Asymptotic series of a matrix-valued function.

Question

Consider the following matrix $$f(\lambda)=\left( \frac{\lambda-1}{\lambda + 1} \right)^{\nu \sigma_3} \ \ \ \lambda \in \mathbb{C} \setminus [-1,1]$$ where $\sigma_3=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ is the third pauli matrix and $\nu $ is a negative real number. Clearly $f(\lambda) \to \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $ as $\lambda \to \infty$. But my question is, what is the asymptotic series of $f(\lambda)$ as $\lambda \to \infty$ ?

Any help is appreciated, Thanks.

Answer 1

$f(\lambda)=I_2-2au+2a^2u^2+(-(4/3)a^3-(2/3)a)u^3+((4/3)a^2+(2/3)a^4)u^4+(-(2/5)a-(4/3)a^3-(4/15)a^5)u^5+((46/45)a^2+(8/9)a^4+(4/45)a^6)u^6+(-(2/7)a-(56/45)a^3-(4/9)a^5-(8/315)a^7)u^7+((88/105)a^2+(44/45)a^4+(8/45)a^6+(2/315)a^8)u^8+(-(2/9)a-(3272/2835)a^3-(76/135)a^5-(8/135)a^7-(4/2835)a^9)u^9+O(u^{10})$

where $u=\dfrac{1}{\lambda}$ and $a^{2p}=\nu^{2p}I_2,a^{2p+1}=\nu^{2p+1}\sigma_3$.

EDIT. (Answer to the8thone) $c_1=-2\nu\sigma_3,c_2=2\nu^2I_2,c_3=(-(4/3)\nu^3-(2/3)\nu)\sigma_3$. Where is the problem ?

Answer 2

Set $a = (\lambda - 1)/(\lambda + 1)$, so that $a \to 1$ as $\lambda \to \infty$. Then $$ \begin{align} a^{-\nu\sigma_3} = \exp(-\nu\ln(a)\sigma_3) & = \sum_{k=0}^\infty{\frac{1}{k!}X^k},\quad X\equiv-\nu\ln(a)\sigma_3\\ & = I\left(\sum_{k=0}^\infty{\frac{\left(\nu\ln(a)\right)^{2k}}{(2k)!}}\right) - \sigma_3\left(\sum_{k=0}^\infty{\frac{\left(\nu\ln(a)\right)^{2k+1}}{(2k+1)!}}\right) \\ & = I\left(1 + \mathcal{O}(\ln^2(a))\right) - \sigma_3\left(\nu\ln(a) + \mathcal{O}(\ln^3(a))\right) \\ & \to I\quad (a\to1) \end{align} $$ where we've used the fact that $\sigma_3^{2k} = I$ and $\sigma_3^{2k + 1} = \sigma_3$.

The identity in the first line equating a matrix exponential to a power series is actually the definition of the matrix exponential, and can be thought of as an analytic continuation of the power series of $e^x$ for $x \in \Bbb{C}$.

Answer 3

Using a combination of ideas from above, I can obtain the first few terms by hand :

Using the same notations, let $a=\displaystyle \frac{\lambda-1}{\lambda+1}$, then $$a^{\nu \sigma_3}=e^{\nu \ln(a)\sigma_3}=\sum^{\infty}_{k=0} \frac{(\nu \ln(a)\sigma_3)^k}{k !} = I + \nu \ln(a)\sigma_3 + \frac{( \ln(a))^2 \nu^2}{2 } I+ \frac{( \ln(a))^3 \nu^3 \sigma_3}{6 } + \cdots \ \ \ \ (*)$$

Now let $\zeta=\frac{1}{\lambda}$, so as $\lambda \to \infty$, $\zeta \to 0$, so we have $$\ln(a)=\ln (\frac{\lambda-1}{\lambda+1}) = \ln (\frac{1-\zeta}{1+\zeta}) = \ln(1-\zeta) - \ln (1+\zeta)= - \sum^{\infty}_{j=0} \frac{\zeta^{j+1}}{j+1} - \sum^{\infty}_{j=0} \frac{(-1)^j\zeta^{j+1}}{j+1} $$ $$ = -\sum^{\infty}_{m=0} \frac{2\zeta^{2m+1}}{2m+1} =-\left[ 2 \zeta + \frac{2\zeta^3}{3} + \frac{2\zeta^5}{5} + \cdots \right] = -\left[ 2 \frac{1}{\lambda} + \frac{2}{3}\frac{1}{\lambda^3} + \frac{2}{5} \frac{1}{\lambda^5} + \cdots \right] \ \ \ (**)$$ combining (*) and (**), It is clear that the coefficient of $\displaystyle \frac{1}{\lambda} $ in the asymptotic series of $a^{\nu \sigma_3}$ is $-2\nu \sigma_3$. the coefficient of $\displaystyle \frac{1}{\lambda^2}$ comes from the term $(\ln(a))^2$, so it is $2\nu^2I$, the coefficient of $\displaystyle \frac{1}{\lambda^3}$ partly comes from the term $(\ln(a))^3$, and partly comes from the term $\ln(a)$, so it is $ \displaystyle(-\frac{2}{3} \nu \sigma_3 - \frac{8}{6} \nu^3 \sigma_3)$ and so on ...