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

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