I am trying to read through a paper and have gotten stuck at the following calculation several times. I've left if for a few days and came back to try it again four different times, but still no luck. So I thought I'd ask if anyone here can help me.
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Long story short, We have two equations
\begin{equation} B = \left[ 2f_{1 + \beta}(x) + 3 \right] C \, . k_* \left( \frac{k_*}{k_0} \right)^{- \frac{5}{3}} \qquad \qquad (1) \end{equation}
and
\begin{equation} L \,. B = \left[ f_{\beta}(x) + \frac{3}{5} \right] C \left( \frac{k_*}{k_0} \right)^{- \frac{5}{3}} \qquad \qquad (2) \end{equation}
where
$$ x = \frac{k_{min}}{k_*} $$
and $f_{\beta}(x)$ is defined as
$$ f_{\beta}(x) = \frac{1}{\beta} \left[ 1 - x^{\beta} \right] $$
with $B$, $L$, and $C$ constants for our purposes.
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Then using (1) to eliminate $B$ by substitution into (2) we can solve for $k_*$ to get
$$ k_* = \frac{1}{L} \frac{ \left[ f_{\beta}(x) + \frac{3}{5} \right] }{ \left[ 2f_{1 + \beta}(x) + 3 \right] } $$
Next, substituting this expression for $k_*$ back into (2) we can solve for $k_0^{\frac{5}{3}} C$ to arrive at
\begin{equation} k_0^{\frac{5}{3}} C = \frac{ \left[ f_{\beta}(x) + \frac{3}{5} \right]^{\frac{2}{3}} }{ \left[ 2f_{1 + \beta}(x) + 3 \right]^{\frac{5}{3}} } \left( \frac{1}{L} \right)^{\frac{2}{3}} B \qquad (3) \end{equation}
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Now, this is not how it is presented in the paper. In the paper the author goes straight from (2) to
\begin{equation} k_0^{\frac{5}{3}} C = \frac{ 5^{\frac{7}{3}}}{2^2} h(x) \left( \frac{1}{L} \right)^{\frac{2}{3}} B \qquad \qquad (4) \end{equation}
where $h(x)$ is defined as
\begin{equation} h(x) = \frac{2^2}{ 5^{\frac{7}{3}}} \frac{ \left[ f_{\beta}(x) + \frac{3}{5} \right]^{\frac{2}{3}} }{ \left[ 2f_{1 + \beta}(x) + 3 \right]^{\frac{5}{3}} } \qquad \qquad (5) \end{equation}
In the next step in the paper the author says that for $ \beta \simeq 0 $ this $h(x)$ becomes
\begin{equation} h(x) \simeq \left[ \frac{5}{8} \ln \left( x \right) + \frac{3}{8} \right]^{\frac{2}{3}} \qquad \qquad (6) \end{equation}
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It is this calculation, getting from equation (5) to equation (6), that I am unable to do.
The bit that I don't understand is why the author seems to take the artifical factor of $ \frac{ 5^{\frac{7}{3}}}{2^2} $ out of the definition of $h(x)$.
I have every bit of faith in the author that she does it for very good reasons that I cannot see, and which if I could see, would likely help to calculate (6) from (5).
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I tried taking the limit of $h(x)$ as $\beta \rightarrow 0$. We find that we need to use L'H$\hat{o}$pital's Rule to take the limit of $f(x)$ as $\beta \rightarrow 0$. Doing this I get
$$ \lim_{\beta \rightarrow 0} f_{\beta}(x) = \lim_{\beta \rightarrow 0} \frac{1}{\beta} \left[ 1 - x^{\beta} \right] \overset{L'H\hat{o}pital}{=} \lim_{\beta \rightarrow 0} \frac{0 - \beta x^{\beta - 1 }}{1} = 0 $$
and
$$ \lim_{\beta \rightarrow 0} f_{1 + \beta}(x) = \lim_{\beta \rightarrow 0} \frac{1}{1 + \beta} \left[ 1 - x^{1 + \beta} \right] \overset{L'H\hat{o}pital}{=} \lim_{\beta \rightarrow 0} \frac{0 - (1 + \beta ) x^{\beta}}{0 + 1} = \frac{-x^0}{1} = -1 $$
however this gives me an exact value for $h(x)$, namely
$$ \lim_{\beta \rightarrow 0} h(x) = \frac{5^{\frac{7}{3}}}{2^2} \frac{ \left[ (0) + \frac{3}{5} \right]^{\frac{2}{3}} }{ \left[ 2(-1) + 3 \right]^{\frac{5}{3}} } = \frac{5^{\frac{5}{3}} 3^{\frac{2}{3}} }{2^2} \simeq 7.6 $$
which is not the desired result of equation (6).
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Can anyone else see how the author gets from equation (5) to equation (6).
I'm afraid my analysis is a little rusty since 2nd year undergrad! If it helps, this calculation is in a Thoeretical Physics paper, so I don't believe that rigor is the most important thing.
Also, apologies for the long equation-full question, but I wanted to try to present the problem as best I could, and what I have tried so far.
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Thanks!
I must confess that I have not been able to identify the path to the approximations. It seems to me that there is a first Taylor expansion around $\beta=0$ .
Nevertheless, I found what I suppose to be a mistake since $$\lim_{\beta \rightarrow 0} f_{\beta}(x) = \lim_{\beta \rightarrow 0} \frac{1}{\beta} \left[ 1 - x^{\beta} \right] = -\log (x)$$ $$\lim_{\beta \rightarrow 0} f_{1+\beta}(x) = \lim_{\beta \rightarrow 0} \frac{1}{1+\beta} \left[ 1 - x^{1+\beta} \right] = 1-x$$.
This is due to the fact that the derivative of $x^{\beta}$ with respect to $\beta$ is $x^{\beta} \log (x)$.
I hope and wish this be of some help to you.
Added later to my answer
If I develop $h(x)$ using Taylor, I find $$h(x) \simeq \frac{4 (3-5 \log (x))^{2/3}}{125 (5-2 x)^{5/3}}$$