laplace701 - 1 Problem 1(10 pts Find the radius and the...

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1 Problem 1 (10 pts) Find the radius and the interval of convergence of the series X n = 0 (2 x - 1) n p n + 3 . Solution: Let a n = (2 x - 1) n p n + 3 . The absolute ratio test gives us: lim n →∞ fl fl fl fl a n + 1 a n fl fl fl fl = lim n →∞ fl fl fl fl fl (2 x - 1) n + 1 p n + 4 · p n + 3 (2 x - 1) n fl fl fl fl fl = lim n →∞ r n + 3 n + 4 | 2 x - 1 | = | 2 x - 1 | . Hence, for this power series to converge absolutely, we must have | 2 x - 1 | < 1, i.e., fl fl fl fl x - 1 2 fl fl fl fl < 1 2 or 0 < x < 1 So, the radius of convergence is R = 1 2 . Now, to determine the interval of convergence, we need to check the convergence at the boundary points, x = 0,1. For x = 0, we have X n = 0 ( - 1) n p n + 3 . This is an alternating series and it satisfies the Alternating Series Test : (i) u n = 1 p n + 3 > 0 for all n N ; (ii) u n + 1 = 1 p n + 4 < 1 p n + 3 = u n for all n N ; (iii) u n 0 as n → ∞ . Hence, this alternating series converges conditionally. On the other hand, for x = 1, we have X n = 0 1 p n + 3 = X n = 3 1 p n , which is a p -series with p = 1 2 < 1. Hence this series diverges. Therefore, the interval of convergence of the given series is: 0 x < 1 or x [0,1) . Score of this page:_______________
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2 Problem 2 (10 pts) Consider the function f ( x ) = p 1 + x . (a) (5 pts) Approximate f ( x ) by a Taylor polynomial of degree 1 centered at 0. Compute the value of p 2 using that approximation. (b) (5 pts) Compute the worst-case approximation error for 0 x 1. Then, confirm that the approximate value of p 2 computed in Part (a) is within this worst-case error. Note that the precise value of p 2 in 3 decimal places is 1.414. Solution to (a): First of all, since f ( x ) = (1 + x ) 1/2 , we have f 0 ( x ) = 1 2 (1 + x ) - 1/2 , and f 00 ( x ) = - 1 4 (1 + x ) - 3/2 . Hence, by Taylor’s formula, we have p 1 + x = P 1 ( x ) + R 1 ( x ) = f (0) + f 0 (0) x + f 00 ( c ) 2 x 2 0 < c < x or x < c < 0 = 1 + 1 2 x - 1 8 (1 + c ) - 3/2 x 2 . Hence, the Taylor polynomial of degree 1 is: P 1 ( x ) = 1 + 1 2 x . So, the approximate value of p 2 is simply p 2 P (1) = 1.5 . Solution to (b): From Taylor’s formula, we have fl fl f ( x ) - P 1 ( x ) fl fl = | R 1 ( x ) | = 1 8 fl fl fl fl x 2 (1 + c ) 3/2 fl fl fl fl 1 8 · (1 + c ) 3/2 since 0 x 1. < 1 8 since 0 < c < 1. Hence the approximation error is bounded by 1 8 = 0.125 , which is the worst-case scenario . Now, f (1) = p 2, so | f (1) - P 1 (1) | = | 1.414 - 1.5 | = 0.086 < 0.125. So, this is within the error bound. Score of this page:_______________
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3 Problem 3 (10 pts) (a) (5 pts) Write proj v u , i.e., the vector projection of u onto v , using the dot product be- tween u and v , the length of the vector v , and the vector v itself. (b) (5 pts) Let u = j + k , v = i + j . Then write u as the sum of a vector parallel to v and a vector orthogonal to v . Solution to (a): The vector projection of u onto v is proportional (i.e., parallel) to v by definition. Let θ be an angle between u and v . Then, from the geometry of these two vectors, we have proj v u = ( | u | cos θ ) v | v | = ( | u || v | cos θ ) v | v | 2 = ( u · v ) v | v | 2 or equivalently u · v | v | 2 v . Solution to (b): From the geometry and the definition of the vector projection, it is clear that proj v u is parallel to v and perpendicular to the residual u - proj v u . Hence, proj v u = ( j + k ) · ( i + j ) | i + j | 2 ( i + j ) = 0,1,1 ⟩·⟨ 1,1,0 |⟨ 1,1,0 ⟩| 2 1,1,0 = 1 2 1,1,0 = ¿ 1 2 , 1 2 ,0 .
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  • Fall '09
  • NUCAMENDI
  • Calculus, lim, pts

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