Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 87245 by redmiiuser last updated on 03/Apr/20

$$expand$$$${(1+x)}^{-1}$$$$usingmaclaurins$$$$theoremandtalyors$$$$formula$$

Commented by jagoll last updated on 03/Apr/20

$$\mathrm{f}\left(0\right)=1$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=-{(1+\mathrm{x})}^{-2}\Rightarrow \mathrm{f}{}^{\prime }\left(0\right)=-1$$$${\mathrm{f}}^{″}\left(\mathrm{x}\right)=2{(1+\mathrm{x})}^{-3}\Rightarrow {\mathrm{f}}^{″}\left(0\right)=2$$$${\mathrm{f}}^{‴}\left(\mathrm{x}\right)=-6{(1+\mathrm{x})}^{-4}\Rightarrow {\mathrm{f}}^{\left(3\right)}\left(0\right)=-6$$$${\mathrm{f}}^{\left(4\right)}\left(\mathrm{x}\right)=24{(1+\mathrm{x})}^{-4}\Rightarrow {\mathrm{f}}^{\left(4\right)}\left(0\right)=24$$$$\mathrm{f}\left(\mathrm{x}\right)=1-\mathrm{x}+{\mathrm{x}}^2-{\mathrm{x}}^3+{\mathrm{x}}^4+...$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{(-\mathrm{x})}^{\mathrm{n}-1}$$

Commented by jagoll last updated on 03/Apr/20

$$\mathrm{x}\ne -1$$

Commented by redmiiuser last updated on 03/Apr/20

$$okmisteriwanttoask$$$$arethereanyconditions$$$$fortheaboveexpansion.$$

Commented by jagoll last updated on 03/Apr/20

$$\mathrm{i}\mathrm{think}\mathrm{no}\mathrm{sir}$$

Commented by redmiiuser last updated on 03/Apr/20

$$Areyou100\mathrm{\%}sure.$$

Commented by Ar Brandon last updated on 03/Apr/20

$$\mid x\mid <1$$

Commented by redmiiuser last updated on 03/Apr/20

$$butwhynot\mid x\mid >1$$

Commented by redmiiuser last updated on 03/Apr/20

$$mrjagollwhy\underset{n=1}{\sum ^n}\ast \ast$$

Commented by jagoll last updated on 03/Apr/20

$$\mathrm{o}\mathrm{yes}\mathrm{it}\mathrm{typo}$$

Commented by redmiiuser last updated on 03/Apr/20

$$thenwhatarethelimits$$$$ofsummation.$$

Commented by redmiiuser last updated on 03/Apr/20

$$cananyonecomment$$

Commented by Joel578 last updated on 03/Apr/20

$$\frac{1}{1+x}=1-x+x^2-x^3+...$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{x}^n$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{x}^n\mathrm{will}\mathrm{converge}\mathrm{if}\mid x\mid <1,\mathrm{otherwise}\mathrm{diverge}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com