27024 - Revision history

Andrei.Horvat at 17:35, 9 June 2024

2024-06-09T17:35:45Z

← Older revision		Revision as of 17:35, 9 June 2024
Line 9:		Line 9:
	<math display="block">=\frac{1}{6(n+1)}\sin(n+1)x\Biggr\|_{0}^{\pi}+\frac{13}{6}I_{n+1},</math>		<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> pentru orice <math>n>0</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>		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>

Andrei.Horvat at 17:35, 9 June 2024

2024-06-09T17:35:10Z

← Older revision		Revision as of 17:35, 9 June 2024
Line 9:		Line 9:
	<math display="block">=\frac{1}{6(n+1)}\sin(n+1)x\Biggr\|_{0}^{\pi}+\frac{13}{6}I_{n+1},</math>		<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>		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>

Andrei.Horvat at 08:28, 16 January 2024

2024-01-16T08:28:35Z

← Older revision		Revision as of 08:28, 16 January 2024
Line 3:		Line 3:
	''Fie '' <math> I_n = \int_{0}^{\pi} \frac{\cos nx}{13-12\cos x}\,dx, n\ge0.</math>'' Să se calculeze '' <math>\lim_{n \to \infty}(I_0+I_1+I_2+\ldots+I_n).</math>		''Fie '' <math> I_n = \int_{0}^{\pi} \frac{\cos nx}{13-12\cos x}\,dx, n\ge0.</math>'' Să se calculeze '' <math>\lim_{n \to \infty}(I_0+I_1+I_2+\ldots+I_n).</math>

	'''Soluție. ''' ''Să observăm că''		'''Soluție. ''' Să observăm că

	<math display="block">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=</math>		<math display="block">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=</math>
Line 9:		Line 9:
	<math display="block">=\frac{1}{6(n+1)}\sin(n+1)x\Biggr\|_{0}^{\pi}+\frac{13}{6}I_{n+1},</math>		<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>.

	În consecință, <math>\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>.		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>

Robert Manc at 15:52, 5 December 2023

2023-12-05T15:52:47Z

← Older revision		Revision as of 15:52, 5 December 2023
Line 5:		Line 5:
	'''Soluție. ''' ''Să observăm că''		'''Soluție. ''' ''Să observăm că''

	<math>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=		<math display="block">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=</math>
	=\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>		<math display="block">=\frac{1}{6(n+1)}\sin(n+1)x\Biggr\|_{0}^{\pi}+\frac{13}{6}I_{n+1},</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>.
			'' 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>.

	În consecință, <math>\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>.		În consecință, <math>\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>.

Robert Manc: Pagină nouă: '''27024 (Gheorghe Szöllösy)''' ''Fie '' $I_n = \int_{0}^{\pi} \frac{\cos nx}{13-12\cos x}\,dx, n\ge0.$ '' Să se calculeze '' $\lim_{n \to \infty}(I_0+I_1+I_2+\ldots+I_n).$ '''Soluție. ''' ''Să observăm că'' $I_{n+2}+I_n = \int_{0}^{\pi}\frac{2\cos x-\cos(n_1)x}{13-12\cos x}\,dx = \frac{1}{6}\int_{0}^{\pi}\frac{(12\cos x-13+13)\cos(n+1)x}{13-12\cos x}\,dx= =\frac{1}{6(n+1)}\sin(n+1)x\Biggr|_{0}^{\pi}+\frac{13}{6}I_{n+1},$ '' oricare...

2023-12-05T15:28:39Z

Pagină nouă: '''27024 (Gheorghe Szöllösy)''' ''Fie '' <math> I_n = \int_{0}^{\pi} \frac{\cos nx}{13-12\cos x}\,dx, n\ge0.</math>'' Să se calculeze '' <math>\lim_{n \to \infty}(I_0+I_1+I_2+\ldots+I_n).</math> '''Soluție. ''' ''Să observăm că'' <math>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= =\frac{1}{6(n+1)}\sin(n+1)x\Biggr|_{0}^{\pi}+\frac{13}{6}I_{n+1}, </math>'' oricare...

New page

'''27024 (Gheorghe Szöllösy)'''

''Fie '' <math> I_n = \int_{0}^{\pi} \frac{\cos nx}{13-12\cos x}\,dx, n\ge0.</math>'' Să se calculeze '' <math>\lim_{n \to \infty}(I_0+I_1+I_2+\ldots+I_n).</math>

'''Soluție. ''' ''Să observăm că''

<math>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=
=\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>''Obținem ''<math>\alpha=\frac{\pi}{5}, \beta=0</math>'' și ''<math>I_n=\frac{\pi}{5}\left(\frac{2}{3}\right)^n</math>.

În consecință, <math>\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>.