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

expand (1+x)^(−1) using maclaurins theorem and talyors formula

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

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

x ≠ −1

$$\mathrm{x}\:\neq\:−\mathrm{1} \\ $$

Commented by redmiiuser last updated on 03/Apr/20

ok mister i want to ask are there any conditions for the above expansion.

$${ok}\:{mister}\:{i}\:{want}\:{to}\:{ask} \\ $$$${are}\:{there}\:{any}\:{conditions} \\ $$$${for}\:{the}\:{above}\:{expansion}. \\ $$

Commented by jagoll last updated on 03/Apr/20

i think no sir

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

Commented by redmiiuser last updated on 03/Apr/20

Are you 100% sure.

$${Are}\:{you}\:\mathrm{100\%}\:{sure}. \\ $$

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

∣x∣<1

$$\mid{x}\mid<\mathrm{1} \\ $$

Commented by redmiiuser last updated on 03/Apr/20

but why not ∣x∣>1

$${but}\:{why}\:{not}\:\:\mid{x}\mid>\mathrm{1} \\ $$

Commented by redmiiuser last updated on 03/Apr/20

mr jagoll why Σ_(n=1) ^n ∗∗

$${mr}\:{jagoll}\:{why}\:\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\ast\ast \\ $$

Commented by jagoll last updated on 03/Apr/20

o yes it typo

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

Commented by redmiiuser last updated on 03/Apr/20

then what are the limits of summation.

$${then}\:{what}\:{are}\:{the}\:{limits} \\ $$$${of}\:{summation}. \\ $$

Commented by redmiiuser last updated on 03/Apr/20

cananyone comment

$${cananyone}\:{comment} \\ $$

Commented by Joel578 last updated on 03/Apr/20

(1/(1 + x)) = 1 − x + x^2 − x^3 + ... = Σ_(n=0) ^∞ (−1)^n x^n Σ_(n=0) ^∞ (−1)^n x^n will converge if ∣x∣<1, otherwise diverge

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

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