584 Real Analysis
Proof.Letg(c)=y. Thenc=g(g(c))=g(y); hencey=g(c)=g(g(y)). Thusyis
a fixed point ofg◦g.Ify=a, thena=g(a)=g(y)=c, leading to a contradiction.
Similarly,y=bforcesc=b.Ify=c, thenc=g(y)=g(c),socis a fixed point of
g, again a contradiction. It follows thaty=d, i.e.,g(c)=d, and similarlyg(d)=c.
Suppose there isf:S→Ssuch thatf◦f=g. Thenf◦g=f◦f◦f=g◦f.
Thenf(a)= f(g(a))=g(f (a)),sof(a)is a fixed point ofg. Examining case by
case, we conclude thatf({a, b})⊂{a, b}andf({a, b, c, d})⊂{a, b, c, d}. Because
f◦f=g, the inclusions are, in fact, equalities.
Considerf(c).Iff(c)=a, thenf(a)=f(f(c))=g(c)=d, a contradiction
sincef(a)is in{a, b}. Similarly, we rule outf(c)=b. Of course,cis not a fixed point
off, since it is not a fixed point ofg. We are left with the only possibilityf(c)=d.
But thenf(d)=f(f(c))=g(c)=d, and this again cannot happen becausedis not a
fixed point ofg. We conclude that such a functionfcannot exist.
In the particular case of our problem,g(x)=x^2 −2 has the fixed points−1 and
2, andg(g(x))=(x^2 − 2 )^2 −2 has the fixed points− 1 , 2 ,−^1 +
√ 5
2 , and− 1 −√ 5
2. This
completes the solution.
(B.J. Venkatachala,Functional Equations: A Problem Solving Approach, Prism
Books PVT Ltd., 2002)
540.The standard approach is to substitute particular values forxandy. The solution
found by the student S.P. Tungare does quite the opposite. It introduces an additional
variablez. The solution proceeds as follows:
f(x+y+z)
=f(x)f(y+z)−csinxsin(y+z)
=f(x)[f(y)f(z)−csinysinz]−csinxsinycosz−csinxcosysinz
=f(x)f(y)f(z)−cf (x)sinysinz−csinxsinycosz−csinxcosysinz.Because obviouslyf(x+y+z)=f(y+x+z), it follows that we must have
sinz[f(x)siny− f(y)sinx]=sinz[cosxsiny−cosysinx].Substitutez=π 2 to obtain
f(x)siny−f(y)sinx=cosxsiny−cosysinx.Forx=πandynot an integer multiple ofπ, we obtain siny[f(π)+ 1 ]=0, and hence
f(π)=−1.
Then, substituting in the original equationx=y=π 2 yields
f(π)=[
f(π
2)]
−c,