3

I have been trying to figure out why we simply don't use Taylor Series everywhere to solve integrals. For instance, I have been reading this article Seized opportunities (Notices of the AMS, April 2010), https://www.ams.org/notices/201004/rtx100400476p.pdf . The author says it is difficult to solve some definite integrals, why don't we just use Taylor Series? I don't mean exclusively definite integrals, I may also point out some well-known differential equations, for instance the damped pendulum, which is:

$$\phi^{(2)}(t) + A \phi^{(1)}(t) + B \sin( \phi(t) ) = 0$$

For real $A$ and $B$. Is it because it does not give an analytical answer, in both situations?

EDIT

For instance, let's suppose I have the following complex integral:

$f(z) = \displaystyle\int_{z_0}^z g(s) ds = \displaystyle\int_\Gamma g(s)ds$

Let's also suppose that $g$ is complex analytic and $\Gamma$ is a continuously differentiable path, then could we either:

  1. Use complex power series to rewrite $g$ and then integrate. (Supposing the sums are interchangeable):

$f(z) = \displaystyle\int_\Gamma \displaystyle\sum_{n=0}^\infty \frac{(s-z_1)^n g^{(n)}(z_1)}{n!} ds$

Or 2) Use the FTC and then rewrite $f$ as a power series:

$f(z) = \displaystyle\sum_{n=0}^\infty \frac{(z-z_1)^n \partial_z^n ( \displaystyle\int_{z_0}^z g(s) ds ) }{n!}$

Are these even possible to do? Additionally, what about the case when it is a improper integral as follows:

$f(z) = \lim_{a \to z_0} \displaystyle\int_a^z g(s) ds$

Could we use analytic power series?

Mr. N
  • 516
  • 1
  • 5
  • 17
  • 4
    Taylor series are not necessarily convergent. Many functions are not analytical. We cannot necessarily interchange the order of integration and summation. – Carl Christian Jan 11 '21 at 18:58
  • I would much rather evaluate $\int_a^b \sin x \ dx$ through the FTC than Taylor series. – Randall Jan 11 '21 at 19:00
  • 1
    Sometimes we do use Taylor/Maclaurin/Binomial/other series expansions to approximate integrals as you describe. In particular, it’s standard to approximate the equation of motion for a (damped) pendulum when the angle $\theta$ is small by it’s small-angle approximation, and small angle approximations come from Taylor series (although there are geometric proofs of the small angle approximations as well). Also, in some books/contexts, (complex-valued functions in particular) are sometimes defined as their Taylor/Maclaurin series. – Adam Rubinson Jan 11 '21 at 19:02
  • Taylor expansion is valid in a neighborhood of some value. What value do you use for the damped pendulum equation? – Raffaele Jan 11 '21 at 19:02
  • 1
    The whole realm of definite integrals appearing in practice is not that easy to grasp. How do you suppose to handle something like $\int_0^1\sqrt{-\log x},dx$ using Taylor series? (Let alone the fact that it has a well-known closed form.) Moreover, even if one has to resort to (high precision!) numerical evaluation, Taylor series are most often not the way to follow. – metamorphy Jan 11 '21 at 19:06
  • @Raffaele, I don't have any in mind. – Mr. N Jan 11 '21 at 19:12
  • @Mr.N $\sqrt{\log(1/x)}$ is quite hard to Taylorize, actually – Raffaele Jan 11 '21 at 19:15
  • @metamorphy and Raffaele, indeed. Just seen its closed form and it is quite surprising. – Mr. N Jan 11 '21 at 19:18
  • @CarlChristian , hm interesting. But I have never seen such kind of non-interchangeable summations. May I ask you an example? – Mr. N Jan 11 '21 at 19:20
  • 2
    As for numerical evaluation, there are very robust methods like this one, outperforming Gaussian methods, let alone any Taylor-series-based idea. But even these have their limits. A sum like $\sum_{n=0}^\infty(-1)^n/p_n$, where $p_n$ denotes the $n$-th prime, has a Mellin-type integral representation which seems good, but there's still no known high-precision method to compute it (up to, say, $1000$ digits). – metamorphy Jan 11 '21 at 19:21
  • @metamorphy I've never seen this method before. Thanks for the example! Still on mathematical series, so the real application is to use Taylor series and similar ones just as means to achieve closed form? – Mr. N Jan 11 '21 at 19:27
  • I wouldn't be that restrictive. The theoretical value is clearly broader (see e.g. this answer of mine, where some analytic conclusions are made, but no closed form obtained). At the same time, most likely, your question won't get a satisfactory answer, for the same reason (it's just too broad). – metamorphy Jan 11 '21 at 20:23
  • good question... – Aatmaj Jan 14 '21 at 12:12

0 Answers0