MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 2 pdf

... substitutions kin and 24 2-n (2. 74) to write (2. 75) n=O C21 (2 - n)!n! (1 - 2n)! Comparing this with the right-hand side of Equation (2. 65), which is we obtain (2. 76) (2. 77) 2. 3.3 Recursion ... write &s+l = - (4s + 3, - p2s (O) U2S +2( avo), s = 0,1 ,2, 2 (2s +2) (2. 1 32) Substituting these in Equation (2. 127 ) we finally obtain the potential outside the sphere as ... obtain (2. 61) (2. 62) (2. 63) (2. 64) thus proving the equivalence of Equations (2. 60) and (2. 59). 2. 3 .2 Generating Function Another way to define the Legendre polynomials is by using...

MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 7 pdf

... div and the grad operators for the fol- i) ds2 = dr2 + p2dB2 + dz2 ii) ds2 = [ ] dr2 + r2a2 + r2 sin2 Odd2, 1 - kr2/R2 where k takes the values k = 0,1, -1. 21 6 ... 10.9): 1 V 2J = J [." - (,) x'] 1 - u2/c2 1 V -1 z - - J [-(,)."+x'] 1 - u2/c2 (10 .25 2) (10 .25 3) (10 .25 4) (10 .25 5) -2 - 2 x -x 2= x. 3 Inverse transformation ... element and the defi- nition of the proper time to find uUuu as ds2 = c2dt2 - (d~’)~ - (d~~)~ - (d~~)~, (10 .29 6) (10 .29 7) c2 = uau,. Thus. (10 .29 8) ( 10 .29 9) U 22 P pU...

MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 13 pdf

... (15 .22 5) Applying this formula with z = x to the sine and cosine functions we obtain 22 00 sinz=zn (I ) nW n=l (15 .22 6) and (15 .22 7) These products are finite for all the finite values ... l1 (2k + 2) (2lc + 1) lim k - - 1 = lim k k-cc [Uztt2 ] k+oo [ (2k )2- n2 6k +2+ n2 3 k-co (21 C)Z - n2 which indicates that the series is convergent at the end points as ... 1.3 1.3.5 . (2n - 3) + (-1)n-l 2x (2x2)"- 1 where (15.190) 454 INFINITE SERIES 4. n 5. 6. (15. 124 ) (15. 125 ) 7. In the interval [O, 11 and for s 2 1, the only...

MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 1 potx

... Generators Normal Forms 25 0 25 1 25 1 25 5 25 5 25 7 25 8 26 0 26 1 26 2 26 4 26 6 26 8 26 9 27 2 27 2 27 3 27 4 27 4 27 4 27 5 27 6 27 7 27 8 27 9 28 1 28 1 28 2 28 5 Problems 28 8 ... Harmonics and Representations of R(3) 11.11.1 Angular Momentum in Quantum Mechanics 20 2 20 4 20 5 20 7 20 8 20 8 20 9 21 1 21 3 21 4 21 5 21 6 22 3 22 4 22 4 22 6 22 7 22 8 23 1 23 2 23 4 ... of Diflerintegrals in Science and Engineering 424 14.7 .2 Fractional Fokker-Planck Equations 427 Problems 429 14.7.1 Continuous Time Random Walk (CTRW) 424 15 INFINITE SERIES 431 15.1...

MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 3 docx

... H: (2) - 22 HL (x) + 2nHn (x) = 0. Hermite Polynomials HO (2) = 1 HI (2) = 22 H2(z) = -2 + 4x2 H&) = - 122 + 823 H4 (2) = 12 - 48x2 + 16x4 HS (2) = 120 2 - ... Frobenius method to i> 2x 3 -++x2-++3y=O d2Y dY dx2 dx ii) iii) iv) v) vi) 3 d2Y 2 dY 8 x -++x -+(x3+-x)y=O 9 dx2 dx 3 d2Y 2 dY 3 x -+fx -+(x3+-x)y=O dx2 dx 4 2d2Y dY ... left-hand side of Equation (4.61) explicitly we get 00 &e-t2+2tx -2+ 2sx - 12 e ex (4. 62) = e2st (s + t) ,h, (4.63) where we have defined u=x-(s+t). (4.64) Expanding this in...

MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 4 pptx

... C(c0sx) with a two-term recursion relation. 5 .2 Using the line element ds2 = c2dt2 - &(t )2[ dX2 + sin2 xdB2 + sin2 Xsin2 13d4~], find the spatial volume of a closed universe. ... functions are given as X1 22 + ), x<1, jdx) + (21 + l)!! (1 - 2( 2l+ 1) (21 - l)!! 22 f - xl+l (1 - 2( 1- 21 ) + ), x<1, 1 LIT jl(x) = ;sin(x - -), x >> ... -1(1+ 2) + (w2 - -)@. % 90 BESSEL FUNCTIONS 6.3 .2 Integral Definitions Bessel function J,(x) also has the following integral definitions: (6.48) and 1 2 (1 - t2)n-6 cosztdt,...

MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 5 ppsx

... (9.138) 1 2 p(1 72) = (m - - )2. (9.139) For convenience we have taken p(0) = a2c2 rather than zero in Equation (9. 92) . 122 STURM-LIOUVILLE SYSTEMS AND THE FACTORIZATION METHOD 9.1 ... write O+(z, 2) O- (2, l)yX"(z) = [-m2 + P]i+(z) (9 .20 2) 0-(z,1+l)O+(z,I+ l)~Xm(z) = [-m2+(1+1 )2] 1/Ixm(.z). (9 .20 3) Using (9 .20 4) GENERALIZED FOURIER SERIES 113 pointwise convergence ... ZL: (9 .24 6) and Ul+l n = [x+(r,l+l)lc- (9 .24 7) The normalized ladder operators are defined by Equation (9.48) as Using (9 .24 0) the energy levels are obtained as mZ2e4 2ii2n2 '...

MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 6 pps

... written as - 21 = allxl f a 122 2 + a13x3 22 = a2 121 + a 222 2 + a2 323 x3 = a31x1 + a 322 2 + (333x3 . (10 .23 ) - - aij are constants independent of 7 and 7. A convenient ... obtained from SinJ-n+l /2 COSJ+n+l /2 - e 2 2 by using 2 1r2 1 YE-’ = (J + - )2 - (m - - )2 L The corresponding eigenvalues are c7 + K + 1/4 = (J + 1 /2) 2 J - Iml and ... = 0,1 ,2, so that (9 .27 0) (9 .27 1) (9 .27 2) (9 .27 3) (9 .27 4) (9 .27 5) (9 .27 6) 9.10 BESSEL FUNCTIONS (TYPE C) Bessel’s equation is given as z2J$(z) +zJL(z) + (A 22 - m2)Jm(.)...

MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 8 pps

... (1 1 .22 7) d icotOsin4- 84 a a ae 34 L* = L, * ZL, = e*;q*- + icot 0-), L2 = L: + L; + Lq (11 .22 8) (11 .22 9) (11 .23 0) (11 .23 1) Using Equations (11 .22 7-11 .22 9) we can ... (1 1 .22 5) Expressing the components of the angular momentum operator in spherical polar coordinates: x + zy = r sin &+a6 and z = rcosf?, (1 1 .22 6) 25 2 CONTINUOUS GROUPS AND ... (11.157) I- 0 -LO1 -LO2 -LO3 -hl 0 ‘5 12 L13 -LO2 -Ll2 0 L23 -LO3 -L13 -L23 0 L= 1 Introducing six independent parameters (PI, 02, j33) and (01, 02, 03), this can also be...

MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 9 pps

... (Fig. 12. 2) as w = f (z) = u (2, y) + iv (z, y) . ( 12. 7) ( 12. 8) ( 12. 9) ( 12. 10) ( 12. 11) ( 12. 12) ( 12. 13) ( 12. 14) ( 12. 15) ( 12. 16) ( 12. 17) 27 6 CONTINUOUS GROUPS AND REPRESENTATIONS ... IZll + 1z21 7 I21z21 = 121 1 122 1 7 arg (21 22 ) = arg z1 + arg 22 are also very useful in calculation with complex numbers. 12. 2 COMPLEX FUNCTIONS We can define a complex ... sin2 6, MAPPINGS 301 Since this means u=z+zo and 2) = y +yo, ( 12. 53) ( 12. 54) a point (z, y) in the z-plane is mapped int,o the translated point (Z + 20 , 9 + yo) in...

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