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

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 to - ~ i + 11 ~ j - 7 ~ k . This is not a unit vector, so we normalize it by dividing by its length, p ( - 1) 2 + 11 2 + ( - 7) 2 = 171(= 3 19).
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern