Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 5 ppt

Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 2 ppt

... is the angle from a to b and n is a unit vector that is orthogonal to a and b and in the direction s uch thatthe ordered triple of vectors a, b and n form a right-handed system.29abbθbFigure ... vectors a and b. We can write b = b⊥+ bwhere b⊥is orthogonal to a and bisparallel to a. Show thata × b = a × b⊥.Finally prove the distributive law for arbitrary b and c.Hint 2 .5 Write ... 0.31Solution 2 .5 We kn ow thati × j = k, j × k = i, k × i = j, and thati × i = j × j = k × k = 0.Now we write a and b in terms of their rectangular components and use the distributive law to expand the...

Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 3 pptx

... 3.11.-1-0 .5 0 .5 10 .5 11 .5 22 .5 -1-0 .5 0 .5 10 .5 11 .5 22 .5 -1-0 .5 0 .5 10 .5 11 .5 22 .5 -1-0 .5 0 .5 10 .5 11 .5 22 .5 Figure 3.11: Four Finite Taylor Series Approximations ofexNote that for ... passes through the points (x, y(x)) and (x + ∆x, y(x + ∆x)). See Figure 3 .5. yx∆y∆xFigure 3 .5: The increments ∆x and ∆y. 56 -10 -5 510-2-1 .5 -1-0 .5 0 .5 1Figure 3.14: Ten Term Taylor Series ... 3 .5. 3 Consider y = x3 and the point x = 0. The function is diﬀerentiable. The derivative, y= 3x2ispositive for x < 0 and positive for 0 < x. Since yis not identically zero and...

Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 4 pptx

... of x and an increasingfunction of δ for positive x and δ. Thus for any ﬁxed δ, the maximum value of√x + δ −√x is bounded by√δ.Therefore on the interval (0, 1), a suﬃcient condition for ... −2)−1/3The ﬁrst derivative exists and is nonzero for x = 2. At x = 2, the derivative does not exist and thus x = 2 is acritical point. For x < 2, f(x) < 0 and for x > 2, f(x) > 0. ... aresin(x) = x −x36+x 5 120−x7 50 40+x9362880+ ··· .The seventh derivative of sin x is −cos x. Thus we have thatsin(x) = x −x36+x 5 120−cos x0 50 40x7,where 0 ≤ x0≤...

Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 5 pdf

... (0 .58 9 755 , 0, 0.34781).The closest point is shown graphically in Figure 5. 10.-1-0 .5 00 .5 1-1-0 .5 00 .5 100 .5 11 .5 2-1-0 .5 00 .5 100 .5 11 .5 2Figure 5. 10: Paraboloid, Tangent Plane and ... particle at time t, then the velocity and acceleration of the particle aredrdt and d2rdt2, 154 Figure 5. 2: The gradient of the distance from the origin.u vθθntvu-tθθFigure 5. 3: ... . . . , xk, . . . , xn)∆x.Partial derivatives have the same diﬀerentiation formulas as ordinary derivatives. 155 Solution 4.16Expanding the integral in partial fractions,x + 1x3+ x2−...

Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 7 ppt

... This is zero if and only if u0= u1= u2= u3= 0. Thus there217-10 -5 510- 15 -10 -5 5Figure 6.23: (z) − (z) = 5 Solution 6.16|eıθ−1| = 2eıθ−1e−ıθ−1= 41 −eıθ−e−ıθ+1 ... real-variablecounterparts.7.1 Curves and RegionsIn this section we introduce curves and regions in the complex plane. This material is necessary for the study ofbranch points in this chapter and later for ... ı√320482.(11 + ı4)2= 1 05 + ı88Solution 6.41.2 + ıı6 − (1 − ı2)2=2 + ı−1 + ı82=3 + ı4−63 − ı16=3 + ı4−63 − ı16−63 + ı16−63 + ı16= − 253 42 25 − ı20442 25 2 15 2.(1 − ı)7=(1...

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

... are an inﬁnite set of rational numbers for which ızhas 1 as one of its values. For example,ı4 /5 = 11 /5 =1,eı2π /5 ,eı4π /5 ,eı6π /5 ,eı8π /5 7.8 Riemann SurfacesConsider the mapping ... SeeFigure 7.18 and Figure 7.19 for plots of the real and imaginary parts of the cosine and sine, respectively. Figure 7.20shows the modulus of the cosine and the sine.The hyperbolic sine and cosine. ... y = 0.This becomes the two equations (for the real and imaginary parts)sin x cosh y = 0 and cos x sinh y = 0.Since cosh is real-valued and positive for real argument, the ﬁrst equation dictates...

Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 9 ppt

... HintsCartesian and Modulus-Argument FormHint 7.1Hint 7.2Trigonometric FunctionsHint 7.3Recall that sin(z) =1ı2(eız−e−ız). Use Result 6.3.1 to convert between Cartesian and modulus-argument form.Hint ... solutions.Solution 7 .5 We write the expressions in terms of Cartesian coordinates.ez2=e(x+ıy)2=ex2−y2+ı2xy=ex2−y2304 25 50 75 100-11Figure ... on which f(0) = ı√6. Write out an explicit formula for thevalue of the function on this branch.Figure 7.33: Four candidate sets of branch cuts for ((z − 1)(z − 2)(z − 3))1/2.Hint, Solution294change...

Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 2 pptx

... the real and imaginary parts.ur=1rvθ, vr= −1ruθur=1rvθ, uθ= −rvrSolution 8. 15 Since w is analytic, u and v satisfy the Cauchy-Riemann equations,ux= vy and uy= ... =1r∂∂rr∂u∂r+1r2∂2u∂θ2= 0.Therefore u is harmonic and is the real part of some analytic function.Example 9.3.9 Find an analytic function f(z) whose real part isu(r, θ) = r (log r cos θ −θ ... Consider analytic functions f1(z) and f2(z) deﬁned on the domains D1 and D2, respectively. Suppose that D1∩ D2is a region or an arc and that f1(z) = f2(z) for allz ∈ D1∩ D2. (See...

Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 3 ppt

... contour and do the integration.z − z0=eıθ, θ ∈ [0 . . . 2π)C(z − z0)ndz =2π0eınθıeıθdθ=eı(n+1)θn+12π0 for n = −1[ıθ]2π0 for n = −1=0 for n = −1ı2π for ... have found a formula for writing the analytic function in termsof its real part, u(r, θ). With the same method, we can ﬁnd how to write an analytic function in terms of its imaginary part, v(r, ... uy) dx dy + ıD(ux− vy) dx dy= 0Since the two integrands are continuous and vanish for all C in Ω, we conclude that the integrands are identically zero.This implies that the Cauchy-Riemann...

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

... ,converges for α > 1 and diverges for α ≤ 1.Hint, Solution 56 4Example 12.3.2 Convergence and Uniform Convergence. Consider the serieslog(1 − z) = −∞n=1znn.This series converges for |z| ... large. Thus this series is not uniformly convergent in the domain|z| ≤ 1, z = 1. The series is uniformly convergent for |z| ≤ r < 1. 54 512.2.2 Uniform Convergence and Continuous Functions.Consider ... necessarycondition for uniform convergence. The Weierstrass M-test can succeed only if the series is uniformly and absolutelyconvergent.Example 12.2.1 The seriesf(x) =∞n=1sin xn(n + 1)is uniformly and...

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