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Question Number 119754 by Bird last updated on 26/Oct/20

$$find\sum _{n=1}^{\mathrm{\infty }}\frac{u_n}{n!}if{u}_n=u_{n+1}+u_{n-1}$$

Commented by Dwaipayan Shikari last updated on 26/Oct/20

$$u_n=u_{n+1}+u_{n-1}$$$$r^n=r^{n+1}+r^{n-1}$$$$1=r+\frac{1}{r}\Rightarrow r^2-r+1=0\Rightarrow r=\frac{1\pm i\sqrt{3}}{2}=e^{\pm \frac{i\pi }{3}}$$$$u_n=\mathrm{{\rm Y}}{e}^{\frac{i\pi n}{3}}+\mathrm{\Lambda }{e}^{-\frac{i\pi n}{3}}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{u_n}{n!}=\mathrm{{\rm Y}}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\left(e^{\frac{i\pi }{3}}\right)}^n}{n!}+\mathrm{\Lambda }\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\left(e^{\frac{-i\pi }{3}}\right)}^n}{n!}$$$$=\mathrm{{\rm Y}}(e^{e^{\frac{i\pi }{3}}}-1)+\mathrm{\Lambda }(e^{e^{\frac{-i\pi }{3}}}-1)$$

Commented by talminator2856791 last updated on 27/Oct/20

$$\mathrm{WOW}\mathrm{that}\mathrm{is}\mathrm{a}\mathrm{very}\mathrm{beautiful}\mathrm{answer}$$$$\mathrm{where}\mathrm{do}\mathrm{you}\mathrm{learn}\mathrm{that}?$$

Commented by Dwaipayan Shikari last updated on 27/Oct/20

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}=e^x-1$$$$Herex=e^{\frac{i\pi }{3}}and{e}^{\frac{-i\pi }{3}}$$$$:)$$$$:)$$

Answered by Olaf last updated on 26/Oct/20

$$u_{n+1}-u_n+u_{n-1}=0$$$$r^2-r+1=0$$$$\mathrm{\Delta }={(-1)}^2-4\left(1\right)\left(1\right)=-3$$$$r=\frac{1\pm \sqrt{3}i}{2}=e^{\pm i\frac{\pi }{3}}$$$$u_n=\lambda e^{+in\frac{\pi }{3}}+\mu e^{-in\frac{\pi }{3}}$$$$\left(\lambda \mathrm{and}\mu \mathrm{depend}\mathrm{on}{u}_0\mathrm{and}{u}_1\right)$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{u_n}{n!}=\lambda \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(e^{i\frac{\pi }{3}}\right)}^n}{n!}+\mu \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(e^{-i\frac{\pi }{3}}\right)}^n}{n!}-u_0$$$$\mathrm{S}=\lambda e^{\frac{1}{2}+i\frac{\sqrt{3}}{2}}+\mu e^{\frac{1}{2}-i\frac{\sqrt{3}}{2}}-u_0$$$$\mathrm{S}=\sqrt{e}(\lambda e^{i\frac{\sqrt{3}}{2}}+\mu e^{-i\frac{\sqrt{3}}{2}})-u_0$$$$\mathrm{with}:$$$$u_0=\lambda +\mu$$$$u_1=\lambda e^{i\frac{\pi }{3}}+\mu e^{-i\frac{\pi }{3}}$$$$\{\begin{array}{l}\end{array}$$

Answered by mathmax by abdo last updated on 26/Oct/20

$${\mathrm{u}}_{\mathrm{n}}={\mathrm{u}}_{\mathrm{n}+1}+{\mathrm{u}}_{\mathrm{n}-1}\Rightarrow {\mathrm{u}}_{\mathrm{n}+1}-{\mathrm{u}}_{\mathrm{n}}+{\mathrm{u}}_{\mathrm{n}-1}=0\Rightarrow {\mathrm{u}}_{\mathrm{n}+2}-{\mathrm{u}}_{\mathrm{n}+1}+{\mathrm{u}}_{\mathrm{n}}=0$$$$\to {\mathrm{r}}^2-\mathrm{r}+1=0\to \mathrm{\Delta }=1-4=-3\Rightarrow {\mathrm{r}}_1=\frac{1+\mathrm{i}\sqrt{3}}{2}={\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}$$$${\mathrm{r}}_2=\frac{1-\mathrm{i}\sqrt{3}}{2}={\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}\Rightarrow {\mathrm{u}}_{\mathrm{n}}={\mathrm{ar}}_1^{\mathrm{n}}+{\mathrm{br}}_1^{\mathrm{n}}={\mathrm{ae}}^{\frac{\mathrm{in}\pi }{3}}+{\mathrm{be}}^{-\mathrm{i}\frac{\mathrm{n}\pi }{3}}$$$$\Rightarrow \sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{u}}_{\mathrm{n}}}{\mathrm{n}}=\mathrm{a}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\frac{\mathrm{in}\pi }{3}}}{\mathrm{n}}+\mathrm{b}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\frac{\mathrm{in}\pi }{3}}}{\mathrm{n}}$$$$\mathrm{let}\mathrm{determine}\mathrm{f}\left(\mathrm{z}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\mathrm{n}}}{\mathrm{n}}\mathrm{we}\mathrm{have}\mathrm{for}\mid \mathrm{z}\mid <1$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{z}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{z}}^{\mathrm{n}-1}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{z}}^{\mathrm{n}}=\frac{1}{1-\mathrm{z}}\Rightarrow \mathrm{f}\left(\mathrm{z}\right)=-\ln (1-\mathrm{z})+\mathrm{c}$$$$\mathrm{c}=\mathrm{f}\left(0\right)=0\Rightarrow \sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\mathrm{n}}}{\mathrm{n}}=-\ln (1-\mathrm{z})\Rightarrow$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{u}}_{\mathrm{n}}}{\mathrm{n}}=-\mathrm{aln}(1-{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}})-\mathrm{bln}(1-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}})\mathrm{we}\mathrm{have}$$$$\ln (1-{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}})=\ln (1-\cos \left(\frac{\pi }{3}\right)-\mathrm{isin}\left(\frac{\pi }{3}\right))$$$$=\ln ({2\sin }^2\left(\frac{\pi }{6}\right)-2\mathrm{isin}\left(\frac{\pi }{6}\right)\cos \left(\frac{\pi }{6}\right))$$$$=\ln (-2\mathrm{isin}\left(\frac{\pi }{6}\right){\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}})=\ln \left(2\right)+\ln (-\mathrm{i})+\ln \left(\sin \frac{\pi }{6}\right)+\frac{\mathrm{i}\pi }{6}$$$$=\ln \left(2\right)-\frac{\mathrm{i}\pi }{2}+\ln \left(\sin \left(\frac{\pi }{6}\right)\right)+\frac{\mathrm{i}\pi }{6}$$$$\ln (1-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}})=\ln (1-\cos \left(\frac{\pi }{3}\right)+\mathrm{isin}\left(\frac{\pi }{3}\right))$$$$=\ln ({2\sin }^2\left(\frac{\pi }{6}\right)+2\mathrm{isin}\left(\frac{\pi }{6}\right)\cos \left(\frac{\pi }{6}\right))$$$$=\ln \left(2\mathrm{isin}\left(\frac{\pi }{6}\right){\mathrm{e}}^{-\frac{\mathrm{i}\pi }{6}}\right)=\ln \left(2\right)+\ln \left(\mathrm{i}\right)+\ln \left(\sin \left(\frac{\pi }{6}\right)\right)-\frac{\mathrm{i}\pi }{6}$$$$=\ln \left(2\right)+\frac{\mathrm{i}\pi }{2}+\ln \left(\sin \left(\frac{\pi }{2}\right)\right)-\frac{\mathrm{i}\pi }{6}\Rightarrow$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{u}}_{\mathrm{n}}}{\mathrm{n}}=-\mathrm{a}\{\ln \left(2\right)-\frac{\mathrm{i}\pi }{2}+\ln \left(\frac{1}{2}\right)+\frac{\mathrm{i}\pi }{6}\}-\mathrm{b}\{\ln \left(2\right)+\frac{\mathrm{i}\pi }{2}+\ln \left(\frac{1}{2}\right)-\frac{\mathrm{i}\pi }{6}\}$$$$=-\mathrm{ai}(\frac{\pi }{6}-\frac{\pi }{2})-\mathrm{bi}(\frac{\pi }{2}-\frac{\pi }{6})$$$$=-\mathrm{ai}(-\frac{\pi }{3})-\mathrm{bi}\left(\frac{\pi }{3}\right)=(\mathrm{a}-\mathrm{b})\mathrm{i}\frac{\pi }{3}$$$$\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{can}\mathrm{be}\mathrm{found}\mathrm{by}\mathrm{initial}\mathrm{conditions}$$

Commented by mathmax by abdo last updated on 27/Oct/20

$$\mathrm{sorry}\mathrm{i}\mathrm{have}\mathrm{solved}\mathrm{\Sigma }\frac{{\mathrm{u}}_{\mathrm{n}}}{\mathrm{n}}\mathrm{not}\mathrm{\Sigma }\frac{{\mathrm{u}}_{\mathrm{n}}}{\mathrm{n}!}...!$$

Commented by mathmax by abdo last updated on 27/Oct/20

$$\mathrm{we}\mathrm{have}{\mathrm{u}}_{\mathrm{n}}={\mathrm{ae}}^{\frac{\mathrm{in}\pi }{3}}+{\mathrm{be}}^{-\frac{\mathrm{in}\pi }{3}}\Rightarrow$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{u}}_{\mathrm{n}}}{\mathrm{n}!}=\mathrm{a}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\left({\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}\right)}^{\mathrm{n}}}{\mathrm{n}!}+\mathrm{b}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\left({\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}\right)}^{\mathrm{n}}}{\mathrm{n}!}$$$$=\mathrm{a}({\mathrm{e}}^{{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}}-1)+\mathrm{b}({\mathrm{e}}^{{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{3}}}-1)$$$$=\mathrm{a}\{{\mathrm{e}}^{\frac{1}{2}+\frac{\mathrm{i}\sqrt{3}}{2}}-1)+\mathrm{b}({\mathrm{e}}^{\frac{1}{2}-\frac{\mathrm{i}\sqrt{3}}{2}}-1)$$$$=\mathrm{a}{\mathrm{e}}^{\frac{1}{2}}(\cos \left(\frac{\sqrt{3}}{2}\right)+\mathrm{isin}\left(\frac{\sqrt{3}}{2}\right))+{\mathrm{be}}^{\frac{1}{2}}(\cos \left(\frac{\sqrt{3}}{2}\right)-\mathrm{isin}\left(\frac{\sqrt{3}}{2}\right))-(\mathrm{a}+\mathrm{b})$$$$=(\mathrm{a}+\mathrm{b})\sqrt{\mathrm{e}}\cos \left(\frac{\sqrt{3}}{2}\right)+\mathrm{i}(\mathrm{a}-\mathrm{b})\sqrt{\mathrm{e}}\sin \left(\frac{\sqrt{3}}{2}\right)-(\mathrm{a}+\mathrm{b})$$

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