# Sequences defined as solutions to equations : $u_{n}=v_{n}^n$

For $n$ a positive integer at least equal to $2$, define the two following functions as follows : \begin{align*} f_{n}(x) & = \pi/4 + \arctan(\sqrt[n]{x})-\arctan(1/x) \text{ for all nonzero }x, \\ g_{n}(x) & = x^{n+1} +x^n+x-1 \text{ for all real }x. \end{align*} We can find, using differentiation for instance, that these two functions have each one unique real root and, using the intermediate value theorem, that it lies in the interval $(0,1)$. If we define the sequence $u_{n}$ as the real root of $f_{n}$ and $v_{n}$ as the real root of $g_{n}$, my question is to prove that the equality $u_{n}=v_{n}^n$ holds for all $n\geq 2$.

EDITED : I corrected the statement from $g_{n}(x) = x^{n+1} +x^4+x-1$ to $g_{n}(x) = x^{n+1} +x^n+x-1$.

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Careful. There is definitely a real root of $g_n$ in $(0,1)$. But if $n$ is odd, then the degree of $g_n$ is even. Since complex roots come in pairs, there will have to be more than one real root in these cases. –  Dimitrije Kostic Nov 23 '11 at 21:19
For example, if $n=3$ then $g_n(x) = 2x^4 +x -1$ has two real roots, at $-1$ and $0.6477988...$ (according to magma). –  Dimitrije Kostic Nov 23 '11 at 21:32

Since $\arctan(1/x)=\pi/2-\arctan(x)$ for every positive $x$, $\arctan(\sqrt[n]{u})+\arctan(u)=\pi/4$. Since $\tan(a+b)=(\tan(a)+\tan(b))/(1-\tan(a)\tan(b))$ for every suitable $a$ and $b$, one gets $1=\tan(\pi/4)=(u+\sqrt[n]{u})/(1-u\sqrt[n]{u})$, that is, $u+\sqrt[n]{u}+u\sqrt[n]{u}=1$. Using $v=\sqrt[n]{u}$, one gets $v^{n+1}+v^n+v=1$.
This proves that $u=v^n$, provided $v$ solves $v^{n+1}+v^n+v=1$.
And, in the question, $x^4$ should read $x^n$. –  Did Nov 24 '11 at 0:47
Thank you, Mr. Piau. I found this problem manuscript so maybe I misread $x^n$. The statement is now edited. –  user20010 Nov 24 '11 at 12:47