Say I have the MacLaurin Series expansion of a function and get \begin{equation} 1+2f(0)c_is_i\theta+o(\theta) \end{equation} and I want to take the log of this function -i.e. $\log[1+2f(0)c_is_i\theta+o(\theta)]$. How would the last term be expressed?
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Use MacLaurin for log. – zkutch Dec 01 '20 at 03:33
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Are we assuming that $\theta$ is small? – paulinho Dec 01 '20 at 03:34
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@paulinho yes! Sorry should have mentioned this. – Carl Dec 01 '20 at 03:34
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Write $a = 2 f(0) c_i s_i$ so your expression is $1 + a \theta + o(\theta)$. We want to compare the log of this expression to $\log (1 + a \theta)$. We have
$$\log (1 + a \theta + o (\theta)) - \log (1 + a \theta) = \log (1 + o \left( \frac{\theta}{1 + a \theta} \right) )$$
and the Taylor series of the logarithm gives $\log (1 + x) = x + O(x^2)$ which gives, altogether,
$$\log (1 + a \theta + o (\theta)) = \log (1 + a \theta) + o \left( \frac{\theta}{1 + a \theta} \right).$$
(This way of writing it is agnostic about whether you want $\theta \to 0$ or $\theta \to \infty$; it works either way. If $\theta \to 0$ it simplifies to $o(\theta)$ and if $\theta \to \infty$ it simplifies to $o(1)$.)
Qiaochu Yuan
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