Mathematical methods for physics and engineering riley k f, hobson m p, bence s j

... collection of mathematical tools for both understanding and solving problems in robotics and computer vision Several classes at Stanford cover the topics presented in this class, and so in much ... want to understand robotics or vision, you should take classes in these subjects, since this course is not on robotics or vision On the other hand, if you plan to study robotics, vision, or other ... following R Courant and D Hilbert, Methods of Mathematical Physics, Volume I and II, John Wiley and Sons, 1989 D A Danielson, Vectors and Tensors in Engineering and Physics, Addison-Wesley, 1992...

... 12 16 12 16 12 19 12 24 12 25 12 48 12 49 12 51 12 51 12 55 12 63 12 70 12 72 12 72 12 78 12 78 12 80 12 81 12 83 12 83 12 84 12 87 12 88 12 94 12 97 13 02 13 03 13 04 25 .14 Solutions ... 10 25 10 27 10 27 10 29 10 32 10 34 10 35 10 41 10 41 10 43 10 45 10 46 10 48 10 50 10 52 10 59 10 59 10 61 10 65 10 65 10 68 10 71 10 74 10 74 10 76 10 77 10 79 10 82 21. 7 .1 Green ... 11 84 11 84 11 88 11 98 12 01 12 03 12 06 10 92 10 95 10 98 11 00 11 04 11 09 11 17 11 23 11 26 11 64 11 65 23.3 23.4 23.5 23.6 23.7 23.8 23.9 Irregular...

... = 10 9 = x0 11 10 = x0 x0 = 10 9 At 7:00 pm the number of bacteria is 10 11 10 60 = 11 60 ≈ 3.04 × 10 11 51 10 At 3:00 pm the number of bacteria was 10 9 11 10 18 0 = 18 10 189 ≈ 35.4 11 180 Figure 1. 13: ... b2 (−i) = (a2 b3 − a3 b2 )i − (a1 b3 − a3 b1 )j + (a1 b2 − a2 b1 )k Next we evaluate the determinant i j k a a a a a a a1 a2 a3 = i − j + k b2 b3 b1 b3 b1 b2 b1 b2 b3 = (a2 b3 − a3 b2 )i − (a1 ... = (a1 i + a2 j + a3 k) × (b1 i + b2 j + b3 k) = a1 i × (b1 i + b2 j + b3 k) + a2 j × (b1 i + b2 j + b3 k) + a3 k × (b1 i + b2 j + b3 k) = a1 b2 k + a1 b3 (−j) + a2 b1 (−k) + a2 b3 i + a3 b1 j...

... = ( 1) n 1 (n − 1) !, for n ≥ By Taylor’s theorem of the mean we have, ln x = (x − 1) − (x − 1) 2 (x − 1) 3 (x − 1) 4 (x − 1) n (x − 1) n +1 + − + · · · + ( 1) n 1 + ( 1) n n n + ξ n +1 72 -1 0.5 1. 5 2.5 ... 6! (2(n − 1) )! (2n)! Here are graphs of the one, two, three and four term approximations 0.5 0.5 -3 -2 -1 -0.5 -1 -3 -2 -1 -0.5 -1 0.5 -3 -2 -1 -0.5 -1 0.5 -3 -2 -1 -0.5 -1 Figure 3 .12 : Taylor ... (x) = + x + f3 (x) = + x + The four approximations are graphed in Figure 3 .11 2.5 1. 5 0.5 -1 -0.5 0.5 2.5 1. 5 0.5 -1 -0.5 2.5 1. 5 0.5 -1 -0.5 0.5 0.5 2.5 1. 5 0.5 -1 -0.5 0.5 Figure 3 .11 : Four Finite...

... approximate sin (1) , 13 15 1 + ≈ 0.8 41 6 67 12 0 10 7 To see that this has the required accuracy, sin (1) ≈ 0.8 41 4 71 Solution 3 .19 Expanding the terms in the approximation in Taylor series, ∆x3 ∆x4 ∆x2 f ... Example 4. 4 .1 Consider the partial fraction expansion of + x + x2 (x − 1) 3 The expansion has the form a0 a1 a2 + + (x − 1) (x − 1) x 1 127 The coefficients are (1 + x + x2 )|x =1 = 3, 0! d a1 = (1 + ... − x→0 10 9 x =0 c ln lim x→+∞ 1+ x x = lim x→+∞ = lim x→+∞ = lim x→+∞ = lim ln + 1/ x 1+ x→+∞ 1+ =1 Thus we have lim x→+∞ 1+ 11 0 x x 1 x − x2 1/ x2 x→+∞ = lim x ln 1+ x x ln + x x = e x 1 d It...

... 87 − 6 1/ 3 1/ 3 6−2/3 + √ 2/3 87 √ + 87 , 0, − 6 1/ 3   ≈ (0 .58 9 755 , 0, 0.347 81) 1/ 3 The closest point is shown graphically in Figure 5 .10 1- 1 -0 .5 0 .5 -1 -0 .5 0 0 .5 1. 5 0 .5 Figure 5 .10 : Paraboloid, ... dx2 = (1 + 2x) x= 1 = 1 x= 1 = x= 1 (2) x= 1 =1 Then we can the integration + x + x2 dx = (x + 1) 3 1 − + (x + 1) (x + 1) x +1 1 + + ln |x + 1| =− 2(x + 1) 2 x + x + 1/ 2 + ln |x + 1| = (x + 1) 2 dx ... x3 x +1 dx = + x2 − 6x 10 Setting x = −3 yields C = − 15 + − dx 6x 10 (x − 2) 15 (x + 3) ln |x − 2| − ln |x + 3| + C = − ln |x| + 10 15 |x − 2|3 /10 = ln 1/ 6 +C |x| |x + 3|2 / 15 − Solution 4 .17 dx...