number of solutions of $f(f(f(f(x))))$

Question

Let $f(x)={x^2+10x+20}$ then the number of real solutions of $f(f(f(f(x))))=0$

My try: I first tried finding the roots of $f(x)=0$ they came out to be

$-5+\sqrt{5}$ and $-5-\sqrt{5}$.

Let these roots be $\alpha$ and $\beta$

thus $f(\alpha)=0$ also if $f(f(f(f(x))))=0$ then $f(f(f(x)))=\alpha$

I was able to go only till this. I skipped the method of actually finding the composite function as it would be too cumbersome. Any ideas, helps will be appreciated.

Answer 1

Hint : $$ f(x) = x^2+10x+20 = (x+5)^2 - 5 $$

so $$ f(f(x)) = ((x+5)^2 - 5 + 5)^2 - 5 = (x+5)^4 - 5 $$ $$f(f(f(x))) = (x+5)^8 - 5 $$ ... and it should be easier to analyse $f^{(4)} (x)$ now.

Answer 2

The minimum of $f(x)$, a concave up quadratic, occurs at $x = -\frac{-10}{2(1)} = -5$, and $f(-5) = -5$. Then the turning point of $f(f(x))$ will occur at $f(f(-5)) = f(-5) = -5$, and so on until $f^4 (x)$ (which denotes $f(f(f(f(x))))$. There is only one minimum of $f(x)$, so there is only one turning point of $f^4 (x)$.

$f^4 (x)$ is also concave up, since the leading term of $f(x), f(f(x)) \cdots$ does not change sign, so the leading term is positive.

Now of a concave up function with only one turning point, how many roots are there in total?

Answer 3

An alternative method to those already presented:

Consider first a simpler case: when is $f^2(x)=0$? (I let $f^2(x)$ denote $f(f(x))$, or, equivalently, $(f \circ f)(x)$.)

It will be whenever $f(x) \in f^{-1}(0)$. (Recall that $f^{-1}(a) = \{ x \mid f(x)=a \}$, i.e. this is a set.)

A similar idea applies to your case. $f^4(x) = 0$ if and only if $f(x) \in (f^{-1})^3(0)$. This is something we can proceed through iteratively.

• First, what elements are in $f^{-1}(0)$? That is, which elements does $f$ map to $0$? As you have determined, these are $-5 \pm \sqrt 5$.

• Next, what elements are in $(f^{-1})^2(0) = f^{-1}(-5 \pm \sqrt 5)$? This means that $$x^2 + 10x + 20 = -5 \pm \sqrt 5$$ so just solve for $x$. You get $-5 \pm \sqrt[4]5$ for the positive root, and $-5 \pm i \sqrt[4]5$ for the negative root. Only the former is relevant.

• Iterate one more time. Then $$x^2 + 10x + 20 = -5 \pm \sqrt[4]5$$ and we solve for $x$. The negative solution gives us more complex numbers, but the positive gives us $-5 \pm \sqrt[8]5$.

Answer 4

The multiple composition of the equation can also be viewed in this way (amounting to a different interpretation of what is described in many of the other posted answers). The function $ \ f(x) \ = \ x^2 + 10x + 20 \ = \ (x + 5)^2 - 5 \ \ $ has the two real zeroes $ \ r_{\pm} \ = \ -5 \ \pm \sqrt{5} \ \ , $ so the next level of composition, $ \ f( \ f(x) \ ) \ = \ 0 \ \ $ requires us to solve $ \ f(x) \ = \ -5 \ \pm \ \sqrt{5} \ \ ; $ further levels of composition seem to call for more daunting calculation.

We might instead look at how the function transforms numbers. The fixed points of the function are given by $$ f(x) \ = \ x \ \ \Rightarrow \ \ x^2 \ + \ 10x \ + \ 20 \ \ = \ \ x \ \ \Rightarrow \ \ x^2 \ + \ 9x \ + \ 20 \ \ = \ \ (x + 5)·(x + 4) \ \ = \ \ 0 $$ $$ \Rightarrow \ \ x \ \ = \ \ -4 \ , \ -5 \ \ . $$

So $ \ f(-4) \ = \ -4 \ \ $ and $ \ f(-5) \ = \ -5 \ \ . $ We could then ask about how other numbers are "mapped" by the function relative to the fixed points. For $ \ x \ = \ -5 + \alpha \ \ , $ we obtain $$ ( \ [-5 + \alpha] + 5)^2 - 5 \ \ = \ \ \alpha^2 \ - \ 5 \ \ , $$ We see from this that $ \ f(x) \ = \ 0 \ \ $ corresponds to the mapping of $ \ -5 + \alpha \ \rightarrow \ 0 \ \ $ such that $ \ \alpha^2 - 5 \ = \ 0 \ \ . $ Hence there are two such points, $ \ -5 \pm \sqrt5 \ \ , $ which are symmetrically located around the axis of symmetry of the parabola representing the quadratic polynomial (as we expect of the zeroes). What we now also see is that $ \ f( \ f(x) \ ) \ = \ 0 \ $ corresponds to $ \ -5 + \alpha' \ \rightarrow \ -5 + \sqrt5 \ \ $ such that $ \ (\alpha')^2 - 5 \ = \ ( \ \alpha^2 \ )^2 - 5 \ = \ 0 \ \ . $

Repeated applications of the function transformation maintain the symmetry of the two points about $ \ x = -5 \ \ $ while "compressing" them successively closer to this fixed point (which would be referred to as an "attractor"). As also described in comments and the other posted answers, we observe that $ \ f^3( x ) \ = \ 0 \ $ is solved by $ \ -5 + \alpha'' \ \rightarrow \ -5 + \alpha' \ \ $ with $ \ (\alpha'')^2 - 5 \ = \ ( \ ( \ \alpha^2 \ )^2 \ ) ^2 - 5 \ = \ 0 \ \ $ and finally $ \ f^4( x ) \ = \ 0 \ $ is solved by $ \ -5 + \alpha''' \ \ $ with $ \ (\alpha''')^2 - 5 \ = \ ( \ ( \ ( \ ( \ \alpha^2\ )^2 \ )^2 \ ) ^2 - 5 \ = \ 0 \ \Rightarrow \ \alpha''' \ = \ \pm 5^{1/16} \ \ . $ At any finite number of compositions then, we continue to obtain just two real zeroes symmetrically placed about $ \ -5 \ $ (and a whole slew of complex zeroes).

What about the other fixed point? If we use $ \ x \ = \ -4 + \beta \ \ , $ we obtain $$ ( \ [-4 + \beta] + 5)^2 - 5 \ \ = \ \ ( \ 1 + \beta \ )^2 \ - \ 5 \ \ = \ \ 0 \ \ \Rightarrow \ \ \beta \ \ = \ \ -1 \pm \sqrt5 \ \ , $$ leading us again to $ \ x \ = \ -5 \pm \sqrt5 \ \ . $ (So the fixed point $ \ -4 \ $ acts as a "repellor".)