Let $f:\mathbb{R}\to \mathbb{R}$ be a function which fulfills for every $n \in \mathbb{N}$ $$f(-n^2+3n+1)=(f(n))^2+1$$

Is it possible that such a function exists?

  • 2
    $\begingroup$ is your function continuous? any thoughts about the problem? where did you face it, the question in the body and in the title are different, is $x\in \mathbb{R}$? I gave a downvote because of the poor formatting (and because the question is not very clear), I will reverse when you fixed it $\endgroup$ – Dominic Michaelis Mar 26 '13 at 7:08
  • 2
    $\begingroup$ By $f^2(n)$, do you mean $f(f(n))$, or $f(n)^2$? $\endgroup$ – Glen O Mar 26 '13 at 7:09
  • $\begingroup$ @math110 i have edited you question, as you did have two different equalites please check if i took the right one (and i reversed the downvote) $\endgroup$ – Dominic Michaelis Mar 26 '13 at 7:36
  • $\begingroup$ Thank you ,my frend, becasuese my English is very poor. $\endgroup$ – math110 Mar 26 '13 at 8:13

The main idea here is writing at first some of the equations you get and look if they have common terms. Indeed here the terms for $n=3$ and $n=1$ are very interessting, as in both only occur $f(1)$ and $f(3)$.

\begin{align*} f(3)&=1+f(1)^2 \tag{$i$}\\ f(1)&=1+f(3)^2 \tag{$ii$}\\ \end{align*} As we don't know that much lets try to get an equation only having $f(1)$.

At first we have this equation: $$f(1)=1+f(3)^2$$ No we use $(i)$ to express $f(3)$ in terms of $f(1)$ $$f(1)= 1+(1+f(1)^2)^2=1+1^2+2f(1)^2+f(1)^4$$ is that possible?

Note that $f(1)$ is a solution of $$0= 2-x+2x^2+x^4$$ but this one has no real solution, hence your function can't exist, as for $x \in [0,1]$ $$2-x+2x^2+x^4\geq 2-x> 0 $$ and for $x\in [1,\infty)$ we know that $x<x^2$ and hence $$2-x+2x^2+x^4 \geq 2+x^2+x^4>0$$ and for $x\in (-\infty,0]$ $$2-x+2x^2+x^4 \geq 2+2x^2 +x^4 >0$$

  • $\begingroup$ @math110 glad i could help, you can accept my answer if you feel so (klicking the hook). $\endgroup$ – Dominic Michaelis Mar 26 '13 at 13:16

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.