IMC 1998/1/3

#1

Έστω

για $x \in \mathbb{R}$ και $f_{n+1}(x) = f_1(f_{n}(x))$ για $n \geqslant 1$ .

(α) Να υπολογιστεί το $\displaystyle{ \lim_{n \to \infty} \int_0^1 f_n(x) \, dx}$
(β) Να υπολογιστούν τα $\displaystyle{ \int_0^1 f_n(x) \, dx}$ για $n=1,2,\ldots$ .

#2

$\displaystyle{{f_1}\left( x \right) = 2x\left( {1 - x} \right)}$ και $\displaystyle{{f_{n + 1}}\left( x \right) = {f_1}\left( {{f_n}\left( x \right)} \right)}$

$\displaystyle{\begin{gathered} \frac{1}{2} - {f_n}\left( x \right) = \frac{1}{2} - 2{f_{n - 1}}\left( x \right)\left( {1 - {f_{n - 1}}\left( x \right)} \right) = 2f_{n - 1}^2\left( x \right) - 2{f_{n - 1}}\left( x \right) + \frac{1}{2} = 2{\left( {\frac{1}{2} - {f_{n - 1}}\left( x \right)} \right)^2} \hfill \\ \begin{array}{*{20}{c}} {\quad \quad \quad \,} & \begin{gathered} = 2{\left( {2{{\left( {\frac{1}{2} - {f_{n - 2}}\left( x \right)} \right)}^2}} \right)^2} = {2^3}{\left( {\frac{1}{2} - {f_{n - 2}}\left( x \right)} \right)^4} = {2^3}{\left( {2{{\left( {\frac{1}{2} - {f_{n - 3}}\left( x \right)} \right)}^2}} \right)^4} = {2^7}{\left( {\frac{1}{2} - {f_{n - 3}}\left( x \right)} \right)^8} = \hfill \\ = {2^{{2^3} - 1}}{\left( {\frac{1}{2} - {f_{n - 3}}\left( x \right)} \right)^{{2^3}}} = {2^{{2^4} - 1}}{\left( {\frac{1}{2} - {f_{n - 4}}\left( x \right)} \right)^{{2^4}}} = .. = {2^{{2^{n - 1}} - 1}}{\left( {\frac{1}{2} - {f_1}\left( x \right)} \right)^{{2^{n - 1}}}} = {2^{{2^n} - 1}}{\left( {\frac{1}{2} - x} \right)^{{2^n}}} \hfill \\ \end{gathered} \\ \end{array} \hfill \\ \end{gathered} }$

Άρα $\displaystyle{\boxed{{f_n}\left( x \right) = \frac{1}{2} - {2^{{2^n} - 1}}{{\left( {\frac{1}{2} - x} \right)}^{{2^n}}}}}$

$\displaystyle{\begin{gathered} {I_n} = \int\limits_0^1 {{f_n}\left( x \right)dx} = \int\limits_0^1 {\left( {\frac{1}{2} - {2^{{2^n} - 1}}{{\left( {\frac{1}{2} - x} \right)}^{{2^n}}}} \right)dx} = \frac{1}{2} - {2^{{2^n} - 1}} \cdot \int\limits_0^1 {{{\left( {\frac{1}{2} - x} \right)}^{{2^n}}}dx} = \mathop = \limits^{1/2 - x = y} = \hfill \\ \begin{array}{*{20}{c}} {} & { = \frac{1}{2} - {2^{{2^n} - 1}} \cdot \int\limits_{ - 1/2}^{1/2} {{y^{{2^n}}}dx} = \frac{1}{2} - \frac{1}{2} \cdot \frac{1}{{{2^n} + 1}} \Rightarrow \boxed{\int\limits_0^1 {{f_n}\left( x \right)dx} = \frac{{{2^{n - 1}}}}{{{2^n} + 1}}}} \\ \end{array} \hfill \\ \end{gathered} }$

.. και φανερά $\displaystyle{\mathop {\lim }\limits_{n \to \infty } {I_n} = \frac{1}{2}}$ .

IMC 1998/1/3

IMC 1998/1/3

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