Why doesn't a Taylor series converge always?

Question

The Taylor expansion itself can be derived from mean value theorems which themselves are valid over the entire domain of the function. Then why doesn't the Taylor series converge over the entire domain? I understand the part about the convergence of infinite series and the various tests. But I seem to be missing something very fundamental here..

Answer 1

It is rather unfortunate that in calc II we teach Taylor series at the same time as we teach Taylor polynomials, all the while not doing a very good job of stressing the distinction between an infinite series and a finite sum. In the process we seem to teach students that Taylor series are a much more powerful tool than they are, and that Taylor polynomials are a much less powerful tool than they are.

The main idea is really the finite Taylor polynomial. The Taylor series is just a limit of these polynomials, as the degree tends to infinity. The Taylor polynomial is an approximation to the function based on its value and a certain number of derivatives at a certain point. The remainder formulae tell us about the error in this approximation. In particular they tell us that higher degree polynomials provide better local approximations to a function.

But the issue is that "local" has a different meaning for different values of the degree $n$. To explain what I mean by that, you can look at the Lagrange remainder, which tells you that the error in an approximation of degree $n$ is

$$\frac{f^{(n+1)}(\xi_n) (x-x_0)^{n+1}}{(n+1)!}$$

where $\xi_n$ is between $x_0$ and $x$. So the ratio of the error with degree $n$ to the error with degree $n-1$ is*

$$\frac{f^{(n+1)}(\xi_n) (x-x_0)}{f^{(n)}(\xi_{n-1}) (n+1)}$$

where similarly $\xi_{n-1}$ is between $x$ and $x_0$. So the error is smaller where this quantity is less than $1$. From this form we can see that if $|x-x_0|<(n+1) \left | \frac{f^{(n)}(\xi_{n-1})}{f^{(n+1)}(\xi_n)} \right |$ then the error with degree $n$ is less than that with degree $n-1$. That looks good because we have that growing factor of $n+1$, but what if $\frac{f^{(n)}(\xi_{n-1})}{f^{(n+1)}(\xi_n)}$ goes to zero, maybe even doing so really fast? Then the interval where the error is reduced by adding another term will shrink, potentially contracting down to just the point of expansion as $n$ tends to infinity.

In other words, if the derivatives near $x_0$ (not necessarily just at $x_0$, because we have to evaluate the derivatives at these $\xi$'s) grow way too fast with $n$, then Taylor expansion has no hope of being successful, even when the derivatives needed exist and are continuous.

*Here I am technically assuming that $f^{(n)}(\xi_{n-1}) \neq 0$. This assumption can fail even when $f$ is not a polynomial; consider $f=\sin,x_0=\pi/2,n=1$. But this is a "degenerate" situation in some sense.

Answer 2

One of the intuitive reasons is that working with functions of real argument we do not care about their singularities in the complex plane. However these do restrict the domain of convergence.

The simplest example is the function $$f(x)=\frac{1}{1+x^2},$$ which can be expanded into Taylor series around $x=0$. The radius of convergence of this series is equal to $1$ because of the poles $x=\pm i$ of $f$ in the complex plane of $x$.

Answer 3

The Taylor expansion is not derived from the mean value theorem. Taylor's expansion is a definition valid for any function which is infinitely differentiable at a point. The various forms for the remainder are derived in various ways. By definition, the remainder function is $R(x)=f(x) - T(x)$ where $f$ is the given function and $T$ is its Taylor expansion (about some point). There is no a-priori guarantee that the Taylor expansion gives any value remotely related to the value of the function, other than at the point of expansion. The various forms for the remainder may be used to obtain bounds on the error which in turn can be used to show convergence at some region, but there is no a-priori reason to expect well-behaved bounds. You simply have a formula for the remainder. The remainder may still be large. Of course, the existence of well-known examples where the Taylor expansion really does not approximate the function at all show that it is hopeless to expect miracles.

Answer 4

As there is few examples I provide one :

$$f(x)=x/(\cosh(x))-xe^{-x}-1/10×x^6/(\sinh(x))^4,g(x)=e^{-x}\frac{\sqrt{1+4x^2}+\sqrt{1+x^2}-2}{2},r(x)=10((f(x)-g(x))+1/12×x^2e^{-2x^2}),t(x)=-0.73(\sqrt{x^2+1/2^{4.85}}-1/2^{4.85/2}),p(x)=-1.5(\sqrt{x^2+1/2^6}-1/8),k(x)=r(x)+t(x)-p(x)$$

I think we have to not separate physics and maths at the 18th century here since the potential wall explain why it fails or converges extremely slowly on the entire positive axis around $x=0$ for $k(x),x>0$