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Question Number 219425 Answers: 2 Comments: 1

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Question Number 219414 Answers: 0 Comments: 0

given the recursive {a_n } define by setting a_(1 ) ∈ (0,1) , a_(n+1) = a_n (1−a_n ) , n≥1 prove that (1) lim_(n→∞) na_n = 1 (2) b_n = n(1−na_n ) is a incresing sequence and diverge to ∞ (3) lim_(n→∞) ((n(1−na_n ))/(ln(n))) = 1

$$\:\:\:\mathrm{given}\:\mathrm{the}\:\mathrm{recursive}\:\left\{\mathrm{a}_{\mathrm{n}} \right\}\:\mathrm{define}\:\mathrm{by}\:\mathrm{setting} \\ $$$$\:\:\mathrm{a}_{\mathrm{1}\:} \:\in\:\left(\mathrm{0},\mathrm{1}\right)\:\:\:,\:\:\:\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} \:=\:\mathrm{a}_{\mathrm{n}} \left(\mathrm{1}−\mathrm{a}_{\mathrm{n}} \right)\:\:\:,\:\mathrm{n}\geqslant\mathrm{1} \\ $$$$\:\:\mathrm{prove}\:\mathrm{that}\:\:\left(\mathrm{1}\right)\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{na}_{\mathrm{n}} =\:\mathrm{1} \\ $$$$\:\:\left(\mathrm{2}\right)\:\:\mathrm{b}_{\mathrm{n}} \:=\:\mathrm{n}\left(\mathrm{1}−\mathrm{na}_{\mathrm{n}} \right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{incresing}\:\mathrm{sequence} \\ $$$$\:\:\:\mathrm{and}\:\mathrm{diverge}\:\mathrm{to}\:\infty \\ $$$$\:\:\:\left(\mathrm{3}\right)\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{n}\left(\mathrm{1}−\mathrm{na}_{\mathrm{n}} \right)}{\mathrm{ln}\left(\mathrm{n}\right)}\:=\:\mathrm{1} \\ $$

Question Number 219408 Answers: 2 Comments: 0

a_1 =(1/2). a_n =(2/(n+3))+(1−(1/(n+3)))^2 a_(n−1) . Find the maximum of a_n .

$${a}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}.\:{a}_{{n}} =\frac{\mathrm{2}}{{n}+\mathrm{3}}+\left(\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{3}}\right)^{\mathrm{2}} {a}_{{n}−\mathrm{1}} . \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{of}\:{a}_{{n}} . \\ $$

Question Number 219404 Answers: 2 Comments: 2

given g(x)= ((x−2023)/(x−1)) find (gogogogogog)(2024)

$$\:\:\mathrm{given}\:\mathrm{g}\left(\mathrm{x}\right)=\:\frac{\mathrm{x}−\mathrm{2023}}{\mathrm{x}−\mathrm{1}} \\ $$$$\:\:\mathrm{find}\:\left(\mathrm{gogogogogog}\right)\left(\mathrm{2024}\right) \\ $$

Question Number 219403 Answers: 1 Comments: 0

Question Number 219388 Answers: 1 Comments: 0

∫_0 ^( ∞) ((sin^2 (u))/u^2 ) du=I I(t)=∫_0 ^( ∞) ((sin^2 (u))/u^2 )e^(−ut) du=∫_0 ^( ∞) (1/u)∙((sin^2 (u))/u)e^(−ut) du= ∫_( t) ^( ∞) L_u {((sin^2 (p))/p)} du=∫_( t) ^( ∞) ∫_0 ^( ∞) ((sin^2 (p))/p)e^(−up) dpdu ∫_( t) ^( ∞) ∫_( w) ^( ∞) L_s {sin^2 (r)}ds dw ∫_( t) ^( ∞) ∫_( w) ^( ∞) ∫_0 ^( ∞) sin^2 (r)e^(−sr) dr ds dw ∫_( t) ^( ∞) ∫_( w) ^( ∞) (2/(s(s^2 +4))) ds dw=∫_( t) ^( ∞) [(1/2)ln(s)−(1/4)ln(s^2 +4)]_(s=w) ^(s=∞) dw ∫_( t) ^( ∞) (1/4)ln((w^2 /(w^2 +4)))dw=[(1/4)w∙ln(1+((2/w))^2 )+tan^(−1) ((1/2)w)]_(w=t) ^(w=∞) (π/2)−(1/4)t∙ln(1+((2/t))^2 )−tan^(−1) ((1/2)t) lim_(t→0) {tan^(−1) ((2/t))−(1/4)t∙ln(1+((2/t))^2 )}=(π/2)

$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}^{\mathrm{2}} \left({u}\right)}{{u}^{\mathrm{2}} }\:\mathrm{d}{u}={I} \\ $$$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}^{\mathrm{2}} \left({u}\right)}{{u}^{\mathrm{2}} }{e}^{−{ut}} \mathrm{d}{u}=\int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{\mathrm{1}}{{u}}\centerdot\frac{\mathrm{sin}^{\mathrm{2}} \left({u}\right)}{{u}}{e}^{−{ut}} \mathrm{d}{u}= \\ $$$$\int_{\:{t}} ^{\:\infty} \:\:\mathcal{L}_{{u}} \left\{\frac{\mathrm{sin}^{\mathrm{2}} \left({p}\right)}{{p}}\right\}\:\mathrm{d}{u}=\int_{\:{t}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{\mathrm{sin}^{\mathrm{2}} \left({p}\right)}{{p}}{e}^{−{up}} \mathrm{d}{p}\mathrm{d}{u} \\ $$$$\int_{\:{t}} ^{\:\infty} \:\int_{\:{w}} ^{\:\infty} \:\mathcal{L}_{{s}} \left\{\mathrm{sin}^{\mathrm{2}} \left({r}\right)\right\}\mathrm{d}{s}\:\mathrm{d}{w} \\ $$$$\int_{\:{t}} ^{\:\infty} \int_{\:{w}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \:\mathrm{sin}^{\mathrm{2}} \left({r}\right){e}^{−{sr}} \mathrm{d}{r}\:\mathrm{d}{s}\:\mathrm{d}{w} \\ $$$$\int_{\:{t}} ^{\:\infty} \int_{\:{w}} ^{\:\infty} \:\:\frac{\mathrm{2}}{{s}\left({s}^{\mathrm{2}} +\mathrm{4}\right)}\:\mathrm{d}{s}\:\mathrm{d}{w}=\int_{\:{t}} ^{\:\infty} \:\:\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left({s}\right)−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left({s}^{\mathrm{2}} +\mathrm{4}\right)\right]_{{s}={w}} ^{{s}=\infty} \mathrm{d}{w} \\ $$$$\int_{\:{t}} ^{\:\infty} \:\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\frac{{w}^{\mathrm{2}} }{{w}^{\mathrm{2}} +\mathrm{4}}\right)\mathrm{d}{w}=\left[\frac{\mathrm{1}}{\mathrm{4}}{w}\centerdot\mathrm{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{{w}}\right)^{\mathrm{2}} \right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}{w}\right)\right]_{{w}={t}} ^{{w}=\infty} \\ $$$$\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}{t}\centerdot\mathrm{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{{t}}\right)^{\mathrm{2}} \right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}{t}\right) \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{{t}}\right)−\frac{\mathrm{1}}{\mathrm{4}}{t}\centerdot\mathrm{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{{t}}\right)^{\mathrm{2}} \right)\right\}=\frac{\pi}{\mathrm{2}} \\ $$

Question Number 219384 Answers: 1 Comments: 0