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Let $\{f_n(x)\}_{n\in \Bbb N}$ be a sequence of polynomials defined inductively as $$\begin{matrix} f_1(x) & = & (x-2)^2 & \\ f_{n+1}(x)& = & (f_n(x)-2)^2, &\text{ for all }n\ge 1.\end{matrix}$$

Let $a_n$ and $b_n$ respectively denote the constant term and the coefficient of $x$ in $f_n(x)$. Then

  1. $(\text A)\,\, a_n=4, b_n=-4^n$
  2. $(\text B)\,\,a_n=4, b_n=-4n^2$
  3. $(\text C)\,\, a_n=4^{(n-1)!}, b_n=-4^n$
  4. $(\text C)\,\, a_n=4^{(n-1)!}, b_n=-4n^2$

Please help me how to solve this problem. I am not able to solve it.

The problem can be found on this link.

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the link of the question is – amitabha Aug 4 '13 at 14:33

3 Answers 3



$f_2(x)=(f_1(x)-2)^2$ $=(x^2-4x+4-2)^2=(2-4x+x^2)^2=2^2+2\cdot2(-4x)+\cdots=4-4^2x+\cdots$

$f_3(x)=(f_2(x)-2)^2$ $=(4-2^4x+\cdots-2)^2=(2-2^4x+\cdots)^2=2^2-2\cdot2\cdot4^2x+\cdots=2^2-4^3x+\cdots$

$f_4(x)=(f_3(x)-2)^2$ $=(2^2-4^3x+\cdots-2)^2=(2-4^3x+\cdots)^2=2^2-2\cdot2\cdot2^6x+\cdots=2^2-4^4x+\cdots$

$f_5(x)=(f_4(x)-2)^2$ $=(4-4^4x+\cdots)^2=(2-4^4x+\cdots)^2=2^2-2\cdot2\cdot2^8x+\cdots=2^2-4^5x+\cdots$

So, $a_n=4$

$b_n=-4^{(n \text{ th term of } 1,2,3,4,5)}=-4^n$

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The following is a proof by induction of the statement $(\forall n\in \Bbb N)(a_n=4\land b_n=-4^n)$.

First start by noting that $\text{deg}(f_n(x))=2^n$, for all $n\in \Bbb N$. Therefore, for certain $c_{2^n}, c_{2^n-1}, \ldots , c_2, c_1 \color{gray}{=b_n}, c_0\color{gray}{=a_n}\in \Bbb Z$ it holds that $$f_n(x)=c_{2^n}x^{2^n}+\cdots + c_2x^2+b_nx+a_n.$$

$\bbox[5px,border:2px solid #0000FF]{(\forall n\in \Bbb N)(a_n=4)}$

$\bullet$ Clearly $a_1=4$.
$\bullet$ Let $n\in \Bbb N$ such that $a_n=4$.
$\,\,\,$ Since $f_{n+1}(x)=(f_n(x)-2)^2=(c_{2^n}x^{(2^n)}+\cdots +b_nx+(a_n-2))^2$, it clearly follows that the
$\,\,\,$ constant term in $f_{n+1}(x)$ is $\color{gray}{(a_n-2)^2=(4-2)^2=}4$.

$\therefore (\forall n\in \Bbb N)(a_n=4)$

$\bbox[5px,border:2px solid #0000FF]{(\forall n\in \Bbb N)(b_n=-4^n)}$

$\bullet$ Clearly $b_1=-4^1$.
$\bullet$ Let $n\in \Bbb N$ and suppose $b_n=-4^n$. Since $$\begin{align}f_{n+1}(x)=(f_n(x)-2)^2&=(c_{2^n}x^{(2^n)}+\cdots +b_nx+(4-2))^2 \\ &=(c_{2^n}x^{(2^n)}+\cdots +b_nx+2))(c_{2^n}x^{(2^n)}+\cdots +b_nx+2),\end{align}$$

it easy to to see that the coefficient of $x$ in $f_{n+1}(x)$ is $\color{gray}{2b_n+2b_n=4b_n=4\cdot (-4^n)=}-4^{n+1}$.

$\therefore (\forall n\in \Bbb N)(b_n=-4^n)$.

The parts regarding the coefficients $a_{n+1}$ and $b_{n+1}$ can be rigorously proved by induction or by applying the multinomial theorem.

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Induction on $\;n\;$ :


$$f_{n+1}(0)=\left(f_n(0)-2\right)^2\stackrel{\text{Ind. Hyp.}}=(4-2)^2=4$$

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