As we know, $$\ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}$$ when $|x|<1$, but what for a function $\log(1+ae^{bx})$? can we use it here?If not, then how'll we expand it?
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You can substitute and then combine like terms. – Ian Feb 06 '15 at 14:30
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$$\ln(1+ae^{bx})=\ln(1+a+abx+ab^2x^2/2+...\\ =\ln((1+a)(1+\frac{abx}{a+1}+\frac{ab^2x^2}{2(a+1)}+...\\ =\ln(1+a)+\ln(1+\frac{abx}{1+a}+\frac{ab^2x^2}{2(1+a)}+...\\ =\ln(1+a)+\left(\frac{abx}{1+a}+\frac{ab^2x^2}{2(1+a)}\right)-\frac{(abx)^2}{2(1+a)^2}+...$$ The last line is because $\ln(1+z)=z-z^2/2+...$
When $x$ is large, $$\ln(1+ae^{bx})=bx+\ln(a)+\ln\left(1+\frac{e^{-bx}}a\right)\\ =bx+\ln(a)+e^{-bx}/a-e^{-2bx}/(2a^2)+...$$
Empy2
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but how did you expand from 2nd to 3rd line, how can you say $(a[e^{bx}-1])/(1+a)$ – RE60K Feb 06 '15 at 14:39
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I wonder if the result could be $$\log (a+1)+\frac{a b x}{a+1}+\frac{a b^2 x^2}{2 (a+1)^2}-\frac{x^3 \left((a-1) a b^3\right)}{6 (a+1)^3}+O\left(x^4\right)$$ – Claude Leibovici Feb 06 '15 at 14:52
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Lol, there are so many dots that you can't even see the close parentheses because they are infinitely far away. – Mario Carneiro Feb 06 '15 at 14:52