solve f ′(x)=f(f(x))


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Question Number 98913 by mr W last updated on 17/Jun/20

$s o l v e$ $f^{'} (x) = f (f (x))$
Commented by john santu last updated on 17/Jun/20
$let f(x) = Kx^β f ′(x) = Kβ x^(β−1) f(f(x))= K(Kx^β )^β = K^(β+1) x^β^2 ⇔ f ′(x) = f(f(x)) Kβ x^(β−1) = K.K^β x^β^2 ⇒β^2 =β−1 ⇔ { ((β = ((1+i(√3))/2))),((β = ((1−i(√3))/2))) :} β = e^(± i(π/3)) . ⇔K^β = β ⇒K = β^(1/β) for β = e^(i(π/3)) ⇒ (1/β) = e^(−i(π/3)) K= e^(i(π/3).−i(π/3)) = e^(((√3)/6)π (((√3)/2)+(1/2)))$
$let f (x) = {Kx}^{β}$ $f^{'} (x) = K β x^{β - 1}$ $f (f (x)) = K {({Kx}^{β})}^{β} = K^{β + 1} x^{β^{2}}$ $\Leftrightarrow f^{'} (x) = f (f (x))$ $K β x^{β - 1} = K . K^{β} x^{β^{2}}$ $\Rightarrow β^{2} = β - 1 \Leftrightarrow {\begin{cases} β = \frac{1 + i \sqrt{3}}{2} \\ β = \frac{1 - i \sqrt{3}}{2} \end{cases}$ $β = e^{\pm i \frac{π}{3}} .$ $\Leftrightarrow K^{β} = β \Rightarrow K = β^{\frac{1}{β}}$ $for β = e^{i \frac{π}{3}} \Rightarrow \frac{1}{β} = e^{- i \frac{π}{3}}$ $K = e^{i \frac{π}{3} . - i \frac{π}{3}} = e^{\frac{\sqrt{3}}{6} π (\frac{\sqrt{3}}{2} + \frac{1}{2})}$
Commented by mr W last updated on 17/Jun/20

$t h a n k s s i r!$ $n o r e a l f u n c t i o n ?$
Commented by john santu last updated on 17/Jun/20

$no sir .$
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