Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 82985 by M±th+et£s last updated on 26/Feb/20

Answered by ~blr237~ last updated on 26/Feb/20

$$letn⩾2,b_n\ne 0causea+b\ne 0and0<a<b$$$$wehave{a}_n=\frac{b_n^2}{b_{n-1}}and{b}_{n-1}=2a_n-a_{n-1}$$$$sowegot{b}_{n-1}=\frac{2b_n^2}{b_{n-1}}-\frac{b_{n-1}^2}{b_{n-2}}\iff 1=2{\left(\frac{b_n}{b_{n-1}}\right)}^2-\left(\frac{b_{n-1}}{b_{n-2}}\right)$$$$letstate{c}_n=\frac{b_n}{b_{n-1}}wecanprovebyinductionthat\mathrm{\forall }n⩾20<a_n⩽b_n$$$$wehave{c}_n=\sqrt{\frac{1+c_{n-1}}{2}}\left(1\right)and{c}_n\in [0,1]cause{c}_n=\frac{b_n}{b_{n-1}}=\frac{a_n}{b_n}$$$$sothereexist\mathrm{\forall }n⩾2{\theta }_n\in [0,\frac{\pi }{2}]suchas{c}_n=cos{\theta }_n$$$$Using\left(1\right)wegetcos{\theta }_n=cos\left(\frac{{\theta }_{n-1}}{2}\right)cause0<x<\frac{\pi }{2}\to cosx>0$$$$suchas\mathrm{\forall }n⩾2{\theta }_n\in [0,\frac{\pi }{2}]weget{\theta }_n=\frac{{\theta }_{n-1}}{2}.After{\theta }_n={\theta }_2{\left(\frac{1}{2}\right)}^{n-2}\mathrm{\forall }n⩾2$$$$,fork⩾2wehave\frac{b_k}{b_{k-1}}=cos\left({\theta }_2{\left(\frac{1}{2}\right)}^{k-2}\right)$$$$b_n=b_1\underset{k=2}{\prod ^n}\left(cos\left(\frac{{\theta }_2}{2^{k-2}}\right)\right)$$$$b_n=b_{1}cos\left({\theta }_2\right)\underset{k=3}{\prod ^n}\left(\frac{sin\left(\frac{{\theta }_2}{2^{k-3}}\right)}{2sin\left(\frac{{\theta }_2}{2^{k-2}}\right)}\right)$$$$usingsin\left(2x\right)=2sinxcosxanditerationwegot$$$$b_n=\frac{b_{1}cos\left({\theta }_2\right)sin\left({\theta }_2\right)}{2^{n-1}sin\left(\frac{{\theta }_2}{2^{n-2}}\right)}forn⩾2$$$$Now\underset{n\to \mathrm{\infty }}{\lim }{b}_n=\frac{b_{1}cos\left({\theta }_2\right)sin\left({\theta }_2\right)}{2{\theta }_2}cause\underset{n\to \mathrm{\infty }}{\lim }\frac{sin\left(\frac{{\theta }_2}{2^{n-2}}\right)}{\frac{1}{2^{n-2}}}={\theta }_2$$$$using{c}_2=cos{\theta }_{2}and{c}_2=\frac{b_2}{b_1}$$$$\frac{b_2}{b_1}=\sqrt{\frac{a_2}{b_1}}=\sqrt{\frac{1}{2}+\frac{a_1}{2b_1}{}_{}}=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{a_1}{b}}}$$$$c_2=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{a}{2b}}}$$$$$$$$$$

Commented by M±th+et£s last updated on 27/Feb/20

$$thankyousomuchsir$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com