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I have a question which asks to find the coefficient of x and the constant term, for $f_n(x)$ given that

$f_1(x) = (x - 2) ^ 2$ and $f_{n+1}(x) = (f_n(x) - 2) ^ 2, n >= 1$

Now I tried to derive the values like solving for $f_2$ and $f_3$ , but its getting too long. What is a shorter way to solve these problems?

I guess that the constant term would be $4$ or $4^{(n-1)!}$ but I am not sure.

I was proceeding like this,

$f_2 = (f_1 - 2) ^ 2 $

so for $-4 * f_1$ the constant term would be -16 and for $f_1^2$ it will be 16 for $2^4$ and $2*f_1$ would be 4 . so in total that's +4 .. so it could be 4 or $4^{0!}$.

shortCircuit
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  • Experimenting with $f_1$, $f_2$ and $f_3$ would yield very natural conjectures. Did you do that? For example, $f_1(x)=4-4x+$higher terms hence $f_2(x)=(2-4x)^2+$higher terms$=4-16x+$higher terms hence $f_3(x)=(2-16x)^2+$higher terms$=\ldots$ – Did Jun 07 '16 at 05:49

1 Answers1

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At $f_2$, const term is $(2^2-2)^2=2^2=4$, So, for $f_3$, you get the same thing, and this goes on for every $n$ .

Now the constant term being fixed, notice $x$ ca be formed only by multiplying a constant term with a $x$ containing term.So we need only to consider the $2\times (term\ with \ x) \times 2$ while squaring.n So, notice the sequence..

$$f_1\ ->(-2)$$ $$f_2\ ->2\times (-2)\times 2=(-2)\times 2^2$$ $$f_3\ ->2\times (-2)2^2\times 2=(-2)\times 2^4$$ $$f_4\ ->2\times (-2)2^4\times 2=(-2)\times 2^6$$ $$\vdots$$ $$f_n\ ->2\times (-2)2^{2n-4}\times 2=(-2)\times 2^{2n-2}$$

Qwerty
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