Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 8 ppt

... arguments: 1. log( 1) = log 1 1 = log (1) − log( 1) = −log( 1) , therefore, log( 1) = 0.2. 1 = 1 1/2= (( 1) ( 1) ) 1/ 2= ( 1) 1/ 2( 1) 1/ 2= ıı = 1, therefore, 1 = 1. Hint, SolutionExercise 7 .11 Write ... Cartesian form. Denote any multi-valuedness explicitly.22/5, 3 1+ ı,√3 − ı 1/ 4, 1 ı/4.Hint, Solution 287 -2 -1 0 1 2x-2 -1 0 1 2y00.5 1 -2 -1 0 1 2x-2 -1 0 1 2x-2 -1 0 1 2y-202-2 -1 0 1 2xFigure ... logarithm.sin 1 z = −ı logız ±√ 1 − z2Example 7.7.4 Consider the equation sin3z = 1. sin3z = 1 sin z = 1 1/3eız−e−ızı2= 1 1/3eız−ı2 (1) 1/ 3−e−ız= 0eı2z−ı2 (1) 1/ 3eız 1 = 0eız=ı2 (1) 1/ 3±−4 (1) 2/3+...

... a and b are orthogonal, (perpendicular), or one of a and b are zero.26 1 2 1 2246 8 10 1 2Figure 1. 11: Plots of f(x) = p(x)/q(x). 1. 6 HintsHint 1. 1area = constant ×diameter2.Hint 1. 2A ... (1, 1) , (4, 2), (3, 7) and (2, 3)?Hint, SolutionExercise 2.7What is the volume of the tetrahedron with vertices at (1, 1, 0), (3, 2, 1) , (2, 4, 1) and (1, 2, 5)?Hint, SolutionExercise 2 .8 What ... from 1 to n and are called free indices.Example 2 .1. 1 Consider the matrix equation: A · x = b. We can write out the matrix and vectors explicitly.a 11 ··· a1n.........an1···...

... ( 1) n 1 (n − 1) !, for n ≥ 1. By Taylor’s theorem of the mean we have,ln x = (x − 1) −(x − 1) 22+(x − 1) 33−(x − 1) 44+ ··· + ( 1) n 1 (x − 1) nn+ ( 1) n(x − 1) n +1 n + 1 1ξn +1 .723 .8 Exercises3 .8 .1 ... aref 1 (x) = 1 f2(x) = 1 + xf3(x) = 1 + x +x22f4(x) = 1 + x +x22+x36The four approximations are graphed in Figure 3 .11 . -1 -0.5 0.5 1 0.5 1 1.522.5 -1 -0.5 0.5 1 0.5 1 1.522.5 -1 -0.5 ... 0.5 1 0.5 1 1.522.5 -1 -0.5 0.5 1 0.5 1 1.522.5 -1 -0.5 0.5 1 0.5 1 1.522.5 -1 -0.5 0.5 1 0.5 1 1.522.5Figure 3 .11 : Four Finite Taylor Series Approximations ofexNote that for the range of x we are looking...

... thumb: 12 1c.lnlimx→+∞ 1 + 1 xx= limx→+∞ln 1 + 1 xx= limx→+∞x ln 1 + 1 x= limx→+∞ln 1 + 1 x 1/ x= limx→+∞ 1 + 1 x 1 − 1 x2 1/ x2= limx→+∞ 1 + 1 x 1 = ... [ 1, 1] and 1 ≤ cos(x0) ≤ 1, the approximationsin x ≈ x −x36+x5 12 0has a maximum error of 1 5040≈ 0.00 019 8. Using this polynomial to approximate sin (1) , 1 − 1 36+ 1 5 12 0≈ 0 .8 416 67. 10 72.f(0) ... = limδ→0+xa +1 a + 1  1 δ= 1 a + 1 − limδ→0+δa +1 a + 1 This limit exists only for a > 1. Now consider the case that a = 1.  1 0x 1 dx = limδ→0+[ln x] 1 δ= ln(0) − limδ→0+ln...

... ı2√3 8 − 8 √32 1 =−2 + ı2√3 12 8 + 12 8 √3 1 =− 512 − ı 512 √3 1 = 1 512 1 1 + ı√3= 1 512 1 1 + ı√3 1 − ı√3 1 − ı√3= − 1 20 48 + ı√320 48 214 exists a unique ... y2.250-2 -1 0 1 2x-2 -1 0 1 2y-505-2 -1 0 1 2x-2 -1 0 1 2yFigure 7 .10 : A few branches of arg(z).-2 -1 0 1 2x-2 -1 0 1 2y0 1 2-2 -1 0 1 2x-2 -1 0 1 2x-2 -1 0 1 2y-202-2 -1 0 1 2xFigure 7 .11 : Plots of |z| and Arg(z).2 51 -1 ... modulus-2 -1 0 1 2x-2 -1 0 1 2y-2 -1 0 1 2-2 -1 0 1 2x-2 -1 0 1 2x-2 -1 0 1 2y-2 -1 0 1 2-2 -1 0 1 2xFigure 7.9: The real and imaginary parts of f(z) = z = x + ıy.argument form the...

... Hint 7 .19 Hint 7.20Hint 7. 21 Hint 7.22Hint 7.23Hint 7.24Hint 7.25 1. (z2+ 1) 1/ 2= (z − ı) 1/ 2(z + ı) 1/ 22. (z3− z) 1/ 2= z 1/ 2(z − 1) 1/ 2(z + 1) 1/ 23. log (z2− 1) = log(z − 1) + ... z 1 = z2= 1. Arg(( 1) ( 1) ) = Arg (1) = 0, Arg( 1) + Arg( 1) = 2πLog(( 1) ( 1) ) = Log (1) = 0, Log( 1) + Log( 1) = ı2π306sign. This will change the value of arccosh(z). The same is true for ... ıı=eı log ı. log ( (1 + ı)ıπ) has a doubly infin ite set of values.log ( (1 + ı)ıπ) = log(exp(ıπ log (1 + ı))).Hint 7 .17 Hint 7 . 18 Branch Points and Branch Cuts2 98 -1 1 -10 10 Figure 7.36: The...