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Let $T (x)= 4x (1 − x)$ be a map from $X = [0, 1]$ into itself. Prove or disprove that this map has a trajectory of period $7$.

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Note that $$ T(\sin^2 t) = 4\sin^2t\cos^2 t=\sin^2(2t)\ , $$ so that $$ \begin{aligned} T^{(7)}(\sin^2 t) &:= ( T\circ T\circ T\circ T\circ T\circ T\circ T) (\sin^2 t) \\ &=\sin (2^7 t)\ , \end{aligned} $$ and we can now easily find a fixed point, by specifying the difference between $t$ and $2^7t$ to be some small multiple of $2\pi$, for instance by requiring $ 2^7 t-t=2\pi$. This gives $t=2\pi/127$. (Or the difference may be $4\pi$, or $6\pi$, or ...) Fixed points of period seven are thus $$ \sin ^2\frac {2k\pi}{127}\ ,\qquad k=1,2,\dots, 126\ . $$


Numerically, pari/gp code:

? T(x) = 4 * (1-x) * x;
? p(x) = T(T(T(T(T(T(T(x))))))) - x;
? p( sin(2*Pi/127)^2 )
%8 = 1.6071211827648462023 E-40
? p( sin(4*Pi/127)^2 )
%9 = 6.428484731059384809 E-40
? p( sin(6*Pi/127)^2 )
%10 = 2.755064884739736347 E-38

Here is also an exact check working in the cyclotomic field of order $4\cdot 127$, which contains $j=\sqrt {-1}$ and a primitive root $z$ of unity of order $127$:

sage: K.<z> = CyclotomicField(127*4)
sage: R.<x> = PolynomialRing(K)
sage: w = z^4    # so w is a primitive root of unity of order 127
sage: j = z^127  # so j is a primitive root of unity of order 4, it is sqrt(-1)
sage: w.multiplicative_order(), j.multiplicative_order()
(127, 4)
sage: T = 4*x*(1-x)
sage: p = T(T(T(T(T(T(T(x))))))) - x
sage: p( ( (w - 1/w) / (2*j) )^2 )    # corresponds to p( sin(2pi/127)^2 )
0
sage: p( ( (w^2 - 1/w^2) / (2*j) )^2 )    # corresponds to p( sin(4pi/127)^2 )
0
sage: set( [ p( ( (w^k - 1/w^k) / (2*j) )^2 ) for k in [1..126] ] )
{0}

The last set containing the only element zero tells us, that all values listed above, $\sin^2(2k\pi/127)$ are roots of $p(x)=T^{(7)}(x)-x$. (Of course, we avoid $0,3/4$, which are the (only) fixed points of $T$

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