Fie\( f:[0,1]\rightarrow\mathbb{R}^+ \) o functie continua cu f(1)=1 si sirul \( (a_n)_{n\geq1} \) definit prin
\( a_n=\int_0^1{\frac{f(x)}{1+x^n}dx} \), \( n\geq1. \)
Aratati ca \( \lim_{n\to\infty}n(\int_0^1{f(x)dx}-a_n)=\ln 2 \).
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Limita unui sir integral 3
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Solutie.
a) Vom arata ca \( a_{n}=\int_0^1\frac{f(x)}{1+x^{n}}dx\to\int_0^1 f(x)dx \). Pentru aceasta vom folosi criteriul clestelui. Intr-adevar, avem:
\( 0\leq\int_0^1f(x)dx- \int_0^1\frac{f(x)}{1+x^{n}}dx=\int_0^1\frac{x^{n}f(x)dx}{1+x^{n}}dx\leq\int_0^1x^{n}f(x)dx\leq\frac{\sup_{x\in [0,1]}f(x)}{n+1} \). Trecand la limita dupa \( n \) tinzand la \( \infty \), avem ca \( \lim_{n\to\infty}a_{n}=\int_0^1f(x)dx \).
b) Avem \( \lim_{n\to\infty}n(\int_0^1f(x)dx-a_{n})=\lim_{n\to\infty}\int_0^1\frac{nx^{n}f(x)}{1+x^{n}}dx=\int_0^1(\log(1+x^{n}))^{\prime} xf(x)dx \)
care va fi egala cu o integrare prin parti, cu:
\( f(1)\log 2-\int_0^1\log(1+x^{n})(xf(x))^{\prime} dx \). Acum, ne vom ocupa de sirul \( (b_{n})_{n\geq 1} \) definit prin \( b_{n}=\int_0^1\log(1+x^{n})(xf(x))^{\prime} dx \). Vom proceda ca mai sus (adica cum am facut cu sirul \( a_{n} \)), i.e.
\( \left | \int_0^1\log(1+x^{n})(xf(x))^{\prime} dx\right| \leq \sup_{x\in [0,1]} (xf(x))^{\prime} \int_0^1\log(1+x^{n})dx\leq\frac{\sup_{x\in [0,1]} (xf(x))^{\prime}}{n+1} \), unde am folosit inegalitatea elementara \( \log(1+t)\leq t, \forall t>0 \). Acum, trecand la limita atunci cand \( n\to\infty \), o sa rezulte ca
\( \lim_{n\to\infty}n(\int_0^1f(x)dx-a_{n})=f(1)\log 2=\log 2 \). \( \qed \)
a) Vom arata ca \( a_{n}=\int_0^1\frac{f(x)}{1+x^{n}}dx\to\int_0^1 f(x)dx \). Pentru aceasta vom folosi criteriul clestelui. Intr-adevar, avem:
\( 0\leq\int_0^1f(x)dx- \int_0^1\frac{f(x)}{1+x^{n}}dx=\int_0^1\frac{x^{n}f(x)dx}{1+x^{n}}dx\leq\int_0^1x^{n}f(x)dx\leq\frac{\sup_{x\in [0,1]}f(x)}{n+1} \). Trecand la limita dupa \( n \) tinzand la \( \infty \), avem ca \( \lim_{n\to\infty}a_{n}=\int_0^1f(x)dx \).
b) Avem \( \lim_{n\to\infty}n(\int_0^1f(x)dx-a_{n})=\lim_{n\to\infty}\int_0^1\frac{nx^{n}f(x)}{1+x^{n}}dx=\int_0^1(\log(1+x^{n}))^{\prime} xf(x)dx \)
care va fi egala cu o integrare prin parti, cu:
\( f(1)\log 2-\int_0^1\log(1+x^{n})(xf(x))^{\prime} dx \). Acum, ne vom ocupa de sirul \( (b_{n})_{n\geq 1} \) definit prin \( b_{n}=\int_0^1\log(1+x^{n})(xf(x))^{\prime} dx \). Vom proceda ca mai sus (adica cum am facut cu sirul \( a_{n} \)), i.e.
\( \left | \int_0^1\log(1+x^{n})(xf(x))^{\prime} dx\right| \leq \sup_{x\in [0,1]} (xf(x))^{\prime} \int_0^1\log(1+x^{n})dx\leq\frac{\sup_{x\in [0,1]} (xf(x))^{\prime}}{n+1} \), unde am folosit inegalitatea elementara \( \log(1+t)\leq t, \forall t>0 \). Acum, trecand la limita atunci cand \( n\to\infty \), o sa rezulte ca
\( \lim_{n\to\infty}n(\int_0^1f(x)dx-a_{n})=f(1)\log 2=\log 2 \). \( \qed \)
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