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

... positive for −π < x < 0 and negative for 0 < x < π. Since the sign of y  goes from positive to negative, x = 0 is a relative maxima. See Figure 3. 7. Example 3. 5 .3 Consider y = x 3 and ... y  = 3x 2 is positive for x < 0 and positive for 0 < x. Since y  is not identically zero and the sign of y  does not change, x = 0 is not a relative extrema. See...

Ngày tải lên: 06/08/2014, 01:21

... minimum. b. f  (x) = 2 3 (x −2) −1 /3 The first derivative exists and is nonzero for x = 2. At x = 2, the derivative does not exist and thus x = 2 is a critical point. For x < 2, f  (x) < 0 and for x > ... for r > 0, it must be a global minimum. The cup has a radius of 4 3 √ π cm and a height of 4 3 √ π . 106 a x x x x x x∆ 1 2 3 i n-2 n-1 b f( ) ξ 1 Fig...

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... = ı and an isolated essential singularity at z = 1. (b) sin(z − 3) (z − 3) (z + ı) 6 has a removable singularity at z = 3, a pole of order 6 at z = −ı and an essential singularity at z ∞ . 436 Result ... = 1 r ∂ ∂r  r ∂u ∂r  + 1 r 2 ∂ 2 u ∂θ 2 = 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...

Ngày tải lên: 06/08/2014, 01:21

... dz =  sin 3 z 3  ıπ π = 1 3  sin 3 (ıπ) −sin 3 (π)  = −ı sinh 3 (π) 3 Again the anti-derivative is single-valued. 491 3. e x (sin x cos y cosh y −cos x sin y sinh y) Exercise 9 .3 For an analytic ... max a≤x≤b |f(x)|. 466 with a, b and c complex-valued constants and d a real constant. Substituting z = x + ıy and expanding products yields, a  x 3 + ı3x 2 y − 3xy...

Ngày tải lên: 06/08/2014, 01:21

... −x)  =  2 π  π 0 x(π −x) sin(nx) dx =  2 π 2 n 3 (1 − (−1) n ) =   2 π 4 n 3 for odd n 0 for even n Thus the expansion is x(π −x) = ∞  n=1 oddn 8 πn 3 sin(nx) for x ∈ [0, π]. In the first graph of Figure ... = 1 √ π x −1 e −x 2 − 1 √ π  ∞ x t −2 e −t 2 dt = 1 √ π x −1 e −x 2 − 1 √ π  − 1 2 t 3 e −t 2  ∞ x + 1 √ π  ∞ x 3 2 t −4 e −t 2 dt = 1 √ π e −x 2  x −1 − 1 2 x...

Ngày tải lên: 06/08/2014, 01:21

... defined h i =   ∂x 1 ∂ξ i  2 +  ∂x 2 ∂ξ i  2 +  ∂x 3 ∂ξ i  2 . The gradient, divergence, etc., follow. ∇u = a 1 h 1 ∂u ∂ξ 1 + a 2 h 2 ∂u ∂ξ 2 + a 3 h 3 ∂u ∂ξ 3 ∇ · v = 1 h 1 h 2 h 3  ∂ ∂ξ 1 (h 2 h 3 v 1 ) + ∂ ∂ξ 2 (h 3 h 1 v 2 ) + ∂ ∂ξ 3 (h 1 h 2 v 3 )  ∇ 2 u ... + ∂ ∂ξ 2 (h 3 h 1 v 2 ) + ∂ ∂ξ 3 (h 1 h 2 v 3 )  ∇ 2 u = 1 h 1 h 2 h 3  ∂ ∂ξ 1  h 2 h 3 h 1...

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... 12   β α  2 − 1       < 0 for |β/α| < 1 = 0 for |β/α| = 1 > 0 for |β/α| > 1 Thus we see that β α x may be a minimum for |β/α| ≥ 1 and may be a maximum for |β/α| ≤ 1. Jacobi Condition. ... closed curve, and P (t) is a given entire function of t. Exercise 48.41 Solve −  1 0 f(t) t − x dt + −  3 2 f(t) t − x dt = x where this equation is to hold for x in...

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... . . . . . . . . . . . . . . 1 630 34 .3. 3 Bessel Functions of Non-Integer Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 633 34 .3. 4 Recursion Formulas . . . . . . . . . . . ... q(x). Hint, Solution 12 33 The Gamma Function 1605 33 .1 Euler’s Formul a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1605 33 .2 Hankel’s For...

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... k a 1 a 2 a 3 b 1 b 2 b 3       = i     a 2 a 3 b 2 b 3     − j     a 1 a 3 b 1 b 3     + k     a 1 a 2 b 1 b 2     = (a 2 b 3 − a 3 b 2 )i − (a 1 b 3 − a 3 b 1 )j ... × (b 1 i + b 2 j + b 3 k) = a 1 b 2 k + a 1 b 3 (−j) + a 2 b 1 (−k) + a 2 b 3 i + a 3 b 1 j + a 3 b 2 (−i) = (a 2 b 3 − a 3 b 2 )i − (a 1 b 3 − a 3 b 1 )j + (a...

Ngày tải lên: 06/08/2014, 01:21

... and dv = e 2x dx. Then du = dx and v = 1 2 e 2x .  x 3 e 2x dx = 1 2 x 3 e 2x − 3 4 x 2 e 2x + 3 2  1 2 x e 2x − 1 2  e 2x dx   x 3 e 2x dx = 1 2 x 3 e 2x − 3 4 x 2 e 2x + 3 4 x e 2x − 3 8 e 2x +C Solution ... integral in partial fractions, x + 1 x 3 + x 2 − 6x = x + 1 x(x − 2)(x + 3) = a x + b x − 2 + c x + 3 x + 1 = a(x −2)(x + 3) + bx(x + 3) + cx(x −2) Se...

Ngày tải lên: 06/08/2014, 01:21