problemset11s

# problemset11s - MAT 137Y 2008-09 Winter Session Solutions...

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Unformatted text preview: MAT 137Y 2008-09 Winter Session, Solutions to Problem Set 11 1 (SHE 11.7) 16. Since Z ∞ e dx x ( ln x ) 2 = lim b → ∞ Z b e dx x ( ln x ) 2 = lim b → ∞- 1 ln x b e = lim b → ∞- 1 ln b + 1 = 1 , it follows the integral converges to 1. 24. The integral involved can be solved using partial fractions (which we omit): Z 4 1 dx x 2- 4 = lim b → 2- Z b 1 dx x 2- 4 + lim a → 2 + Z 4 a dx x 2- 4 = lim b → 2- 1 4 ln x- 2 x + 2 b 1 + lim a → 2 + 1 4 ln x- 2 x + 2 4 a = lim b → 2- 1 4 ln b- 2 b + 2- 1 4 ln 1 3 + lim a → 2 + 1 4 ln 1 3- 1 4 ln b- 2 b + 2 = ∞ , so the improper integral diverges. 2 (i) Let u = ln x , so e u = x = ⇒ e u du = dx . Then Z ∞ e dx ( ln x ) 2 = lim b → ∞ Z ln b 1 e u u 2 du = lim c → ∞ Z c 1 e u u 2 du , where c = ln b . But clearly the integral on the right side diverges as lim x → ∞ e x x 2 = ∞ (which can be verified by L’Hˆopital’s Rule), so the integral must diverge. (ii) By comparison we see that for all x , 1 x 2 + sin x ≤ 1 x 2- 1 . The integral Z ∞ 2 dx x 2- 1 = lim b → ∞ 1 2 Z b 2 1 x- 1- 1 x + 1 dx = lim b → ∞ 1 2 ln x- 1 x + 1 b 2 = 1 2 ln3 , so by the comparison test, the integral in question also converges. 3 We need to find the first five derivatives of arcsin x . Doing so gives us, f ( x ) = 1 √ 1- x 2 , f 00 ( x ) = x ( 1- x 2 ) 3 / 2 , f 000 ( x ) = 3 x 2 ( 1- x 2 ) 5 / 2 + ( 1- x 2 )- 3 / 2 = 2 x 2 + 1 ( 1- x 2 ) 5 / 2 Differentiating twice more yields f ( 4 ) ( x ) = 3 x ( 2 x 2 + 3 ) ( 1- x 2 ) 7 / 2 , f ( 5 ) ( x ) = 3 ( 8 x 4 + 24 x 2 + 3 ) ( 1- x 2 ) 9 / 2 . This gives f ( ) = 0, f ( ) = 1, f 00 ( ) = 0, f 000 ( ) = 1, f ( 4 ) ( ) = 0, and f ( 5 ) ( ) = 9. Therefore, the fifth Taylor polynomial P 5 ( x ) is P 5 ( x ) = f ( )+ f ( ) x + f 00 ( ) 2!...
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## This note was uploaded on 04/09/2010 for the course MAT 137 taught by Professor Uppal during the Spring '08 term at University of Toronto.

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problemset11s - MAT 137Y 2008-09 Winter Session Solutions...

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