Adv Alegbra HW Solutions 17

Adv Alegbra HW Solutions 17 - 17 The argument concludes as...

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: 17 The argument concludes as that of the base step. (n + 1)2 = (n ω + εn +1 ) · (n ω + εn +1 ) = n 2 + 2n ω · εn +1 + 1, so that ω · εn +1 = 1. By the base step, ω = εn +1 , and the proof is complete. 1.36 (Star of David) Prove, for all n > r ≥ 1, that n−1 r −1 n r +1 n−1 n+1 = r r n r −1 n+1 . r +1 n−1 n−1 ii r − 1 UUUUUU r i UUUU ii iiii UUU iiii UUUU iii UUUU iiii UUUU iiii UUU iii i n n n ii r + 1 r − 1 UUUUUU r iii UUUU iiii UUUU iiii i UUUU UUUU iiiii i iiii UUUUUU ii n + 1 ii n+1 r r +1 Solution. Using Pascal’s formula, one sees that both sides are equal to (n − 1)!n !(n + 1)! . (r − 1)!r !(r + 1)!(n − r − 1)!(n − r )!(n − r + 1)! 1.37 For all odd n ≥ 1, prove that there is a polynomial gn (x ), all of whose coefficients are integers, such that sin(nx ) = gn (sin x ). Solution. From De Moivre’s theorem, cos nx + i sin nx = (cos x + i sin x )n , we have sin nx = Im (cos x + i sin x )n n = Im r =0 nr r i sin x cosn −r x . r ...
View Full Document

This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

Ask a homework question - tutors are online