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Question Number 83864 by jagoll last updated on 07/Mar/20

$$\mathrm{what}\mathrm{Maclaurin}\mathrm{series}\mathrm{of}\mathrm{function}$$$$\tan \left(\mathrm{x}\right)?$$

Commented by niroj last updated on 07/Mar/20

$$\mathrm{Solution}:$$$$\mathrm{let},\mathrm{f}\left(\mathrm{x}\right)=\tan \mathrm{x},{\mathrm{f}}_0\left(0\right)=0$$$${\mathrm{f}}_1\left(\mathrm{x}\right)={\sec }^2\mathrm{x},{\mathrm{f}}_1\left(0\right)=1$$$${\mathrm{f}}_2\left(\mathrm{x}\right)=2\mathrm{secx}.\mathrm{secx}.\tan \mathrm{x},{\mathrm{f}}_2\left(0\right)=0$$$$={2\sec }^2\mathrm{xtanx}$$$${\mathrm{f}}_3\left(\mathrm{x}\right)={4\sec }^2\mathrm{x}.{\tan }^2\mathrm{x}+{2\sec }^4\mathrm{x},{\mathrm{f}}_3\left(0\right)=2$$$${\mathrm{f}}_4\left(\mathrm{x}\right)={8\sec }^2{\mathrm{xtan}}^3\mathrm{x}+{8\sec }^4\mathrm{x}.\mathrm{tanx}.,{\mathrm{f}}_4\left(0\right)=0$$$${\mathrm{f}}_5\left(\mathrm{x}\right)={16\sec }^2{\mathrm{xtan}}^4\mathrm{x}.+{8\sec }^2\mathrm{x}.{3\tan }^2\mathrm{x}.{\sec }^2\mathrm{x}+$$$${32\sec }^3\mathrm{x}.\sec \mathrm{x}{\tan }^2\mathrm{x}+{8\sec }^6\mathrm{x},{\mathrm{f}}_5\left(0\right)=8$$$$\mathrm{now},\mathrm{we}\mathrm{know}{\mathrm{maclaurin}}^{\prime }\mathrm{s}\mathrm{series}\mathrm{of}$$$$\mathrm{expansion}:$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(0\right)+\frac{\mathrm{x}}{1!}{\mathrm{f}}_1\left(0\right)+\frac{{\mathrm{x}}^2}{2!}{\mathrm{f}}_2\left(0\right)+\frac{{\mathrm{x}}^3}{3!}{\mathrm{f}}_3\left(0\right)+\frac{{\mathrm{x}}^4}{4!}{\mathrm{f}}_4\left(0\right)+\frac{{\mathrm{x}}^5}{5!}{\mathrm{f}}_5\left(0\right)$$$$\tan \mathrm{x}=0+\mathrm{x}.\left(1\right)+\frac{{\mathrm{x}}^2}{2}\left(0\right)+\frac{{\mathrm{x}}^3}{6}.\left(2\right)+\frac{{\mathrm{x}}^4}{24}\left(0\right)+\frac{{\mathrm{x}}^5}{120}.\left(8\right)$$$$=\mathrm{x}+\frac{1}{3}{\mathrm{x}}^3+\frac{2}{15}{\mathrm{x}}^5+.....$$$$\mathrm{hence}\mathrm{proved}{\mathrm{maclaurin}}^{\prime }\mathrm{s}\mathrm{function}$$$$\mathrm{of}\tan \mathrm{x}.$$$$$$$$$$$$$$

Commented by jagoll last updated on 07/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by niroj last updated on 07/Mar/20

$$youmustwelcomesir.$$

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