Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

Question Number 197359 by sniper237 last updated on 14/Sep/23

$$\underset{x\to 0}{\lim }\frac{1-cosxcos2x...cos\left(nx\right)}{x^2}=\frac{n(n+1)(2n+1)}{12}$$

Commented by universe last updated on 16/Sep/23

Answered by witcher3 last updated on 14/Sep/23

$$\mathrm{if}\mathrm{n}=0$$$$\cos \left(\mathrm{x}\right)\cos \left(2\mathrm{x}\right).....\cos \left(\mathrm{nx}\right)={\mathrm{p}}_{\mathrm{n}}\left(\mathrm{x}\right)$$$$1-{\mathrm{p}}_0=0\Rightarrow \underset{x\to 0}{\lim }\frac{1-{\mathrm{p}}_{\mathrm{n}}\left(\mathrm{x}\right)}{{\mathrm{x}}^2}=0=\frac{0\left(1\right)\left(1\right)}{12}$$$$\mathrm{\forall }\mathrm{n}\in \mathbb{N}\mathrm{suppose}\frac{1-{\mathrm{p}}_{\mathrm{n}}\left(\mathrm{x}\right)}{{\mathrm{x}}^2}=\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{12}$$$$\mathrm{show}\mathrm{That}\frac{1-\cos \left(\mathrm{x}\right)\cos \left(2\mathrm{x}\right)....\cos \left((\mathrm{n}+1)\mathrm{x}\right)}{{\mathrm{x}}^2}$$$$=\frac{(\mathrm{n}+1)(\mathrm{n}+2)(2\mathrm{n}+3)}{12},$$$$\mathrm{t}\to \cos \left(\mathrm{t}\right)\mathrm{over}[0,(\mathrm{n}+1)\mathrm{x}]$$$$\cos \left((\mathrm{n}+1)\mathrm{y}\right)=1-\frac{\cos \left(\mathrm{c}\right)}{2}{(\mathrm{n}+1)}^2{\mathrm{y}}^2$$$$\iff \underset{x\to 0}{\lim }\frac{1-{\mathrm{p}}_{\mathrm{n}}\left(\mathrm{x}\right)(1-\frac{\cos \left(\mathrm{c}\right)}{2}{(\mathrm{n}+1)}^2{\mathrm{x}}^2)}{{\mathrm{x}}^2}$$$$=\underset{x\to 0}{\lim }\frac{1-{\mathrm{p}}_{\mathrm{n}}\left(\mathrm{x}\right)+\frac{{\mathrm{p}}_{\mathrm{n}}\left(\mathrm{x}\right)}{2}\cos \left(\mathrm{c}\right){(\mathrm{n}+1)}^2{\mathrm{x}}^2}{{\mathrm{x}}^2}$$$$=\underset{x\to 0}{\lim }\frac{1-{\mathrm{p}}_{\mathrm{n}}\left(\mathrm{x}\right)}{{\mathrm{x}}^2}+{(\mathrm{n}+1)}^2\cos \left(\mathrm{c}\right)\frac{{\mathrm{p}}_{\mathrm{n}}}{2}$$$$\cos \left((\mathrm{n}+1)\mathrm{y}\right)⩽\cos \left(\mathrm{c}\right)⩽1...$$$$\cos \left(\mathrm{c}\right)\frac{{\mathrm{p}}_{\mathrm{n}}}{2}\to \frac{1}{2}$$$$=\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{12}+\frac{{(\mathrm{n}+1)}^2}{2}=\frac{(\mathrm{n}+1)(\mathrm{n}(2\mathrm{n}+1)+6\mathrm{n}+6)}{12}$$$$=\frac{(\mathrm{n}+1)(2\mathrm{n}+3)(\mathrm{n}+2)}{12}$$$$\mathrm{We}\mathrm{can}\mathrm{Show}\mathrm{This}\mathrm{without}$$$$\mathrm{recursion}\mathrm{elementry}\mathrm{way}$$$${\mathrm{U}}_{\mathrm{n}}=\underset{x\to 0}{\lim }\frac{1-{\mathrm{p}}_{\mathrm{n}}\left(\mathrm{x}\right)}{{\mathrm{x}}^2},{\mathrm{U}}_{\mathrm{n}+1}={\mathrm{U}}_{\mathrm{n}}+\frac{{(\mathrm{n}+1)}^2}{2}$$$$\Rightarrow \underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}({\mathrm{u}}_{\mathrm{k}+1}-{\mathrm{U}}_{\mathrm{k}})=\frac{1}{2}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}{\mathrm{k}}^2=\frac{1}{2}.\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+3)}{6}$$$$\Rightarrow {\mathrm{U}}_{\mathrm{n}}-{\mathrm{U}}_0=\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+3)}{12}={\mathrm{U}}_{\mathrm{n}},{\mathrm{U}}_0=0$$$$$$

Commented by sniper237 last updated on 15/Sep/23

$$Welldone$$

Answered by MM42 last updated on 15/Sep/23

$$hop\to {lim}_{x\to 0}\frac{sinx\times cos2x\times cos3x\times ...\times cosnx+2cosxsin2x\times cos3x\times ...\times cosnx+3cosxcos2x\times sin3x+...+ncosx\times cos2x\times ...\times sinnx}{2x}$$$$\stackrel{lawofequivalence}{=}{lim}_{x\to 0}\frac{x+4x+9x+...+n^{2}x}{2x}$$$$=\frac{1+2^2+3^2+...+n^2}{2}=\frac{n(n+1)(2n+1)}{12}✓$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com