Why does my solution only work for $|x|\le1$?

Question

In my research, I experimented with the following set of equations: $$\sqrt{f(x)}+\sqrt{x+f(x)}=k_n$$ $$\sqrt{x+f(x)}=k_{n+1}$$ Upon solving this, it then follows that: $$\sqrt{g(x)}+\sqrt{x+g(x)}=k_{n+1}$$ $$\sqrt{x+g(x)}=k_{n+2}$$ And so forth, with $g(x)>f(x)\therefore |k_{n+1}-\sqrt{x}|<|k_n-\sqrt{x}|$ (i.e. successive k's are closer to the value of $\sqrt{x}$ - this is an approximation formula)

Attempting to find a general solution for the first equation, I attempted to solve it for $k_0=1$.The result I got was that: $$f(x)=\frac{1}{4}(1-x)^2$$ (The algebra to get to this is quite easy to figure out, square both sides and drag the non-roots over to the RHS, square again and cancel down)

Upon checking this by graphing it, I noticed this equation only solves my questions for $|x|\le1$

I would like to know why this occurs, and if there is a way I can find a solution to satisfy my equation for a broader range of x?

Answer 1

Note that the solution to $\sqrt{f(x)} + \sqrt{f(x)+x} = k_n$ is, if it exists, $f(x) = \frac{(|x|-k_n^2)^2}{4k_n}$.

In addition, in order to take the square root of $f(x)$, you need $f(x) \geq 0$. Therefore :

• if $x \geq 0$, you immediately recover $k_n = \sqrt{f(x)} + \sqrt{f(x)+x} \geq \sqrt{x}$;
• if $x \leq 0$ (i.e. $x = -|x|$, then $f(x)-|x|$ must be non-negative in order to take its square root, so $f(x) \geq |x|$, and $k_n = \sqrt{f(x)} + \sqrt{f(x)-|x|} \geq \sqrt{|x|}$.

Finally, you get that a necessary condition for your equation to possess a solution $f(x)$ is $|x| \leq k_n^2$. In particular, you will not be able to find a solution for $k_n = 1$ if $|x| > 1$.