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Question Number 58309 by Smail last updated on 21/Apr/19
Prove that lim_(x→0) ((1−cos(x)cos(x/2)cos(x/3)...)/x^2 )=(π^2 /(12))
$${Prove}\:{that}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}−{cos}\left({x}\right){cos}\left({x}/\mathrm{2}\right){cos}\left({x}/\mathrm{3}\right)…}{{x}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$
Commented by maxmathsup by imad last updated on 21/Apr/19
let use hospital theorem first we have by chang.x=6t lim_(x→0) ((1−cos(x)cos((x/2))cos((x/3)))/x^2 ) =lim_(t→0) ((1−cos(6t) cos(3t)cos(2t))/(36 t^2 )) but we have cos(6t)cos(3t) =(1/2) {cos(9t)+cos(3t)} ⇒ cos(6t)cos(3t)cos(2t) =(1/2){cos(9t)cos(2t) +cos(3t)cos(2t)} =(1/4){cos(11t)+cos(7t) +cos(5t) +cos(t)} let f(t) =1−(1/4){cos(t) +cos(5t) +cos(7t) +cos(11t)} and g(t)=36x^2 ⇒ f^′ (t) =(1/4){sin(t)+5sin(5t)+7sin(7t)+11sin(11t)} ⇒ f^(′′) (t) =(1/4){cost +25 cos(t) +49 cos(7t) +121 cos(11t)} ⇒ lim_(t→0) f^(′′) (t) =(1/4){1+25 +49 +121} =(1/4){26 +49 +121} =(1/4){121 +75} =(1/4)(196) =((98)/2) =49 also we have g^′ (t) =72 x and g^(′′) (x) =72 ⇒ lim_(x→0) ((f^(′′) (t))/(g^(′′) (t))) =((49)/(72)) so there is a error in the question ...
$${let}\:{use}\:{hospital}\:{theorem}\:\:{first}\:{we}\:{have}\:\:{by}\:{chang}.{x}=\mathrm{6}{t}\: \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}−{cos}\left({x}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{3}}\right)}{{x}^{\mathrm{2}} }\:={lim}_{{t}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}−{cos}\left(\mathrm{6}{t}\right)\:{cos}\left(\mathrm{3}{t}\right){cos}\left(\mathrm{2}{t}\right)}{\mathrm{36}\:{t}^{\mathrm{2}} } \\ $$$${but}\:{we}\:{have}\:{cos}\left(\mathrm{6}{t}\right){cos}\left(\mathrm{3}{t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:\left\{{cos}\left(\mathrm{9}{t}\right)+{cos}\left(\mathrm{3}{t}\right)\right\}\:\Rightarrow \\ $$$${cos}\left(\mathrm{6}{t}\right){cos}\left(\mathrm{3}{t}\right){cos}\left(\mathrm{2}{t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\mathrm{9}{t}\right){cos}\left(\mathrm{2}{t}\right)\:+{cos}\left(\mathrm{3}{t}\right){cos}\left(\mathrm{2}{t}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left\{{cos}\left(\mathrm{11}{t}\right)+{cos}\left(\mathrm{7}{t}\right)\:+{cos}\left(\mathrm{5}{t}\right)\:+{cos}\left({t}\right)\right\}\:{let}\: \\ $$$${f}\left({t}\right)\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\left\{{cos}\left({t}\right)\:+{cos}\left(\mathrm{5}{t}\right)\:+{cos}\left(\mathrm{7}{t}\right)\:+{cos}\left(\mathrm{11}{t}\right)\right\}\:{and}\:{g}\left({t}\right)=\mathrm{36}{x}^{\mathrm{2}} \:\Rightarrow \\ $$$${f}^{'} \left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}\left\{{sin}\left({t}\right)+\mathrm{5}{sin}\left(\mathrm{5}{t}\right)+\mathrm{7}{sin}\left(\mathrm{7}{t}\right)+\mathrm{11}{sin}\left(\mathrm{11}{t}\right)\right\}\:\Rightarrow \\ $$$${f}^{''} \left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}\left\{{cost}\:+\mathrm{25}\:{cos}\left({t}\right)\:+\mathrm{49}\:{cos}\left(\mathrm{7}{t}\right)\:+\mathrm{121}\:{cos}\left(\mathrm{11}{t}\right)\right\}\:\Rightarrow \\ $$$${lim}_{{t}\rightarrow\mathrm{0}} {f}^{''} \left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}\left\{\mathrm{1}+\mathrm{25}\:+\mathrm{49}\:+\mathrm{121}\right\}\:=\frac{\mathrm{1}}{\mathrm{4}}\left\{\mathrm{26}\:+\mathrm{49}\:+\mathrm{121}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left\{\mathrm{121}\:+\mathrm{75}\right\}\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{196}\right)\:=\frac{\mathrm{98}}{\mathrm{2}}\:=\mathrm{49}\:\:\:{also}\:{we}\:{have}\:{g}^{'} \left({t}\right)\:=\mathrm{72}\:{x}\:{and} \\ $$$${g}^{''} \left({x}\right)\:=\mathrm{72}\:\Rightarrow\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{f}^{''} \left({t}\right)}{{g}^{''} \left({t}\right)}\:=\frac{\mathrm{49}}{\mathrm{72}}\:\:\:\:\:\:\:{so}\:{there}\:{is}\:{a}\:{error}\:{in}\:{the}\:{question}\:… \\ $$
Commented by mr W last updated on 21/Apr/19
question is correct sir. it′s lim_(x→0) ((1−cos (x) cos (x/2) cos (x/3).....)/x^2 ) not lim_(x→0) ((1−cos (x) cos (x/2) cos (x/3))/x^2 )
$${question}\:{is}\:{correct}\:{sir}. \\ $$$${it}'{s}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\left({x}\right)\:\mathrm{cos}\:\left({x}/\mathrm{2}\right)\:\mathrm{cos}\:\left({x}/\mathrm{3}\right)…..}{{x}^{\mathrm{2}} } \\ $$$${not}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\left({x}\right)\:\mathrm{cos}\:\left({x}/\mathrm{2}\right)\:\mathrm{cos}\:\left({x}/\mathrm{3}\right)}{{x}^{\mathrm{2}} } \\ $$
Commented by Mr X pcx last updated on 21/Apr/19
i hav nt seen the point but nrvermind i have treated a special case ...
$${i}\:{hav}\:{nt}\:{seen}\:{the}\:{point}\:{but}\:{nrvermind} \\ $$$${i}\:{have}\:{treated}\:{a}\:{special}\:{case}\:… \\ $$
Answered by tanmay last updated on 21/Apr/19
cosx=1−(x^2 /(2!))+(x^4 /(4!))−(x^6 /(6!))+... here we take cosx≈1−(x^2 /2) other terms ignored because those terms contain x^r when r>2 f(x)=cosxcos((x/2))cos((x/3))cos((x/4))... f(x)=(1−(x^2 /(2×1^2 )))(1−(x^2 /(2×2^2 )))(1−(x^2 /(2×3^2 )))... f(x)=Π_(r=1) ^∞ (1−(x^2 /(2×r^2 ))) f(x)=1−(x^2 /2)((1/1^2 )+(1/2^2 )+(1/3^2 )+...)+others terms ignored so value of lim_(x→0) ((1−f(x))/x^2 ) =lim_(x→0) (((x^2 /2)((1/1^2 )+(1/2^2 )+(1/3^2 )+...))/x^2 ) =(1/2)((1/1^2 )+(1/2^2 )+(1/3^2 )+...)=(1/2)×(π^2 /6)=(π^2 /(12))
$${cosx}=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}−\frac{{x}^{\mathrm{6}} }{\mathrm{6}!}+… \\ $$$${here}\:{we}\:{take}\:{cosx}\approx\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\:{other}\:{terms}\:{ignored} \\ $$$${because}\:{those}\:{terms}\:{contain}\:{x}^{{r}} \:\:{when}\:{r}>\mathrm{2} \\ $$$$\: \\ $$$${f}\left({x}\right)={cosxcos}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{3}}\right){cos}\left(\frac{{x}}{\mathrm{4}}\right)… \\ $$$${f}\left({x}\right)=\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}×\mathrm{1}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}×\mathrm{3}^{\mathrm{2}} }\right)… \\ $$$${f}\left({x}\right)=\underset{{r}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}×{r}^{\mathrm{2}} }\right) \\ $$$${f}\left({x}\right)=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…\right)+{others}\:{terms}\:{ignored} \\ $$$${so}\:{value}\:{of}\: \\ $$$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{\mathrm{1}−{f}\left({x}\right)}{{x}^{\mathrm{2}} } \\ $$$$={li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\:\frac{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…\right)}{{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…\right)=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$
Commented by Tawa1 last updated on 21/Apr/19
Wow great sir. But how is, (1/1^2 ) + (1/2^2 ) + (1/3^2 ) + ... = (π^2 /6)
$$\mathrm{Wow}\:\mathrm{great}\:\mathrm{sir}. \\ $$$$\:\:\:\:\mathrm{But}\:\mathrm{how}\:\mathrm{is},\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:…\:\:\:\:\:\:=\:\:\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$
Commented by tanmay last updated on 21/Apr/19
Commented by tanmay last updated on 21/Apr/19
pls check srl no19.19
$${pls}\:{check}\:{srl}\:{no}\mathrm{19}.\mathrm{19}\: \\ $$
Commented by Smail last updated on 21/Apr/19
thanks
$${thanks} \\ $$
Commented by tanmay last updated on 21/Apr/19
the question created ripples in thought process later i found the light to reach goal... i have solved by expansion method..
$${the}\:{question}\:{created}\:{ripples}\:{in}\:{thought}\:{process} \\ $$$${later}\:{i}\:{found}\:{the}\:{light}\:{to}\:{reach}\:{goal}… \\ $$$${i}\:{have}\:{solved}\:{by}\:{expansion}\:{method}.. \\ $$
Commented by Tawa1 last updated on 21/Apr/19
How can we prove that (1/1^2 ) + (1/2^2 ) + (1/3^2 ) + (1/4^2 ) + ... ∞ = (π^2 /6)
$$\mathrm{How}\:\mathrm{can}\:\mathrm{we}\:\mathrm{prove}\:\mathrm{that}\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }\:+\:\:…\:\:\infty\:\:=\:\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$
Commented by peter frank last updated on 21/Apr/19
thank you
$${thank}\:{you} \\ $$
Commented by maxmathsup by imad last updated on 22/Apr/19
sir Tawa take a look at the plateform this result is proved by different methods...
$${sir}\:{Tawa}\:{take}\:{a}\:{look}\:\:{at}\:{the}\:{plateform}\:{this}\:{result}\:{is}\:{proved}\:{by}\:{different} \\ $$$${methods}… \\ $$
Commented by Tawa1 last updated on 22/Apr/19
I don′t know anyone sir
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{anyone}\:\mathrm{sir} \\ $$
Commented by Smail last updated on 22/Apr/19
Check a channel called on Youtube called 3Blue 1Brown. They used a brilliant way to proof that identity.
$${Check}\:{a}\:{channel}\:{called}\:{on}\:{Youtube}\:{called}\: \\ $$$$\mathrm{3}{Blue}\:\mathrm{1}{Brown}.\:{They}\:{used}\:{a}\:{brilliant}\:\: \\ $$$${way}\:{to}\:{proof}\:{that}\:{identity}. \\ $$$$ \\ $$
Answered by mr W last updated on 21/Apr/19
let f(x)=cos (x) cos (x/2) cos (x/3) ... lim_(x→0) f(x)=1 ln f(x)=cos (x)+cos (x/2)+cos (x/3) ... ((f′(x))/(f(x)))=−sin (x)−(1/2)sin (x/2)−(1/3)sin (x/3)−.... f′(x)=−f(x)[sin (x)+(1/2)sin (x/2)+(1/3)sin (x/3)+....] lim_(x→0) f′(x)=0 f′′(x)=−f′(x)[sin (x)+(1/2)sin (x/2)+(1/3)sin (x/3)+....]−f(x)[cos x+(1/2^2 )cos (x/2)+(1/3^2 )cos (x/3)+...] =f(x){[sin (x)+(1/2)sin (x/2)+(1/3)sin (x/3)+....]^2 −[cos x+(1/2^2 )cos (x/2)+(1/3^2 )cos (x/3)+...]} lim_(x→0) f′′(x)=1×{[0+0+0+...]^2 −[1+(1/2^2 )+(1/3^2 )+...]}=−(1+(1/2^2 )+(1/3^2 )+...)=−(π^2 /6) lim_(x→0) ((1−cos(x)cos(x/2)cos(x/3)...)/x^2 ) =lim_(x→0) ((1−f(x))/x^2 ) (=(0/0)) =lim_(x→0) ((−f′(x))/(2x)) (=(0/0)) =lim_(x→0) ((−f′′(x))/2) =(1/2)(1+(1/2^2 )+(1/3^2 )+...) =(1/2)×(π^2 /6) =(π^2 /(12))
$${let}\:{f}\left({x}\right)=\mathrm{cos}\:\left({x}\right)\:\mathrm{cos}\:\left({x}/\mathrm{2}\right)\:\mathrm{cos}\:\left({x}/\mathrm{3}\right)\:… \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left({x}\right)=\mathrm{1} \\ $$$$ \\ $$$$\mathrm{ln}\:{f}\left({x}\right)=\mathrm{cos}\:\left({x}\right)+\mathrm{cos}\:\left({x}/\mathrm{2}\right)+\mathrm{cos}\:\left({x}/\mathrm{3}\right)\:… \\ $$$$\frac{{f}'\left({x}\right)}{{f}\left({x}\right)}=−\mathrm{sin}\:\left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\frac{{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}\:\frac{{x}}{\mathrm{3}}−…. \\ $$$${f}'\left({x}\right)=−{f}\left({x}\right)\left[\mathrm{sin}\:\left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}\:\frac{{x}}{\mathrm{3}}+….\right] \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}'\left({x}\right)=\mathrm{0} \\ $$$$ \\ $$$${f}''\left({x}\right)=−{f}'\left({x}\right)\left[\mathrm{sin}\:\left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}\:\frac{{x}}{\mathrm{3}}+….\right]−{f}\left({x}\right)\left[\mathrm{cos}\:{x}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\mathrm{cos}\:\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\mathrm{cos}\:\frac{{x}}{\mathrm{3}}+…\right] \\ $$$$={f}\left({x}\right)\left\{\left[\mathrm{sin}\:\left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}\:\frac{{x}}{\mathrm{3}}+….\right]^{\mathrm{2}} −\left[\mathrm{cos}\:{x}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\mathrm{cos}\:\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\mathrm{cos}\:\frac{{x}}{\mathrm{3}}+…\right]\right\} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}''\left({x}\right)=\mathrm{1}×\left\{\left[\mathrm{0}+\mathrm{0}+\mathrm{0}+…\right]^{\mathrm{2}} −\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…\right]\right\}=−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$ \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}−{cos}\left({x}\right){cos}\left({x}/\mathrm{2}\right){cos}\left({x}/\mathrm{3}\right)…}{{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}−{f}\left({x}\right)}{{x}^{\mathrm{2}} }\:\:\:\left(=\frac{\mathrm{0}}{\mathrm{0}}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{−{f}'\left({x}\right)}{\mathrm{2}{x}}\:\:\:\:\left(=\frac{\mathrm{0}}{\mathrm{0}}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{−{f}''\left({x}\right)}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$
Commented by tanmay last updated on 21/Apr/19
bah darun sir...LHospital rule excellent...
$${bah}\:{darun}\:{sir}…{LHospital}\:{rule}\:{excellent}… \\ $$
Commented by Mr X pcx last updated on 21/Apr/19
sir mrw have played a kriket match with this limit ...
$${sir}\:{mrw}\:{have}\:{played}\:{a}\:{kriket}\:{match} \\ $$$${with}\:{this}\:{limit}\:… \\ $$
Commented by tanmay last updated on 29/Apr/19
f(x)=cosxcos((x/2))cos((x/3))... lnf(x)=ln(cosx)+lncos((x/2))+lncos((x/3))+... ((f^′ (x))/(f(x)))=−[tanx+(1/2)tan((x/2))+(1/3)tan((x/3))+...] sir pls check...
$${f}\left({x}\right)={cosxcos}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{3}}\right)… \\ $$$${lnf}\left({x}\right)={ln}\left({cosx}\right)+{lncos}\left(\frac{{x}}{\mathrm{2}}\right)+{lncos}\left(\frac{{x}}{\mathrm{3}}\right)+… \\ $$$$\frac{{f}^{'} \left({x}\right)}{{f}\left({x}\right)}=−\left[{tanx}+\frac{\mathrm{1}}{\mathrm{2}}{tan}\left(\frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{3}}{tan}\left(\frac{{x}}{\mathrm{3}}\right)+…\right] \\ $$$${sir}\:{pls}\:{check}… \\ $$

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