Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $f(4)$ ?

share|cite|improve this question
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
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 $f(4) = 1 + 4^3 = 65$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.