Functie continua care are puncte fixe

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Cezar Lupu
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Functie continua care are puncte fixe

Post by Cezar Lupu »

Fie \( f:[0,1]\to\mathbb{R} \) o functie continua astfel incat sa avem \( \int_0^1f(x)dx=\frac{1}{2} \). Demonstrati ca exista \( x_{0}\in (0,1) \) astfel incat \( f(x_{0})=x_{0} \).

ONM 1977
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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Vlad Matei
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Post by Vlad Matei »

Fie \( F \) o primitiva al lui \( f \). Avem ca \( \displaystyle \int_{0}^{1}f(x)dx=F(1)-F(0)=\frac{1}{2} \). Fie \( g(x)=F(x)-\frac{x^2}{2} \). Avem ca \( g(1)=g(0) \) de unde din teorema lui Rolle avem ca exista \( x_{0} \) astfel incat \( g'(x_{0})=0 \) de unde \( f(x_{0})=x_{0} \).
Gelfand Valentin
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Post by Gelfand Valentin »

Cred ca poti evita teorema lui Role. Este destul de clar ca \( \int_{0}^{1}(f(x)-x) dx=0 \), iar aici poti aplica teorema de medie.
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Cezar Lupu
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Post by Cezar Lupu »

Da Valentin, ai dreptate. Se poate evita teorema lui Rolle, folosind prima teorema de medie, dar sa stii ca teorema de medie se poate demonstra si folosind teorema lui Rolle.
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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