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Question Number 95786 by abdomathmax last updated on 27/May/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}\ln (1+2\mathrm{x})$$$$1)\mathrm{calculate}{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)\mathrm{and}{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(1\right)$$$$2)\mathrm{developp}\mathrm{f}\mathrm{at}\mathrm{integr}\mathrm{serie}\mathrm{at}{\mathrm{x}}_0=1$$$$3)\mathrm{developp}\mathrm{f}\mathrm{at}\mathrm{integr}\mathrm{serie}\mathrm{at}{\mathrm{x}}_0=0$$

Commented by Rio Michael last updated on 28/May/20

$$\mathrm{sir}\mathrm{when}\mathrm{you}\mathrm{talk}\mathrm{about}\mathrm{integer}\mathrm{series}$$$$\mathrm{do}\mathrm{you}\mathrm{mean}\mathrm{taylors}\mathrm{series}?$$

Commented by mathmax by abdo last updated on 29/May/20

$$\mathrm{yes}$$

Answered by mathmax by abdo last updated on 02/Jun/20

$$1){\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\left(\ln (2\mathrm{x}+1)\right)}^{\left(\mathrm{k}\right)}{\left(\frac{1}{\mathrm{x}}\right)}^{(\mathrm{n}-\mathrm{k})}$$$$={\mathrm{C}}_{\mathrm{n}}^0\ln (2\mathrm{x}+1)\frac{{(-1)}^{\mathrm{n}}\mathrm{n}!}{{\mathrm{x}}^{\mathrm{n}+1}}+\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\left(\ln (2\mathrm{x}+1)\right)}^{\left(\mathrm{k}\right)}\times \frac{{(-1)}^{\mathrm{n}-\mathrm{k}}(\mathrm{n}-\mathrm{k})!}{{\mathrm{x}}^{\mathrm{n}-\mathrm{k}+1}}$$$$\mathrm{we}\mathrm{have}\frac{\mathrm{d}}{\mathrm{dx}}\ln (2\mathrm{x}+1)=\frac{2}{2\mathrm{x}+1}=\frac{1}{\mathrm{x}+\frac{1}{2}}\Rightarrow \frac{{\mathrm{d}}^{\mathrm{k}}}{{\mathrm{dx}}^{\mathrm{k}}}\ln (2\mathrm{x}+1)={\left(\frac{1}{\mathrm{x}+\frac{1}{2}}\right)}^{(\mathrm{k}-1)}$$$$=\frac{{(-1)}^{\mathrm{k}-1}(\mathrm{k}-1)!}{{(\mathrm{x}+\frac{1}{2})}^{\mathrm{k}}}\Rightarrow$$$${\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)=\frac{{(-1)}^{\mathrm{n}}\mathrm{n}!}{{\mathrm{x}}^{\mathrm{n}+1}}\ln (2\mathrm{x}+1)+\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}\frac{{(-1)}^{\mathrm{k}-1}(\mathrm{k}-1)!}{{(\mathrm{x}+\frac{1}{2})}^{\mathrm{k}}}\times \frac{{(-1)}^{\mathrm{n}-\mathrm{k}}(\mathrm{n}-\mathrm{k})!}{{\mathrm{x}}^{\mathrm{n}-\mathrm{k}+1}}$$$$=\frac{{(-1)}^{\mathrm{n}}\mathrm{n}!}{{\mathrm{x}}^{\mathrm{n}+1}}\ln (2\mathrm{x}+1)+\sum _{\mathrm{k}=1}^{\mathrm{n}}{(-1)}^{\mathrm{n}-1}\frac{\mathrm{n}!}{\mathrm{k}!(\mathrm{n}-\mathrm{k})!}\times \frac{(\mathrm{k}-1)!(\mathrm{n}-\mathrm{k})!}{{\mathrm{x}}^{\mathrm{n}-\mathrm{k}+1}{(\mathrm{x}+\frac{1}{2})}^{\mathrm{k}}}$$$${\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)=\frac{{(-1)}^{\mathrm{n}}\mathrm{n}!}{{\mathrm{x}}^{\mathrm{n}+1}}\ln (2\mathrm{x}+1)+\mathrm{n}!{(-1)}^{\mathrm{n}-1}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}{\mathrm{x}}^{\mathrm{n}-\mathrm{k}+1}{(\mathrm{x}+\frac{1}{2})}^{\mathrm{k}}}$$

Commented by mathmax by abdo last updated on 02/Jun/20

$${\mathrm{f}}^{\left(\mathrm{n}\right)}\left(1\right)=\mathrm{n}!{(-1)}^{\mathrm{n}}\ln \left(3\right)+\mathrm{n}!{(-1)}^{\mathrm{n}-1}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}{\left(\frac{3}{2}\right)}^{\mathrm{k}}}$$$$=\mathrm{n}!{(-1)}^{\mathrm{n}}\ln \left(3\right)+\mathrm{n}!{(-1)}^{\mathrm{n}-1}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}{\left(\frac{2}{3}\right)}^{\mathrm{k}}$$

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