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If $f(x)$ is a polynomial satisfying $ f(x)f(\frac 1x) = f(x)+f(\frac 1x)$ and $f(3)=28$, then how could we find $f(4)$ ?

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How about re-writing it as $$\big(f(x)-1\big)\cdot\big(f(1/x)-1\big)=1$$ –  GEdgar Nov 22 '11 at 16:24
    
@GEdgar:and then ...? –  Quixotic Nov 22 '11 at 18:58
    
... and then we know $f(x)-1 = x^m$ for some $m$ or $f(x)-1 = -x^m$ for some $m$ as the only ways to get it. Assuming INTEGER COEFFICIENTS (which is not stated), from $\pm 3^m = 27$ we have a good guess for $m$ and the sign. –  GEdgar Nov 22 '11 at 20:47
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1 Answer

up vote 7 down vote accepted

Solving the functional equation for $f\left(\frac{1}{x} \right) = \frac{f(x)}{f(x)-1}$. This means that $f(x)-1$ must be a monomial. Let $f(x) = 1 + c x^d$. Then $$ c \left( \frac{1}{x} \right)^d +1 = \frac{1}{c} \left( \left( \frac{1}{x} \right)^d + c \right) $$ This, implies $c^2 = 1$. Now use $f(3) = 28$ to determined $c$ and $d$. Since $28 = 1 + 1 \times 3^3$, we conclude $c=1$ and $d=3$.

Thus $f(4) = 1 + 4^3 = 65$.

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I never studied functional equations before, is it algebra pre-calculus? –  Quixotic Nov 22 '11 at 16:34
    
Never mind, replace the phrase functional equation with equation. As a side note, the problem asks you to solve the functional equation (i.e. equation for a function, satisfied for every $x$) in polynomials. You may glance though wiki page on functional equations. Another famous functional equation is $f(x+1) = 2 f(x)$, which is solved by $f(x) = f_0 2^x$. –  Sasha Nov 22 '11 at 16:41
    
Two things (1)why $f(x)-1$ must be a monomial ? (2) what is $f_0$ in your comment? is it f(0)? –  Quixotic Nov 22 '11 at 18:57
    
@MaX Rational function $f(x)/(f(x)-1)$ is a polynomial in $\frac{1}{x}$. Set $d$ be a degree of this polynomial. This $x^d f\left( \frac{1}{x} \right)$ is a polynomial in $x$ of degree $d$, denote it $p(x)$. Thus, $x^d f(x) = (f(x)-1) p(x)$, and $(-x^d + p(x)) f(x) = p(x)$. Since $f(x)$ has degree $d$ and $p(x)$ has degree $d$, it follows that $p(x)-x^d$ is a constant, thus $p(x) = c + x^d$ and $f(x) = p\left(\frac{1}{x} \right) x^d = c x^d + 1$. –  Sasha Nov 22 '11 at 19:21
    
@MaX Yes, $f_0$ is an arbitrary constant, and $f(0) = f_0$. I think GEdgar's hint is the most elegant way to arrive at $f(x) = 1 + c x^d$, with $c^2 = 1$. –  Sasha Nov 22 '11 at 19:22
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