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Question Number 145339 by qaz last updated on 04/Jul/21

$$\mathrm{Let}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}\cos \mathrm{x},\mathrm{Find}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)}{2^{\mathrm{n}}}=?$$

Answered by mathmax by abdo last updated on 04/Jul/21

$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}\Rightarrow {\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\left(\mathrm{cosx}\right)}^{\left(\mathrm{k}\right)}{\left({\mathrm{e}}^{\mathrm{x}}\right)}^{(\mathrm{n}-\mathrm{k})}$$$$=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}\cos (\mathrm{x}+\frac{\mathrm{k}\pi }{2}){\mathrm{e}}^{\mathrm{x}}\Rightarrow$$$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)}{2^{\mathrm{n}}}={\mathrm{e}}^{\mathrm{x}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{2^{\mathrm{n}}}\left(\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}\cos (\mathrm{x}+\frac{\mathrm{k}\pi }{2})\right)$$$$\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}\cos (\mathrm{x}+\frac{\mathrm{k}\pi }{2})=\mathrm{Re}\left(\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\mathrm{e}}^{\mathrm{i}(\mathrm{x}+\frac{\mathrm{k}\pi }{2})}\right)$$$$\sum _{\mathrm{k}=0}^{\mathrm{n}}(...)={\mathrm{e}}^{\mathrm{ix}}\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\left(\mathrm{i}\right)}^{\mathrm{k}}={\mathrm{e}}^{\mathrm{ix}}{(1+\mathrm{i})}^{\mathrm{n}}$$$$={\mathrm{e}}^{\mathrm{ix}}{\left(\sqrt{2}\right)}^{\mathrm{n}}{\left({\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}\right)}^{\mathrm{n}}={\left(\sqrt{2}\right)}^{\mathrm{n}}{\mathrm{e}}^{\mathrm{ix}}{\mathrm{e}}^{\frac{\mathrm{in}\pi }{4}}={\left(\sqrt{2}\right)}^{\mathrm{n}}{\mathrm{e}}^{\mathrm{i}(\mathrm{x}+\frac{\mathrm{n}\pi }{4})}$$$$\Rightarrow \sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}\cos (\mathrm{x}+\frac{\mathrm{k}\pi }{2})={\left(\sqrt{2}\right)}^{\mathrm{n}}\cos (\mathrm{x}+\frac{\mathrm{n}\pi }{4})\Rightarrow$$$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)}{2^{\mathrm{n}}}={\mathrm{e}}^{\mathrm{x}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\left(\frac{\sqrt{2}}{2}\right)}^{\mathrm{n}}\cos (\mathrm{x}+\frac{\mathrm{n}\pi }{4})$$$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\left(\frac{1}{\sqrt{2}}\right)}^{\mathrm{n}}\cos (\mathrm{x}+\frac{\mathrm{n}\pi }{4})=\mathrm{Re}\left(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\left(\frac{1}{\sqrt{2}}\right)}^{\mathrm{n}}{\mathrm{e}}^{\mathrm{i}(\mathrm{x}+\frac{\mathrm{n}\pi }{4}}\right))$$$$\mathrm{\Sigma }(...)={\mathrm{e}}^{\mathrm{ix}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\left(\frac{1}{\sqrt{2}}\right)}^{\mathrm{n}}{\left({\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}\right)}^{\mathrm{n}}$$$$={\mathrm{e}}^{\mathrm{ix}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\left(\frac{{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}}{\sqrt{2}}\right)}^{\mathrm{n}}={\mathrm{e}}^{\mathrm{ix}}\times \frac{1}{1-\frac{1}{\sqrt{2}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}}$$$$=\frac{\sqrt{2}{\mathrm{e}}^{\mathrm{ix}}}{\sqrt{2}-\frac{1}{\sqrt{2}}-\frac{\mathrm{i}}{\sqrt{2}}}=\frac{{2\mathrm{e}}^{\mathrm{ix}}}{1-\mathrm{i}}=\frac{{2\mathrm{e}}^{\mathrm{ix}}(1+\mathrm{i})}{2}$$$$={\mathrm{e}}^{\mathrm{ix}}\sqrt{2}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}=\sqrt{2}{\mathrm{e}}^{\mathrm{i}(\mathrm{x}+\frac{\pi }{4})}\Rightarrow \mathrm{Re}(....)=\sqrt{2}\cos (\mathrm{x}+\frac{\pi }{4})\Rightarrow$$$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)}{2^{\mathrm{n}}}=\sqrt{2}{\mathrm{e}}^{\mathrm{x}}\cos (\mathrm{x}+\frac{\pi }{4})$$

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