MAT 1322M3 Winter2005

# MAT 1322M3 Winter2005 - MAT 1322B Winter 2005 Professor...

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Unformatted text preview: MAT 1322B, Winter 2005 Professor: Venanzio Capretta MIDTERM TEST 3, version A Solutions 1. [2 points, 8.6 #8] Find a power series representation of the function: f ( x ) = 3 x 1- 2 x 2 A. ∑ ∞ n =0 ( 3 2 ) n x 2 n B. ∑ ∞ n =0 6 n x n +1 C. ∑ ∞ n =0 2 3 n x 2 n +2 D . ∑ ∞ n =0 3 · 2 n x 2 n +1 E. ∑ ∞ n =0 3 2 n x 2 n- 1 F. ∑ ∞ n =0 2 · 3 n x 2 n +1 1 1- x = ∞ X n =0 x n 1 1- 2 x 2 = ∞ X n =0 (2 x 2 ) n = ∞ X n =0 2 n x 2 n 3 x 1- 2 x 2 = ∞ X n =0 3 x · 2 n x 2 n = ∞ X n =0 3 · 2 n x 2 n +1 2. [2 points, 8.8 #13] Let f ( x ) = 3 √ 1 + x 2 . Evaluate f (6) (0). [Hint: Use the binomial series to find the Maclaurin series of f ( x ).] A. 400 B. 400 9 C. 2 D.- 90 E. 105 243 F.- 800 9 f ( x ) = 3 p 1 + x 2 = (1 + x 2 ) 1 / 3 By the Binomial Series we have: (1 + x ) 1 / 3 = ∞ X n =0 1 / 3 n x n So we have: f ( x ) = (1 + x 2 ) 1 / 3 = ∞ X n =0 1 / 3 n x 2 n But we also have, by the Maclaurin series: f ( x ) = ∞ X n =0 f ( n ) (0) n ! x n The coefficient of the term x 6 = x 2 · 3 from the two formulas must be the same: f (6) (0) 6! = 1 / 3 3 f (6) (0) = 6! 1 / 3 3 = 6! 1 / 3(1 / 3- 1)(1 / 3- 2) 3! = 400 9 3. [2 points, 11.3 #55] Let f ( x, y, z ) = xe 2 z sin 3 y . Determine the partial derivative: ∂ 3 f ∂x∂y∂z . A . 6 e 2 z cos 3 y B. x 3 cos 3 y C. 5 e 2 z cos 3 y D. 5 e 3 z cos 2 y E.- 6 e 3 z sin 2 y F.- xe 3 z sin 2 y ∂f ∂z = 2 xe 2 z sin 3 y ∂ 2 f ∂y∂z = 6 xe 2 z cos 3 y ∂ 3 f ∂x∂y∂z = 6 e 2 z cos 3 y 4. [2 points, 8.7#23, 25 ] Find the Maclaurin series for the function: f ( x ) = Z x t 2 e 3 t dt [Hint: First find the Maclaurin series for...
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MAT 1322M3 Winter2005 - MAT 1322B Winter 2005 Professor...

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