27024: Difference between revisions

Revision as of 17:35, 9 June 2024

27024 (Gheorghe Szöllösy)

Fie Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_n = \int_{0}^{\pi} \frac{\cos nx}{13-12\cos x}\,dx, n\ge0.} Să se calculeze Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty}(I_0+I_1+I_2+\ldots+I_n).}

Soluție. Să observăm că

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_{n+2}+I_n = \int_{0}^{\pi}\frac{2\cos x-\cos(n_1)x}{13-12\cos x}\,dx = \frac{1}{6}\int_{0}^{\pi}\frac{(12\cos x-13+13)\cos(n+1)x}{13-12\cos x}\,dx=}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{1}{6(n+1)}\sin(n+1)x\Biggr|_{0}^{\pi}+\frac{13}{6}I_{n+1},}

oricare ar fiFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n\in \mathbb{N}.} Atunci Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_n=\alpha\left(\frac{2}{3}\right)^n+\beta\left(\frac{3}{2}\right)^n} , unde Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha+\beta=I_0=\frac{\pi}{5}} și Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{2}{3}\alpha+\frac{3}{2}\beta=I_1=\frac{2\pi}{15}.}

Obținem Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha=\frac{\pi}{5}, \beta=0} și Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_n=\frac{\pi}{5}\left(\frac{2}{3}\right)^n} pentru orice Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n>0} .

În consecință, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty}(I_0+I_1+I_2+\ldots+I_n)= \frac{\frac{\pi}{5}}{1-\frac{2}{3}}=\frac{3\pi}{5}.}

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 <math display="block">=\frac{1}{6(n+1)}\sin(n+1)x\Biggr|_{0}^{\pi}+\frac{13}{6}I_{n+1},</math>
-oricare ar fi<math>n\in \mathbb{N}.</math> Atunci<math>I_n=\alpha\left(\frac{2}{3}\right)^n+\beta\left(\frac{3}{2}\right)^n</math>, unde <math>\alpha+\beta=I_0=\frac{\pi}{5}</math>și <math> \frac{2}{3}\alpha+\frac{3}{2}\beta=I_1=\frac{2\pi}{15}.</math>
+oricare ar fi<math>n\in \mathbb{N}.</math> Atunci <math>I_n=\alpha\left(\frac{2}{3}\right)^n+\beta\left(\frac{3}{2}\right)^n</math>, unde <math>\alpha+\beta=I_0=\frac{\pi}{5}</math>și <math> \frac{2}{3}\alpha+\frac{3}{2}\beta=I_1=\frac{2\pi}{15}.</math>
-Obținem <math>\alpha=\frac{\pi}{5}, \beta=0</math>și<math>I_n=\frac{\pi}{5}\left(\frac{2}{3}\right)^n</math>.
+Obținem <math>\alpha=\frac{\pi}{5}, \beta=0</math> și <math>I_n=\frac{\pi}{5}\left(\frac{2}{3}\right)^n</math> pentru orice <math>n>0</math>.
 În consecință, <math display="block">\lim_{n \to \infty}(I_0+I_1+I_2+\ldots+I_n)= \frac{\frac{\pi}{5}}{1-\frac{2}{3}}=\frac{3\pi}{5}.</math>