7x 3 = 9; upto D places with Initial root (5,5,5) iv. x 2y + = 8; x + y + 2 = 9; 3x y + = 6 with Initial root (3,3,2); Ans:... 6. Verify that each of the following equations has a root on the interval (0,). Next, perform the bisection method to determine p 3, the third approximation to the location of the root (a) ln( + x) cos x = 0 (b) x 5 + 2x = 0 (c) e x x = 0 (d) cos x x = 0 7. It was noted that the function f(x) = x 3 + 2x 2 3x has a ero on the interval ( 3, 2) and another on the interval (-,0). Approximate both of these eroes to within an absolute tolerance of 5 0 5. 8. Approximate 3 3 to three decimal places by applying the bisection method to the equation x 3 3 = 0. 9. Approximate /37 to five decimal places by applying the bisection method to the equation /x 37 = 0. 0. onsider the function g(x) = cos x. (a) Graphically verify that this function has a unique fixed point on the real line. (b) an we prove that the fixed point is unique using the theorems of this section? Why or why not?
. onsider the function g(x) = + x 8 x3. (a) Analytically verify that this function has a unique fixed point on the real line. (b) an we prove that the fixed point is unique using the theorems of this section? Why or why not? 2. Each of the following equations has a root on the interval (0,). Perform Newton s method to determine p 4, the fourth approximation to the location of the root. (a) ln( + x) cos x = 0 (b) x 5 + 2x = 0 (c) e x x = 0 (d) cos x x = 0 3. The equation x 3 + x 2 3x 3 = 0 has a root on the interval (,2), namely x = 3. 4. The equation x 7 = 3 has a root on the interval (,2), namely x = 7 3. 5. The equation /x 37 = 0 has a ero on the interval (0.0,0.) namely x = /37. 6. Show that when Newton s method is applied to the equation x 2 a = 0, the resulting iteration function is g(x) = 2 (x + a x ). 7. Show that when Newton s method is applied to the equation /x a = 0, the resulting iteration function is g(x) = x(2 ax). 8. For each of the functions given below, use Newton s method to approximate all real roots. Use an absolute tolerance of 0 6 as a stopping condition. (a) f(x) = e x + x 2 x 4 (b) f(x) = x 3 x 2 0x + 7 (c) f(x) =.05.04x + ln x 9. Each of the following equations has a root on the interval (0,). Perform the secant method to determine p 4, the fourth approximation to the location of the root. (a) ln( + x) cos x = 0 (b) x 5 + 2x = 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
(i) A = 2 3 4 2 3 2 5, b = 0 5 3 4, b 2 = 4 5 3 4, b 3 = 0 2 0 3 (ii) A = 4 3 6 0 2 5 3, b = 2 0, b 2 = 6 4, b 3 = 3 0 5 3 2 2 7 5 4 4 3 (iii) A = 6 20 0, b = 36, b 2 = 6, b 3 = 2 4 3 0 7 7 6 2. In Exercises -3, use the Gauss-Seidel method to solve the indicated linear system of equations. Take x (0) = 0, and terminate iteration when x (k+) x (0) falls below 5 0 6.. 2. 3. 4x + x 2 + x 3 + x 4 = 5 x + 8x 2 + 2x 3 + 3x 4 = 23 x + 2x 2 5x 3 = 9 x + 2x 3 + 4x 4 = 4 4x x 2 = 2 x + 4x 2 x 3 = 4 x 2 + 4x 3 = 0 7x + 3x 2 + = 4 3x + 9x 2 + x 3 = 6 x 2 + 3x 3 x 4 = 3 x 3 + 0x 4 4x 5 = 7 4x 4 + 6x 5 = 2 22. Solve the following system of linear equations by Triangulariation /Factoriation or rout s method: (a) 2x + 3y + = 9; 3x + y + 2 = 8; x + 2y + 3 = 6; Ans:... (b) x + y + = 3; 2x y + 3 = 6; 3x + y = 3; Ans: x =, y = 2, = 4 (c) x+2y+3+4w = 20; 3x 2y+8+4w = 26; 2x+y 4+7w = 0, 4x+2y 8 4w = 2; Ans: x = 4, y = 3, = 2, w = 2 3 8 3 8 0 3
Tutorial 2, B. Tech. Sem III, 26 July, 206 ( Difference Operators and Interpolation). Prove the following operator relations: (i). E = + = ( ), (ii) = µδ + δ2, 2 (iii). log f(x) = log( + f(x) ) f(x) (iv). [f(x)g(x)] = f(x) g(x) + g(x + h) f(x), (v). = = = δ 2, (vi). = 2 δ2 + δ + δ2, (vii). 4 δ3 y = y 2 3y + 3y 0 y, 2 (viii). δ(f k g k ) = µf k δ(g k ) + µg k δ(f k ), 2+ (ix). + = 2, (x). = e hd, (xi). µδ = sinh(hd), (xii). e hd =, (xiii). f[x, y, ] = x + y +, f = x 3, (xiv). f[a, b, c] =, f =, abc a (xv). f[a, b, c,...(n symbols)...l] = n f(a) 2. Find for (h = ) (i). 2 (ab ex ), (ii). 2 x 3, Ex 3 (iii). tan (ax), (iv). n ( ), x ( (v). (vi). (x + cosx). ( ( 3. Prove that: e x = 2 E e x and E ex ). 2 e x ) n!h n f 0 + xf + x2 f 2! 2 +... = e x (f 0 + x f 0 + x2 f 2! 0 +...) 2 E ) x 3, 4. Let P n (x) = (x x 0 )(x x )(x x 2 )...(x x n ), where x i = x 0 + ih, i being integer; show that P n = nhp n, hence show that r P n = n! (n r)! hr P n r, r = () n, and n P n = n!h n. 5. Find the third divided difference with argument 2, 4, 9, 0 of the function f(x) = x 3 2x. 6. Form a divided difference table for f(x) = x 4 + 6x 2 + x 2 for values of x = 3 () 3, show that 5th order diferences are ero. 7. Given that f(0) = 8, f() = 68, and f(5) = 23 determine f(2); calculate the error also. 8. Tabulate sin x for x 0 = 30 (2) 40 and interpolate sin 3 0 and sin 33 0. ompare with exact values. 9. Tabulate e x for x =.7 (0.) 2.2 and interpolate at x =.7, 2.5. 0. Find log 0 52.5 and log 0 6.3 using the following data: x 50 55 60 65 70 75 80 log 0 (x) 3.06069 3.06258 3.06445 3.06632 3.0688 3.07003 3.0788 and error in the result.. Use Lagrange s formula to interpolate the values of f(5) from x 2 3 4 7 f(x) 2 4 8 6 28 How much it deviates from 2 5.? 4
2. A third degree polynomial passes through the points (0, ), (, ), (2, ), (3, 2). Find the polynomial. Ans: 6 x3 2 x2 + 8 3 x 3. 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 3 42 5 35 3 Ans: approx. 48 4. Find the form of f(x), given that f(0) = 8, f() =, f(4) = 68 and f(5) = 23 also determine f(2). Ans: x 3 x 2 + 3x + 8, 8 5. Given that f(0) = 8, f() = 0 = f(3) = f(6), f(5) = 248 and f(9) = 304 find the form of f(x) assuming it to be a polynomial of degree 5th. {Hint: f(x) = (x )(x 3)(x 6)φ(x), φ(x) is a polynomial of degree 2 with φ(0) =, φ(5) = 3, φ(9) = 9 } Ans: x 5 9x 4 + 8x 3 x 2 + 9x 8 6. Find log 0 30 using Newton s Divided difference interpolation formula from the data: x 300 304 305 307 Ans: 2.4786 log 0 (x) 2.477 2.4829 2.4843 2.487 7. The values of y = x are listed below: x 4 6 7 0 y 2 2.449 2.646 3.62 ompute x corresponding to y = 2.5. Ans: 6.2548 8. 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 /2 0 6? Ans: The number of subdivisions required n = 57. 5
Tutorial 3, B. Tech. Sem III, 22 August, 206 (Numerical Differentiation, Integration and Differential Equations). 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 : 0 300 600 900 200 500 800 y : 35 49 57 83 20 205 93 2. Find the first three derivatives of the function at x =.5 from the data x :.5 2.0 2.5 3.0 3.5 4.0 y : 3.375 7.0 3.625 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 =. t :.0..2.3.4 v : 43. 47.7 52. 56.4 60.8 Ans: 44.97 4. Derive Newton ote s quadrature formula. Hence deduce (i) Trapeoidal rule, One third Simpson s and 3/8 Simpson s rule of Numerical integrations. (ii) alculate Truncation error as well as Max. Global error in the rules. 5. Evaluate dx, with h = 0.2 (up to 3D) by using Trapeoidal rule of Numerical 0 +x 2 integration. Hence find an approximate value of π. Give your decision about the statement an we apply One third Simpson s and 3/8 Simpson s rule of Numerical integrations for this problem Ans: 3.35, No 6. Evaluate 0 ex dx by One third Simpson s correct to 5 D places with proper choice of h. Ans:.7828 7. Evaluate 6 dx, up to 3D by using Trapeoidal rule and Simpson s rules of Numerical 0 +x 2 integrations. Also check your results by actual integration. Ans:.4,.366,.357, Actual value is.406 8. onstruct the divided difference table for the following data set, and then write out the Newton form of the interpolating polynomial. x - 0 2 y 3 - -3 9. onstruct the divided difference table for the following data set, and then write out the Newton form of the interpolating polynomial. x -7-5 -4 - y 0 5 2 0 0. 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
. Apply Picard s Method with 4-iterations upto 4D places to find the values of y at x = 0.(0.3)0.3, given that dy = y x, y(0) = 2. Ans: 2.2, 2.42, 2.65 dx 2. Apply Picard s Method with 3-iterations to find the values of y at x = 0., 0.2, given that dy dx = y2 + x 2, y(0) = 0. Ans: 0.00033, 0.00267 3. By using Picard s Method with 5-iterations, find y at x = 0., 0.2 upto 4D places from the differential equation dy dx = x + x2 y, y(0) =. Ans:.0053,.0227 4. Write out the Newton form of the interpolating polynomial for f(x) = e x that passes through the points (, e ), (0, e 0 ), and (, e ). 5. Use Runge Kutta method of fourth order to find a numerical solution at x = 0.5 for dy (x y), y(0) = taking h = 0.25. dx = 2 6. Solve the differential Equation dy = x+y+xy, y(0) = by using Taylor s series expansion dx to get y at x = 0.(0.)0.5 (use terms up to x 5 in the expansion ) by shifting the origin for each value. 7. Evaluate y upto 4D places for x = 0.(0.)0.5 by using Taylor s series method for dy dx = x + y 2, y(0) = 0. 8. Solve dy dx = x + y2, y(0) = 0 for x = 0.(0.)0.5 by using modified Euler s method correct to 4D places. 9. Solve dy = x + y + xy, y(0) = by using modified Euler s method correct to 4D places to dx obtain y at x = 0.(0.)0.5 20. Using Runge Kutta fourth order method compute y at x = 0.2(0.2)0.6 for dy dx = + y 2, y(0) = 0.9 correct to 3D places. 2. Using fourth order Runge Kutta method solve dy = xy, y(0) = in the interval [0,0.6] dx 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(.), y(.2) for dy dx = x2 +y 2, y() = 2. Ans 2.6384, 3.7080 23. Find y at x = 0.2 by using Euler s method correct to 3D places to solve dy dx = x y2, y(0) = taking h = 0.. 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 Types of Newton-otes Formulae:. Trapeoidal Rule (Two point formula) b a x x 0 x x 2... x n f(x) f(x 0 ) f(x ) f(x 3 )... f(x n ) f(x)dx = h 2 [f(x 0) + 2 (f(x ) + f(x 2 ) f(x n )) + f(x n )] Error in composite Trapeoidal rule Error in the ith interval [x i, x i ] is E T (I) = (x i x i ) 3 f (ξ i ) = h3 2 2 f (ξ i ) where ξ i (x i, x i )). Hence, the Max. error in composite rule (Global Error) is E T = h3 2 n f (ξ i ) or (b a)h2 = f (ξ), 2 where a < ξ < b, 2. Simpson s /3 Rule (Three Point formula); Then b a f(x)dx = x2 x 0 f(x)dx + x4 i= x 2 f(x)dx + + x2i x 2i 2 f(x)dx + + x2n = h 3 {f(x 0) + 4 [f(x ) + f(x 3 ) + f(x 5 ) + + f(x 2n )]+ +2 [f(x 2 ) + f(x 4 ) + f(x 6 ) + + f(x 2n 2 )] + f(x 2n )} x 2n 2 f(x)dx The Error in the omposite Simpson s 3 Rule i.e. error in the ith interval (x 2i 2, x 2i ) is E cs (I) = h5 90 f (4) (ξ i ), ξ i (x 2i 2, x 2i ) Max. Error in composite Simpson s /3-rule is given by E S = (b a) h4 80 f (4) (ξ), 8
where ξ [a, b] 3. omposite Simpson s 3 8 rule or (Four point formula); The [a, b] is divided into 3n equal subintervals. (h = b a.) and we apply 3 rule on each 3n 8 of the n intervals [x 3i 3, x 3i ] for i =, 2, 3,, n. Hence, Remember: b a f(x)dx x3 x 0 =a f(x)dx + x6 x 3 f(x)dx + + x3n =b x 3n 3 f(x)dx = 3h 8 [f 0 + 3f + 3f 2 + 2f 3 + 3f 4 + 3f 5 + 2f 6 + 3f 7 + + 3f 3n + f 3n ] f with suffices of multiple 3 are multiplied by 2. Others by 3, except the end points. T. Error G. Error E s = 3h4 f (4) (ξ) where ξ [a, b] 80 E s = 3h5 80 f (4) (ξ i ), ξ i (x 3i 3, x 3i ) 9
Tutorial 4, B. Tech. Sem III, 5 Sep., 206 (Introduction of omplex numbers & Analytic functions). Find the locus of in each of the following relations: (i) 5 = 6, (ii) + 2i, (iii)re( + 2) =, (iv) i = + i, (v) + 3 + + = 4, (vi) 3 2, (vii) + 3 + =. 2. For which complex number following are true, justify in each (i) =, (ii) =, (iii) =, (iv) = 3. Define an analytic function at a point and in a domain. 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() = u + iv to be analytic. 8. Define an analytic function at a point. Illustrate such a function. 9. If f() = ( )2, 0; f(0) = 0 then f() satisfies auchy Riemann equations (R) at origin. 0. Using Milne Thomson method construct an analytic function f() = u + iv for which 2u + 3v = 3(x 2 y 2 ) + 2x + 3y.. State and Prove auchy Riemann equations in polar coordinate system. 2. Let f(x, y) = ( xy ), then (a) f x, f y do not exist at (0, 0); (b) f x (0, 0) = ; (c) f y (0, 0) = 0; (d) f is differentiable at (0, 0). Ans: c ( ) 2, 3. If function f() = when 0, f() = 0 for = 0, Then f() (a) satisfies.r. equations at = 0; (b) is not continuous at = 0; (c) is differentiable at = 0; (d) is analytic at = Ans: b 4. Show that f() = ( )2, when 0, f() = 0 for = 0, satisfies.r. equations at = 0; but is not differentiable at = 0. 5. The harmonic conjugate of u = x 2 y 2 + xy is (a) x 2 y 2 xy; (b) x 2 + y 2 xy; (c) /2( x 2 + y 2 ) + 2xy. (d) 2( x 2 + y 2 ) + /2 Ans: c 6. f() = ( ) 2 is (a) continuous everywhere but nowhere differentiable; (b) continuous at = 0 but differentiable everywhere; (c) continuous nowhere; (d) none of these. Ans: d 7. Examine the nature of the function f() = x2 y 5 (x+iy) ; if 0, otherwise 0, in a region x 4 +y 0 including the origin. 0
8. Prove that the function f() = u + iv, where f() = x3 (+i) y 3 ( i) ; if 0, otherwise x 2 +y 2 0 is continuous and that auchy Riemann equations are satisfied at the origin, yet f () does not exists there. 9. If f() = x3 y(y ix) ; if 0 and f(0) = 0, show that f() f(0) 0 as 0 along any x 6 +y 2 radius vector but not as 0 in any manner. 20. If f() = xy2 (x+iy) ; if 0 and f(0) = 0, prove that f() f(0) 0 as 0 along any x 2 +y 4 radius vector but not as 0 in any manner. 2. The function f() = is (a) analytic at = 0, (b) differentiable only at = 0;(c) satisfies.r. equations everywhere; (d) nowhere analytic. Ans: d 22. Derive the.r. equations for an analytic function f(r, θ) = u(r, θ) + i v(r, θ) and deduce that u rr + r u r + r 2 u θθ = 0. 23. Find the point where the.r. equations are satisfied for the function f() = xy 2 + ix 2 y. In which region f () exists? 24. f() = ( ) 2 is (a) continuous everywhere but nowhere differentiable except at 0; (b) continuous at = 0 but differentiable everywhere (c) continuous nowhere; (d) none of these. Ans: a 25. Prove that the function f() = is nowhere analytic. 26. If f() is an analytic function such that Ref () = 3x 2 4y 3y 2 and f( + i) = 0 then f() is (a) 3 + 6 2i, (b) 3 + 2i 2 + 6 2i, (c) 3 + 2i 2 2i, (d) 3 + 2 2 + 6 2i Ans: b Hint: u x = 3x 2 4y 3y 2 = φ (x, y)(say), integrating partially w.r.t. y we get u = x 3 4xy 3xy 2 + g(y) Therefore, u y = 4x 6xy + g (y) or u y = v x = φ 2 (x, y)(say) = 4x + 6xy g (y) Thus φ (, 0) = 3 2, φ 2 (, 0) = 4 + g (0) Now, applying Milne Thomson we get f() = (3 2 + i 4)d + constt. or f() = 3 + 2i 2 + constt. and applying f( + i) = 0 implies onstt = 6 2i 27. The orthogonal trajectory of u = e x (x cos y y sin y) is (a) e x (cos y + x sin y) + c; (b) e x x sin y + c ; (d) e x (y cos y + x sin y) + c. Ans: c 28. Find the locus of points in the plane satisfying the relation + 5 2 + 5 2 = 75. Ans: circle 29. The function f() = is (a) analytic at = 0, (b) continuous at = 0;(c) differentiable only at = 0; (d) analytic anywhere. Ans: b 30. If f() = u + iv is an analytic function and u v = (x y)(x 2 + 4xy + y 2 ) then f() is (a) 3 + c, (b) i 2 + ic, (c) i 3 + β, (d) 3 ic, Ans: c cos x+sin x e y 3. If f() = u + iv, is an analytic function of and u v = ; find f() if f() 2 cos x e y e y subject to the condition f(π/2) = 0. Ans: /2{ cot(/2)}
32. If f() = u(r, θ) + iv(r, θ) is an analytic function and u = r 3 sin 3θ then construct the analytic function f(). 2 sin 2x 33. If f() = u + iv, is analytic function of and u + v = ; find f() in terms 2 cos 2x+e 2y e 2y of. Ans: ( + i) cot + d 2 34. hoose the correct code for matching list A and B. i ii iii (a) p r q (b) r p q (c) p q r A (u is given) B (f() = u + iv is an analytic function) p. x 3 3xy 2 + 3x + (i). sin + ci q. y 3 3x 2 y (ii). 3 + 3 + + ci r. sin x cosh y (iii).i( 3 + c) Ans: b 35. if sin(α + iβ) = x + iy prove that (a) x 2 cosh 2 β + y2 =, (b) x 2 sinh 2 β cos 2 α y2 =. sin 2 α 36. Prove that ( 2 x 2 + 2 y 2 ) f() 2 = 4 f () 2 for an analytic function f() = u + iv. 37. Find the harmonic conjugate of u = x 3 3xy 2 + 3x 2 3y 2 + 2x + and corresponding analytic function f() = u + iv. Ans: v = 3x 2 y y 3 + 6xy + 2y + d, f() = 3 + 3 2 + 2 + + id. 38. Find orthogonal trajectory of v = e 2x (x cos 2y y sin 2y) Ans: e 2x (x sin 2y+y cos 2y)+d 2
Tutorial 5, B. Tech. Sem III, 4 Sep., 206 (omplex Integrations). Define simply and multiply connected regions? State and prove auchys theorem for an analytic function? Is it true for multiply connected regions? 2. State and prove auchy integral formula for nth derivative? Is it true for multiply connected regions? If yes, give your explanation with necessary proofs. 3. If f() is analytic in a simply connected region D and a, are two points in D, then a f()d is independent of the path in D joining a and. 4. Evaluate (2,4) (0,3) d 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? 5. Evaluate d from = 0 to = 4 + 2i along the curve given by (i) = t2 + it; (ii) the line from = 0 to = 2i and then the line from = 2i to = 4 + 2i. Ans: (i) 0 8i 3 ; (ii)0 8i. 6. If F (a) = a F (4.5). Ans: Zero 2 +2 5 d, where is the ellipse (x/2) 2 + (y/3) 2 =. Find the value of 7. Evaluate d, where is any simple closed curve and = a is (i) outside ; (ii) a inside. Ans: (i) 0 ; (ii) 2πi. 8. Evaluate d ; n, n Z +, where is any simple closed curve and = a is ( a) n inside. Ans: Zero 9. Suppose f() is analytic inside and on a simple closed curve and = a is inside. Prove that d = d where a a is a circle center at a and totally contained in simple closed curve. 0. Evaluate ( ) 2 d ; where is the circle (i) =, (ii) = Ans: 4πi, 0.. Find d around (a) the circle 2 = 3, (b) the ellipse 3 + + 3 = 0, (c) the square with vertices 0, 2, 2i and 2 + 2i. Ans: 8πi, 40πi, 8i 2. Suppose f() is integrable along a curve having finite length l and there exists a positive real number M such that f() M on. Prove that f()d Ml. Remark: This result is helpful to evaluate the upper bound of an integral without evaluating it. 3. Work out the following integrals around the contour prescribed against it (i) cos π 2 +sin π 2 d; : = 3. Ans: 4πi. ( )( 2) (ii) d; : = 6. Ans: 0 2 +6 3
(iii) (iv) 2πi (v) e 5 2i sin π d; : 2 + + 2 = 6. Ans: 2πie0i e d; : = 3. Ans: e2 2 2 d; is the rectangle with vertices 2 + i, 2 i, 2 + i, 2 i; is the rectangle with vertices i, 2 i, 2 + i, i. Ans: 0, 0. (vi) (vii) e πi 2 4+5 e +cos ( 5)(+5i) Ans: 2πi(πi 5) π 2 +25 sin 6 π/6 d; : 2i = 2. Ans: πeiπ d; where is the boundary of a triangle with vertices:,, 7/2i. (2 cos πi sin πi) (viii) d; : = 2. Ans: πi2 5 (ix) Show that e 2 d = 2 ( i) πe2i ; where consists of = 2 anticlockwise and = clockwise. (x) e d; Ans: e 2 2πi (+2) 2 : =4 4. Evaluate d around = i + 2 5eit. Ans: 0 5. Evaluate ( 2 + 3 + 4/)d, : = 4. Ans: 8πi 6. Evaluate the upper bound of the integral without evaluating it? (a) (4 + )d, : Line segment from 0 to + i. Ans: 5 2. (b) d, : Line segment from 0 to i. Ans:. (c) 2d, : Line segment from i to 2 + i. Ans: 4 5. (d) (x2 + iy 2 )d, : = e it, π/2 < t < π/2. Ans: π 7. Which of the following integrals are compatible to apply the auchy theorem?. 2. sin +2i sin +2i d, : =. d, : + 3i =. 3. e d, : 3i = 6. 4. 5. =b =3 2 +b+ d. e d, 0 < b <. 4
6. +i 0 3 d, along y = x.. Evaluate 2. Evaluate 0 =3 2 2 d, Ans: 0 3 d where is the boundary of a triangle with vertices, 0 and 2i. Ans: 3. Evaluate the following integrals around the contour prescribed against it (i) +e d; : = 7e it, 0 t 2π. Ans: πi. (+iπ) 3 (ii) d; : 3 = 2, Ans: 3πi. 3 ( 2) 2 8 (iii) +e d; : = 7e it, 0 t 2π. Ans: πi. (+iπ) 3 (iv) d; : = 3, Ans: 0. 3 ( 2) 3 (v) 3 4 d. Ans: 3 64 2πi ( 6i) =0 (vi) d. Ans: 0. 4 =5 (vii) d. Ans: 4πi. 4 (+i) i =3/2 e 2 (viii) d. Ans: 2πi(2e 4 ); ( 2) 3 =3 sin (ix) d. Ans: 2πi cos ; ( ) 2 =2 cos (x) d. Ans: 0; (+3i) 6 +i =3/2 3 (xi) d; : = 5, Ans: 0. 2 (+i) 2 e (xii) d. Ans: 2πi sin a; 2 +a 2 a =2a e (xiii) d. Ans: eia π 2 +a 2 a (xiv) ia =a + e (+3) d; : i = 2, Ans: 0. 4. Evaluate (x 2 + iy 2 )ds; : = 2 where s is the arc length. Ans: 8π( + i) 5
Tutorial 6, B. Tech. Sem III, 5 October, 206 (Laurent s Expansions). State and prove Laurents series expansion of a function f(). 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. e (i). 2 ; = ( ) 3 Ans: e 2 [ + 2 + 2 + 4 + 2( ) +...; = is a pole of order 3 and Series ( ) 3 ( ) 2 ( ) 3 3 converges for all. (ii). ( 3) sin ; = 2; (+2) Ans: 5 + 5 +...; = 2 is an essential singularity and +2 6(+2) 2 6(+2) 3 20(+2) 4 Series converges for all 2. (iii). sin ; = 0; Z 3 Ans: 2 4..; = 0 is a removable singularity. Series converges for all. 3! 5! 7! (iv). ; = 2; (+)(+2) 2 Ans: + + ( + 2) + ( + +2 2)2 +..; = 2 a simple pole and Series converges for all in 0 < + 2 <. (v). ; = 3 2 ( 3) 2 Ans: 2 + ( 3) +...; = 3 is a pole of order 2 and Series converges 9( 3) 2 27( 3) 27 243 for all in 0 < 3 < 3. 3. Expand f() = in a Laurent s series valid for expansion (a). < < 3, (+)(+3) (b). > 3, (c) 0 < + < 2, (d) <, Ans: (a). + + + 2 + 3 ; 2 4 2 3 2 2 2 6 8 54 62 (b). 4 + 3 40 + ; 2 3 4 5 (c). + (+) ( + 2(+) 4 8 6 )2 + ; (d). 4 + 3 3 9 27 2 +. 4. Expand f() = in a Laurents series valid for (a). < < 2; (b). > 2; ( )(2 ) (c). 0 < 2 < ; (d). < ; (e). >. Ans: (a). + + 2 + 3 + + + ; 2 4 8 2 (b). 3 7 5 ; 2 2 3 4 (c). 2 ( 2) + ( 2 2)2 ( 2) 3 + ( 2) 4 ; (d). 3 2 4 2 7 8 3 5 9 4. (e). 2 2 ; ( ) ( ) 2 ( ) 3 5. Expand f() = in a Laurents series valid for (a). < 3; (b). > 3 3 Ans: (a). (i) /3 /9 /27 2 /8 3 ; (ii) + 3 2 + 9 3 + 27 4 +. 6. Expand f() = ( 2) Ans: (a). 2 n=0 in a Laurent s series valid for (a). 0 < < 2; (b). > 2 n 2 ; (b). n ; 2 n+2 n+ n= 6
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). ; = 0; e + sin (ii). ; = 0 2 +4 Ans: (i) < π; (ii) < 2 7
Tutorial 7, B. Tech. Sem IV, 24 Feb, 206 (Singularities and Residues). Discuss the types of each singularity for the following functions: (i). ; = π/4 is a simple pole. cos sin cot π (ii). ; = a is a pole of order 2. ( a) 2 e (iii). ; = ±i is a simple pole and = is an essential singularity. 2 + (iv). ; =, 2 are simple poles. (+)(+2) e (v). ; = kπ, k I is a simple pole. sin (vi). arcsin ; = /nπ, (n I+ ) simple pole and = 0 an isolated essential singularity. (vii). arctan π; = /n, (n I + ) is simple pole and = 0 isolated essential singularity. (viii). sin ; = 0 is a removable singularity. + 3 (ix). ; = 0, 2 are simple poles. 2 2 (x). 2 ; = 2 is a simple pole and = is a pole of order 2. (xi). (xii). ( ) 2 (+2) ; 4 + e i ; 2 +a 2 e (iπ+2kπ) 4, k = 0,, 2, 3, 4 each is a simple pole. = ±ai are simple poles. (xiii). e2 ; 3 = 0, a pole of order 2. e (xiv). 2 ; +e = iπ is a simple pole. e (xv). 2 ; = 0 is a simple pole. e (xvi). sin ; = 0 is a pole of order 2. 5 (xvii). ; = 0 is a pole of order 3. 2 (e ) (xviii). sinh ; = 0 is a simple pole. 2 ( (xix). 2 +)e = is a pole of order 3 and = i is a removable singularity. ( ) 3 ( i) ; 2. Examine the nature of singularity for the following functions and find its residue: Hint: residue is defined as coefficient of ( 0 in Laurent s series expansion of a function ) f() about a point 0. Thus, d residue for a pole of order m is lim m (( (m )! 0 d m 0 ) m f()) residue for a removable singularity=0. residue for an essential singularity 0 can be find by finding coefficient of 0 in its Laurent s series expansion about a point 0. (i). e +sin ; = 0 is a removable singularity, Residue= 0 4 (ii). ; = i is a a pole of order 3. Residue = 3. (+i) 3 6i e (iii). i ; = ±ai are simple poles. Residues are = e a. and= ea. 2 +a 2 2 2 (iv). ; = 0 a pole of order 3. Residue= 2 sinh 6 3. Using Residue theorem evaluate the following integrals: (i). 2π dθ, 2π Ans: 0 2+cos θ. 3 (ii). 2π dθ, a > Ans:... 0 2a cos θ+a 2 (iii). π +2 cos θ dθ, Ans: π. 0 5+3 cos θ 24 (iv). 2π cos 2θ dθ, Ans: π 0 5+4 cos θ 6 (v). 2π 0 cos 2 3θ 5 4 cos 2θ dθ, Ans: 3π 8 8
(Bays rule) Tutorial 9, B. Tech. Sem III, 22 October, 206. 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:... 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 the probability that his disease was diagnose correctly. Ans: 6 3 3. A manufacturing firm produces steel pipes in three plants with daily production volumes of 500, 000 & 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, 0.0. What is the probability that a pipe is selected at random from a day s output and found to be defective? Ans:... 4. In a university 4% of male students and % 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 000 parts with 0% defectives and the second shipment contains 500 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 0 urns of which each of 3 contains W, 9B balls, each others 3 contains 9W, B 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 B 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 the bag A was used to draw the ball is 6, find out the value of n. Ans: n = 4. 7 9