# Prove identity involving the Tsallis q-logarithm

The natural logarithm and the exponential can both be generalized to a called q-logarithms and q-exponentials.those functions are defined as follows: \begin{eqnarray} \log_q(x) &:=& \frac{x^{1-q} - 1}{1 - q} \\ \exp_q(x) &:=& \left(1 + (1-q) x\right)^{\frac{1}{1-q}} \end{eqnarray} Here $q>0$. In the limit $q\rightarrow 1$ we retrieve the natural log and the exponential respectively. The $q$-functions are mutual inverses, ie it holds $\exp_q(\log_q(x)) = x$ and $\log_q(\exp_q(x)) = x$. However, what happens if we feed the $q$-log into an ordinary exponential or, vice versa, feed the $q$-exponential into an ordinary log? If $q$ is close '' to unity we should be getting something that is close to identity. How close? In order to quantify that I ask to prove the following series expansion in $q$ around unity:

$$\exp\left(\imath \omega \log_q(x)\right) = x^{\imath \omega} \left[ 1 + \sum\limits_{p=2}^\infty \frac{((1-q) \log(x))^p}{p!} \left( \left.\frac{1}{1!} \left(\frac{\imath \omega}{1-q}\right)^1\right|_{p\ge 2} + \left.\frac{1}{2!} \left(\frac{\imath \omega}{1-q}\right)^2(2^p-2-2 p)\right|_{p\ge 4} + \left.\frac{1}{3!} \left(\frac{\imath \omega}{1-q}\right)^3(3^p-3(1+p/2)2^p+3(p^2+p+1))\right|_{p\ge 6} + \cdots \right) \right]$$

I do not fully understand the expansion in the OP; in absence of further details I would like to consider the limit $q\rightarrow 1$ as follows.

Let $x>0$; then $\ln_q(x)=\frac{e^{(1-q)\ln x}-1}{1-q}$ by definition.

To consider the limit $q\rightarrow 1$, we expand $\ln_q x$ around $q=1$, i.e.

$$\ln_q(x)=\frac{e^{(1-q)\ln x}-1}{1-q}=\frac{1}{1-q}\left[(1+(1-q)\ln x+\frac{(1-q)^2}{2!}\ln x+\dots)-1\right]=\\ \ln x + O(1-q).$$

In other words, in the above expansion the variable is $q$. Then

$$\exp(i\omega\ln_q x)=\exp(i\omega\ln x+ O(1-q))=\exp(O(1-q))\exp(i\omega\ln x)\rightarrow \exp(i\omega\ln x)$$

when $q\rightarrow 1$. We used $\exp(i\omega O(1-q))=\exp(O(1-q))\rightarrow 1$, in the limit $q\rightarrow 1$.