0
$\begingroup$

When I read the paper I met the problem in the step expansion in power. We have \begin{align} s(\epsilon)=\frac{A\epsilon^{a}}{bB|\dot\epsilon|}e^{-Be^{b}} \left[1+\frac{a}{bB}\epsilon^{-b}+\frac{a(a-b)}{b^2B^2}\epsilon^{-2b}+..... \right]\\ \end{align} Defining $q=Be^{b}$ \begin{align} q= Z + \frac{a}{b}lnq + \frac{a}{b} \frac{1}{q} + \frac{a(a-2b)}{2b^2} \frac{1}{q^2} + O(q^3) \end{align} Where \begin{align} Z=ln\frac{A}{bB^{1+a/b}|\dot\epsilon|s}=ln(X/s) \end{align} Then solve for q when $Z>>1$ done as an expansion in power of $1/Z$ and $(lnZ)/Z$ - I do not understand how can we do that, can someone do it step by step for me please, we have the result: \begin{align} q \approx &Z+\frac{a}{b}lnZ+\frac{a}{bZ} \left( \frac{a}{b}lnZ+1 \right)\\ & - \frac{a}{2bZ^2} \left[ \frac{a^2}{b^2}ln^2Z +2\frac{a}{b} \left(1- \frac{a}{b} \right)lnZ +\left( 2-3\frac{a}{b} \right) \right]\\&+O(Z^{-1}lnZ)^3 \end{align} The picture is the article and my problem is go from A3 to A5 and from A5 to A6: enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.