Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 9 docx

... 1)cn+2+ cn−2= 0, for n ≥ 2cn +4 = −cn(n + 4) (n + 3) For our first solution we have the difference equationa0= 1, a1= 0, a2= 0, a3= 0, an +4 = −an(n + 4) (n + 3). For our second solution,b0= ... polynomial areα1=1 +1 − 3 /4 2=3 4 , α2=1 −1 − 3 /4 2=1 4 .Thus our two series solutions will be of the formw1= z3 /4 ∞n=0anzn, w2= z1 /4 ∞n=0bnzn.Substituting ... qN−jaj(α2) and dn= limα→α2ddα(α − α2)an(α) for n ≥ 0.12080.2 0 .4 0.6 0.811.21 .4 0.70.80 .9 1.11.2Figure 23.1: Plot of the Numerical Solution and the First Three...

... =16−1π2∞n=11n2∞n=11n2=π26 141 8 29. 5 HintsHint 29. 1Hint 29. 2Hint 29. 3Hint 29 .4 Write the problem in Sturm-Liouville form to show that the eigenfunctions are orthogonal ... πx)2dx=π3/3π5/30=10π2≈ 1.013 142 9 29 .4 ExercisesExercise 29. 1Find the eigenvalues and eigenfunctions ofy+ 2αy+ λy = 0, y(a) = y(b) = 0,where a < b.Write the problem in Sturm Liouville form. Verify ... Find the eigenvalues and eigenfunctions for: d 4 φdx 4 = λφ,φ(0) = φ(0) = 0,φ(1) = φ(1) = 0.Hint, Solution 144 2(b) We substitute x = 1/2 into the cosine series.1 4 =16−1π2∞n=1cos(nπ)n2∞n=1(−1)nn2=...

... ı1 /4 =2e−ıπ/61 /4 = 4 √2e−ıπ/ 24 11 /4 = 4 √2eı(πn/2−π/ 24) , n = 0, 1, 2, 31ı /4 =e(ı /4) log 1=e(ı /4) (ı2πn)=e−πn/2, n ∈ Z3087.11 HintsCartesian and Modulus-Argument FormHint ... coordinates.ez2=e(x+ıy)2=ex2−y2+ı2xy=ex2−y23 04 25 5075100-11Figure 7 .43 : The values of ıı.See Figure 7 .44 for a plot. -40 -20 20-10-5510Figure 7 .44 : The values of log ((1 + ı)ıπ).317Solution 7.13cot z = 47 (eız+e−ız) ... /2(eız−e−ız) /(ı2)= 47 eız+e−ız= 47 eız−e−ız 46 eı2z 48 = 0ı2z = log 24 23z = −ı2log 24 23z = −ı2ln 24 23+ ı2πn, n ∈ Zz = πn −ı2ln 24 23, n ∈ ZSolution 7. 14 1.log(−ı)...

... are,Res1z 4 + 1,eıπ /4 = limz→eıπ /4 z −eıπ /4 z 4 + 1= limz→eıπ /4 14z3=1 4 e−ı3π /4 =−1 − ı 4 √2,Res1z 4 + 1,eı3π /4 =1 4( eı3π /4 )3=1 4 e−ıπ /4 =1 − ı 4 √2,We evaluate ... ζj) for j = 1, . . . , n.7 04 we can apply Result 13 .4. 1.∞−∞1x 4 + 1dx = ı2πRes1z 4 + 1,eıπ /4 + Res1z 4 + 1,eı3π /4 The appropriate residues are,Res1z 4 + 1,eıπ /4 = ... integral converges for (a) < 1.725Figure 13.8: The real and imaginary part of the integrand for several values of α.Note that the integral exists for all nonzero real α and thatlimα→0+1−11x...

... the problem for u, (and hence the problem for y).As a check, then general solution for y isy = −13cos 2x + c1cos x + c2sin x.1115We guess a particular solution of the formyp= te−t(a ... v +baG(x|ξ)f(ξ) dξ.1 099 A particular solution isyp=t36et +4. The general solution of the differential equation isy = c1et+c2tet+t36et +4. We use the initial conditions ... y(0) = 1c1+ 4 = 1, c1+ c2= 1c1= −3, c2= 4 The solution of the initial value problem isy =t36+ 4t − 3et +4. 2. We consider the problemy+ 2y+ 5y = 4 e−tcos(2t),...

... =x331−1=231−1P2(x)P2(x) dx =1−11 4 9x 4 − 6x2+ 1dx =1 4 9x55− 2x3+ x1−1=251−1P3(x)P3(x) dx =1−11 4 25x6− 30x 4 + 9x2dx =1 4 25x77− 6x5+ 3x31−1=27Solution ... equation isα2+1 4 α −18= 0α +12α −1 4 = 0.Since the roots are distinct and do not differ by an integer, there will be two solutions in the Frobenius form.w1= z1 /4 ∞n=0an(α1)zn, ... criterion for series?SolutionProblem 23.3Define absolute convergence and uniform convergence. What is the relationship between the two?SolutionProblem 23 .4 Write the geometric series and the...

... x3, x 4 } is independent, but not orthogonal in the interval [−1, 1]. Using Gramm-Schmidt orthogo-12 84 123 4 56-60 -40 -20Figure 24. 3: log(error in approximation)In Figure 24. 4 we see ... 21)π 4 P 4 (x)=1058π 4 [(315 − 30π2)x 4 + ( 24 2− 270)x2+ (27 − 2π2)]The cosine and this polynomial are plotted in the second graph in Figure 25.1. The le ast squares fit method usesinformation ... rule of thumb predicts the actual optimal series.12 69 5101520 25-18-16- 14 -12Figure 24. 4: The logarithm of the error in using n terms. 24. 4 Asymptotic SeriesA function f(x) has an asymptotic...

... theformula to obtain information about the eigenvalues before we solve a problem.Example 27 .4. 2 Consider the self-adjoint eigenvalue problem−y= λy, y(0) = y(π) = 0.1322Example 27 .4. 1 ... eigenfunctionφ. Green’s formula statesφ|L[φ] − L[φ]|φ = 0φ|λφ − λφ|φ = 0(λ − λ)φ|φ = 0Since φ ≡ 0, φ|φ > 0. Thus λ = λ and λ is real.13 19 27.7 HintsHint 27.113272. For k = n, φk, ... Heaviside function. On the interval x ∈ (−π, π)x−πδ(t) dt = H(x) =1 for 0 < x < π0 for − π < x < 0.1 299 26.3 Exercises1311...

... solution and discuss in as much detail as possible what goes wrong.13 64 Note that this formula is valid for m = 0, 1, 2, . . Similarly, we can multiply by sin(mx) and integrate to solve for bm. ... ∼∞n=1oddn 4 nx−πsin(nt) dt=∞n=1oddn 4 n−1ncos(nt)x−π=∞n=1oddn 4 n2(−cos(nx) + (−1)n)= 4 ∞n=1oddn−1n2− 4 ∞n=1oddncos(nx)n2Since this series converges uniformly, 4 ∞n=1oddn−1n2− ... =2π0xdydtdt.Express x(t) and y(t) as Fourier series and use the completeness and orthogonality relations to show that L2 4 A ≥ 0.)Exercise 28.201. Find the Fourier sine series expansion and the Fourier...