Remainder for Laurent expansion The error of a finite-order Taylor expansion can be estimated by various remainders, Cauchy, Lagrange, etc. Is there any similar remainder for Laurent expansion? I tried to search in google, but did not find a result
 A: Yes, there is a remainder for the Laurent expansion, but it's actually the sum of two remainders. Let's see how it's derived:
Let two concentric circles $\mathcal{C}_1, \mathcal{C}_2$ have center $a$, and let $f(z)$ be holomorphic in the ring bounded by these two circles. Using Cauchy's Integral Formula:
\begin{align}
f(z) &= \frac{1}{2{\pi}i} \left(\oint_{\mathcal{C}_2} {\frac{f(w)\,dw}{w-z}} - \oint_{\mathcal{C}_1} {\frac{f(w)\,dw}{w-z}}\right) \\
& = \frac{1}{2{\pi}i} \left(\oint_{\mathcal{C}_2} {\frac{f(w)\,dw}{(w-a)-(z-a)}} + \oint_{\mathcal{C}_1} {\frac{f(w)\,dw}{(z-a)-(w-a)}}\right) \\
& = \frac{1}{2{\pi}i} \left(\oint_{\mathcal{C}_2} {\frac{1}{w-a}}\cdot\frac{1}{1-\frac{z-a}{w-a}}f(w)\,dw \; + \; \oint_{\mathcal{C}_1} {\frac{1}{z-a}}\cdot\frac{1}{1-\frac{w-a}{z-a}}f(w)\,dw\right) \\
& = \frac{1}{2{\pi}i} \left(\oint_{\mathcal{C}_2}\frac{f(w)}{w-a}\cdot \sum_{n=0}^\infty\left(\frac{z-a}{w-a}\right)^n\,dw \; + \; \oint_{\mathcal{C}_1}\frac{f(w)}{z-a}\cdot \sum_{n=0}^\infty\left(\frac{w-a}{z-a}\right)^n\,dw\right)
\end{align}
Now an infinite convergent power series starting at an arbitrary integar $m$ is $\sum_{n=m}^{\infty}ar^n = \frac{ar^m}{1-r}$, So $$\sum_{n=m+1}^{\infty}\left(\frac{z-a}{w-a}\right)^n = \frac{(z-a)^{m+1}}{(w-a)^m(w-z)} \qquad \sum_{n=m+1}^{\infty}\left(\frac{w-a}{z-a}\right)^n = \frac{(w-a)^{m+1}}{(z-a)^m(z-w)}$$
And your remainder is $$\mathrm{R}_{n+1} = \frac{1}{2{\pi}i} \left((z-a)^{n+1}\oint_{\mathcal{C}_2}\frac{f(w)\,dw}{(w-a)^{n+1}(w-z)} \; + \; (z-a)^{-(n+1)}\oint_{\mathcal{C}_1}\frac{f(w)(w-a)^{n+1}\,dw}{z-w}\right)$$
Where the first term is the remainder for the series with exponents greater than or equal to $0$, and the second term is the remainder for the series with negative exponents. It shouldn't be too hard to figure out that $\mathrm{R}_{n+1}(z)$ converges to $0$ for sufficiently large $n$.
