Find all $f(x)$ such that $$(f(x))^2-f(x^2)=constant$$
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I Think this proble is very nice, and Now I can't solution.Thank you everyone help – math110 Jul 05 '13 at 12:26
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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 ;-> – Jyrki Lahtonen Jul 05 '13 at 13:55
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Thank you ,what is $F_{2}[x,1/x]?$ – math110 Jul 05 '13 at 14:01
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$$\mathbb{F}2[x,1/x]= {\sum{i=-n}^ma_ix^i\mid n,m\in\mathbb{Z}, a_i\in\mathbb{F}_2,\text{for all $i$}}.$$ – Jyrki Lahtonen Jul 05 '13 at 14:02
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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. – Jyrki Lahtonen Jul 05 '13 at 14:05
2 Answers
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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 ...
OR.
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The constant should be $0$ or $1$. Oh, in the recapitulation forgot to mention $1$. Adding it. – OR. Jul 05 '13 at 13:04
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Okay. You are solving $f(x^2) - (f(x))^2 = 0$. It doesn't really matter. – hot_queen Jul 05 '13 at 13:08
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1In 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 $x^k + 1/x^k$. – hot_queen Jul 05 '13 at 13:14
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I don't know what you mean by that. If $f(x) = x + 1/x$, then $g(x) = f(1/x) = f(x)$ so pole/s of $g$ is a zero of $f$. And you don't need to impress me with your speed - I really like your argument. – hot_queen Jul 05 '13 at 13:25
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You are right. Sorry about that. And I should have said "f and g can only have 0 as their poles implies $f(x)=ax^k$"above. – hot_queen Jul 05 '13 at 13:30
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I see, somehow I thought it says $f(x)^2=f(x^2)$. I'll remove my comment. – S.B. Jul 05 '13 at 14:15
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Yes, I missed that. There is where I am missing the solutions from before, $f(x)=x^k$. – OR. Jul 05 '13 at 15:21
