5
$\begingroup$

Find all $f(x)$ such that $$(f(x))^2-f(x^2)=constant$$

$\endgroup$
5
  • $\begingroup$ I Think this proble is very nice, and Now I can't solution.Thank you everyone help $\endgroup$ – math110 Jul 5 '13 at 12:26
  • $\begingroup$ What is the underlying field of constants? If it is of characteristic two, then, for example, all the functions $f(x)\in\mathbb{F}_2[x,1/x]$ are solutions ;-> $\endgroup$ – Jyrki Lahtonen Jul 5 '13 at 13:55
  • $\begingroup$ Thank you ,what is $F_{2}[x,1/x]?$ $\endgroup$ – math110 Jul 5 '13 at 14:01
  • $\begingroup$ $$\mathbb{F}_2[x,1/x]= \{\sum_{i=-n}^ma_ix^i\mid n,m\in\mathbb{Z}, a_i\in\mathbb{F}_2\,\text{for all $i$}\}.$$ $\endgroup$ – Jyrki Lahtonen Jul 5 '13 at 14:02
  • $\begingroup$ In characteristic two we have, for example $$ (x^3+x)^2=x^6+x^2,$$ so $f(x)=x^3+x$ is a solution in that case. I do expect that you mean that the field of constants is some subfield of the complex numbers. That's why I added the evil grin emoticon. But I do have the point that the term "rational function" does not specify the ground field. $\endgroup$ – Jyrki Lahtonen Jul 5 '13 at 14:05
9
$\begingroup$

Assume that $a$ is a pole of $f$ or order $n$ then $f^2$ has $a$ as a pole of order $2n$ and therefore so does $f(x^2)$. Therefore $a^2$ is a pole of $f$ of order $2n$. Therefore $f$ has no poles different from zero.

Let $g(x):=f(1/x)$. Notice that $g(x)$ also satisfy the equation. Subtract the two equations to get $$(f(x)-g(x))(f(x)+g(x))=f(x^2)-g(x^2).$$

Therefore, for the function $h(x):=f(x)-g(x)$ we have $h(x)(f(x)+g(x))=h(x^2)$. If $a$ is a zero of $h(x)$ of order $n$, then $a^2$ is a zero of $h(x)$ too. Therefore, either $a=0$ or $h(x)=constant$.

Now we have $f(x)=P(x)/x^k$ for some polynomial $P$. Then $P(x)/x^k-P(1/x)/x^{-k}=C$.

If $n$ is the degree of $P$, comparing degrees we get that $n=2k$. Comparing coefficients we get that $P$ is a symmetric polynomial (the coefficients can be put in reverse order), i.e. $P(x)=a_{2k}x^{2k}+a_{2k-1}x^{2k-1}+\ldots+a_{2k-1}x+a_{2k}$.

So, so far we have that $f$ is a linear combination of $x^k+1/x^k$. Plugging a linear combination into $f(x)^2-f(x^2)=constant$,

... to be continued, but from there you should get the solutions ...

$\endgroup$
15
  • $\begingroup$ Nice! You should also add $f(x) =$ constant under $k = 0$. $\endgroup$ – hot_queen Jul 5 '13 at 12:59
  • $\begingroup$ The constant should be $0$ or $1$. Oh, in the recapitulation forgot to mention $1$. Adding it. $\endgroup$ – OR. Jul 5 '13 at 13:04
  • $\begingroup$ Okay. You are solving $f(x^2) - (f(x))^2 = 0$. It doesn't really matter. $\endgroup$ – hot_queen Jul 5 '13 at 13:08
  • 1
    $\begingroup$ In view of Michael's comment, you probably want to fix the part that says "f can only have 0 as its pole and/or zero implies $f(x) = ax^k$" to account for solutions of form $x^k + 1/x^k$. $\endgroup$ – hot_queen Jul 5 '13 at 13:14
  • 1
    $\begingroup$ I saw that poles of $g$ are not the zeros of $f$ long ago. Typing. $\endgroup$ – OR. Jul 5 '13 at 13:18
2
$\begingroup$

The earlier answers have forgotten $f(x)=x+1/x$

$\endgroup$
2
  • $\begingroup$ yes, I think this problem have other $\endgroup$ – math110 Jul 5 '13 at 13:10
  • 1
    $\begingroup$ @S.B. that wasn't the question. Look again. $\endgroup$ – Empy2 Jul 5 '13 at 13:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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