Taylor series has bounds, but what about Laurent series?

Question

For a Taylor series, we have Lagrange remainders:

$$\left|f(x)-\sum_{k=0}^n\frac{f^{(k)}(c)}{k!}(x-a)^k\right|\le\frac{|f^{(n+1)}(\xi)|}{(n+1)!}|x-c|^{n+1},\ |\xi-c|\le x$$

But do we have a similar thing for Laurent series?

$$\left|f(x)-\sum_{k=-n}^na_k(x-c)^k\right|\le~?$$

$$\left|f(x)-\sum_{k=-\infty}^na_k(x-c)^k\right|\le~?$$

$$\left|f(x)-\sum_{k=-n}^{+\infty}a_k(x-c)^k\right|\le~?$$

Such would be fairly useful for solving limits with a squeeze theoerem, for example.

Answer 1

Yes, there are extensions of this idea, although they rely on converting the function to an analytic one (although given that Taylor's Theorem is a quite complicated extension of the Mean Value Theorem, we should perhaps not be too concerned about this). Two possible approaches:

• If $P(z)$ is the principal part of the Laurent series, then $g(z)=f(z)-P(z)$ is analytic, and so has the ordinary sort of Taylor polynomial and remainder, $$ g(z) = \sum_{k=0}^n \frac{g^{(k)}(a)}{k!}(z-a)^k + R_n[g](z), $$ where $R_n[g](z)$ has your favourite expression.
• If $f(z)$ has a pole of order $n$ at $a$, then $h(z)=(z-a)^n f(z)$ is an analytic function, so it has the usual Taylor formula, $$ h(z) = \sum_{k=0}^N \frac{h^{(k)}(a)}{k!}(z-a)^k + R_N[h](z). $$ One may then divide both sides of this by $(z-a)^n$ to find a Taylor-like formula for $f$.

Neither of these are especially useful for a couple of reasons: firstly, complex-analytic/holomorphic functions come with infinitely many derivatives for free, and have Taylor series that converge absolutely inside the disc or annulus of convergence, so there is not much need to having a finite formula that applies to more limiting circumstances. Secondly, integral estimates based on the Residue Theorem tend to be much more useful in complex analysis since one can deform the contour in helpful ways with much more freedom. Such estimates also only need to use $f$, not its derivatives.