4.2 Trigonometry 241(c) Prove there is a positive constantCsuch that one has 0<bn−an< 8 Cn for
alln.684.Two real sequencesx 1 ,x 2 ,...,andy 1 ,y 2 ,...are defined in the following way:
x 1 =y 1 =√
3 ,xn+ 1 =xn+√
1 +x^2 n,yn+ 1 =yn
1 +√
1 +y^2 n, forn≥ 1.Prove that 2<xnyn<3 for alln>1.685.Leta, b, cbe real numbers different from±√^13. Prove that the equalityabc=
a+b+cholds only if
3 a−a^3
3 a^2 − 1·
3 b−b^3
3 b^2 − 1·
3 c−c^3
3 c^2 − 1=
3 a−a^3
3 a^2 − 1+
3 b−b^3
3 b^2 − 1+
3 c−c^3
3 c^2 − 1.
The parametrization of the hyperbolax^2 −y^2 =1byx=cosht,y=sinhtgives
rise to the hyperbolic substitutionx=acoshtin expressions containing
√
a^2 −1. We
illustrate this with an example by the second author.
Example.Leta 1 =a 2 =97 and
an+ 1 =anan− 1 +√
(a^2 n− 1 )(a^2 n− 1 − 1 ), forn> 1.Prove that
(a) 2+ 2 anis a perfect square;
(b) 2+
√
2 + 2 anis a perfect square.Solution.We are led to the substitutionan=coshtnfor some numbertn(which for the
moment might be complex). The recurrence relation becomes
coshtn+ 1 =an+ 1 =coshtncoshtn− 1 +sinhtnsinhtn− 1 =cosh(tn+tn− 1 ).We deduce that the numberstnsatisfyt 0 =t 1 , andtn+ 1 =tn+tn− 1 (in particular they
are all real). And sotn=Fnt 0 , where(Fn)nis the Fibonacci sequence. Consequently,
an=cosh(Fnt 0 ),n≥1.
Using the identity 2(cosht)^2 − 1 =cosh 2t, we obtain
2 + 2 an=(
2 coshFnt 0
2) 2
.
The recurrence relation
2 cosh(k+ 1 )t=(2 cosht)(2 coshkt)−2 cosh(k− 1 )t