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# sC - MAT 137Y 2008-09 Winter Session Solutions to...

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MAT 137Y 2008-09 Winter Session, Solutions to Supplementary Problem Set #3 1 (SHE 11.7) 10. Z 1 0 dx 1 - x = lim a 0 + Z 1 a dx 1 - x = lim a 0 + h - 2 1 - x i 1 a = lim a 0 + 0 + 2 1 - a = 2, so the integral con- verges to 2. 26. Z 1 x ( 1 + x 2 ) 2 dx = lim b Z b 1 x ( 1 + x 2 ) 2 dx = lim b - 1 2 ( 1 + x 2 ) b 1 = lim b - 1 2 ( 1 + b 2 ) + 1 4 = 1 4 , so the integral converges to 1 4 . 54. Since 0 Z π sin 2 2 x x 2 dx Z π 1 x 2 dx , and Z π 1 x 2 dx converges (10.7.1), it follows by the integral comparison test (10.7.2) that Z π sin 2 2 x x 2 dx also converges. 2 (a) We prove by induction on n that lim x 0 + x ( ln x ) n = 0. For n = 1, we have lim x 0 + x ln x = lim x 0 + ln x 1 / x H = lim x 0 + 1 / x - 1 / x 2 = lim x 0 + ( - x ) = 0 , so the base case is true. Now suppose the statement is true for n = k , i.e. lim x 0 + x ( ln x ) k = 0. Then lim x 0 + x ( ln x ) k + 1 = lim x 0 + ( ln x ) k + 1 1 / x H = lim x 0 + ( k + 1 )( ln x ) k · ( 1 / x ) - 1 / x 2 = - ( k + 1 ) lim x 0 + x ( ln x ) k = 0 , so the statement is true for all positive integers n . (b) Once again we apply induction to show that Z 1 0 ( ln x ) n dx = ( - 1 ) n n !. If n = 1, then Z 1 0 ln x dx = lim a 0 + Z 1 a ln x dx = lim a 0 + h x ln x - x i 1 a = lim a 0 + - 1 - a ln a + a = - 1 = ( - 1 ) 1 1! since lim a 0 + a ln a = 0 by part (a). This proves the base case. Now assume Z 1 0 ( ln x ) k dx = ( - 1 ) k k ! for some positive integer k . Integrating by parts and applying the result fro part (a), we have Z 1 0 ( ln x ) k + 1 dx = lim a 0 + Z 1 a ( ln x ) k + 1 dx = lim a 0 + h x ( ln x ) k + 1 i 1 a - ( k + 1 ) lim a 0 + Z 1 a ( ln x ) k dx = lim a 0 + a ( ln a ) k + 1 - ( k + 1 ) · ( - 1 ) k k ! = 0 +( - 1 ) k + 1 ( k + 1 ) ! = ( - 1 ) k + 1 ( k + 1 ) ! , so the statement is true for n = k + 1, and hence the statement is true for all positive integers n . 3 (SHE 12.5) 4. Computing derivatives, f ( x ) = sec x = f 0 ( x ) = sec x tan x = f 00 ( x ) = sec x tan 2 x + sec 3 x = f 000 ( x ) = sec x tan 3 x + 5sec 3 x tan x = f ( 4 ) ( x ) = sec x tan 4 x + 18sec 3 x tan 2 x + 5sec 5 x ,

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so f ( 0 ) = 1 , f 0 ( 0 ) = 0 , f 00 ( 0 ) = 1 , f 000 ( 0 ) = 0 , f ( 4 ) ( 0 ) = 5 . Hence, the fourth Taylor polynomial for sec x is P 4 ( x ) = f ( 0 )+ f 0 ( 0 ) x + f 00 ( 0 ) 2! x 2 + f 000 ( 0 ) 3! x 3 + f ( 4 ) ( 0 ) 4! x 4 = 1 + x 2 2 + 5 x 4 24 . 8. Computing derivatives, if f ( x ) = x cos ( x 2 ) , then f 0 ( x ) = cos ( x 2 ) - 2 x 2 sin ( x 2 ) = f 00 ( x ) = - 6 x sin ( x 2 ) - 4 x 3 cos ( x 2 ) = f 000 ( x ) = - 24 x 2 cos ( x 2 ) - 6sin ( x 2 )+ 8 x 4 sin ( x 2 ) = f ( 4 ) ( x ) = - 60 x cos ( x 2 )+ 80 x 3 sin ( x 2 )+ 16 x 5 cos ( x 2 ) = f ( 5 ) ( x ) = - 60cos ( x 2 )+ 360 x 2 sin ( x 2 )+ 240 x 4 cos ( x 2 ) - 32 x 6 sin ( x 2 ) , so all derivatives at 0 are zero except f 0 ( 0 ) = 1 and
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sC - MAT 137Y 2008-09 Winter Session Solutions to...

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