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)


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

$Write down a series expansion for$ $\ln [\frac{1 - 2 x}{{(1 + 2 x)}^{2}}] in ascending powers of x$ $up to and including the term in x^{4} .$ $if x is small that terms in x^{2} and higher powers$ $are negleted show that {(\frac{1 - 2 x}{1 + 2 x})}^{\frac{1}{2 x}} ≅ (1 + x) e^{- 3}$
Commented by mr W last updated on 29/Feb/20

$q u e s t i o n i s w r o n g a g a i n . p l e a s e c h e c k!$ $m a y b e i t i s {(\frac{1 - 2 x}{{(1 + 2 x)}^{2}})}^{\frac{1}{2 x}} ≅ (1 + x) e^{- 3}$
	Answered by mr W last updated on 29/Feb/20

	$\ln [\frac{1 - 2 x}{{(1 + 2 x)}^{2}}] = \ln (1 - 2 x) - 2 \ln (1 + 2 x)$ $= (- 2 x) - \frac{{(- 2 x)}^{2}}{2} + \frac{{(- 2 x)}^{3}}{3} - . . .$ $- 2 [(2 x) - \frac{{(2 x)}^{2}}{2} + \frac{{(2 x)}^{3}}{3} - . . .]$ $= - 6 x + 2 x^{2} - 8 x^{3} + o (x^{3})$ $\frac{1}{2 x} \ln [\frac{1 - 2 x}{{(1 + 2 x)}^{2}}] = - 3 + x - 4 x^{2} + o (x^{2})$ ${(\frac{1 - 2 x}{{(1 + 2 x)}^{2}})}^{\frac{1}{2 x}} = e^{\frac{1}{2 x} \ln \frac{1 - 2 x}{{(1 + 2 x)}^{2}}} = e^{- 3 + x - 4 x^{2} + o (x^{2})}$ $= e^{- 3} e^{x - 4 x^{2} + o (x^{2})}$ $= e^{- 3} (1 + x + o (x))$ $\approx e^{- 3} (1 + x)$
	Commented by Rio Michael last updated on 29/Feb/20

	$t h a n k y o u s i r, y o u a r e r i g h t$
	Commented by peter frank last updated on 29/Feb/20

	$t h a n k y o u$
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