So I got this function and I'm looking for an asymptotic expansion for different values of$\ \beta > 1 $ $\ f(x)=\left(1- \beta \frac{\log \left( \log(x) \right)}{\log(x)} \right)^{\beta}$ as $\ x \to \infty $

Does anyone have any idea on how to go about this? Obviously for very large $\ x$, we go to $\ 1$ very quickly and the first order is obviously $\ 1$, but what about the other orders?

  • $\begingroup$ Use the binomial theorem for $(1+y)^\beta$ then set $y = -\beta \log\log x/\log x$. $\endgroup$ Jul 28 '15 at 2:59

Let the function $f(x)$ be given by

$$f(x)=\left(1-\beta\frac{\log \log x}{\log x}\right)^{\beta}$$

where $x>1$. We wish to find the Taylor series for $f(x)$.

To that end, let $g(t)$ be the function given by


and note that $f(x)=g\left(\beta\frac{\log \log x}{\log x}\right)$.

Straightforward calculation shows that the $n$'th order derivative $g^{(n)}(0)$ of $g(t)$ at $t=0$ can be written as

$$\begin{align} g^{(n)}(0)&=(-1)^n\beta(\beta-1)(\beta-2)\cdots(\beta-n+1)\\\\ &=(-1)^n\frac{\Gamma(\beta+1)}{\Gamma(\beta-n+1)} \tag 1 \end{align}$$

where $\Gamma(\beta)$ is the Gamma Function. Note that we exploited the identity $\Gamma(\beta+1)=\beta\Gamma(\beta)$ repeatedly ($n\,\,$times) to arrive at $(1)$.

Now, the Taylor series for $g(t)$ is

$$g(t)=\sum_{n=0}^{\infty}\frac{\Gamma(\beta+1)}{\Gamma(\beta-n+1)} \frac{(-1)^nt^n}{n!}$$

Simple application of the ratio test show that the series is convergent for all $t$.

Finally, letting $t=\beta\,\frac{\log \log x}{\log x}$, $x>1$ reveals

$$\bbox[5px,border:2px solid #C0A000]{f(x)=\sum_{n=0}^{\infty}\frac{\Gamma(\beta+1)}{\Gamma(\beta-n+1)}\frac{(-1)^n\beta^n}{n!}\left(\frac{\log\log x}{\log x}\right)^n} \tag 2$$

For convenience, we provide the first $4$ terms of $(2)$ as

$$\bbox[5px,border:2px solid #C0A000]{f(x)\sim 1-\beta^2\left(\frac{\log\log x}{\log x}\right)+\frac{\beta^3(\beta-1)}{2!}\left(\frac{\log\log x}{\log x}\right)^2-\frac{\beta^4(\beta-1)(\beta-2)}{3!}\left(\frac{\log\log x}{\log x}\right)^3}$$



$$ \lim_{x \rightarrow \infty} \left( 1 - \beta \frac{\log(\log(x))}{\log(x)} \right)^\beta = \lim_{y \rightarrow \infty} \left( 1 - \beta \frac{\log(y)}{y} \right)^\beta = \lim_{z \rightarrow \infty} \Big( 1 - \beta z \exp(-z) \Big)^\beta $$

  • $\begingroup$ I still don't see how that helps me get different orders. The first order is $\ 1$, but the other orders are the issue. I thought about trying to expand $\ e^{-z}$ as a Taylor Series, but that doesn't hold at $\ x=\infty$ so it doesn't help in the asymptotic expansion. $\endgroup$
    – Suvy
    Jul 28 '15 at 0:42
  • $\begingroup$ Okay so I kept playing around with it and this is what I got: $\ f(x)=(1-\beta ln(ln(x))+\beta (ln(ln(x))^{2}- \beta (ln(ln(x))^{3}/2!+...)^{\beta}$ $\endgroup$
    – Suvy
    Jul 28 '15 at 1:05

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