advanced mathematics pdf

www.EngineeringBooksPDF.com www.EngineeringBooksPDF.com This page intentionally left blank www.EngineeringBooksPDF.com www.EngineeringBooksPDF.com Copyright © 2009, New Age International (P) Ltd., Publishers Published by New Age International (P) Ltd., Publishers All rights reserved No part of this ebook may be reproduced in any form, by photostat, microfilm, xerography, or any other means, or incorporated into any information retrieval system, electronic or mechanical, without the written permission of the publisher All inquiries should be emailed to rights@newagepublishers.com ISBN (13) : 978-81-224-2719-6 PUBLISHING FOR ONE WORLD NEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS 4835/24, Ansari Road, Daryaganj, New Delhi - 110002 Visit us at www.newagepublishers.com www.EngineeringBooksPDF.com Preface We feel happy and honoured while presenting this book “Advanced Mathematics” for engineering students studying in B Tech IV Semester (EE and EC Branch) of Rajasthan Technical University and all Indian Universities In this book we have presented the subject matter in very simple and precise manner The treatment of the subject is systematic and the exposition easily understandable All standard examples have been included and their model solutions have also been given This book falls into five units: In first and second unit we have discussed the Numerical Analysis The unit I deals with Finite Difference—Forward, Backward and Central difference, Newton’s formula for Forward and Backward differences, Interpolation, Stirling’s formula and Lagrange’s interpolation formula Solution of nonlinear equations in one variable by Newton-Raphson method, Simultaneous algebraic equation by Gauss and Regula-Falsi method, Solution of simultaneous equations by Gauss elimination and Gauss Seidel methods, Fitting of curves (straight line and parabola of second degree) by method of least squares are also discussed In unit II, we have discussed Numerical differentiation, Numerical Integration, Trapezoidal rule, Simpson’s one-third and three-eighth rules Numerical solution of ordinary differential equations of first order, Picard’s method, Euler’s and modified Euler’s methods Miline’s method and Runga-Kutta fourth order method, Simple linear difference equations with constant coefficients are also discussed in the unit Unit III deals with the special functions, Bessel’s functions of first and second kind, Simple recurrence relations, Orthogonal property of Bessel’s transformation and generating functions Legendre’s function of first kind, Simple recurrence relations, Orthogonal property and generating function are also discussed In unit IV, the basic principles of probability theory is given in order to prepare the background for its application to various fields Baye’s theorem with simple applications, Expected value, Theoretical probability distributions—Binomial, Poisson and Normal distributions are discussed Unit V deals with Lines of regression, concept of simple Co-relation and Rank correlation Ztransforms, its inverse, simple properties and application to difference equations are also discussed We are grateful to New Age International (P) Limited, Publishers and the editorial department for their commitment and encouragement in bringing out this book within a short span of period AUTHORS www.EngineeringBooksPDF.com This page intentionally left blank www.EngineeringBooksPDF.com Acknowledgement We are thankful to Prof L K Maheshwari, Vice-Chancellor, Prof R K Mittal, Deputy Director (Administration), Prof G Raghurama, Deputy Director (Academic) of Birla Institute of Technology & Science (BITS), Pilani for their encouragement and all over support in completing the book The authors are also highly thankful to Dr P S Bhatnagar, Director, BK Birla Institute of Engineering & Technology (BKBIET), Pilani for his motivation, time to time support and keen interest in the project Dr S R Singh Pundir, Reader, D N College Meerut, deserves for special thanks.Thanks are also due to Mr Anil and Mr Rahul of BKBIET for providing necessary help during the project We also place our thanks on record to all those who have directly or indirectly helped us in completion of the project At the last but not in the least we are very much indebted to our family members without whom it was not possible for us to complete this project in time Thanks are also due to M/s NEW AGE INTERNATIONAL (P) LTD PUBLISHERS and their editorial department AUTHORS www.EngineeringBooksPDF.com This page intentionally left blank www.EngineeringBooksPDF.com Contents Preface v Acknowledgement vii UNIT I : NUMERICAL ANALYSIS-I CHAPTER CALCULUS OF FINITE DIFFERENCES 3–20 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 Finite Differences Forward Differences Backward Differences Central Differences Shift Operator E Relations Between the Operators Fundamental Theorem of the Difference Calculus Factorial Function To Show that Δn x(n) = n ! hn and Δn+1 x(n) = To Show that f(a + nh) = f(a) + nC1 Δf(a) + nC2 Δ2 f(a) + + nCn Δn f(a) Solved Examples 10 Exercise 1.1 18 Answers 20 CHAPTER INTERPOLATION 21–41 2.1 2.2 2.3 2.4 2.5 To Find One Missing Term 21 Newton-Gregory’s Formula for Forward Interpolation with Equal Intervals 22 Newton-Gregory’s Formula for Backward Interpolation with Equal Intervals 23 Lagrange’s Interpolation Formula for Unequal Intervals 24 Stirling’s Difference Formula 25 Solved Examples 25 Exercise 2.1 39 Answers 41 *CHAPTER SOLUTION OF LINEAR SIMULTANEOUS EQUATIONS 42–51 3.1 3.2 3.3 Linear Equations 42 Gauss Elimination Method 43 Gauss-Seidel Method 44 *Not for EC branch students www.EngineeringBooksPDF.com 310 ADVANCED MATHEMATICS Solution: (i) We have ∞ Z{un} = ∑uz n −n n=−∞ + + + z z z z = 15z4 + 10z3 + 7z2 + 4z + – (ii) We have Z{un} = ∑ n=−2 ∞ (iii) We have Z{un} = ∑ −∞ Ans −n 1 1 z = 4z2 + 2z + + + + + Ans n 4z 8z 16 z 2z 1 1 + + + Ans n = + 9z + 3z + + 3z z 27 z 3 RS a UV , n ≥ Tn !W n Example 2: Find the Z-transform of Solution: We have Ra U ZS V = ∑ T n !W ∑ FH IK a F aI F aI + =1+ + H K z 2! z 3!H zK Ra U Example 3: Find the Z-transform of S n V , n ≥ T W n ∞ n=0 a n −n z = n! ∞ n=0 1n a n! z n a z = e Ans n Solution: We have RS a UV TnW=∑ n Z ∞ n =1 ∑ FH IK ∞ a a n −n z = n n n =1 F H = – log − I K FG H Z 2n + sin nπ − 3a 4 = F I H K a a + z z + a Ans z Example 4: Find the Z-transform of 2n + sin Solution: We know that n nπ – 3a4 IJ = 2Z(n) + 5Z FG sin nπ IJ H 4K K – 3a4 Z(1) π z sin z z − 3a =2 +5 π ( z − 1) z −1 z − z cos + www.EngineeringBooksPDF.com FI HK a z 311 Z-TRANSFORM 5z 2z z = + − 3a z −1 ( z − 1) z − 2z +1 = 2z 5z 3a z Ans + − 2 ( z − 1) ( z − z + 1) z − Example 5: Find the Z-transform of (n – 1)2 Solution: We know that Z[(n – 1)2] = Z(n2 – 2n + 1) = Z(n2) – 2Z(n) + Z(1) = z2 + z z z −2 + ( z − 1) ( z − 1) ( z − 1) = z + z − z( z − 1) + z( z − 1) z − 3z + z = ( z − 1) ( z − 1) Ans Example 6: Find the Z-transform of sin (an + b), where a and b are constants Solution: We know that Z [sin (an + b)] = Z(sin an cos b + cos an sin b) = cos b Z (sin an) + sin b Z (cos an) = cos b = Thus z − z cos a z sin a + sin b z − z cos a + z − z cos a + z(sin a cos b + z sin b − cos a sin b) z − z cos a + z[sin ( a − b) + z sin b] Ans z − z cos a + Z [sin (an + b)] = Example 7: Find the Z-transform of the following: (i) 2n (ii) sin 2n Solution: (i) We have Z(an) = z z−a Z(2n) = z z−2 Then ∞ ∞ Aliter: We have Z(an) = ∑ a n z −n = n=0 ∞ ∴ Z(2n) = ∑ n=0 (iii) an + ∑ n=0 an zn F I + F 2I H K H zK 2n 2 + n = + z z z www.EngineeringBooksPDF.com 312 ADVANCED MATHEMATICS = (ii) We have Then (iii) We have = 1− z z Ans z−2 Z(sin nθ) = z sin θ z − z cos θ + Z(sin 2n) = z sin Ans z − z cos + 2 Z(an + 3) = z (ana3) = a3 z (an) = Thus, z (an + 3) = a3 z z−a a3 z Ans z−a Example 8: Find the Z-transform of the following: (i) nan (ii) n2 an Solution: (i) We have Z(n) = U(z) = Using damping rule, we get Z(nan) = U (ii) We have z ( z − 1) F z I = z/a H aK F z I H a − 1K Z(n2) = U (z) = = az ( z − a) Ans z2 + z ( z − 1) Using damping rule, we get F zI + F zI H a K H a K az + a z z a ) = UF I = H a K F z − 1I = (z − a) Ha K 2 Z(n2 n 3 Ans Example 9: Find the Z-transform of the following: (ii) an cos nθ (i) cos nθ Solution: (i) We have Then Z(1) = U (z) = Z(e–in θ) = Z = z z −1 LMde i 1OP = U (e N Q iθ − n iθ z) z( z − e iθ ) e iθ z z = = ( z − e − iθ ) ( z − e i θ ) e iθ z − z − e −iθ www.EngineeringBooksPDF.com (by damping rule) 313 Z-TRANSFORM = or Z(cos nθ – i sin nθ) = z − ze iθ z − z(e iθ + e − iθ ) + z − z(cos θ + i sin θ) z − z cos θ + Compair on both sides real parts, we get Z(cos nθ) = (ii) Let Then z − z cos θ Ans z − z cos θ + Z (cos nθ) = U(z) F zI H aK F z I − F z I cos θ H aK H aK = F z I − 2F z I cos θ + H aK H aK Z(an cos nθ) = U [by damping rule] 2 Z(an cos nθ) = z − az cos θ Ans z − az cos θ + a Example 10: Find the Z-transform of (ii) an sinh nθ (i) sinh nθ Solution: (i) We have FG e − e IJ = [Z(e ) − Z(e ) ] H K 1L z z O z( z − e ) − z( z − e ) = M − P= Nz − e z − e Q ( z − e )( z − e ) nθ − nθ θ n Z(sinh nθ) = Z −θ n −θ θ (ii) We have where −θ −θ = z sin h θ z (e θ − e − θ ) = θ −θ 2 z − z(e + e ) + z − z cos h θ + = z sin h θ Ans z − z cos h θ + Z(an sinh nθ) = Z[(a–1)–n sinh nθ] = U (a–1z) U(z) = Z(sinh nθ) ∴ θ θ Z(an sinh nθ) = a −1 z sin h θ ( a z) − 2( a −1 z ) cos h θ + −1 www.EngineeringBooksPDF.com (By damping rule) 314 ADVANCED MATHEMATICS (an sinh nθ) = za sin h θ Ans z − az cos θ + a 2 FG IJ = e H n !K F IJ (ii) Z F I Hence, evaluate (i) Z G GH n + ! JK H n + !K 1/z Example 11: Show that Z Solution: We have FG IJ = ∑ H n !K ∞ Z n=0 −n 1 1 1 z =1+ + + n! 1! z ! z ! z3 = e1/z Hence proved (i) Shifting one unit to the left, we get n! Z (ii) Shifting FG IJ = z LMZFG IJ − 1OP = z(e H (n + 1) !K N H n !K Q 1/ z − 1) one unit to the left, we get ( n + 1) ! Z FG IJ = z LMZFG IJ − 1OP = z [z(e H (n + 2) !K N H (n + 1) !K Q 1/z – 1) – 1] = z2 [e1/z – – z–1] Ans Example 12: If U(z) = 2z + 3z + then find u1 , u2 and u3 (z − 3) Solution: We have U (z) = z + 3z + (2 + 3z −1 + z −2 ) = ( z − 3) z (1 − 3z −1 )3 Then by initial value theorem u0 = lim U ( z ) = lim z→∞ z→0 + z −1 + z −2 =0 z (1 − 3z −1 ) u1 = lim z[U(z) – u0] = lim z z→∞ z→0 + 3z −1 + z −2 =2 (1 − 3z −1 ) z u2 = lim z2[U(z) – u0 – u1 z–1] z→∞ = lim z z→∞ LM1 × + 3z + 4z N z (1 − 3z ) −1 −2 −1 www.EngineeringBooksPDF.com −0− OP Q = 21 z 315 Z-TRANSFORM u3 = lim z3 [U (z) – u0 – u1z–1 – u2 z –2] z→∞ = lim z z→∞ LM × + 3z + 4z N z (1 − 3z ) −1 −2 −1 −0− OP Q 21 = 139 − z z2 EXERCISE 2.1 Find the Z-transform of the following: (i) 5n (iii) (n + 1)2 (ii) sin 3n (iv) (cos θ + i sin θ)n (vii) sin (n + 1) θ nπ + 5a (v) 3n – sin (viii) sinh (vi) F nπ I H2K ane − a n! (ix) cos F kπ + π I H 4K Find the Z-transform of the following: (i) e an (ii) ne an Find the Z-transform of (i) sin nθ Find the Z-transform of (i) cosh nθ Find the Z-transform of (i) cos F nπ + π I H 4K (iii) n2ean (ii) an sin (ii) an nθ cosh nθ (ii) cosh F nπ + θI H2 K Find the Z-transform of n sin nθ If U(z) = z + z + 14 , find u0, u1, u2 and u3 ( z − 1)4 Find the Z-transform of et sin 2t Using Z(n) = 10 Using Z(n2) = z ( z + z ) cos θ − z , show that Z(n cos nθ) = ( z − 1) ( z − z cos θ + 1)2 (z3 + z2 ) (z + z) 2] = , show that Z[(n + 1) ( z − 1)3 ( z − 1)3 11 Find Z (– e–an sin nθ) 12 If Z{un}= z z , then find Z{un + 2} + z −1 z +1 13 Find the Z-transform of unit impulse function which is given by RS T if n = δn = if n ≠ www.EngineeringBooksPDF.com 316 ADVANCED MATHEMATICS 14 Find the Z -transform of {un} where {un} = RS4 T3 n n if n < if n ≥ 15 Find the Z-transform of the discrete unit step function given by un = RS1 T0 if n ≥ if n < ANSWERS z (2 z + 1) ( z − 1)3 z z−5 (ii) z sin z − z cos + (iii) (iv) z z − eiθ (v) 2z (3 − 5a )z + 5az – z − 2z + ( z − 1)2 a( z (vi) e (vii) z sin θ z − z cos θ + z sin h π / 2 z − z cos h π / + (ix) ea z ( z − e a )2 (iii) (i) (viii) (i) z z − ea (i) z sin θ z − z cos θ + (ii) az sin θ z − az cos θ + (i) z − z cos h θ z − z cos h θ + (ii) z ( z − a cos h θ ) z − az cos h θ + a (i) 2 z( z − 1) 2( z + 1) (ii) − 1) z2 − z ( z + 1) ze a ( z + e a ) ( z − e a )3 2 F π − θI H2 K πI F − z cos h H 2K +1 z cos h θ − z cos h (ii) z2 z ( z − 1) sin θ ( z − z cos θ + 1)2 ez sin z − ez + e 11 12 z ( z − z + 2) ( z − 1) ( z + 1) 13 14 −z z − z + 12 −1 u0 = 0, u1 = 0, u2 = and u3 = 13 e a z sin θ e z − e a z cos θ + 2a www.EngineeringBooksPDF.com 317 Z-TRANSFORM 2.11 INVERSE Z-TRANSFORM LM MN ∑ u ∞ It is denoted by z–1U (z) = un and defined as un = Z–1 [U(z)] = Z–1 n OP PQ z −n n=0 2.12 CONVOLUTION THEOREM ∞ If Z–1[U (z)] = un and Z–1 [V(z)] = then Z–1[U (z) V(z)] = ∑u m m=0 − m = u ∗ v n n Where, the symbol ∗ denotes the convolution operation ∞ Proof: We have ∑uz Z (un) = U (z) = n ∞ −n and Z(vn) = V(z) = n=0 ∞ ∴ ∞ ∑ u z ∑ v z U (z) V(z) = −n n n n=0 = (u0 + ∑vz n −n m=0 −n n=0 u1z–1 + u2 z–2 + + un z–n ) × (v0 + v1z –1 + v2z –2 + + z –n ) ∞ ∑ (u v = n + u1vn −1 + + un −1v1 + un v0 ) z − n n=0 F GG ∑ u H n = Z(u0vn + u1vn–1 + + un – v1 + unv0) = Z m=0 n Hence ∑u Z–1 [U(z) V(z)] = m − m Hence proved m=0 SOLVED EXAMPLES LM z OP by convolution theorem N (z − 2) Q Example 1: Find Z–1 LM z OP = Nz − 2Q LM z OP = Z LMFG z IJ FG z IJ OP N (z − 2) Q NH z − K H z − K Q Solution: We have Z–1 n ∴ Z–1 −1 www.EngineeringBooksPDF.com m − m I JJ K 318 ADVANCED MATHEMATICS n = ∑2 m 2n−m (By Convolution theorem) m=0 n ∑ = 2n m − m = 2n n=0 n ∑ = 2n (n + 1) Ans n=0 LM z OP N (z − a) (z − b) Q L z OP = a and Z LM z OP = b Solution: We have Z M Nz − bQ Nz − aQ L z OP = Z LM z × z OP ∴ Z M N (z − a)(z − b) Q N z − a z − b Q Example 2: Using convolution theorem evaluate Z −1 –1 n –1 n −1 –1 n = ∑a m n− m b (By Convolution theorem) m=0 L a a a a O ∑ FH b IK = b MN1 + b + FH b IK + + FH b IK PQ R| F a I − 1U| |V = b RS a − b UV |S H b K a || b − || T b (a − b) W W T n = bn m n n m=0 n +1 = bn n n+ n LM z OP = a − b N (z − a)(z − b) Q a − b L 2z(2z − 1) OP Example 3: Find Z M N z − 5z + 8z − Q n +1 n+1 n+ Z–1 Ans –1 Solution: We have z( z − 1) z( z − 1) = z − 5z + 8z − ( z − 1) ( z − 2) = Then, Z–1 LM Nz 2z 2z 6z − + ( z − 1) ( z − 2) ( z − 2) OP Q (By partial fraction) LM N 2z 2z 6z z( z − 1) − + = Z–1 z − z − (z − 2)2 − 5z + 8z − = 2Z–1 LM z OP – 2Z L z O + 3Z MN z − PQ N z − 1Q –1 = 1n – 2.2n + 3n 2n –1 Ans www.EngineeringBooksPDF.com OP Q LM 2z OP N ( z − 2) Q 319 Z-TRANSFORM 2z + 3z (z + 2) (z − 4) Examples 4: Find the inverse Z-transform of FG H z + 3z z 11 z = + ( z + 2) (z − 4) ( z + 2) z − Solution: We have IJ K LM 2z + 3z OP = Z LM z + 11 z OP N ( z + 2) ( z − 4) Q N z + z − Q Then Z–1 –1 = 11 (–2)n + (4)n 6 Ans FG z IJ by power series method H z + 1K F z IJ = log = − log F1 + I U(z) = log G H zK H z + 1K F + I H zK L 1 + OP =– M − N z z 3z Q Example 5: Find the inverse Z-transform of log Solution: We have ∞ =– 1 ( − 1) n − n + − = z z 2z n 3z n=0 ∑ LM MN ∑ ∞ ∴ Z–1 [U(z)] = Z–1 n=0 ⇒ un = ( − 1) n − n z n OP PQ ( − 1) n if n ≥ Ans n Example 6: Find Z–1 LM z OP for | z | > N (z − 1) (z − 2) Q Solution: We have U(z) = Since, | z | > ⇒ | z | > and Now z = − ( z − 1) ( z − 2) z − z − 1 2