+ 8x + 2, find f(X): Immediately, I took the X




Дата канвертавання19.04.2016
Памер16.77 Kb.
For f(f(x)) = x4 – 4x3 + 8x + 2, find f(x):
Immediately, I took the x4 as a clue that f(x) must be a quadratic (since squaring a square yields the forth power).
Assuming f(x) = ax2 + bx + c (it is obvious that a=1, but we will take the long route anyway).
f(f(x)) = a(ax2 + bx + c)2 + b(ax2 + bx + c) + c
= a(a2x4 + 2abx3 + 2acx2 + b2 x2 + 2bcx + c2) + abx2 + b2x + bc + c
= a3x4 + 2a2bx3 + 2a2cx2 + ab2x2 + 2abcx + ac2 + abx2 + b2x + bc + c
= a3x4 + 2a2bx3 + (ab + ab2 + 2a2c)x2 + (b2 + 2abc)x + (ac2 + bc + c)

with a3 = 1, a = 1


f(f(x)) = x4 + 2bx3 + (b2 + b + 2c)x2 + (b2 + 2bc)x + (c2 + bc + c)
with 2b = -4, b = -2
f(f(x)) = x4 - 4x3 + (2c + 2)x2 + (4 - 4c)x + (c2 - c)
with 2c + 2 = 0, c = -1
f(f(x)) = x4 - 4x3 + 8x + 2
Thus
f(x) = x2 - 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) = ax3 + bx2 + cx + d
Then

f(f(x)) = a(ax3 + bx2 + cx + d)3 + b(ax3 + bx2 + cx + d)2 + c(ax3 + bx2 + cx + d) + d


= a4x9 + 3a3bx8 + (3a3c+3a2b2)x7 + (3a3d+6a2bc+ab3)x6 + (6a2bd+3a2c2+2ab2c)x5 + (6a2cd+3ab2d+3abc2)x4 + (3a2d2+6abcd+ac3)x3 + (3abd2+3ac2d)x2 + 3acd2x + ad3 + a2bx6 + 2ab2x5 + (b3+2abc)x4 + (2abd+2b2c)x3 + (bc2+2b2d)x2 + 2bcdx + bd2 + acx3 + bcx2 + c2x + cd + d
= a4x9 + 3a3bx8 + (3a3c+3a2b2)x7 + (3a3d+6a2bc+ab3+a2b)x6 + (6a2bd+3a2c2+2ab2c+2ab2x5)x5 + (6a2cd+3ab2d+3abc2+b3+2abc)x4 + (3a2d2+6abcd+ac3+ac+2abd+2b2c)x3 + (3abd2+3ac2d+bc2+2b2d+bc)x2 + (3acd2+c2+2bcd)x + (cd+d+bd2+ad3)

Now suppose our target is

f(f(x)) = ax4 + bx3 + cx2 + dx + f
Then a = 0 and

f(f(x)) = b3x4 + 2b2cx3 + (bc2+2b2d+bc)x2 + (c2+2bcd)x + (cd+d+bd2)

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)) = ax2+bx+c (where a is non-zero) and guess that f(x) = mx2+nx+p. Then,


f(f(x)) = m(mx2+nx+p)2 + n(mx2+nx+p) + p
= m3x4+2m2nx3+(2m2p+mn2)x2 + 2mnpx + mp2 + mnx2 + n2x + np + p
= m3x4 + 2m2nx3 + (2m2p+mn2+mn)x2 + (2mnp+n2)x + (mp2+np+p)
m must be zero. Then

f(f(x)) = n2x + (np+p)


Now there is no x2 term, which tells us that f(f(x)) = ax2+bx+c cannot be derived from a simple system of powers of x.
What are some alternative functions we can try?

More to come…


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