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

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Question Number 83297 by Rio Michael last updated on 29/Feb/20

$$\mathrm{Write}\mathrm{down}\mathrm{a}\mathrm{series}\mathrm{expansion}\mathrm{for}$$$$\ln \left[\frac{1-2x}{{(1+2x)}^2}\right]\mathrm{in}\mathrm{ascending}\mathrm{powers}\mathrm{of}\mathrm{x}$$$$\mathrm{up}\mathrm{to}\mathrm{and}\mathrm{including}\mathrm{the}\mathrm{term}\mathrm{in}{x}^4.$$$$\mathrm{if}\mathrm{x}\mathrm{is}\mathrm{small}\mathrm{that}\mathrm{terms}\mathrm{in}{\mathrm{x}}^2\mathrm{and}\mathrm{higher}\mathrm{powers}$$$$\mathrm{are}\mathrm{negleted}\mathrm{show}\mathrm{that}{\left(\frac{1-2x}{1+2x}\right)}^{\frac{1}{2x}}\cong (1+x)e^{-3}$$$$$$

Commented by mr W last updated on 29/Feb/20

$$questioniswrongagain.pleasecheck!$$$$maybeitis{\left(\frac{1-2x}{{(1+2x)}^2}\right)}^{\frac{1}{2x}}\cong (1+x)e^{-3}$$

Answered by mr W last updated on 29/Feb/20

$$\ln \left[\frac{1-2x}{{(1+2x)}^2}\right]=\ln (1-2x)-2\ln (1+2x)$$$$=(-2x)-\frac{{(-2x)}^2}{2}+\frac{{(-2x)}^3}{3}-\dots$$$$-2[\left(2x\right)-\frac{{\left(2x\right)}^2}{2}+\frac{{\left(2x\right)}^3}{3}-\dots ]$$$$=-6x+2x^2-8x^3+o\left(x^3\right)$$$$\frac{1}{2x}\ln \left[\frac{1-2x}{{(1+2x)}^2}\right]=-3+x-4x^2+o\left(x^2\right)$$$$$$$${\left(\frac{1-2x}{{(1+2x)}^2}\right)}^{\frac{1}{2x}}=e^{\frac{1}{2x}\ln \frac{1-2x}{{(1+2x)}^2}}=e^{-3+x-4x^2+o\left(x^2\right)}$$$$=e^{-3}{e}^{x-4x^2+o\left(x^2\right)}$$$$=e^{-3}(1+x+o\left(x\right))$$$$\approx e^{-3}(1+x)$$

Commented by Rio Michael last updated on 29/Feb/20

$$thankyousir,youareright$$

Commented by peter frank last updated on 29/Feb/20

$$thankyou$$

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