Why don't taylor series represent the entire function?

Question

Say, I have a continuos function that is infinitely differentiate on the interval $I$.

It can then be written as a taylor series. However, taylor series aren't always completely equal to the function - in other words, they don't necessarily converge for all $x$ in $I$.

Why? The way I think of taylor series is that if you know the position , velocity, acceleration, jolt etc. of a particle at one moment in time, you can calculate its position at any time. Taylor series not converging for all $x$ suggests there's a limitation on this analogy.

So why do taylor series "not" work for some $x$?

Using the particle analogy, described above shouldn't taylor series allow you to find the "location" of the function at any "time"?

Please note, I am not looking for a proof - I'm looking for an intuitive explanation of why taylor series don't always converge for all $x$.

Answer 1

There are actually three cases to consider:

• the taylor series might not converge at all, or
• it may converge but will converge to a function different from the one from which you calculated the coefficients
• Finally it might converge to the function you used to get the coeficients, but will only converges on a small interval, this is then due to the fact that the radius of convergence is smaller than the domain of definition.

For each of these phenomena you should find examples in each textbook on Real Analysis.

Asking why it not always converges is a funny question in my opinion, I think it is way more astounding that the series sometimes actually does converge to the function. In general it is a rather stringent statement to have a power series representation for a function, these functions are called analytic and are just a relatively small set, even in the space of $C^{\infty}$ functions. If they are defined on (an open subset of) $\mathbb{C}$ they are holomrphic, which is a very special property. They are easy to detect, it is sufficient that the function is complex differentiable. The real case is much more complicated, it is usually very hard to show that a given real function is analytic.

Answer 2

May be an example clarify the situation:

Consider $f(x)=e^{-\frac{1}{x^2}}$

$f'(x)=\frac{1}{2x^3}e^{-\frac{1}{x^2}}$ which converges to $0$ as $x\to 0$.

Similarly you can check that $f$ is infintely many times differentiable and that all the derivatives of $f$ at $0$ vanishes.

Therefore, the Taylor series fo $f$ is... the zero function!

The explanation here is that $f$ is "too flat" at zero.

The functions that satisfy your "intuition" are called analytic. So $f$ is not analytic.

Even without this complications: consider a function which is zero for $x<0$ and then start increasing. Like a particle that stay on site for $x<0$ and then suddenly move.

This is not an "analytic" behaviour, but it may happen.

Answer 3

Because the behaviour of a function on an very small interval around a point $x$, i.e. on $[e-\epsilon,e+\epsilon]$ doesn't necessarily determine the whole function.

Take a function $f \,:\, \mathbb{R} \to \mathbb{R}$ which is represented by its taylor series on the whole real line - for example, $\sin x$. Then take a second function, which is differentiable arbitrarily often, and which is zero outside some interval. Such functions are called smooth functions with compact support. An example for such a function is $$ g(x) = \begin{cases} e^{-\frac{1}{1-x^2}} &\text{if $|x| < 1$} \\ 0 &\text{otherwise} \end{cases} $$

Now look at $$ h(x) = f(x) + g(x) $$ Outside of the interval $[-1,1]$, $h(x) = f(x)$, since $g(x)$ is zero. Thus, if you compute the taylor series of $h$ around, say, $2$, you'll find exactly the same taylor series as $f$ yields around $2$. This happens since $h(x) = f(x)$ around $2$ implies that $h^{(n)}(x) = f^{(n)}(x)$ around $2$, and in particular that $$ h^{(n)}(2) = f^{(n)}(2) \text{.} $$ That taylor series cannot, therefore, represent $h$ faithfully on the interval $[-1,1]$ - the taylor series will model $f$, not $h$, on that interval.

Answer 4

A much more basic example than the ones shown already:

Let $f(x) = \begin{cases} 0 \text{ if |x|<1}\\ 1\text{ else}\end{cases}$. We have absolutely no reason to suspect that the taylor series at the origin has anything to do with the values of the function when $|x|>1$. This example just shows that not all functions act like you would expect of a "physical process". We would hope that if we knew everything about the motion of an object in a small time interval, we could predict what it would do in the future: this is basically the goal of physics. But the function above models a completely nonphysical motion: teleportation. Other functions like $e^{\frac{-1}{x^2}}$ have a similarly "nonphysical" feel to them.

Answer 5

My professor used to say:

You might want to do calculus in $\Bbb{R}$, but the functions themselves naturally live in $\Bbb{C}$. Euler was the first to discover that if you don't look at what they do everywhere in the complex plane, you don't really understand their habits.

This is as subjective as it gets, but it has always helped my intuition. In particular, you might think that some function is doing nothing wrong, so it should be analytic. Well, if it does nothing wrong in $\Bbb{R}$, look at what it does in $\Bbb{C}$! If also in $\Bbb{C}$ it does nothing wrong, then it is analytic. If in $\Bbb{C}$ it makes some mess, then you have to be careful also in $\Bbb{R}$. To quote my professor again:

Even in $\Bbb{R}$, and in the most practical and applied problems, you can hear distant echos of the complex behavior of the functions. It's their nature, you can't change it.