onequestion
TRANSCRIPT
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8/7/2019 OneQuestion
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1. Given that tan is, where i = 1, 2, . . . , n, are n zeros ofxn a1x
n1 + a2xn2 + . . .+ (1)nan = 0.
Prove that
tan
n
i=1i
=
a1a3+a5...+(1)ka(2k+1)
1a2+a4...(1)k1a(2k), ifn = 2k + 1
a1a3+a5...+(1)k1a(2k1)
1a2+a4...(1)k1a2k , ifn = 2k
Solution We need to find tan (n
i=1 i). Start with a polynomial x2 a1x + a2 = 0 with roots tan 1
and tan 2:
tan(1 + 2) =tan 1 + tan 2
1 tan 1 tan 2=
a1
1 a2,
and a polynomial x3 a1x2 + a2x a3 = 0 with roots tan 1, tan 2 and tan 3:
tan(1 + 2 + 3) =
tan 1+tan 21tan 1 tan 2
+ tan 3
1 tan 1+tan 21tan 1 tan 2 tan 3
=tan 1 + tan 2 + tan 3(1 tan 1 tan 2)
1 tan 1 tan 2 (tan 1 + tan 2)tan 3
=
3i=1 tan i
3i=1 tan i
1
1i
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Notice that for u + 1 v 1, we have
Au+1(v 1) + tan vAu(v 1)
= (tan 1 tan 2 . . . tan u tan u+1 u+1) + . . . + (tan 1 tan 2 . . . tan u tan v1 u+1
) + . . .
+(tan1 tan vu . . . tan v2 tan v1 u+1
) + . . . + (tan vu1 tan vu . . . tan v2 tan v1 u+1
)
+tan v
(tan 1 tan 2 . . . tan u1 tan u
u
) + . . . + (tan 1 tan 2 . . . tan u1 tan v1 u
) + . . .
+(tan1 tan vu+1 . . . tan v2 tan v1 u
) + . . . + (tan vu tan vu+1 . . . tan v2 tan v1 u
)
=( vu+1)
1t1
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This indicates that the identity is true when m is odd. Notice that we use the result 2 except thelast term in the denominator in which we use 3.
2. m = 2k
For n = m + 1, i.e. n = 2k + 1. then
tan(m+1 +m
i=1 i)
=tan 2k+1+
A1(2k)A3(2k)+A5(2k)...+(1)k2
A(2k3)(2k)+(1)k1
A(2k1)(2k)
1A2(2k)+A4(2k)...(1)k2A(2k2)(2k)(1)
k1A(2k)(2k)
1tan 2k+1A1(2k)A3(2k)+A5(2k)...+(1)
k2A(2k3)(2k)+(1)k1A(2k1)(2k)
1A2(2k)+A4(2k)...(1)k2A(2k2)(2k)(1)
k1A(2k)(2k)
=tan 2k+1[1A2(2k)+A4(2k)...(1)k2A(2k2)(2k)(1)k1A(2k)(2k)]
1A2(2k)+A4(2k)...(1)k2A(2k2)(2k)(1)k1A(2k)(2k)to be continued
+A1(2k)A3(2k)+A5(2k)...+(1)k2A(2k3)(2k)+(1)
k1A(2k1)(2k)
tan 2k+1[A1(2k)A3(2k)+A5(2k)...+(1)k2A(2k3)(2k)+(1)k1A(2k1)(2k)]
=A1(2k+1)A3(2k+1)+A5(2k+1)...+(1)
k2A(2k1)(2k+1)+(1)k1A(2k+1)(2k+1)
1A2(2k+1)+A4(2k+1)...(1)k2A(2k2)(2k+1)(1)k1A(2k)(2k+1)
One may compare this to the identity and found it matching exactly. It is indeed this proves theidentity as the degree increases by one and becomes the case of odd m.
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