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What are the solutions of the functional equation $f(x) + f(x^2) = 2$? Will they be one to one or many to one? Will they be periodic or not?

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closed as off-topic by LutzL, Crostul, Thomas Andrews, Servaes, user223391 Nov 9 '15 at 0:12

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  • $\begingroup$ Yes sir! One solution $f(x)=1.$ But if you want full answer, I would recommend to show some effort, because it is unlikely that someone here will do your homework completely for you. $\endgroup$ – iiivooo Oct 28 '15 at 17:13
  • $\begingroup$ $f(-x)=f(x)$ (they are both equal to $2-f(x^2)$) therefore $f$ is not one to one. $\endgroup$ – Iulia Oct 28 '15 at 17:17
  • $\begingroup$ @iiivooo Sir, you don't need to be negative always. I have tried this question many times. Been working on this since past three days, couldn't get satifactory answer, like other i also did get f(x)=1 as one of the solution but i wanted to know if there are other possibilities. It is just like i wanted to discuss ideas, no telling you all to do my homework. Thank you. $\endgroup$ – Tyrion Lannister Oct 28 '15 at 18:18
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Select some positive $q\ne 1$ and $g(t)=f(q^{2^t})-1$. Then $$ g(t+1)=f((q^{2^t})^2)-1=-g(t)=g(t-1) $$ Thus $g$ is $2$-periodic. But note that $q$ may be arbitrarily close to $1$.


In the end this means that you can divide the positive numbers in equivalence classes $\{q^{4^n}:n\in\Bbb Z\}$ where $f$ is necessarily constant, $f(q^{4^n})=f(q)$. The classes of $q$ and $q^2$ are connected by the functional equation and in general the function values of every equivalence class are determined by the values over $(1/4,1/2]$ and $[2,4)$ for the positive axis and thus for the full real line, adding the trivial values $f(\pm1)=1=f(0)$.


Only by demanding some continuity condition do you get the unique solution which is the constant function $1$.

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  • $\begingroup$ Can you explain, Sir? $\endgroup$ – Tyrion Lannister Oct 28 '15 at 18:20
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Let $x=2^{2^t}$ and $f(x)=g(t)$.

Then,

$$g(t)+g(t+1)=2,$$ which is solved by $$g(t)=1+C(-1)^t.$$

As only values one unit apart are related to each other, we can write

$$g(t)=1+c(t-\lfloor t\rfloor)(-1)^{\lfloor t\rfloor}$$ where $c(t)$ is arbitrary in $[0,1)$.

Thus $$f(x)=1+h(\sqrt[\lfloor \text{lg}(\text{lg}(x))\rfloor]x)(-1)^{\lfloor \text{lg}(\text{lg}(x))\rfloor},$$

where $h(x)$ is arbitrary in $[2,4)$.

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  • $\begingroup$ That $x = 2^{2^t}$ substitution can be quite useful. $\endgroup$ – marty cohen Oct 29 '15 at 17:40
  • $\begingroup$ @martycohen: yep, $x=2^t$ handles $x\to x/2$ and $2^{2^t}$ is for $x\to \sqrt x$. $\endgroup$ – Yves Daoust Oct 29 '15 at 17:51
  • $\begingroup$ Same reasoning applies for the interval $(0,1)$ using $x=2^{-2^t}$ and the negative half-axis. $\endgroup$ – LutzL Oct 30 '15 at 22:20

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