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# 527plsolns - 642:527 PREREQUISITE QUIZ SOLUTIONS 2n x3n n2...

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Unformatted text preview: 642:527 PREREQUISITE QUIZ: SOLUTIONS 2n x3n . n2 n=0 ∞ Fall 2009 1. Find the radius of convergence of the power series Solution: We use the ratio test: if bn = 2n x3n /n2 is a typical term of the series then n→∞ lim 2n+1 x3(n+1) /(n + 1)2 2|x|3 n2 bn+1 = lim = lim = 2|x|3 . n→∞ n→∞ (n + 1)2 bn 2n x3n /n2 The series converges if this limit is less than 1, i.e., if |x| < 2−1/3 , so the radius of convergence is 2−1/3 . 2. Find the eigenvalues of the matrix A = 3 6 , and ﬁnd an eigenvector for one of them. −1 −4 3−λ 6 = λ2 + λ − 6 = 0: −1 −4 − λ u1 3+3 6 λ1 = −3, λ2 = 2. To ﬁnd an eigenvector u for λ1 we solve = 0, ﬁnding u2 −1 −4 + 3 −1 −6 . . Similarly, an eigenvector v for λ2 is v = u= 1 1 Solution: The eigenvalues are the roots of the equation 3. Suppose that f (x, y ) = x2 y ln(xy 2 ). Find fy and fxy . Solution: By the product and chain rules: fy = fxy = ∂ ∂x ∂f 1 = x2 ln(x2 y ) + x2 y 2xy = x2 ln(x2 y ) + 2x2 , ∂y xy 2 ∂f 1 y 2 + 4x = 2x ln(xy 2 ) + 5x. = 2x ln(x2 y ) + x2 ∂y xy 2 Notice that we never have to compute fx , since always fxy = fyx . 4. Find the general solution y (x) of the equation y ′ = ye2x . Solution: We separate variables (remember that y ′ = dy/dx) to obtain dy = e2x dx. y Integrating gives ln y = 1 e2x + C or y = ee 2 2x /2+C = Kee 2x /2 , where K = eC . 5. Find the solution y (x) of the initial value problem y ′′ − 4y = 0, y (0) = 3, y ′ (0) = 2. Solution: Since this is a constant coeﬃcient linear equation we expect solutions of the form y = erx ; substituting this form into the equation we ﬁnd that r must satisfy r 2 − 4 = 0, so r = ±2. The general solution is y = c1 e2x + c2 e−2x ; imposing the initial conditions gives y = 2e2x + e−2x . 6. Does sin x dx converge or diverge? Justify your answer. x2 π Solution: The function is ﬁnite for all x in the range of integration so the only question is ∞ π ∞ dx π x2 ∞ convergence at x = ∞. Now converges: ∞ dx π x2 1 = −x = 1 . π But the sin x factor just oscillates between 1 and −1, which will not destroy this convergence; we conclude that the ∞ 1 integral converges. Technically we are using a comparison test: sin2x ≤ x2 , π dx converges, so x x2 ∞ sin x 2 dx converges x π 1 ...
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