Fie \( f:[0,1]\to\mathbb{R} \) o functie integrabila astfel incat \( \int_0^1 f(x)dx=\int_0^1 xf(x)dx=1 \).
Sa se arate ca \( \int_0^1 f^{2}(x)dx\geq 4 \).
I. Rasa, ONM 2004
O problema data ONM 2004
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Cezar Lupu
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O problema data ONM 2004
Last edited by Cezar Lupu on Thu Jan 31, 2008 2:18 am, edited 2 times in total.
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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Alin Galatan
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Cauchy:
\( \int_{0}^{1}f^2(x)dx\cdot\int_{0}^{1}(3x-1)^2dx\geq(\int_{0}^{1}(3x-1)f(x)dx)^2 \).
Bineinteles, se stie ca \( (3x-1)f(x) \) e integrabila, deoarece e produs dintre o continua si o integrabila.
Dezvoltand parantezele, iese imediat
\( \int_{0}^{1}f^2(x)dx\geq (3-1)^2=4 \).
\( \int_{0}^{1}f^2(x)dx\cdot\int_{0}^{1}(3x-1)^2dx\geq(\int_{0}^{1}(3x-1)f(x)dx)^2 \).
Bineinteles, se stie ca \( (3x-1)f(x) \) e integrabila, deoarece e produs dintre o continua si o integrabila.
Dezvoltand parantezele, iese imediat
\( \int_{0}^{1}f^2(x)dx\geq (3-1)^2=4 \).
Last edited by Alin Galatan on Wed Sep 26, 2007 5:31 pm, edited 1 time in total.
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Cezar Lupu
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Da, asta este si solutia mea. De fapt se pot da vreo 9 solutii la problema asta. Marea "provocare" ca sa zic asa este gasirea unui polinom care sa verifice egalitatea din ipoteza si acel polinom este intr-adevar \( p(x)=3x-1 \).
P.S. Surprinzator sau nu, acesta tehnica si bineinteles problema in sine provine din Analiza Numerica
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P.S. Surprinzator sau nu, acesta tehnica si bineinteles problema in sine provine din Analiza Numerica
.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.