Series and Useful Formulas

Series and Useful Formulas

Appendix A Series and Useful Formulas A.1 Series and Useful Formulas Xn1 n .a C id/ D .a C l/ 2 iD0 where l D a C .n 1/d A.2 Geometric Series Xn1 a.1 rn/ ari D 1 r iD0 X1 a ari D 1 r iD0 where r ¤ 1 in the above two equations. © Springer International Publishing Switzerland 2015 541 F. Gebali, Analysis of Computer Networks, DOI 10.1007/978-3-319-15657-6 542 A Series and Useful Formulas A.3 Arithmetic-Geometric Series Xn1 a.1 rn/ rd 1 nrn1 C .n 1/ rn / .a C id/ ri D C 1 r .1 r/2 iD0 where r ¤ 1 in the above two equations. If 1<r<1, the series converge and we get X1 a rd .a C id/ ri D C 1 r .1 r/2 iD0 X1 r C r2 i 2 ri D .1 r/3 iD0 A.4 Sums of Powers of Positive Integers Xn n.n C 1/ i D 2 (A.1) iD1 Xn n.n C 1/.2n C 1/ i 2 D 6 (A.2) iD1 Xn n2.n C 1/2 i 3 D 4 (A.3) iD1 A.5 Binomial Series ! Xn n ani bi D .a C b/n i iD0 special cases ! Xn n D 2n i iD0 A Series and Useful Formulas 543 ! Xn n .1/i D 0 i iD0 !2 ! Xn n 2n D i n iD0 q.q 1/ q.q 1/.q 2/ .1 C x/q D 1 C qx C x2 C x3 2Š 3Š .1 C x/1 D 1 x C x2 x3 C x4 where x<1in the above equations. A.5.1 Properties of Binomial Coefficients ! ! n n D i n i ! ! n n n 1 D i i i 1 ! ! ! n n n C 1 C D i i 1 i ! ! n n D D 1 n 0 p nC 1 n nŠ 2n 2 e Stirling’s formula A.6 Other Useful Series and Formulas ! Xn n i ai1bni D na.a C b/n1 i iD0 ! Xn n i 2 ai bni D n.n 1/a2.a C b/n2 C na.a C b/n1 i iD0 544 A Series and Useful Formulas ! Xn n 1 1 1 ai .1 a/ni D Œ1 .1 a/n i 1 i n iD1 X1 ai D ea iŠ iD0 X1 ai .i C 1/ D ea .a C 1/ iŠ iD0 x lim .1 ˙ /n D e˙x n!1 n ! n lim xn D 0 jxj <1 n!1 i ! ! ! n n n C 1 C D i i 1 i a Án lim 1 D ea n!1 n ! i a n ni a e lim ai .1 a/ D a<1 n!1 i iŠ The last equation is used to derive the Poisson distribution from the binomial distribution. A.7 Integrals of Exponential Functions Z 1 ecx x D ecx d c Z 1 xecx2 x D ecx2 d 2c Z 1 1 eax dx D 0 a r Z 1 2 eax dx D ;a>0 0 a Z 1  à b 1 n C 1 xneax dx D a.nC1/=b 0 b b Appendix B Solving Difference Equations B.1 Introduction Difference equations describe discrete-time systems just like differential equations describe continuous-time systems. We encounter difference equations in many fields in telecommunications, digital signal processing, electromagnetics, civil engineering, etc. In Markov chains, and many queuing models, we often get a special structure for the state transition matrix that produces a recurrence relation between the system states. This appendix is based on the results provided in [1] and [2]. We start first by exploring simple approaches for simple situations, then we deal with the more general situation. B.2 First-Order Form Assume we have the simple recurrence equation si D asi1 C b (B.1) where a and b are given. Our task is to find values for the unknown si for all values of i D 0; 1; that satisfy the above equation. This is a first-order form since each sample depends on the immediate past value only. Since this is a linear relationship, we assume that the solution for si is composed of two components, a constant component c and a variable component v that depends on i. Thus, we write the trial solution for si as si D vi C c (B.2) © Springer International Publishing Switzerland 2015 545 F. Gebali, Analysis of Computer Networks, DOI 10.1007/978-3-319-15657-6 546 B Solving Difference Equations Substitute this into our recursion to get vi C c D a .vi1 C c/ C b (B.3) We can group the constant parts and the variable parts together to get c D acC b (B.4) vi D avi1 (B.5) and the value of the constant component of s is b c D 1 a (B.6) Assume a solution for vi in (B.5)oftheform i vi D (B.7) Substitute this solution in the recursion formula (B.5)forvi to get i D ai1 (B.8) which gives D a (B.9) and vi is given by i i vi D D a (B.10) Thus si is given from (B.2)as b s D ai C i 1 a (B.11) This is the desired solution to the difference equations. B.3 Second-Order Form Assume we have the simple recurrence equation siC1 C asi C bsi1 D 0 (B.12) B Solving Difference Equations 547 This is a second-order form since each sample depends on the two most recent past samples. Assume a solution for si of the form i si D (B.13) The recursion formula gives 2 C aC b D 0 (B.14) There are two possible solutions (roots) for ; which we denote as ˛ and ˇ; and there are three possible situations. B.3.1 Real and Different Roots ˛ ¤ ˇ When the two roots are real and different, si becomes a linear combination of these two solutions i i si D A˛ C Bˇ (B.15) where A and B are constants. The values of A and B are determined from any restrictions on the solutions for si such as given initial conditions. For example, if si represent the different components of the distribution vector in a Markov chain, then the sum of all the components must equal unity. B.3.2 Real and Equal Roots ˛ D ˇ When the two roots are real and equal, si is given by i si D .A C iB/ ˛ (B.16) B.3.3 Complex Conjugate Roots In that case we have ˛ D C j (B.17) ˇ D j (B.18) si is given by i si D ŒA cos.i / C B sin.i / (B.19) 548 B Solving Difference Equations B.4 General Approach Consider the N -order difference equations given by XN si D ak sik;i>0 (B.20) kD0 whereweassumedsi D 0 when i<0. We define the one-sided z-transform [3]ofsi as X1 i S.z/ D si z (B.21) iD0 Now take the z-transform of both sides of (B.20) to obtain " # XN X1 k .ik/ S.z/ s0 D ak z sik z (B.22) kD0 iD0 We assume that ak D 0 when k>Nand we also assume that si D 0 when i<0. Based on these two assumptions, we can change the upper limit for the summation over k and we can change the variable of summation of the term in square brackets as follows. " # X1 X1 k m S.z/ s0 D ak z sm z (B.23) kD0 mD0 where we introduced the new variable m D i k. Define the z-transform of the coefficients ak as X1 k A.z/ D ak z (B.24) kD0 Thus we get " # X1 m S.z/ s0 D A.z/ sm z mD0 D A.z/ S.z/ (B.25) Notice that the two summations on the RHS are now independent. Thus we finally get s0 S.z/ D (B.26) 1 A.z/ B Solving Difference Equations 549 We can write the above equation in the form s0 S.z/ D (B.27) D.z/ where the denominator polynomial is D.z/ D 1 A.z/ (B.28) MATLAB allows us to find the inverse z-transform of S.z/ using the command RESIDUE.a; b/ where a and b are the coefficients of the nominator and denominator polynomials A.z/ and B.z/, respectively, in descending powers of z1. The function RESIDUE returns the column vectors r, p, and c which give the residues, poles, and direct terms, respectively. The solution for si is given by the expression Xm .i1/ si D ci C rj .pj / i>0 (B.29) j D1 where m is the number of elements in r or p vectors. References 1. J.R. Norris, Markov Chains (Cambridge University Press, New York, 1997) 2. V.K. Ingle, J.G. Proakis, Digital Signal Processing Using MATLAB (Brooks/Cole, Pacific Grove, 2000) 3. A. Antoniou, Digital Filters: Analysis, Design, and Applications (McGraw-Hill, New York, 1993) Appendix C Finding s.n/ Using the Z-Transform When the transition matrix P of a Markov chain is not diagonalizable, we could use the z-transform technique to find the value of the distribution vector at any time instance s.n/.

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