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Question Number 94110 by Jidda28 last updated on 16/May/20

Commented by mathmax by abdo last updated on 17/May/20

$$a)lettakeatryletf\left(x\right)=e^{cosx}maclaurinat0$$$$f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{f^{\left(n\right)}\left(0\right)}{n!}{x}^{n}wecanfind{f}^{\left(n\right)}\left(0\right)byrecurrence$$$$fiseven\Rightarrow f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{f^{\left(2n\right)}\left(0\right)}{\left(2n\right)!}{x}^{2n}$$$$wehave{f}^{{}^{\prime }}\left(x\right)=-sinxf\left(x\right)\Rightarrow f^{\left(n\right)}\left(x\right)=-{\left(sinxf\left(x\right)\right)}^{(n-1)}$$$$=-\sum _{k=0}^{n-1}{C}_n^k{f}^{\left(k\right)}\left(x\right){\left(sinx\right)}^{(n-1-k)}$$$$=-\sum _{k=0}^{n-1}{C}_n^k{f}^{\left(k\right)}\left(x\right)sin(x+\frac{(n-1-k)\pi }{2})\Rightarrow$$$$f^{\left(2n\right)}\left(x\right)=-\sum _{k=0}^{2n-1}{C}_{2n}^k{f}^{\left(k\right)}\left(x\right)sin(x+\frac{(2n-1-k)\pi }{2})\Rightarrow$$$$f^{\left(2n\right)}\left(0\right)=-\sum _{k=0}^{2n-1}{C}_{2n}^k{f}^{\left(k\right)}\left(0\right)sin\left(\frac{(2n-1)\pi }{2}\right)$$$$=-\sum _{k=0}^{2n-1}{C}_{2n}^k{f}^{\left(k\right)}\left(0\right)sin(n\pi -\frac{\pi }{2})=\sum _{k=0}^{2n-1}{C}_{2n}^k{f}^{\left(k\right)}\left(0\right)$$$$forexemplen=2\Rightarrow f^{\left(4\right)}\left(0\right)=\sum _{k=0}^3{C}_4^k{f}^{\left(k\right)}\left(0\right)$$$$=C_4^{0}f\left(0\right)+C_4^1{f}^{\left(1\right)}\left(0\right)+C_4^2{f}^{\left(2\right)}\left(0\right)+C_4^3{f}^{\left(3\right)}\left(0\right)=...andwecan$$$$writef\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\left(\sum _{k=0}^{2n-1}{C}_{2n}^k{f}^{\left(k\right)}\left(0\right)\right)x^{2n}$$$$b)g\left(x\right)=e^{{cos}^{2}x}\Rightarrow g\left(x\right)=e^{\frac{1+cos\left(2x\right)}{2}}$$$$=\sqrt{e}\times e^{\frac{1}{2}cos\left(2x\right)}andweaplly{Q}_{n}1....$$

Commented by abdomathmax last updated on 17/May/20

$$ifyouhaveanotherwaypostit...$$

Answered by niroj last updated on 17/May/20

$$\left(\mathrm{a}\right){\mathrm{e}}^{\cos \mathrm{x}}$$$$\mathrm{let},\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\cos \mathrm{x}},\mathrm{f}\left(0\right)=\mathrm{e}$$$${\mathrm{f}}_1\left(\mathrm{x}\right)=-{\mathrm{e}}^{\cos \mathrm{x}}.\sin \mathrm{x},{\mathrm{f}}_1\left(0\right)=0$$$${\mathrm{f}}_2\left(\mathrm{x}\right)=-[-{\mathrm{e}}^{\mathrm{cosx}}.{\sin }^2\mathrm{x}+{\mathrm{e}}^{\cos \mathrm{x}}.\mathrm{cosx})={\mathrm{e}}^{\cos \mathrm{x}}{\sin }^2\mathrm{x}-{\mathrm{e}}^{\cos \mathrm{x}}.\cos \mathrm{x}$$$${\mathrm{f}}_2\left(0\right)=-\mathrm{e}$$$${\mathrm{f}}_3\left(\mathrm{x}\right)=-{\mathrm{e}}^{\cos \mathrm{x}}.{\sin }^3\mathrm{x}+{\mathrm{e}}^{\cos \mathrm{x}}\sin 2\mathrm{x}$$$${\mathrm{f}}_3\left(0\right)=0$$$${\mathrm{f}}_4\left(\mathrm{x}\right)={\mathrm{e}}^{\cos \mathrm{x}}{\sin }^4\mathrm{x}-{\mathrm{e}}^{\cos \mathrm{x}}.{3\sin }^2\mathrm{x}.\cos \mathrm{x}-{\mathrm{e}}^{\cos \mathrm{x}}.\mathrm{sinx}.\sin 2\mathrm{x}+{2\mathrm{e}}^{\mathrm{cosx}}.\cos 2\mathrm{x}$$$${\mathrm{f}}_4\left(0\right)=2\mathrm{e}$$$${\mathrm{maclaurin}}^{\prime }\mathrm{s}\mathrm{series}\mathrm{of}\mathrm{expansion}.$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(0\right)+\frac{\mathrm{x}}{1!}{\mathrm{f}}_1\left(0\right)+\frac{{\mathrm{x}}^2}{2!}{\mathrm{f}}_2\left(0\right)+\frac{{\mathrm{x}}^3}{3!}{\mathrm{f}}_3\left(0\right)+\frac{{\mathrm{x}}^4}{4!}{\mathrm{f}}_4\left(0\right)+...=.$$$${\mathrm{e}}^{\cos \mathrm{x}}=\mathrm{e}+\mathrm{x}.0+\frac{{\mathrm{x}}^2}{2!}(-\mathrm{e})+\frac{{\mathrm{x}}^3}{3!}.\left(0\right)+\frac{{\mathrm{x}}^4}{4!}\left(2\mathrm{e}\right)+....$$$${\mathrm{e}}^{\cos \mathrm{x}}=\mathrm{e}-\frac{1}{2}{\mathrm{x}}^2\mathrm{e}+\frac{2}{4.3.2.1}{\mathrm{x}}^4\mathrm{e}+....$$$${\mathrm{e}}^{\cos \mathrm{x}}=\mathrm{e}-\frac{1}{2}{\mathrm{x}}^2\mathrm{e}+\frac{1}{12}{\mathrm{x}}^4\mathrm{e}+....$$

Commented by mathmax by abdo last updated on 17/May/20

$$but{f}^{\left(n\right)}\left(0\right)stillunknown...!$$

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