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Question Number 83296 by Rio Michael last updated on 29/Feb/20

$$\mathrm{Obtain}\mathrm{a}\mathrm{maclaurin}\mathrm{expansion}\mathrm{for}$$$$\mathrm{a}){\mathrm{e}}^{\cos x}\mathrm{b}){\mathrm{e}}^{\cos {}^{2}x}$$

Commented by niroj last updated on 29/Feb/20

$$$$$$\left(\mathrm{a}\right).{\mathrm{e}}^{\cos \mathrm{x}}$$$$\mathrm{let},\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\cos \mathrm{x}},{\mathrm{f}}_0\left(0\right)={\mathrm{e}}^{\cos 0}=\mathrm{e}$$$${\mathrm{f}}_1\left(\mathrm{x}\right)={\mathrm{e}}^{\cos \mathrm{x}}.(-\mathrm{sinx}),{\mathrm{f}}_1\left(0\right)=0$$$${\mathrm{f}}_2\left(\mathrm{x}\right)=\sin {}^2\mathrm{x}.{\mathrm{e}}^{\cos \mathrm{x}}-\cos \mathrm{x}.{\mathrm{e}}^{\cos \mathrm{x}}$$$${\mathrm{f}}_2\left(0\right)=-\mathrm{e}$$$${\mathrm{f}}_3\left(\mathrm{x}\right)=2\sin \mathrm{x}\cos \mathrm{x}.{\mathrm{e}}^{\cos \mathrm{x}}-{\sin }^2\mathrm{x}.{\mathrm{e}}^{\cos \mathrm{x}}.\mathrm{sinx}+\sin \mathrm{x}.{\mathrm{e}}^{\cos \mathrm{x}}+\cos \mathrm{x}.{\mathrm{e}}^{\cos \mathrm{x}}.\sin \mathrm{x}$$$${\mathrm{f}}_3\left(0\right)=0$$$$\mathrm{we}\mathrm{know}{\mathrm{maclaurine}}^{\prime }\mathrm{s}\mathrm{series}..$$$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{f}}_0\left(0\right)+\frac{\mathrm{x}}{1!}{\mathrm{f}}_1\left(0\right)+\frac{{\mathrm{x}}^2}{2!}{\mathrm{f}}_2\left(0\right)+\frac{{\mathrm{x}}^3}{3!}{\mathrm{f}}_3\left(0\right)+...$$$${\mathrm{e}}^{\cos \mathrm{x}}=\mathrm{e}+\frac{\mathrm{x}}{1\times 1}\left(0\right)+\frac{{\mathrm{x}}^2}{2\times 1}(-\mathrm{e})+\frac{{\mathrm{x}}^3}{3\times 2\times 1}\left(0\right)+...$$$${\mathrm{e}}^{\cos \mathrm{x}}=\mathrm{e}-{\mathrm{x}}^2\mathrm{e}+....$$$$$$$$$$$$$$

Commented by Rio Michael last updated on 29/Feb/20

$$thankssir$$

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