2

Let $\{f_n(x)\}$ be a sequence of polynomials defined inductively as \begin{align*} f_1(x) &= (x-2)^2, \\ f_{n+1}(x) &= (f_n(x)-2)^2, \quad n \geq 1. \end{align*} Let $a_n$ and $b_n$ respectively denote the constant term and the coefficient of $x$ in $f_n(x)$. Then

  1. $a_n = 4$, $b_n = -4^n$,
  2. $a_n = 4$, $b_n = -4n^2$,
  3. $a_n = 4^{(n-1)!}$, $b_n = -4^n$,
  4. $a_n = 4^{(n-1)!}, b_n = -4n^2$.

I see that general or $n$-th term of this will be $((x-2)^2 - 2)^2-2)^2...$ But how do I simplify this to get to the constant term

Jendrik Stelzner
  • 16,401
Taylor Ted
  • 3,408

2 Answers2

2

The trick is to ignore $x^2$ terms because they will never affect $a_n$ or $b_n$ in later values.
$f_1(x)=x^2g_1(x)-4x+4$
$f_2(x)=...+(-4x+2)^2=...-16x+4$
$f_3(x)=...+(-16x+2)^2=...-64x+4$
$f_4(x)=...+(-64x+2)^2=...-256x+4$

Empy2
  • 50,853
1

To expand on the comment by Vinod, for any polynomial, we get the constant term by plugging 0 in for $x$. In fact, that's pretty much the definition of the "constant term". Doing this we see that we will get

$$\begin{align*} (\cdots (0-2)^2 - 2)^2 - \cdots - 2)^2 &= (\cdots ((-2)^2-2)^2 - 2)^2 - \cdots - )^2 \\ &= (\cdots (4-2)^2 - 2)^2 - \cdots - 2)^2 \\ &= (\cdots (2)^2 - 2)^2 - \cdots - 2)^2 \\ &= (\cdots (4 - 2)^2 - \cdots - 2)^2 \\ &= \cdots \\ &= 4 \end{align*}$$

(We continually remove a set of parentheses in each step, and eventually end up with $4$ as our constant term.)

Now onto the coefficient of $x$, we can perform a similar analysis. We first note that the coefficient of $x$ for $n=1$ is $-4$:

$$f_1(x) = (x-2)^2 = x^2 - 4x + 4$$

Now we prove by induction that the coefficient is $-4^n$.

Assume that the coefficient of $x$ is $-4^n$ for $f_n$. Then we know $f_{n+1}(x)$ is defined as

$$ f_{n+1}(x) = (f_n(x) - 2)^2$$

We know three very important things about this equation. First, by the inductive hypothesis, we know that the coefficient of $x$ for $f_n$ is $-4^n$, we know by the previous step that the constant term of $f_n$ is $4$, and we know that we don't care about the other terms. Observe:

$$\begin{align*} (f_n(x) - 2)^2 &= (c_n x^n + \cdots + c_2 x^2 - 4^n x + 4 - 2)^2 \\ &= (c_n x^n + \cdots + c_2 x^2 - 4^n x + 2)^2 \\ &= (c_n x^n + \cdots + c_2 x^2 - 4^n x + 2) \cdot (c_n x^n + \cdots + c_2 x^2 - 4^n x + 2) \\ &= \cdots - 4^n x \cdot 2 - 4^n x \cdot 2 + 2 \cdot 2 \\ &= \cdots - 4 \cdot 4^n x + 4 \\ &= \cdots - 4^{n+1} x + 4 \end{align*}$$

as desired. Note that the $\cdots$ in the last few equations are all higher order terms, i.e. when multiplying $(c_n x^n + \cdots + c_2 x^2 - 4^n x + 2)$ by itself, the $\cdots$ represent all of the terms whose exponents are 2 or greater.

Finally, it is worth noting that once you know the constant term is 4, you can manually calculate the $f_1$, $f_2$, and $f_3$ by hand and see that the coefficients for $x$ are $-4^1 = -4$, $-4^2 = -16$, and $-4^3 = -64$ respectively, and then conclude (since it is a multiple choice problem), that the answer is $a_n = 4$ and $b_n = -4^n$, but where's the fun in that?

kbrose
  • 286