Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 2 docx

... +∞n=1ndnRn−1sin(nθ)18 15 m=3,n=1 m=3,n =2 m=3,n=3m =2, n=1 m =2, n =2 m =2, n=3m=1,n=1 m=1,n =2 m=1,n=3m=3,n=1 m=3,n =2 m=3,n=3m =2, n=1 m =2, n =2 m =2, n=3m=1,n=1 m=1,n =2 m=1,n=3Figure 37.4: The modes of ... the sum and difference of these solutions to obtainφ = expα 2 − β 2 t ±αaxcossin 2 βt ±βax1817m =2, n=1m =2, n =2 m =2, n=3m=1, n=1m=1, n =2 m=1, n=3m=0, n=1m=0, n =2 m=0, ... ıβ) 2 ,XX=(α + ıβ) 2 a 2 T = e xp(α + ıβ) 2 t, X = exp±α + ıβaxT = e xpα 2 − β 2 t + 2 βt, X = exp±αax ±ıβaxφ = expα 2 − β 2 t ±αax + ı 2 βt...

... =2x 2 x 2 + βx + χNow we use the fact that f(x) is even to conclude that q(x) is even and thus β = 0.f(x) =2x 2 x 2 + χFinally, we use that f(1) = 1 to determine χ.f(x) =2x 2 x 2 + 1 20 Einstein ... there and that χ = 0.f(x) =ax 2 x 2 + βx + χWe n ote that f (x) → 2 as x → ∞. This d etermine s the parameter a.limx→∞f(x) = limx→∞ax 2 x 2 + βx + χ= limx→∞2ax2x + β= limx→∞2a 2 = ... a 2 j + a3k) × (b1i + b 2 j + b3k)= a1i × (b1i + b 2 j + b3k) + a 2 j × (b1i + b 2 j + b3k) + a3k × (b1i + b 2 j + b3k)= a1b 2 k + a1b3(−j) + a 2 b1(−k) + a 2 b3i...

... A = 0, 2 ln t +1 2 t 2 + c, A = 0−1 2 t 2 + ln t + c, A = 2 y =1A+t 2 A +2 + ct−A, A = 2 ln t +1 2 t 2 + c, A = 0−1 2 + t 2 ln t + ct 2 , A = 2 For positive A, the ... are:y =1A+t 2 A +2 , A > 01A+t 2 A +2 + ct−A, A < 0, A = 2 −1 2 + t 2 ln t + ct 2 , A = 2 Equations in the Complex PlaneSolution 14. 15 1. Consider the equation w+sin ... 5 c1c 2 =13c1=1 2 , c 2 =1 2 The solution subject to the initial condition isx =1 2 11e−t+1 2 1 5 e3t For large t, the solution looks likex ≈1 2 1 5 e3t.Both coordinates...

... −1−√ 5 2 1+√ 5 2 √ 5 =1+√ 5 2 1+√ 5 2 √ 5 =1√ 5 Substitute this result into the equation for c 2 .c 2 =1r 2 1 −1√ 5 r1= 2 1 −√ 5 1 −1√ 5 1 +√ 5 2 = − 2 1 −√ 5 1 −√ 5 2 √ 5 = −1√ 5 Thus ... solutionbn=nj=0−4j 2 − 2j + 1(2j + 2) (2j + 1).Thus we have thatan=n /2 j=0−4j 2 −2j+1(2j +2) (2j+1) for even n,0 for odd n.1193c1=1 − r 2 r 2 1− r1r 2 =1 −1−√ 5 2 1+√ 5 2 √ 5 =1+√ 5 2 1+√ 5 2 √ 5 =1√ 5 Substitute ... isan= c11 +√ 5 2 n+ c 2 1 −√ 5 2 n.From the initial conditions we havec1r1+c 2 r 2 = 1c1r 2 1+c 2 r 2 2= 1.Solving for c 2 in the first equation,c 2 =1r 2 (1 − c1r1).We...

... cases for even and odd n separately.a2n= −a2n 2 2n=a2n−4(2n)(2n − 2) = (−1)na0(2n)(2n − 2) ···4 · 2 = (−1)na0nm=12m, n ≥ 0a2n+1= −a2n−12n + 1=a2n−3(2n + 1)(2n − 1)= ... =x331−1= 2 31−1P 2 (x)P 2 (x) dx =1−1149x4− 6x 2 + 1dx =149x 5 5− 2x3+ x1−1= 2 5 1−1P3(x)P3(x) dx =1−114 25 x6− 30x4+ 9x 2 dx =14 25 x77− 6x 5 + ... polynomial.y 2 =2n+1m=1odd m1m!m 2 k=1odd kk(k + 1) − α(α + 1)xm 122 9Calculating the bn’s,b1= −b01 ·3 2 b 2 =b01 · 2 ·3 2 · 5 2 bn=(−1)n 2 nb0n! · 3 · 5...

... v]x=a 27 .2 Formally Self-Adjoint OperatorsExample 27 .2. 1 The linear operatorL[y] = x 2 y+ 2xy+ 3yhas the adjoint operatorL∗[y] =d 2 dx 2 (x 2 y) −ddx(2xy) + 3y= x 2 y+ ... 4xy+ 2y −2xy− 2y + 3y= x 2 y+ 2xy+ 3y.In Example 27 .2. 1, the adjoint operator is the same as the operator. If L = L∗, the operator is said to be formallyself-adjoint.13 15 Here ... eigenfunctions for each of the positive eigenvalues.We choose the eigenvalues and eigenfunctions.λ0= 0, φ0=1 2 λn= n 2 , φ2n−1= cos(nx), φ2n= sin(nx), for n = 1, 2, 3, . . .13 32 The eigenvalues...

... that∞n=1na 2 n+ b 2 n+ c 2 n+ d 2 n=∞n=1n 2 a 2 n+ b 2 n+ c 2 n+ d 2 n14 15 -1-0 .5 0 .5 1-0 .2 -0.10.10 .2 -1-0 .5 0 .5 1-0 .2 -0.10.10 .2 Figure 28 .13: The odd and even periodic ... inequality |ab| ≤1 2 |a 2 + b 2 |, which follows from expanding(a − b) 2 ≥ 0.A ≤π 2 ∞n=1na 2 n+ b 2 n+ c 2 n+ d 2 n≤π 2 ∞n=1n 2 a 2 n+ b 2 n+ c 2 n+ d 2 nWe can express ... function.λ1≤π0(u) 2 dxπ0u 2 dx=π0(2x − π) 2 dxπ0(x 2 − πx) 2 dx=π3/3π 5 /30=10π 2 ≈ 1.0131 429 29 .4 ExercisesExercise 29 .1Find the eigenvalues and eigenfunctions ofy+ 2 y+...

... transform of ˆy1.y1=t36−t 2 2+1 2 e−t+1 2 sin t −1 2 cos t.We can find y 2 and y3by differentiating the expression for y1.y 2 =t 2 2− t −1 2 e−t+1 2 cos t +1 2 sin ... −1 2 ∞0e−rt√π−ı√re 2 √a√r(−dr)=1 2 √π∞0e−rt1√re− 2 √a√r+e 2 √a√rdr=1 2 √π∞01√re−rt 2 cos 2 √a√rdr 1 52 8Hint 31.17Hint 31.18Hint 31.19Hint 31 .20 Hint 31 .21 Hint 31 .22 Hint 31 .23 150 6Exercise 31. 12 Find the inverse Laplace transform ofˆf(s) ... =1s3− 2s 2 + s − 2 with the following methods. 1. Expandˆf(s) using partial fractions and then use the table of Laplace transforms. 2. Factor the denominator into (s − 2) (s 2 + 1) and then...