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Question Number 199666 by sonukgindia last updated on 07/Nov/23

	Answered by witcher3 last updated on 07/Nov/23

	$I = \int_{0}^{\infty} \frac{x^{a}}{1 + x^{b}} dx x \to 0 \frac{x^{a}}{1 + x^{b}} \sim x^{a} cv if only if ? a > - 1 . . true$ $a \in N a ⩾ 0$ $x \to \infty \sim x^{a - b} integrabe if b - a > 1 \Leftrightarrow b - a ⩾ 2$ $b - a \in] 1, \infty [\cap N \Rightarrow b - a ⩾ 2 \Rightarrow b ⩾ 2 + a ⩾ 2$ $x^{b} = t \Rightarrow t^{\frac{1}{b}} \Rightarrow dx = \frac{t^{\frac{1}{b} - 1}}{b}$ $I = \frac{1}{b} \int_{0}^{\infty} \frac{t^{\frac{a + 1}{b} - 1}}{(1 + t)} dt$ $β (a, b) = \int_{0}^{\infty} \frac{t^{a - 1}}{{(1 + t)}^{b + a}} dt$ $I = β (\frac{a + 1}{b}; 1 - \frac{1 + a}{b})$ $\int_{0}^{\infty} \frac{x^{a}}{1 + x^{b}} dx = \frac{1}{b} \frac{π}{\sin (π (\frac{a + 1}{b})}$ ${(4 + \sqrt{15})}^{n} + {(4 - \sqrt{15})}^{n} = U_{n}$ $u_{n} = \underset{k = 0}{\sum^{n}} {(\sqrt{15})}^{k} 4^{n - k} + {(- 1)}^{k} {(\sqrt{15})}^{k} 4^{n - k}$ $= 2 \underset{k = 0}{\sum^{[\frac{n}{2}]}} . (\begin{matrix} n \\ k \end{matrix}) 15^{k} 4^{n - 2 k}; 201 = 2.100 + 1 for n = 2 m + 1$ $U_{n} = 2 \underset{k = 0}{\sum^{m}} (\begin{matrix} 2 m + 1 \\ k \end{matrix}) 15^{k} 4^{2 m - 2 k + 1}$ $U_{n} \equiv 2 . 4^{2 m + 1} [15]$ $15^{k} \equiv 1 [7], \forall k \in N$ $U_{n} \equiv 2 \underset{k = 0}{\sum^{m}} (\begin{matrix} 2 m + 1 \\ 2 k \end{matrix}) 4^{2 m - 2 k + 1} [7]$ $U_{n} = \underset{k = 0}{\sum^{2 m + 1}} (\begin{matrix} 2 m + 1 \\ k \end{matrix}) (4^{2 m + 1 - k} {(1)}^{k} + {(- 1)}^{k} 4^{2 m + 1 - 2 k})$ $Missing \left or extra \right$ $= 5 . 5^{2 m} + 3^{2 m + 1} [7]$ $= 5 . {(25)}^{m} + 3 . {(9)}^{m} \equiv 5 . 4^{m} . 3 . 2^{m} [7]$ $easy from here$
	Answered by deleteduser1 last updated on 07/Nov/23

	$225 \equiv 15 (m o d 105) \Rightarrow \sqrt{15} \equiv \sqrt{225} \equiv 15 (m o d 105)$ $\Rightarrow {(4 + \sqrt{15})}^{201} + {(4 - \sqrt{15})}^{201} \overset{105}{\equiv} 19^{201} - 11^{201}$ $ϕ (105) = 48 \Rightarrow {(19^{48})}^{4} {(19)}^{9} - {(11^{48})}^{4} {(11)}^{9} \overset{105}{\equiv} 19^{9} - 11^{9}$ $\equiv {(46)}^{4} (19) - {(16)}^{4} 11 \equiv {(16)}^{2} (19) - {(46)}^{2} (11)$ $\equiv (46) (19) - (16) (11) \overset{105}{\equiv} 68$
	Answered by deleteduser1 last updated on 07/Nov/23

	$x = {(4 + \sqrt{15})}^{201} + {(4 - \sqrt{15})}^{201} \equiv {(1)}^{201} + {(1)}^{201} \overset{3}{\equiv} 2$ $x \equiv {(- 1)}^{201} + {(- 1)}^{201} \equiv 3 (m o d 5)$ $x \equiv {(4 + 1)}^{201} + {(4 - 1)}^{201} \overset{7}{\equiv} (5^{198}) (5^{3}) + (3^{198}) (3^{3})$ $\equiv 6 - 1 = 5 (m o d 7)$ $x = 3 k + 2 \equiv 3 (m o d 5) \Rightarrow - 2 (k - 1) \overset{5}{\equiv} 8 \Rightarrow k \overset{5}{\equiv} 2$ $\Rightarrow x = 3 (5 q + 2) + 2 = 15 q + 8 \overset{7}{\equiv} 5 \Rightarrow q \equiv 4 (m o d 7)$ $\Rightarrow x = 15 (7 l + 4) + 8 = 105 l + 68 \Rightarrow x \equiv 68 (m o d 105)$
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