Question Number 83297 by Rio Michael last updated on 29/Feb/20
![Write down a series expansion for ln [((1−2x)/((1+2x)^2 ))] in ascending powers of x up to and including the term in x^4 . if x is small that terms in x^2 and higher powers are negleted show that (((1−2x)/(1+2x)))^(1/(2x)) ≅ (1 + x)e^(−3)](https://www.tinkutara.com/question/Q83297.png)
$$\mathrm{Write}\:\mathrm{down}\:\mathrm{a}\:\mathrm{series}\:\mathrm{expansion}\:\mathrm{for}\: \\ $$$$\:\mathrm{ln}\:\left[\frac{\mathrm{1}−\mathrm{2}{x}}{\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} }\right]\:\mathrm{in}\:\mathrm{ascending}\:\mathrm{powers}\:\mathrm{of}\:\mathrm{x}\: \\ $$$$\mathrm{up}\:\mathrm{to}\:\mathrm{and}\:\mathrm{including}\:\mathrm{the}\:\mathrm{term}\:\mathrm{in}\:{x}^{\mathrm{4}} .\: \\ $$$$\mathrm{if}\:\mathrm{x}\:\mathrm{is}\:\mathrm{small}\:\mathrm{that}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{x}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{higher}\:\mathrm{powers} \\ $$$$\mathrm{are}\:\mathrm{negleted}\:\mathrm{show}\:\mathrm{that}\:\:\:\left(\frac{\mathrm{1}−\mathrm{2}{x}}{\mathrm{1}+\mathrm{2}{x}}\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} \:\cong\:\left(\mathrm{1}\:+\:{x}\right){e}^{−\mathrm{3}} \\ $$$$ \\ $$
Commented by mr W last updated on 29/Feb/20
$${question}\:{is}\:{wrong}\:{again}.\:{please}\:{check}! \\ $$$${maybe}\:{it}\:{is}\:\left(\frac{\mathrm{1}−\mathrm{2}{x}}{\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} \:\cong\:\left(\mathrm{1}\:+\:{x}\right){e}^{−\mathrm{3}} \\ $$
Answered by mr W last updated on 29/Feb/20
![ln [((1−2x)/((1+2x)^2 ))]=ln (1−2x)−2 ln (1+2x) =(−2x)−(((−2x)^2 )/2)+(((−2x)^3 )/3)−... −2[(2x)−(((2x)^2 )/2)+(((2x)^3 )/3)−...] =−6x+2x^2 −8x^3 +o(x^3 ) (1/(2x))ln [((1−2x)/((1+2x)^2 ))]=−3+x−4x^2 +o(x^2 ) (((1−2x)/((1+2x)^2 )))^(1/(2x)) =e^((1/(2x))ln ((1−2x)/((1+2x)^2 ))) =e^(−3+x−4x^2 +o(x^2 )) =e^(−3) e^(x−4x^2 +o(x^2 )) =e^(−3) (1+x+o(x)) ≈e^(−3) (1+x)](https://www.tinkutara.com/question/Q83308.png)
$$\mathrm{ln}\:\left[\frac{\mathrm{1}−\mathrm{2}{x}}{\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} }\right]=\mathrm{ln}\:\left(\mathrm{1}−\mathrm{2}{x}\right)−\mathrm{2}\:\mathrm{ln}\:\left(\mathrm{1}+\mathrm{2}{x}\right) \\ $$$$=\left(−\mathrm{2}{x}\right)−\frac{\left(−\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(−\mathrm{2}{x}\right)^{\mathrm{3}} }{\mathrm{3}}−… \\ $$$$−\mathrm{2}\left[\left(\mathrm{2}{x}\right)−\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\mathrm{2}{x}\right)^{\mathrm{3}} }{\mathrm{3}}−…\right] \\ $$$$=−\mathrm{6}{x}+\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}{x}^{\mathrm{3}} +{o}\left({x}^{\mathrm{3}} \right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{x}}\mathrm{ln}\:\left[\frac{\mathrm{1}−\mathrm{2}{x}}{\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} }\right]=−\mathrm{3}+{x}−\mathrm{4}{x}^{\mathrm{2}} +{o}\left({x}^{\mathrm{2}} \right) \\ $$$$ \\ $$$$\left(\frac{\mathrm{1}−\mathrm{2}{x}}{\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} ={e}^{\frac{\mathrm{1}}{\mathrm{2}{x}}\mathrm{ln}\:\frac{\mathrm{1}−\mathrm{2}{x}}{\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} }} ={e}^{−\mathrm{3}+{x}−\mathrm{4}{x}^{\mathrm{2}} +{o}\left({x}^{\mathrm{2}} \right)} \\ $$$$={e}^{−\mathrm{3}} {e}^{{x}−\mathrm{4}{x}^{\mathrm{2}} +{o}\left({x}^{\mathrm{2}} \right)} \\ $$$$={e}^{−\mathrm{3}} \left(\mathrm{1}+{x}+{o}\left({x}\right)\right) \\ $$$$\approx{e}^{−\mathrm{3}} \left(\mathrm{1}+{x}\right) \\ $$
Commented by Rio Michael last updated on 29/Feb/20
$${thank}\:{you}\:{sir},\:{you}\:{are}\:{right} \\ $$
Commented by peter frank last updated on 29/Feb/20
$${thank}\:{you} \\ $$