Define the function f(x)=\frac{2x}{1-x^2}. Find the number of distinct real solutions of the equation f^{(5)} (x) =x. Let x = tan(y). Then f(x) = \frac{2\tan y}{1-\tan^2 y} = \tan(2y). Then, f^5(x) = f^5(\tan y) = tan(32y). Thus, we are asked to find the number of distinct solutions to \tan y = \tan(32y). Since tangent is periodic with period \pi, we have 32y = y + n\pi, for some integer n. Thus, y = \frac{n \pi}{31}. Since we only find the \tan y values and not the actual y values, we see clearly that there are only [Math Processing Error] solutions.

If you expanded the polynomial form, you do NOT get a degree 31 polynomial, but a degree 33 polynomial. You can check that the 'base' case " f(x) =x" yields a cubic x(1-x^2) = 2x, which gives 0, i, -i as the fixed points. Even though we know that i, -i are 2 complex roots of the degree 33 polynomial, it is difficult to justify that this polynomial must have exactly 33-2 =31 real roots, as you have to show that there are no other complex roots, or repeated real roots.

We can extend the domain of the tangent function to the complex numbers, using the definition \tan z = \frac { \sin z} { \cos z} = \frac {i (e^{-iz} - e^{iz}) } {e^{-iz} + e^{iz} }. This accounts for the 2 extra roots that we get. It is slightly interesting that the additional roots of f^{(n)} (x)=x are all real valued, which is highly unlikely for a general polynomial / rational function.

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TopNewestshow that there is no posible integers for when√n+1+√n+1is rational

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