expand-1-x-1-using-maclaurins-theorem-and-talyors-formula-

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

$${expand}\: \\ $$$$\left(\mathrm{1}+{x}\right)^{−\mathrm{1}} \\ $$$${using}\:{maclaurins} \\ $$$${theorem}\:{and}\:{talyors} \\ $$$${formula} \\ $$

Commented by jagoll last updated on 03/Apr/20

f(0) = 1 f ′(x)=−(1+x)^(−2) ⇒f ′(0) = −1 f′′(x)= 2(1+x)^(−3) ⇒f′′(0) = 2 f′′′(x)=−6(1+x)^(−4) ⇒f^((3)) (0)=−6 f^((4)) (x)=24(1+x)^(−4) ⇒f^((4)) (0)=24 f(x) = 1 −x+x^2 −x^3 +x^(4 ) +... = Σ_(n=1) ^∞ (−x)^(n−1)

$$\mathrm{f}\left(\mathrm{0}\right)\:=\:\mathrm{1} \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)=−\left(\mathrm{1}+\mathrm{x}\right)^{−\mathrm{2}} \:\Rightarrow\mathrm{f}\:'\left(\mathrm{0}\right)\:=\:−\mathrm{1} \\ $$$$\mathrm{f}''\left(\mathrm{x}\right)=\:\mathrm{2}\left(\mathrm{1}+\mathrm{x}\right)^{−\mathrm{3}} \:\Rightarrow\mathrm{f}''\left(\mathrm{0}\right)\:=\:\mathrm{2} \\ $$$$\mathrm{f}'''\left(\mathrm{x}\right)=−\mathrm{6}\left(\mathrm{1}+\mathrm{x}\right)^{−\mathrm{4}} \:\Rightarrow\mathrm{f}^{\left(\mathrm{3}\right)} \:\left(\mathrm{0}\right)=−\mathrm{6} \\ $$$$\mathrm{f}^{\left(\mathrm{4}\right)} \left(\mathrm{x}\right)=\mathrm{24}\left(\mathrm{1}+\mathrm{x}\right)^{−\mathrm{4}} \Rightarrow\mathrm{f}^{\left(\mathrm{4}\right)} \left(\mathrm{0}\right)=\mathrm{24} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{1}\:−\mathrm{x}+\mathrm{x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{4}\:} +… \\ $$$$=\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{x}\right)^{\mathrm{n}−\mathrm{1}} \\ $$

Commented by jagoll last updated on 03/Apr/20