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 Dec8 awarded Caucus Dec1 revised Function such that $f(x) f(\pi/2 - x) = 1$ edited tags Dec1 awarded Scholar Dec1 accepted Function such that $f(x) f(\pi/2 - x) = 1$ Dec1 revised Function such that $f(x) f(\pi/2 - x) = 1$ added 454 characters in body Dec1 comment Function such that $f(x) f(\pi/2 - x) = 1$ @StevenStadnicki okay, I see. However, I have to choose the $C^\infty$ function such that its power series works at the midpoint. Setting all the derivatives to zero is like making it locally the constant function, and I could also locally make it match a $\tan$ function. Maybe I should have asked for how many analytic functions satisfy the requirement, since each analytic function would characterize the type of power series the $C^\infty$ function can have at the midpoint. Dec1 revised Function such that $f(x) f(\pi/2 - x) = 1$ added 13 characters in body Dec1 comment Function such that $f(x) f(\pi/2 - x) = 1$ by smooth I mean $C^\infty$. Looking at the second derivative at $x=pi/4$ (or I guess $1$ in your example) I think gives a nontrivial restriction on the function at that point, $f''-(f')^2=0$. Higher derivatives give still more restrictions, and I'm wondering if $tan(x)$ and the constant function are the unique solutions satisfying all these smoothness conditions. Nov30 comment Function such that $f(x) f(\pi/2 - x) = 1$ I changed the interval to \$0