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Question Number 191205 by ajfour last updated on 20/Apr/23

Commented by ajfour last updated on 20/Apr/23

$$Find\underset{i=1}{\sum ^{\mathrm{\infty }}}{r}_i.Take{r}_1=a,\theta =\alpha$$

Commented by mr W last updated on 20/Apr/23

$$verynicequestion!$$

Answered by mr W last updated on 21/Apr/23

$${\theta }_1=\alpha$$$${\theta }_2=\frac{{\theta }_1}{2}=\frac{\alpha }{2}$$$$...$$$${\theta }_{n-1}=\frac{\alpha }{2^{n-2}}$$$$fromQ191192:$$$$\frac{r_{n-1}}{\tan {\theta }_{n-1}}+2\sqrt{r_n{r}_{n-1}}=\frac{r_n}{\tan \frac{{\theta }_{n-1}}{2}}$$$$let\lambda =\sqrt{\frac{r_n}{r_{n-1}}},t=\tan \frac{{\theta }_{n-1}}{2}=\tan \frac{\alpha }{2^{n-1}}$$$${\lambda }^2-2t\lambda -\frac{1-t^2}{2}=0$$$$\Rightarrow \lambda =t+\sqrt{\frac{1+t^2}{2}}=\frac{1+\sqrt{2}\sin \frac{\alpha }{2^{n-1}}}{\sqrt{2}\cos \frac{\alpha }{2^{n-1}}}$$$$\frac{r_n}{r_{n-1}}={\lambda }^2={\left(\frac{\frac{1}{\sqrt{2}}+\sin \frac{\alpha }{2^{n-1}}}{\cos \frac{\alpha }{2^{n-1}}}\right)}^2$$$$\Rightarrow r_n=a\underset{k=1}{\prod ^n}{\left(\frac{\frac{1}{\sqrt{2}}+\sin \frac{\alpha }{2^{k-1}}}{\cos \frac{\alpha }{2^{k-1}}}\right)}^2$$$$or{r}_n={\left[\frac{(\frac{1}{\sqrt{2}}+\sin \alpha )(\frac{1}{\sqrt{2}}+\sin \frac{\alpha }{2})...(\frac{1}{\sqrt{2}}+\sin \frac{\alpha }{2^{n-1}})}{\cos \alpha \cos \frac{\alpha }{2}...\cos \frac{\alpha }{2^{n-1}}}\right]}^{2}a$$$$-----------$$$$\cos \alpha \cos \frac{\alpha }{2}\cos \frac{\alpha }{2^2}...\cos \frac{\alpha }{2^{n-1}}$$$$=\frac{1}{2\sin \frac{\alpha }{2^{n-1}}}\cos \alpha \cos \frac{\alpha }{2}\cos \frac{\alpha }{2^2}...\cos \frac{\alpha }{2^{n-1}}2\sin \frac{\alpha }{2^{n-1}}$$$$=\frac{1}{2^2\sin \frac{\alpha }{2^{n-1}}}\cos \alpha \cos \frac{\alpha }{2}\cos \frac{\alpha }{2^2}...\cos \frac{\alpha }{2^{n-2}}2\sin \frac{\alpha }{2^{n-2}}$$$$...$$$$=\frac{1}{2^n\sin \frac{\alpha }{2^{n-1}}}\cos \alpha 2\sin \alpha$$$$=\frac{\sin 2\alpha }{2^n\sin \frac{\alpha }{2^{n-1}}}$$$$--------------$$$$S=\underset{n=1}{\sum ^{\mathrm{\infty }}}{r}_n=a\underset{n=1}{\sum ^{\mathrm{\infty }}}{\left[\frac{(\frac{1}{\sqrt{2}}+\sin \alpha )(\frac{1}{\sqrt{2}}+\sin \frac{\alpha }{2})...(\frac{1}{\sqrt{2}}+\sin \frac{\alpha }{2^{n-1}})}{\cos \alpha \cos \frac{\alpha }{2}...\cos \frac{\alpha }{2^{n-1}}}\right]}^2$$$$examples:$$$$\alpha =60°:S\approx 79.619654909509a$$$$\alpha =45°:S\approx 21.061652923501a$$$$\alpha =30°:S\approx 7.043218497988a$$$$\alpha =10°:S\approx 1.907032259467a$$

Commented by mr W last updated on 21/Apr/23

$$i{can}^{\prime }tfindaclosedformulafor$$$$r_{n}and\underset{n=1}{\sum ^{\mathrm{\infty }}}{r}_n.$$

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