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Question Number 147467 by mathmax by abdo last updated on 21/Jul/21

$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{x}}$$$$1)\mathrm{calculate}{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(0\right)\mathrm{and}{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(1\right)$$$$2)\mathrm{developp}\mathrm{f}\mathrm{at}\mathrm{integr}\mathrm{serie}$$$$3)\mathrm{calculate}{\int }_0^1\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$$

Answered by mathmax by abdo last updated on 22/Jul/21

$$1)\mathrm{leibniz}\mathrm{give}{\mathrm{f}}^{\left(\mathrm{p}\right)}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{p}}{\mathrm{C}}_{\mathrm{p}}^{\mathrm{k}}{\left({\mathrm{x}}^{\mathrm{n}}\right)}^{\left(\mathrm{k}\right)}{\left({\mathrm{e}}^{-\mathrm{x}}\right)}^{(\mathrm{p}-\mathrm{k})}\Rightarrow$$$$=\sum _{\mathrm{k}=0}^{\mathrm{p}}{\mathrm{C}}_{\mathrm{p}}^{\mathrm{k}}{(-1)}^{\mathrm{p}-\mathrm{k}}{\mathrm{e}}^{-\mathrm{x}}{\left({\mathrm{x}}^{\mathrm{n}}\right)}^{\left(\mathrm{k}\right)}$$$$\mathrm{if}\mathrm{k}>\mathrm{n}\Rightarrow {\left({\mathrm{x}}^{\mathrm{n}}\right)}^{\left(\mathrm{k}\right)}=0$$$$\mathrm{if}\mathrm{k}⩽\mathrm{n}\Rightarrow {\left({\mathrm{x}}^{\mathrm{n}}\right)}^{\left(\mathrm{k}\right)}=\mathrm{n}(\mathrm{n}-1)...(\mathrm{n}-\mathrm{k}+1){\mathrm{x}}^{\mathrm{n}-\mathrm{k}}$$$$=\frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{k})!}{\mathrm{x}}^{\mathrm{n}-\mathrm{k}}\mathrm{we}\mathrm{have}$$$${\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{(-1)}^{\mathrm{n}-\mathrm{k}}{\mathrm{e}}^{-\mathrm{x}}\times \frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{k})!}{\mathrm{x}}^{\mathrm{n}-\mathrm{k}}$$$$=\sum _{\mathrm{k}=0}^{\mathrm{n}-1}(...){\mathrm{x}}^{\mathrm{n}-\mathrm{k}}+{\mathrm{e}}^{-\mathrm{x}}\mathrm{n}!\Rightarrow {\mathrm{f}}^{\left(\mathrm{n}\right)}\left(0\right)=\mathrm{n}!$$

Commented by mathmax by abdo last updated on 22/Jul/21

$${\mathrm{f}}^{\left(\mathrm{n}\right)}\left(1\right)={\mathrm{e}}^{-1}\sum _{\mathrm{k}=0}^{\mathrm{n}}{(-1)}^{\mathrm{n}-\mathrm{k}}\frac{\mathrm{n}!}{\mathrm{k}!(\mathrm{n}-\mathrm{k})!}\times \frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{k})!}$$$$={(-1)}^{\mathrm{n}}{\mathrm{e}}^{-1}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}!}\times \frac{{(\mathrm{n}!)}^2}{{((\mathrm{n}-\mathrm{k})!)}^2}$$

Answered by mathmax by abdo last updated on 22/Jul/21

$$2)\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{x}}={\mathrm{x}}^{\mathrm{n}}\sum _{\mathrm{p}=0}^{\mathrm{\infty }}\frac{{(-\mathrm{x})}^{\mathrm{p}}}{\mathrm{p}!}$$$$=\sum _{\mathrm{p}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{p}}}{\mathrm{p}!}{\mathrm{x}}^{\mathrm{n}+\mathrm{p}}{=}_{\mathrm{n}+\mathrm{p}=\mathrm{k}}\sum _{\mathrm{k}=\mathrm{n}}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{k}-\mathrm{n}}}{(\mathrm{k}-\mathrm{n})!}{\mathrm{x}}^{\mathrm{k}}$$

Answered by mathmax by abdo last updated on 22/Jul/21

$$3){\mathrm{A}}_{\mathrm{n}}={\int }_0^1{\mathrm{x}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}\mathrm{by}\mathrm{parts}\mathrm{we}\mathrm{get}$$$${\mathrm{A}}_{\mathrm{n}}={\left[\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}{\mathrm{e}}^{-\mathrm{x}}\right]}_0^1+{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}{\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}=\frac{{\mathrm{e}}^{-1}}{\mathrm{n}+1}+\frac{1}{\mathrm{n}+1}{\mathrm{A}}_{\mathrm{n}+1}\Rightarrow$$$$(\mathrm{n}+1){\mathrm{A}}_{\mathrm{n}}=\frac{1}{\mathrm{e}}+{\mathrm{A}}_{\mathrm{n}+1}\Rightarrow {\mathrm{A}}_{\mathrm{n}+1}=(\mathrm{n}+1){\mathrm{A}}_{\mathrm{n}}-\frac{1}{\mathrm{e}}\Rightarrow$$$${\mathrm{A}}_{\mathrm{n}}={\mathrm{nA}}_{\mathrm{n}-1}-\frac{1}{\mathrm{e}}\mathrm{let}{\mathrm{u}}_{\mathrm{n}}=\frac{{\mathrm{A}}_{\mathrm{n}}}{\mathrm{n}!}\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}+1}=\frac{{\mathrm{A}}_{\mathrm{n}+1}}{(\mathrm{n}+1)!}=\frac{(\mathrm{n}+1){\mathrm{A}}_{\mathrm{n}}-\frac{1}{\mathrm{e}}}{(\mathrm{n}+1)!}=\frac{{\mathrm{A}}_{\mathrm{n}}}{\mathrm{n}!}-\frac{1}{\mathrm{e}(\mathrm{n}+1)!}$$$$={\mathrm{u}}_{\mathrm{n}}-\frac{1}{\mathrm{e}(\mathrm{n}+1)!}\Rightarrow {\mathrm{u}}_{\mathrm{n}+1}-{\mathrm{u}}_{\mathrm{n}}=-\frac{1}{\mathrm{e}(\mathrm{n}+1)!}\Rightarrow$$$$\sum _{\mathrm{k}=0}^{\mathrm{n}}({\mathrm{u}}_{\mathrm{k}+1}-{\mathrm{u}}_{\mathrm{k}})=-\frac{1}{\mathrm{e}}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{1}{(\mathrm{k}+1)!}\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}}-{\mathrm{u}}_0=-\frac{1}{\mathrm{e}}\sum _{\mathrm{k}=1}^{\mathrm{n}+1}\frac{1}{\mathrm{k}!}\Rightarrow {\mathrm{u}}_{\mathrm{n}}={\mathrm{u}}_{\mathrm{o}}-\frac{1}{\mathrm{e}}\sum _{\mathrm{k}=1}^{\mathrm{n}+1}\frac{1}{\mathrm{k}!}$$$${\mathrm{u}}_{\mathrm{o}}={\mathrm{A}}_0={\int }_0^1{\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}={[-{\mathrm{e}}^{-\mathrm{x}}]}_0^1=1-\frac{1}{\mathrm{e}}\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}}=1-\frac{1}{\mathrm{e}}\sum _{\mathrm{k}=0}^{\mathrm{n}+1}\frac{1}{\mathrm{k}!}$$$${\mathrm{A}}_{\mathrm{n}}=\mathrm{n}!{\mathrm{u}}_{\mathrm{n}}=\mathrm{n}!(1-\frac{1}{\mathrm{e}}\sum _{\mathrm{k}=0}^{\mathrm{n}+1}\frac{1}{\mathrm{k}!})$$

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