Consider $$\log(1-x)$$ where $x=x(n)\rightarrow 1$ when $n\rightarrow \infty$.
What is the asymptotics of this function?
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2This looks a lot simpler than the recent related question whose answer you have accepted. – Ethan Bolker Apr 18 '24 at 14:49
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1$\log(1-x)$ or $\log(1+x)$ ? – Claude Leibovici Apr 18 '24 at 15:09
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@ClaudeLeibovici minus. I am sorry – chloe Apr 18 '24 at 15:18
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@ClaudeLeibovici Thank you very much! – chloe Apr 18 '24 at 17:44
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You are supposed to know the rules of this site. Please show some of your working first, please. – Claude Leibovici Apr 19 '24 at 03:59
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https://math.stackexchange.com/questions/4307784/approximation-of-lnx-in-the-vicinity-of-x-0 – Claude Leibovici Apr 19 '24 at 04:18
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It is not possible to describe the singular behaviour of the logarithm at the origin by simpler functions. – Gary Apr 19 '24 at 04:56
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@Gary I see! I alternatively used the upper and lower bound of $\log x$ near $x=0$. https://rgmia.org/papers/v7n2/pade.pdf – chloe Apr 19 '24 at 07:06
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It is true that $\lim _{x \to 0^ + } x^\varepsilon \log x \to 0$ for any $\varepsilon>0$. – Gary Apr 19 '24 at 07:08
2 Answers
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$$ \log(1-x) = x+ \frac{x^2}{2} + \frac{x^3}{3} + \cdots . $$
At $x=-1$ the alternating harmonic series converges to $\log 2$. The error is bounded by the absolute value of the first omitted term in the partial sum.
The analysis harder at $x= 1$ where you want to know how fast the logarithm approaches $-\infty$. Sorry I'm no help there.
Gary
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Ethan Bolker
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1but Taylor expansion works for $x\in(-1,1)$ and at boundary $\pm 1$ it might be inaccurate? I am not sure – chloe Apr 18 '24 at 14:53
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@chloe I didn't read the question carefully enough. See my edit. I can delete my answer if it's no help. – Ethan Bolker Apr 18 '24 at 15:36
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Since you look at the behaviour of $\log(x)$ when $x \to 0^+$, have a look at this question where I proposed to use $$x\log(x)\sim \frac{5 (x-1) x \left(5 x^3+37 x^2+37 x+5\right)}{6 \left(x^4+16 x^3+36 x^2+16 x+1\right)}$$
If you want a simpler approximation of mine $$x\log(x)\sim x^{\log (e-1)} \, (x-1)$$
Claude Leibovici
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