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Problem Of The Week #390 Nov 1st, 2019

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Congratulations to the following members for their correct solution!(Cool)

1. castor28
2. MegaMoh

Partial credit goes to lfdahl for his partially correct solution!

Solution from castor28 :
Let us write $f^n$ for the n-fold composition of $f$, so that $f^0(x) = x$ and $f^{n+1}(x) = f(f^n(x))$. We must solve $f^5(x)=0$.

The graph of $f$ has a minimum at $(-6,-6)$. This shows that $-6$ is a fixed point of $f$, and makes it worthwhile to try the substitution $x=z-6$. Under that substitution, we get:
$$
f(z-6) = z^2 - 6
$$
and, by induction,
$$
f^n(z-6) = z^{2^n}-6
$$
We must now solve:
$$
f^5(x) = f^5(z-6) = z^{32}-6 = 0
$$
giving the real roots $z=\pm\sqrt[32]{6}$ and $x=-6\pm\sqrt[32]{6}$.

The complex solutions are $x=-6 + \sqrt[32]{6}\,e^{\frac{2\pi ni}{32}}$, with $0\le n<32$.
 
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