Question Number 197359 by sniper237 last updated on 14/Sep/23
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{1}−{cosxcos}\mathrm{2}{x}…{cos}\left({nx}\right)}{{x}^{\mathrm{2}} }\:=\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}}\: \\ $$
Commented by universe last updated on 16/Sep/23
Answered by witcher3 last updated on 14/Sep/23
![if n=0 cos(x)cos(2x).....cos(nx)=p_n (x) 1−p_0 =0⇒lim_(x→0) ((1−p_n (x))/x^2 )=0=((0(1)(1))/(12)) ∀n∈N suppose ((1−p_n (x))/x^2 )=((n(n+1)(2n+1))/(12)) show That ((1−cos(x)cos(2x)....cos((n+1)x))/x^2 ) =(((n+1)(n+2)(2n+3))/(12)), t→cos(t) over [0,(n+1)x] cos((n+1)y)=1−((cos(c))/2)(n+1)^2 y^2 ⇔lim_(x→0) ((1−p_n (x)(1−((cos(c))/2)(n+1)^2 x^2 ))/x^2 ) =lim_(x→0) ((1−p_n (x)+((p_n (x))/2)cos(c)(n+1)^2 x^2 )/x^2 ) =lim_(x→0) ((1−p_n (x))/x^2 )+(n+1)^2 cos(c)(p_n /2) cos((n+1)y)≤cos(c)≤1... cos(c)(p_n /2)→(1/2) =((n(n+1)(2n+1))/(12))+(((n+1)^2 )/2)=(((n+1)(n(2n+1)+6n+6))/(12)) =(((n+1)(2n+3)(n+2))/(12)) We can Show This without recursion elementry way U_n =lim_(x→0) ((1−p_n (x))/x^2 ),U_(n+1) =U_n +(((n+1)^2 )/2) ⇒Σ_(k=0) ^(n−1) (u_(k+1) −U_k )=(1/2)Σ_(k=1) ^n k^2 =(1/2).((n(n+1)(2n+3))/6) ⇒U_n −U_0 =((n(n+1)(2n+3))/(12))=U_n ,U_0 =0](https://www.tinkutara.com/question/Q197361.png)
$$\mathrm{if}\:\mathrm{n}=\mathrm{0} \\ $$$$\mathrm{cos}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{2x}\right)…..\mathrm{cos}\left(\mathrm{nx}\right)=\mathrm{p}_{\mathrm{n}} \left(\mathrm{x}\right) \\ $$$$\mathrm{1}−\mathrm{p}_{\mathrm{0}} =\mathrm{0}\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{p}_{\mathrm{n}} \left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} }=\mathrm{0}=\frac{\mathrm{0}\left(\mathrm{1}\right)\left(\mathrm{1}\right)}{\mathrm{12}} \\ $$$$\forall\mathrm{n}\in\mathbb{N}\:\mathrm{suppose}\:\frac{\mathrm{1}−\mathrm{p}_{\mathrm{n}} \left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{12}} \\ $$$$\mathrm{show}\:\mathrm{That}\:\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{2x}\right)….\mathrm{cos}\left(\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{2n}+\mathrm{3}\right)}{\mathrm{12}}, \\ $$$$\mathrm{t}\rightarrow\mathrm{cos}\left(\mathrm{t}\right)\:\mathrm{over}\:\left[\mathrm{0},\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}\right] \\ $$$$\mathrm{cos}\left(\left(\mathrm{n}+\mathrm{1}\right)\mathrm{y}\right)=\mathrm{1}−\frac{\mathrm{cos}\left(\mathrm{c}\right)}{\mathrm{2}}\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \\ $$$$\Leftrightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{p}_{\mathrm{n}} \left(\mathrm{x}\right)\left(\mathrm{1}−\frac{\mathrm{cos}\left(\mathrm{c}\right)}{\mathrm{2}}\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{p}_{\mathrm{n}} \left(\mathrm{x}\right)+\frac{\mathrm{p}_{\mathrm{n}} \left(\mathrm{x}\right)}{\mathrm{2}}\mathrm{cos}\left(\mathrm{c}\right)\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{p}_{\mathrm{n}} \left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} }+\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} \mathrm{cos}\left(\mathrm{c}\right)\frac{\mathrm{p}_{\mathrm{n}} }{\mathrm{2}} \\ $$$$\mathrm{cos}\left(\left(\mathrm{n}+\mathrm{1}\right)\mathrm{y}\right)\leqslant\mathrm{cos}\left(\mathrm{c}\right)\leqslant\mathrm{1}… \\ $$$$\mathrm{cos}\left(\mathrm{c}\right)\frac{\mathrm{p}_{\mathrm{n}} }{\mathrm{2}}\rightarrow\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{12}}+\frac{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}=\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}\left(\mathrm{2n}+\mathrm{1}\right)+\mathrm{6n}+\mathrm{6}\right)}{\mathrm{12}} \\ $$$$=\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{3}\right)\left(\mathrm{n}+\mathrm{2}\right)}{\mathrm{12}} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{Show}\:\mathrm{This}\:\mathrm{without} \\ $$$$\mathrm{recursion}\:\mathrm{elementry}\:\mathrm{way} \\ $$$$\mathrm{U}_{\mathrm{n}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{p}_{\mathrm{n}} \left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} },\mathrm{U}_{\mathrm{n}+\mathrm{1}} =\mathrm{U}_{\mathrm{n}} +\frac{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\left(\mathrm{u}_{\mathrm{k}+\mathrm{1}} −\mathrm{U}_{\mathrm{k}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{3}\right)}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{U}_{\mathrm{n}} −\mathrm{U}_{\mathrm{0}} =\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{3}\right)}{\mathrm{12}}=\mathrm{U}_{\mathrm{n}} ,\mathrm{U}_{\mathrm{0}} =\mathrm{0} \\ $$$$ \\ $$
Commented by sniper237 last updated on 15/Sep/23
$${Well}\:{done} \\ $$
Answered by MM42 last updated on 15/Sep/23
$${hop}\rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{sinx}×{cos}\mathrm{2}{x}×{cos}\mathrm{3}{x}×…×{cosnx}+\mathrm{2}{cosxsin}\mathrm{2}{x}×{cos}\mathrm{3}{x}×…×{cosnx}+\mathrm{3}{cosxcos}\mathrm{2}{x}×{sin}\mathrm{3}{x}+…+{ncosx}×{cos}\mathrm{2}{x}×…×{sinnx}}{\mathrm{2}{x}} \\ $$$$\overset{{law}\:{of}\:{equivalence}} {=}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}+\mathrm{4}{x}+\mathrm{9}{x}+…+{n}^{\mathrm{2}} {x}}{\mathrm{2}{x}}\:\:\:\:\:\: \\ $$$$=\frac{\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +…+{n}^{\mathrm{2}} }{\mathrm{2}}=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}}\:\checkmark \\ $$$$ \\ $$