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- Title: M461solassD
- Type: Notes
- School: ASU
- Course: MAT 461
- Term: Fall
Coursehero >> Arizona >> ASU >> MAT 461
Path: ASU >> MAT >> 461 Fall, 2007
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Path: ASU >> PHY >> 252 Spring, 2008
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Path: ASU >> PHY >> 252 Spring, 2008
Path: ASU >> PHY >> 252 Spring, 2008
Path: ASU >> PHY >> 252 Spring, 2008
Path: ASU >> PHY >> 252 Spring, 2008
Path: ASU >> PHY >> 252 Spring, 2008
Path: ASU >> PHY >> 252 Spring, 2008
Path: ASU >> PHY >> 252 Spring, 2008
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Path: ASU >> PHY >> 252 Spring, 2008
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Path: ASU >> PHY >> 132 Fall, 2007
Path: ASU >> PHY >> 132 Fall, 2007
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Path: Penn State >> MATH >> 415 Spring, 2008
Path: Penn State >> MATH >> 415 Spring, 2008
Path: Penn State >> MATH >> 415 Spring, 2008
Path: Penn State >> MATH >> 415 Spring, 2008
Path: Penn State >> MATH >> 415 Spring, 2008
Path: Penn State >> MATH >> 250 Fall, 2007
Path: Penn State >> MATH >> 415 Spring, 2008
Path: Penn State >> MATH >> 415 Spring, 2008
Path: Penn State >> MATH >> 415 Spring, 2008
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461 MAT Assignment D - Solutions Posted November 12 (1) Assume f (z) is analytic on and inside the circle |z| = 4. Show that if f (z) = 0 for every z such that |z| = 4, then f (z) = 0 for every z such that |z| < 4 as well. (That is, show that if f (z) = 0 on the circle, then f (z) = 0 inside too.) By the Maximum Modulus Principle (or more specifically, the corollary on pg 171), the maximum value of |f (z)| in the circle |z| 4 must occur on the boundary |z| = 4. Since f (z) = 0 on the boundary, |f (z)| 0 inside the circle. This implies that f (z) = 0 inside the circle. Another way: Let C denote the circle and let z0 be a point interior to the circle. Then the conditions necessary for the Cauchy Integral Formula hold, and we can write f (z0 ) = 1 2i f (z) dz z - z0 C Since f (z) = 0 on C (and z - z0 = 0 on C), the integral on the right is zero and so f (z0 ) = 0. This works for all point inside C, so f (z) = 0 in the whole inside of C. (2) Find the Taylor series of the given function, expanded about the given point z0 . (a) f (z) = ez , z0 = i 2 ez = ez- 2 + 2 (z - i )n i 2 2 = e n! n=0 i i = n=0 i(z - i )n 2 n! (b) f (z) = z , 4 + z2 z0 = 0 z z 1 = +4 4 1 + z42 z = 4 = = n=0 z2 (-1) n=0 n z2 4 n z 4 n=0 (-1)n 2n z 4n (-1)n 2n+1 z 4n+1 1 NOTE: For questions (3),(4) and (5), C represents the circle radius 2, center 0, positively oriented. (3) (a) Find the Laurent series which represents f (z) = z sin in the region 0 < |z| < . sin( 1 ) = z2 = n=0 1 z2 (-1)n n=0 ( z12 )2n+1 (2n + 1)! 1 1 4n+2 (2n + 1)! z (-1)n 1 1 1 1 - + - + 2 6 10 z 3!z 5!z 7!z 14 1 1 1 (-1)n z sin( 2 ) = 4n+1 z (2n + 1)! z n=0 = = (b) Evaluate the following integral: z sin C 1 1 1 1 - + - + z 3!z 5 5!z 9 7!z 13 1 z2 Since dz the contour C is simple, closed, positive, around z = 0 and in the region where the Laurent Series from part 3(a) is valid, we can evaluate this integral by taking b1 1 (the coefficient of z ) and multiplying it by 2i. We see that b1 = 1. So z sin C 1 z2 1 dz = 2i (4) Evaluate the integral C ze z dz 1-z 1 by finding b2 (coefficient of 1 ) z2 ez valid for 1 < |z| < . in the Laurent expansion of 1-z e 1 z = n=0 1 1 n! z n 1 1 1 + + + 2 z 2!z 3!z 3 1 1 1 = 1-z z 1 -1 z = 1+ 2 = = 1 -1 1 z1- z 1 z n=0 -1 zn for 1 <1 z 1 1 1 - 2 - 3 - z z z 1 1 1 1 = - - 2 - 3 - 4 - z z z z 1 1 1 1 1 1 1 ez = (1 + + 2 + 3 + ) (- - 2 - 3 - ) 1-z z 2z 6z z z z -1 1 1 1 = + (-1 - 1) 2 + (-1 - 1 - ) 3 + z z 2 z 1 = z -1 - 1 By multiplying the series for e z and 1-z above, we see that b2 = -2 for the Laurent series valid for 1 < |z| < . Since C is simple, closed, positive, around |z| = 1 and in the region where our series is valid, we know that: 1 b2 1 z 1 = 2i C ez 1-z 1 (z - 0)1 dz C ze dz = 2i(-2) = -4i 1-z (5) Find the residue at z = 0 for the following function f (z) = Use this residue to evaluate the integral z 2 (3 1 - z) 1 z 2 (3 - z) C 1 1 1 = 3-z 31- 1 = 3 z 3 n=0 1 n z 3n z z2 z3 1 + 2 + 3 + 4 + = 3 3 3 3 1 1 1 1 z = + 2 + 3 + 4 + 2 (3 - z) 2 z 3z 3z 3 3 The above Laurent Series is valid for 0 < |z| < 3, so the residue at z = 0 is 1 . By 9 Cauchy's Residue Theorem, we know an integral around a closed simple positive contour 3 C is equal to 2i times the sum of the residues at the singularities inside C. z = 0 is the only singularity inside C (the other, at z = 3, is outside). So the integral z 2 (3 1 2i dz = - z) 9 C 4
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M461solassE
Path: ASU >> MAT >> 461 Fall, 2007
Description: MAT 461 Assignment E - Solutions Posted November 30th 2007 Evaluate the following integrals, where C is the circle |z| = 5 in the positive sense. (1) sin z dz - 1)(z + 10) C The function has singularities at z = 1, -1, -10 of which 1, -1 are inside t...
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Description: MAT 461 Assignment A - Solutions (1) Sketch the set of points detemined by the given condition. a. filled in circle, radius 5, center 1 - 2i. The circle should pass through the origin. b. line y = x c. circle radius 1, center -2i d. all below the l...
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HW4Path: Penn State >> MATH >> 415 Spring, 2008
Description: 8.59 B = 2, = 10, 8.108 p = 0.67, n = 415 ^ 95% CI for p is p 1.96 ^ 9.2 n= 4 2 B2 = 100 p(1-^) ^ p n = 0.832 0.0153 = (0.817, 0.847) a E(^1 ) = 1 2 = , 2 E(^2 ) = 1 + 4 (n-2) 2(n-2) + 1 , 4 (n-2) 2 4(n-2)2 E(^3 ) = n n = 1 b V (^...
HW9Path: Penn State >> MATH >> 415 Spring, 2008
Description: Stat 415 HW 9 i Stat 415 HW 9 ii ...
HW3Path: Penn State >> MATH >> 415 Spring, 2008
Description: 8.39 yn unif (0, ), a fy = 1 , y (0, ) Let Y(n) = max(Y1 , , Yn ), U = 1 Y(n) FU (u) = P ( 1 Y(n) u) = P (Y(n) u) = P (Y1 u, Yn u) () = P (Y u)n = u n un () = Because (*) : Y(n) is a maximum in Y1 , , Yn () : Y is random sampl...
HW1Path: Penn State >> MATH >> 415 Spring, 2008
Description: 4.129 Find y0 satisfying P (Y < y0 ) = 0.9. P (Y < y0 ) = 0.9 P (Y y0 ) = 0.1. -70 Since Y N (70, 122 ), Z = Y 12 N (0, 1). 0.1 = P (Z > 1.28) = P (Y > 85.36) by Normal table. 4.151 a Since 0 a 1, 1. f (y) = af1 (y) + 1 - af2 (y) 0 2. f (y)dy...
415Mid1Solution08Path: Penn State >> MATH >> 415 Spring, 2008
Description: This confidence interval means that, with 90% chance, the true difference in proportion between the two groups is covered by this interval. Since 0 is not covered by this interval, with 90% confidence we can say that eating saturated fats do change r...
HW10Path: Penn State >> MATH >> 415 Spring, 2008
Description: Stat 415 HW 10 14.1 If no lane is preferred over another, the probability that a car will be driven in lane 1, 2, 3, 4 is 1 H0 : p1 = p2 = p3 = p4 = , 4 4 1 4. H1 : Some lanes were preffered over another X = i=1 2 (ni - 250)2 [ni - E(ni )]2 = =...
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Description: \'. Math 250 Fdl 2007 Exa,nr I K\"l 5 ofl\"\\1 \"^\" NAME: ID\'No: SECTION: This ocam contains l0 questionson g pages (including this title page). This exa,nris worth a totd of 100 points. The exasr is broken into two parts. There are six multiple choice...
HW6Path: Penn State >> MATH >> 415 Spring, 2008
Description: Stat 415 HW 6 10.2 The test statistic Y has a binomial distribution with n=20 and p a A type I error occurs if the experimenter concluded that the drug dosage level induces sleep in less that 80% of the people suffering form insomnia when, in fact, ...
HW2Path: Penn State >> MATH >> 415 Spring, 2008
Description: 8.2 ^ a Show that E(3 ) = ^ ^ ^ ^ ^ E(3 ) = E(a1 + (1 - a)2 ) = aE(1 ) + (1 - a)E(2 ) = a + (1 - a) = ^ ^ b Find a = arg min V ar()3 under 1 and 2 are independent. ^ ^ ^ ^ ^ Let L = var(3 ) = V ar(a1 + (1 - a)2 ) = a2 V ar(1 ) + (1 - a)2 var(2 ) =...
HW5Path: Penn State >> MATH >> 415 Spring, 2008
Description: 9.35 L(y1 , , yn |, ) = n -1 yi n = h(y1 , , yn )g( n Yi , ) By theorem 9.4, where h(y1 , , yn ) = 1, Yi is sufficient for 1 9.38The exponential distribution is given by f (y) = L(y1 , , yn |) = 1 e y1 / ey/ . Yi is sufficient...
exam2_solPath: Penn State >> MATH >> 250 Fall, 2007
Description: !\"1* [l*^ s Math 250 Fall 2007 Exam 2 NAME: ID No: SECTION: This exarn contains 10 questionson 10 pages(including this title page). This exam is worth a total of 100 points. The exam is broken into two parts. There are six multiple choicequestions,e...