Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 152601 by mnjuly1970 last updated on 30/Aug/21

     solve....    lim_( n→∞) {  Π_(k=1) ^n (1 −(k/n)+(k^( 2) /n^( 2) ) )^( (1/n)) }=?    m.n...

$$ \\ $$$$\:\:\:{solve}.... \\ $$$$\:\:{lim}_{\:{n}\rightarrow\infty} \left\{\:\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}\:−\frac{{k}}{{n}}+\frac{{k}^{\:\mathrm{2}} }{{n}^{\:\mathrm{2}} }\:\right)^{\:\frac{\mathrm{1}}{{n}}} \right\}=? \\ $$$$\:\:{m}.{n}... \\ $$$$ \\ $$

Commented by mindispower last updated on 30/Aug/21

hello sir i can′tfind one post of you  ∫((sin(x))/x^x )dx somthig lik that  have a good day god bless you

$${hello}\:{sir}\:{i}\:{can}'{tfind}\:{one}\:{post}\:{of}\:{you} \\ $$$$\int\frac{{sin}\left({x}\right)}{{x}^{{x}} }{dx}\:{somthig}\:{lik}\:{that} \\ $$$${have}\:{a}\:{good}\:{day}\:{god}\:{bless}\:{you} \\ $$

Answered by puissant last updated on 30/Aug/21

U_n =Π_(k=1) ^n (1−(k/n)+(k^2 /n^2 ))^(1/n)     V_n =ln(U_n )  ⇒ lim_(n→∞) V_n = lim_(n→∞) (1/n)Σ_(k=1) ^n ln(1−(k/n)+(k^2 /n^2 ))  =∫_0 ^1 ln(1−x+x^2 )dx   { ((u=ln(1−x+x^2 ))),((v′=1)) :} ⇒  { ((u′=((2x−1)/(1−x+x^2 )))),((v=x)) :}  =[xln(1−x+x^2 )]_0 ^1 −∫_0 ^1 ((2x^2 −x)/(1−x+x^2 ))dx  To be continued...

$${U}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}−\frac{{k}}{{n}}+\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$ \\ $$$${V}_{{n}} ={ln}\left({U}_{{n}} \right) \\ $$$$\Rightarrow\:{lim}_{{n}\rightarrow\infty} {V}_{{n}} =\:{lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{ln}\left(\mathrm{1}−\frac{{k}}{{n}}+\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right){dx} \\ $$$$\begin{cases}{{u}={ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)}\\{{v}'=\mathrm{1}}\end{cases}\:\Rightarrow\:\begin{cases}{{u}'=\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{1}−{x}+{x}^{\mathrm{2}} }}\\{{v}={x}}\end{cases} \\ $$$$=\left[{xln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}^{\mathrm{2}} −{x}}{\mathrm{1}−{x}+{x}^{\mathrm{2}} }{dx} \\ $$$${To}\:{be}\:{continued}... \\ $$

Commented by SANOGO last updated on 30/Aug/21

merci bien mon prof

$${merci}\:{bien}\:{mon}\:{prof} \\ $$

Commented by puissant last updated on 30/Aug/21

je suis pas prof broo je suis e^� le^� ve..

$${je}\:{suis}\:{pas}\:{prof}\:{broo}\:{je}\:{suis}\:\acute {{e}l}\grave {{e}ve}.. \\ $$

Commented by Ar Brandon last updated on 30/Aug/21

∗etudiant   😜

$$\ast\mathrm{etudiant}\: \\ $$😜

Answered by Olaf_Thorendsen last updated on 30/Aug/21

∫_0 ^1 ln(1−x(1−x)) dx = lim_(n→∞) (1/n)Σ_(k=0) ^n ln(1−(k/n)+(k^2 /n^2 ))  ∫_0 ^1 ln(1−x(1−x)) dx =  [xln(1−x+x^2 )−2x−(1/2)ln(1−x+x^2 )  +(√3)arctan(((2x−1)/( (√3))))]_0 ^1   = 2(√3)arctan((1/( (√3))))−2  lim_(n→∞) Π_(k=1) ^n ln(1−(k/n)+(k^2 /n^2 ))^(1/n)  = e^(2(√3)arctan((1/( (√3))))−2)

$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−{x}\left(\mathrm{1}−{x}\right)\right)\:{dx}\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{ln}\left(\mathrm{1}−\frac{{k}}{{n}}+\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−{x}\left(\mathrm{1}−{x}\right)\right)\:{dx}\:= \\ $$$$\left[{x}\mathrm{ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)−\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)\right. \\ $$$$\left.+\sqrt{\mathrm{3}}\mathrm{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\:\mathrm{2}\sqrt{\mathrm{3}}\mathrm{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)−\mathrm{2} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{ln}\left(\mathrm{1}−\frac{{k}}{{n}}+\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{{n}}} \:=\:{e}^{\mathrm{2}\sqrt{\mathrm{3}}\mathrm{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)−\mathrm{2}} \\ $$

Commented by SANOGO last updated on 30/Aug/21

merci bien mon prof

$${merci}\:{bien}\:{mon}\:{prof}\: \\ $$

Commented by mnjuly1970 last updated on 30/Aug/21

 thanks alot mr olaf...

$$\:{thanks}\:{alot}\:{mr}\:{olaf}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com