Vol. 11 No 1 - Pi Mu Epsilon

2 JACK SAMUEL CALCUT III
power terms in binomial expansion with alternating signs. That is:
Ln/2J - 1
Ln/2J
-1)i (n.)x2i,
t=O i=O 2~
(1) Pn(x) = ~ ( - 1)i (2i: 1)x2i+1 and qn(x) = L (
where (~) = k!(:~ k)! (the binomial coefficients).
(n) + ( n ) a
Before we prove that equations (1) hold recall that (n+ 1 ) =
. . . ' . k k k- 1 '
f undamental mgred1ent of the Pascal tnangle, which can be easily proven by induction
[2].
THEOREM 2. tan(narctan(x)) is defined by equations (1) for all natural n and
real x.
Proof We will proceed by induction on n. We have seen that equations (1) hold
true for n = 1, 2, 3, 4. Assume (1) is true for all n:::; k. We must show (1) is true for
k + 1. There are two cases to consider: k even and k odd. Suppose k is even. Then
k + 1 is odd. So
ARCTANGENT IDENTITIES 3
The only possible rational roots of this polynomial are x = ± 1 by the rational
root theorem. 0
Theorem 3 characterizes all single rational arctangent identities for rr /4.
Before we characterize all single rational arctangent identities for all rational
multiples of rr, it is necessary that we make some observations. First, we look at
Pn(1) and qn(1). From (1) we get: P1 (1) = 1, P2(1) = 2, P3 (1) = 2, P4(1) = 0,
q1(1) = 1, q2(1) = 0, q3(1) = -2 and q4(1) = - 4.
Further inspection leads one to the following conjecture (where n = 4d+ r, r < 4):
(2)
Pn(1) = ~
0, n = O(mod 4)
( - 4)d, n = 1(mod 4)
2( - 4)d, n = 2(mod 4)
2( - 4)d, n = 3(mod 4)
n = O(mod 4)
n = 1(mod 4)
n = 2(mod 4)
n = 3(mod 4)
Proving equations (2) hold is straightforward after we prove the next proposition.
PROPOSITION 4. Pk+1(1) = qk(1) + Pk(1) and qk+1(1) = qk(1)- Pk(1).
Proof Case 1. (k is even) Then k + 1 is odd and
k
Pk .( 1 ) = ~ {-1 )' Gi: ~) = ~( - 1 )' [ ci ~ 1) + (;i) J
k
k- 1 k
= t(- 1)i ( . k ) + t(- 1)i (k.)
i=O 2 ~ + 1 i=O 2 ~
Also,
1 + t(- 1)'+' [( . k ) + ( . k )]x2i+2
i=O 2~ + 2 2~ + 1
k
qk +l( 1 ) = ~( - 1 )' e ~ 1 ) = ~( - 1 )'
k
[ Gi) + (2i ~ 1) J
k
k
k
t (_1) i (k ~ 1) x2i
i=O
as desired. The proof for k odd is virtually identical, as the reader may wish to verify.
0
Note that Theorem 2 applies to all nonzero integral n, since if n < 0 then -n > 0
and tan(n arctan(x)) =- tan( -n arctan(x)) = -P- n(x)/q- n(x).
It is transparent from Theorem 2 that Pn(x) and qn(x) are rational for all integral
nand rational x. Clearly then, tan(narctan(x)) is rational for rational x provided
qn(x) # 0.
We now apply Theorem 2 to rr/4.
THEOREM 3. lfwe have narctan(x) = rr/4forsome nonzero integraln, then the
only possible rational values for x are x = ± 1.
Proof. narctan(x) = "i => tan(narctan(x)) = 1 => Pn(x)/qn(x) = 1 => Pn(x) =
qn(x) => qn(x)- Pn(x) = 0 => 1- (7)x- G)x 2 + · · · ± (n~ 1 )xn - 1 ± xn = 0.
2 ~
= ~(- 1 )'(~) - ~( - 1 ) ' (2i~ 1)
= qk(1) - Pk(1).
Case 2. (k is odd) The proof is virtually identical to Case 1, as the reader may
wish to verify. 0
THEOREM 5. Equations (2) hold for all natural n.
Proof We will proceed by induction on n. We have seen that equations (2)
hold true for n = 1, 2, 3, 4. Assume equations (2) hold for all n :::; k and write
k = 4 * d + r, r < 4.
Case 1. k = O(mod4) implies Pk(1) = 0 and qk(1) = ( - 4)d. Proposition 4 implies
Pk +1(1) = (-4)d and qk+1(1) = (-4)d.
Case 2. k = 1(mod4) implies Pk(1) = ( - 4)d and qk(1) = ( - 4)d. Proposition 4
implies Pk+1(1) = 2( -4)d and qk+1(1) = 0.
Case 3. k = 2(mod4) implies Pk(1) = 2( -4)d and qk(1) = 0. Proposition 4
impliespk+1(1) = 2(-4)d and qk+1(1) = -2(- 4)d.