Question Number 112083 by bemath last updated on 06/Sep/20
$$\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{4}^{{n}} \:\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\alpha}{\mathrm{2}^{{n}} }\right)\right)\:?\: \\ $$$$\:\:\sqrt{{bemath}} \\ $$
Answered by bobhans last updated on 06/Sep/20
$$\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{4}^{\mathrm{n}} \left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\alpha}{\mathrm{2}^{\mathrm{n}} }\right)\right)\:=\:\mathrm{L} \\ $$$$\mathrm{setting}\:\frac{\alpha}{\mathrm{2}^{\mathrm{n}} }\:=\:\mathrm{r}\:;\:\mathrm{2}^{\mathrm{n}} \:=\:\frac{\alpha}{\mathrm{r}}\:\mathrm{with}\:\mathrm{r}\rightarrow\mathrm{0} \\ $$$$\mathrm{L}\:=\:\underset{\mathrm{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\alpha^{\mathrm{2}} \left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{r}}{\mathrm{r}^{\mathrm{2}} }\right)\:=\:\alpha^{\mathrm{2}} \:\underset{\mathrm{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{r}}{\mathrm{2}}\right)}{\mathrm{r}^{\mathrm{2}} } \\ $$$$\mathrm{L}\:=\:\frac{\alpha^{\mathrm{2}} }{\mathrm{2}}\:. \\ $$
Answered by Dwaipayan Shikari last updated on 06/Sep/20
$$\mathrm{4}^{{n}} \left(\mathrm{2}{sin}^{\mathrm{2}} \frac{\alpha}{\mathrm{2}^{{n}+\mathrm{1}} }\right)=\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} \left(\frac{\alpha}{\mathrm{2}^{{n}+\mathrm{1}} }\right)^{\mathrm{2}} =\frac{\alpha^{\mathrm{2}} }{\mathrm{2}}\:\:\:\:\:\:\:{sin}\frac{\alpha}{\mathrm{2}^{{n}} }\rightarrow\frac{\alpha}{\mathrm{2}^{{n}} } \\ $$