For f(f(x)) = x^{4} – 4x^{3} + 8x + 2, find f(x):
Immediately, I took the x^{4 }as a clue that f(x) must be a quadratic (since squaring a square yields the forth power).
Assuming f(x) = ax^{2 }+ bx + c (it is obvious that a=1, but we will take the long route anyway).
f(f(x)) = a(ax^{2 }+ bx + c)^{2} + b(ax^{2 }+ bx + c) + c
= a(a^{2}x^{4} + 2abx^{3} + 2acx^{2} + b^{2} x^{2} + 2bcx + c^{2}) + abx^{2 }+ b^{2}x + bc + c
= a^{3}x^{4} + 2a^{2}bx^{3} + 2a^{2}cx^{2} + ab^{2}x^{2} + 2abcx + ac^{2} + abx^{2 }+ b^{2}x + bc + c
= a^{3}x^{4} + 2a^{2}bx^{3} + (ab + ab^{2} + 2a^{2}c)x^{2} + (b^{2} + 2abc)x + (ac^{2 }+ bc + c)
with a^{3 }= 1, a = 1
f(f(x)) = x^{4} + 2bx^{3} + (b^{2} + b + 2c)x^{2} + (b^{2} + 2bc)x + (c^{2 }+ bc + c)
with 2b = 4, b = 2
f(f(x)) = x^{4}  4x^{3} + (2c + 2)x^{2} + (4  4c)x + (c^{2 } c)
with 2c + 2 = 0, c = 1
f(f(x)) = x^{4}  4x^{3} + 8x + 2
Thus
f(x) = x^{2}  2x  1
As far as a discussion regarding “these types of problems”…
Did I get lucky with the assumption that f(x) was a quadratic.
Suppose f(x) = ax^{3} + bx^{2} + cx + d
Then
f(f(x)) = a(ax^{3} + bx^{2} + cx + d)^{3} + b(ax^{3} + bx^{2} + cx + d)^{2} + c(ax^{3} + bx^{2} + cx + d) + d
= a^{4}x^{9} + 3a^{3}bx^{8} + (3a^{3}c+3a^{2}b^{2})x^{7} + (3a^{3}d+6a^{2}bc+ab^{3})x^{6} + (6a^{2}bd+3a^{2}c^{2}+2ab^{2}c)x^{5} + (6a^{2}cd+3ab^{2}d+3abc^{2})x^{4} + (3a^{2}d^{2}+6abcd+ac^{3})x^{3} + (3abd^{2}+3ac^{2}d)x^{2} + 3acd^{2}x + ad^{3 }+ a^{2}bx^{6} + 2ab^{2}x^{5} + (b^{3}+2abc)x^{4} + (2abd+2b^{2}c)x^{3} + (bc^{2}+2b^{2}d)x^{2} + 2bcdx + bd^{2 }+ acx^{3} + bcx^{2} + c^{2}x + cd^{ }+ d
= a^{4}x^{9} + 3a^{3}bx^{8} + (3a^{3}c+3a^{2}b^{2})x^{7} + (3a^{3}d+6a^{2}bc+ab^{3}+a^{2}b)x^{6} + (6a^{2}bd+3a^{2}c^{2}+2ab^{2}c+2ab^{2}x^{5})x^{5} + (6a^{2}cd+3ab^{2}d+3abc^{2}+b^{3}+2abc)x^{4} + (3a^{2}d^{2}+6abcd+ac^{3}+ac+2abd+2b^{2}c)x^{3} + (3abd^{2}+3ac^{2}d+bc^{2}+2b^{2}d+bc)x^{2} + (3acd^{2}+c^{2}+2bcd)x + (cd+d+bd^{2}+ad^{3})
Now suppose our target is
f(f(x)) = ax^{4} + bx^{3} + cx^{2} + dx + f
Then a = 0 and
f(f(x)) = b^{3}x^{4} + 2b^{2}cx^{3} + (bc^{2}+2b^{2}d+bc)x^{2} + (c^{2}+2bcd)x + (cd+d+bd^{2})
Now f(f(x)) has a maximum power of four and with a=0, f(x) is a quadratic.
What about other maximum powers for f(f(x))

Max power f(f(x))

Max power f(x)?

1

1

2

?

3

?

4

2

5

?

6

3



Let’s try f(f(x)) = ax^{2}+bx+c (where a is nonzero) and guess that f(x) = mx^{2}+nx+p. Then,
f(f(x)) = m(mx^{2}+nx+p)^{2} + n(mx^{2}+nx+p) + p
= m^{3}x^{4}+2m^{2}nx^{3}+(2m^{2}p+mn^{2})x^{2} + 2mnpx + mp^{2} + mnx^{2 }+ n^{2}x + np + p
= m^{3}x^{4 }+ 2m^{2}nx^{3 }+ (2m^{2}p+mn^{2}+mn)x^{2} + (2mnp+n^{2})x + (mp^{2}+np+p)
m must be zero. Then
f(f(x)) = n^{2}x + (np+p)
Now there is no x^{2 }term, which tells us that f(f(x)) = ax^{2}+bx+c cannot be derived from a simple system of powers of x.
What are some alternative functions we can try?
More to come… 