Solution A3

Solution A3 - MATH 222, Calculus 3, Fall 2011 Solutions to...

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MATH 222, Calculus 3, Fall 2011 Solutions to Assignment 3 1. Find the Maclaurin series first for (1 - t 2 ) - 1 2 and then for arcsin x = sin - 1 x . (Hint: integrate.) Use this give another formula for π 4 . Solution: From the binomial series (1 + x ) m = n =0 ± m n x n , we see that (1 - t 2 ) - 1 2 = X n =0 ± - 1 2 n ( - 1) n t 2 n = 1 - 1 2 t 2 + 1 2 · 1 2 · 3 2 t 4 - 1 6 · 1 2 · 3 2 · 5 2 t 6 ··· . Since arcsin x = R x 0 (1 - t 2 ) - 1 2 , we integrate this series term-by-term to get arcsin x = X n =0 ± - 1 2 n ( - 1) n 2 n + 1 x 2 n +1 . This series expansion is valid at least for - 1 < x < 1, and arcsin( 1 2 ) = π 4 , so we get that π 4 = X n =0 ± - 1 2 n ( - 1) n (2 n + 1) 2 2 n +1 . [This can be tidied up a little.] 2. Find a unit vector perpendicular to both h- 3 , 1 , 2 i and h 1 , 2 , 3 i . Solution: The vector h- 3 , 1 , 2 i × h 1 , 2 , 3 i is perpendicular to both the given vectors. This cross product is det ~ i ~ j ~ k - 3 1 2 1 2 3 , which works out
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This note was uploaded on 01/11/2012 for the course MATH 100 taught by Professor Loveys during the Fall '11 term at McGill.

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Solution A3 - MATH 222, Calculus 3, Fall 2011 Solutions to...

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