lim-x-0-x-sin-sin-sin-sin-x-n-times-x-3-

Question Number 211295 by efronzo1 last updated on 05/Sep/24

$$\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left.\mathrm{x}−\underset{\mathrm{n}\:\mathrm{times}} {\underbrace{\mathrm{sin}\:\left(\mathrm{sin}\:\left(\mathrm{sin}\:\left(….\left(\mathrm{sin}\:\mathrm{x}\right)\right)\right)\right)\right)}}}{\mathrm{x}^{\mathrm{3}} } \\ $$

Answered by mahdipoor last updated on 05/Sep/24

get f_n =f(f(f(....f(x))...) n time if f_1 (x)=f(x)=a_1 x+b_1 x^3 +R_1 (R_1 =c_1 x^5 +...) and f_m (x)=a_m x+b_m x^3 +R_m ⇒f_(m+1) (x)=f(f_m (x))=a_1 (f_m )+b_1 (f_m )^3 +R_1 =(a_1 a_m )x+(a_1 b_m +a_m ^3 b_1 )x^3 +R_(m+1) ⇒ { ((a_(m+1) =a_1 a_m )),((b_(m+1) =a_1 b_m +a_m ^3 b_1 )) :}⇒a_1 =1 b_1 =−(1/6) f(x)=sin(x)=x−(1/6)x^3 +R ⇒ { ((a_(m+1) =1 )),((b_(m+1) =b_m +b_1 =b_m −(1/6) ⇒ b_(m+1) =((−(m+1))/6))) :} ⇒f_n =sin(sin(....sin(x)...))=x−(n/6)x^3 +R_n lim_(x→0) ((x−f_n )/x^3 )=(n/6)+(R_n /x^3 )=(n/6)+(c_n x^2 +...)=(n/6)

$$ \\ $$$${get}\:{f}_{{n}} ={f}\left({f}\left({f}\left(….{f}\left({x}\right)\right)…\right)\:\:\:\:{n}\:{time}\:\right. \\ $$$${if}\:\:{f}_{\mathrm{1}} \left({x}\right)={f}\left({x}\right)={a}_{\mathrm{1}} {x}+{b}_{\mathrm{1}} {x}^{\mathrm{3}} +{R}_{\mathrm{1}} \:\:\:\:\:\left({R}_{\mathrm{1}} ={c}_{\mathrm{1}} {x}^{\mathrm{5}} +…\right) \\ $$$${and}\:\:\:{f}_{{m}} \left({x}\right)={a}_{{m}} {x}+{b}_{{m}} {x}^{\mathrm{3}} +{R}_{{m}} \\ $$$$\Rightarrow{f}_{{m}+\mathrm{1}} \left({x}\right)={f}\left({f}_{{m}} \left({x}\right)\right)={a}_{\mathrm{1}} \left({f}_{{m}} \right)+{b}_{\mathrm{1}} \left({f}_{{m}} \right)^{\mathrm{3}} +{R}_{\mathrm{1}} \\ $$$$=\left({a}_{\mathrm{1}} {a}_{{m}} \right){x}+\left({a}_{\mathrm{1}} {b}_{{m}} +{a}_{{m}} ^{\mathrm{3}} {b}_{\mathrm{1}} \right){x}^{\mathrm{3}} +{R}_{{m}+\mathrm{1}} \\ $$$$\Rightarrow\begin{cases}{{a}_{{m}+\mathrm{1}} ={a}_{\mathrm{1}} {a}_{{m}} }\\{{b}_{{m}+\mathrm{1}} ={a}_{\mathrm{1}} {b}_{{m}} +{a}_{{m}} ^{\mathrm{3}} {b}_{\mathrm{1}} }\end{cases}\Rightarrow{a}_{\mathrm{1}} =\mathrm{1}\:\:\:\:{b}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${f}\left({x}\right)={sin}\left({x}\right)={x}−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} +{R} \\ $$$$\Rightarrow\begin{cases}{{a}_{{m}+\mathrm{1}} =\mathrm{1}\:}\\{{b}_{{m}+\mathrm{1}} ={b}_{{m}} +{b}_{\mathrm{1}} ={b}_{{m}} −\frac{\mathrm{1}}{\mathrm{6}}\:\Rightarrow\:{b}_{{m}+\mathrm{1}} =\frac{−\left({m}+\mathrm{1}\right)}{\mathrm{6}}}\end{cases} \\ $$$$\Rightarrow{f}_{{n}} ={sin}\left({sin}\left(….{sin}\left({x}\right)…\right)\right)={x}−\frac{{n}}{\mathrm{6}}{x}^{\mathrm{3}} +{R}_{{n}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{x}−{f}_{{n}} }{{x}^{\mathrm{3}} }=\frac{{n}}{\mathrm{6}}+\frac{{R}_{{n}} }{{x}^{\mathrm{3}} }=\frac{{n}}{\mathrm{6}}+\left({c}_{{n}} {x}^{\mathrm{2}} +…\right)=\frac{{n}}{\mathrm{6}} \\ $$

Answered by mr W last updated on 05/Sep/24

sin x=x−(x^3 /(3!))+(x^5 /(5!))−...=x−(x^3 /(3!))+O(x^3 ) sin (sin x)=x−(x^3 /(3!))−(x^3 /(3!))+O(x^3 )=x−((2x^3 )/(3!))+O(x^3 ) sin (sin (sin ...(sin x)))=x−((nx^3 )/(3!))+O(x^3 ) lim_(x→0) ((x−sin (sin (...(sin x))))/x^3 )=(n/(3!))=(n/6)

$$\mathrm{sin}\:{x}={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−…={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+{O}\left({x}^{\mathrm{3}} \right) \\ $$$$\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+{O}\left({x}^{\mathrm{3}} \right)={x}−\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{3}!}+{O}\left({x}^{\mathrm{3}} \right) \\ $$$$\mathrm{sin}\:\left(\mathrm{sin}\:\left(\mathrm{sin}\:…\left(\mathrm{sin}\:{x}\right)\right)\right)={x}−\frac{{nx}^{\mathrm{3}} }{\mathrm{3}!}+{O}\left({x}^{\mathrm{3}} \right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−\mathrm{sin}\:\left(\mathrm{sin}\:\:\left(…\left(\mathrm{sin}\:{x}\right)\right)\right)}{{x}^{\mathrm{3}} }=\frac{{n}}{\mathrm{3}!}=\frac{{n}}{\mathrm{6}} \\ $$

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