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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=∫_0 ^∞ (x^a /(1+x^b ))dx x→0 (x^a /(1+x^b ))∼x^a cv if only if?a>−1..true a∈N a≥0 x→∞ ∼x^(a−b) integrabe if b−a>1⇔b−a≥2 b−a∈]1,∞[∩N⇒b−a≥2⇒b≥2+a≥2 x^b =t⇒t^(1/b) ⇒dx=(t^((1/b)−1) /b) I=(1/b)∫_0 ^∞ (t^(((a+1)/b)−1) /((1+t)))dt β(a,b)=∫_0 ^∞ (t^(a−1) /((1+t)^(b+a) ))dt I=β(((a+1)/b);1−((1+a)/b)) ∫_0 ^∞ (x^a /(1+x^b ))dx=(1/b)(π/(sin(π(((a+1)/b)))) (4+(√(15)))^n +(4−(√(15)))^n =U_n u_n =Σ_(k=0) ^n ((√(15)))^k 4^(n−k) +(−1)^k ((√(15)))^k 4^(n−k) =2Σ_(k=0) ^([(n/2)]) . ((n),(k) )15^k 4^(n−2k) ;201=2.100+1 for n=2m+1 U_n =2Σ_(k=0) ^m (((2m+1)),(k) )15^k 4^(2m−2k+1) U_n ≡2.4^(2m+1) [15] 15^k ≡1[7],∀k∈N U_n ≡2Σ_(k=0) ^m (((2m+1)),((2k)) )4^(2m−2k+1) [7] U_n =Σ_(k=0) ^(2m+1) (((2m+1)),(k) ) (4^(2m+1−k) (1)^k +(−1)^k 4^(2m+1−2k) ) =(4+1)^(2m+1) +(4−1)^(2m+1) =5.5^(2m) +3^(2m+1) [7] =5.(25)^m +3.(9)^m ≡5.4^m .3.2^m [7] easy from here
$$\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{a}} }{\mathrm{1}+\mathrm{x}^{\mathrm{b}} }\mathrm{dx}\:\mathrm{x}\rightarrow\mathrm{0}\:\frac{\mathrm{x}^{\mathrm{a}} }{\mathrm{1}+\mathrm{x}^{\mathrm{b}} }\sim\mathrm{x}^{\mathrm{a}} \:\mathrm{cv}\:\mathrm{if}\:\mathrm{only}\:\mathrm{if}?\mathrm{a}>−\mathrm{1}..\mathrm{true} \\ $$$$\mathrm{a}\in\mathbb{N}\:\mathrm{a}\geqslant\mathrm{0} \\ $$$$\mathrm{x}\rightarrow\infty\:\sim\mathrm{x}^{\mathrm{a}−\mathrm{b}} \:\mathrm{integrabe}\:\mathrm{if}\:\mathrm{b}−\mathrm{a}>\mathrm{1}\Leftrightarrow\mathrm{b}−\mathrm{a}\geqslant\mathrm{2} \\ $$$$\left.\mathrm{b}−\mathrm{a}\in\right]\mathrm{1},\infty\left[\cap\mathbb{N}\Rightarrow\mathrm{b}−\mathrm{a}\geqslant\mathrm{2}\Rightarrow\mathrm{b}\geqslant\mathrm{2}+\mathrm{a}\geqslant\mathrm{2}\right. \\ $$$$\mathrm{x}^{\mathrm{b}} =\mathrm{t}\Rightarrow\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{b}}} \Rightarrow\mathrm{dx}=\frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{b}}−\mathrm{1}} }{\mathrm{b}} \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{b}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\frac{\mathrm{a}+\mathrm{1}}{\mathrm{b}}−\mathrm{1}} }{\left(\mathrm{1}+\mathrm{t}\right)}\mathrm{dt} \\ $$$$\beta\left(\mathrm{a},\mathrm{b}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} }{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{b}+\mathrm{a}} }\mathrm{dt} \\ $$$$\mathrm{I}=\beta\left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{b}};\mathrm{1}−\frac{\mathrm{1}+\mathrm{a}}{\mathrm{b}}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{a}} }{\mathrm{1}+\mathrm{x}^{\mathrm{b}} }\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{b}}\frac{\pi}{\mathrm{sin}\left(\pi\left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{b}}\right)\right.} \\ $$$$\left(\mathrm{4}+\sqrt{\mathrm{15}}\right)^{\mathrm{n}} +\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)^{\mathrm{n}} =\mathrm{U}_{\mathrm{n}} \\ $$$$\mathrm{u}_{\mathrm{n}} =\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\left(\sqrt{\mathrm{15}}\right)^{\mathrm{k}} \mathrm{4}^{\mathrm{n}−\mathrm{k}} +\left(−\mathrm{1}\right)^{\mathrm{k}} \left(\sqrt{\mathrm{15}}\right)^{\mathrm{k}} \mathrm{4}^{\mathrm{n}−\mathrm{k}} \\ $$$$=\mathrm{2}\underset{\mathrm{k}=\mathrm{0}} {\overset{\left[\frac{\mathrm{n}}{\mathrm{2}}\right]} {\sum}}.\begin{pmatrix}{\mathrm{n}}\\{\mathrm{k}}\end{pmatrix}\mathrm{15}^{\mathrm{k}} \mathrm{4}^{\mathrm{n}−\mathrm{2k}} ;\mathrm{201}=\mathrm{2}.\mathrm{100}+\mathrm{1}\:\mathrm{for}\:\mathrm{n}=\mathrm{2m}+\mathrm{1} \\ $$$$\mathrm{U}_{\mathrm{n}} =\mathrm{2}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{m}} {\sum}}\begin{pmatrix}{\mathrm{2m}+\mathrm{1}}\\{\mathrm{k}}\end{pmatrix}\mathrm{15}^{\mathrm{k}} \mathrm{4}^{\mathrm{2m}−\mathrm{2k}+\mathrm{1}} \\ $$$$\mathrm{U}_{\mathrm{n}} \equiv\mathrm{2}.\mathrm{4}^{\mathrm{2m}+\mathrm{1}} \left[\mathrm{15}\right] \\ $$$$\mathrm{15}^{\mathrm{k}} \equiv\mathrm{1}\left[\mathrm{7}\right],\forall\mathrm{k}\in\mathbb{N} \\ $$$$\mathrm{U}_{\mathrm{n}} \equiv\mathrm{2}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{m}} {\sum}}\begin{pmatrix}{\mathrm{2m}+\mathrm{1}}\\{\mathrm{2k}}\end{pmatrix}\mathrm{4}^{\mathrm{2m}−\mathrm{2k}+\mathrm{1}} \left[\mathrm{7}\right] \\ $$$$\mathrm{U}_{\mathrm{n}} =\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{2m}+\mathrm{1}} {\sum}}\begin{pmatrix}{\mathrm{2m}+\mathrm{1}}\\{\mathrm{k}}\end{pmatrix}\:\left(\mathrm{4}^{\mathrm{2m}+\mathrm{1}−\mathrm{k}} \left(\mathrm{1}\right)^{\mathrm{k}} +\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{4}^{\mathrm{2m}+\mathrm{1}−\mathrm{2k}} \right) \\ $$$$=\left(\mathrm{4}+\mathrm{1}\overset{\mathrm{2m}+\mathrm{1}} {\right)}+\left(\mathrm{4}−\mathrm{1}\overset{\mathrm{2m}+\mathrm{1}} {\right)} \\ $$$$=\mathrm{5}.\mathrm{5}^{\mathrm{2m}} +\mathrm{3}^{\mathrm{2m}+\mathrm{1}} \left[\mathrm{7}\right] \\ $$$$=\mathrm{5}.\left(\mathrm{25}\right)^{\mathrm{m}} +\mathrm{3}.\left(\mathrm{9}\right)^{\mathrm{m}} \equiv\mathrm{5}.\mathrm{4}^{\mathrm{m}} .\mathrm{3}.\mathrm{2}^{\mathrm{m}} \left[\mathrm{7}\right] \\ $$$$\mathrm{easy}\:\mathrm{from}\:\mathrm{here} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by AST last updated on 07/Nov/23
225≡15(mod 105)⇒(√(15))≡(√(225))≡15(mod 105) ⇒(4+(√(15)))^(201) +(4−(√(15)))^(201) ≡^(105) 19^(201) −11^(201) φ(105)=48⇒(19^(48) )^4 (19)^9 −(11^(48) )^4 (11)^9 ≡^(105) 19^9 −11^9 ≡(46)^4 (19)−(16)^4 11≡(16)^2 (19)−(46)^2 (11) ≡(46)(19)−(16)(11)≡^(105) 68
$$\mathrm{225}\equiv\mathrm{15}\left({mod}\:\mathrm{105}\right)\Rightarrow\sqrt{\mathrm{15}}\equiv\sqrt{\mathrm{225}}\equiv\mathrm{15}\left({mod}\:\mathrm{105}\right) \\ $$$$\Rightarrow\left(\mathrm{4}+\sqrt{\mathrm{15}}\right)^{\mathrm{201}} +\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)^{\mathrm{201}} \overset{\mathrm{105}} {\equiv}\mathrm{19}^{\mathrm{201}} −\mathrm{11}^{\mathrm{201}} \\ $$$$\phi\left(\mathrm{105}\right)=\mathrm{48}\Rightarrow\left(\mathrm{19}^{\mathrm{48}} \right)^{\mathrm{4}} \left(\mathrm{19}\right)^{\mathrm{9}} −\left(\mathrm{11}^{\mathrm{48}} \right)^{\mathrm{4}} \left(\mathrm{11}\right)^{\mathrm{9}} \overset{\mathrm{105}} {\equiv}\mathrm{19}^{\mathrm{9}} −\mathrm{11}^{\mathrm{9}} \\ $$$$\equiv\left(\mathrm{46}\right)^{\mathrm{4}} \left(\mathrm{19}\right)−\left(\mathrm{16}\right)^{\mathrm{4}} \mathrm{11}\equiv\left(\mathrm{16}\right)^{\mathrm{2}} \left(\mathrm{19}\right)−\left(\mathrm{46}\right)^{\mathrm{2}} \left(\mathrm{11}\right) \\ $$$$\equiv\left(\mathrm{46}\right)\left(\mathrm{19}\right)−\left(\mathrm{16}\right)\left(\mathrm{11}\right)\overset{\mathrm{105}} {\equiv}\mathrm{68} \\ $$
Answered by AST last updated on 07/Nov/23
x=(4+(√(15)))^(201) +(4−(√(15)))^(201) ≡(1)^(201) +(1)^(201) ≡^3 2 x≡(−1)^(201) +(−1)^(201) ≡3(mod 5) x≡(4+1)^(201) +(4−1)^(201) ≡^7 (5^(198) )(5^3 )+(3^(198) )(3^3 ) ≡6−1=5(mod 7) x=3k+2≡3(mod 5)⇒−2(k−1)≡^5 8⇒k≡^5 2 ⇒x=3(5q+2)+2=15q+8≡^7 5⇒q≡4(mod 7) ⇒x=15(7l+4)+8=105l+68⇒x≡68(mod 105)
$${x}=\left(\mathrm{4}+\sqrt{\mathrm{15}}\right)^{\mathrm{201}} +\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)^{\mathrm{201}} \equiv\left(\mathrm{1}\right)^{\mathrm{201}} +\left(\mathrm{1}\right)^{\mathrm{201}} \overset{\mathrm{3}} {\equiv}\mathrm{2} \\ $$$${x}\equiv\left(−\mathrm{1}\right)^{\mathrm{201}} +\left(−\mathrm{1}\right)^{\mathrm{201}} \equiv\mathrm{3}\left({mod}\:\mathrm{5}\right) \\ $$$${x}\equiv\left(\mathrm{4}+\mathrm{1}\right)^{\mathrm{201}} +\left(\mathrm{4}−\mathrm{1}\right)^{\mathrm{201}} \overset{\mathrm{7}} {\equiv}\left(\mathrm{5}^{\mathrm{198}} \right)\left(\mathrm{5}^{\mathrm{3}} \right)+\left(\mathrm{3}^{\mathrm{198}} \right)\left(\mathrm{3}^{\mathrm{3}} \right) \\ $$$$\equiv\mathrm{6}−\mathrm{1}=\mathrm{5}\left({mod}\:\mathrm{7}\right) \\ $$$${x}=\mathrm{3}{k}+\mathrm{2}\equiv\mathrm{3}\left({mod}\:\mathrm{5}\right)\Rightarrow−\mathrm{2}\left({k}−\mathrm{1}\right)\overset{\mathrm{5}} {\equiv}\mathrm{8}\Rightarrow{k}\overset{\mathrm{5}} {\equiv}\mathrm{2} \\ $$$$\Rightarrow{x}=\mathrm{3}\left(\mathrm{5}{q}+\mathrm{2}\right)+\mathrm{2}=\mathrm{15}{q}+\mathrm{8}\overset{\mathrm{7}} {\equiv}\mathrm{5}\Rightarrow{q}\equiv\mathrm{4}\left({mod}\:\mathrm{7}\right) \\ $$$$\Rightarrow{x}=\mathrm{15}\left(\mathrm{7}{l}+\mathrm{4}\right)+\mathrm{8}=\mathrm{105}{l}+\mathrm{68}\Rightarrow{x}\equiv\mathrm{68}\left({mod}\:\mathrm{105}\right) \\ $$

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