f(f(x)) = x² - x + 1
f(0) = ?
(@edsaperia inflicted this on me earlier :P )
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Replying to @increpare @edsaperia
because + 1 is the only output that doesn't depend on x, so I assume that applying f once does + 1/2
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Replying to @undefdev @edsaperia
Take a simplified version of the problem (equivalent in your analysis :P ) g(g(x))=1 . defining g(x)=1 is a solution to this, giving g(0)=1
1:48 PM - 16 Mar 2017
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