Proof that convergent Taylor Series converge to actual value of function

Question

Taylor series (or Maclaurin Series) are the only way to get values for some functions, such as

$$\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{t^2} dt = \frac{2}{\sqrt{\pi}}\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{n!(2n+1)}.$$

However, while it is easy to show the students convergence of any such series (using the ratio or root tests) and have them find the radius and interval of convergence, the students might legitimately ask: "it converges to a number (if inside the interval of convergence), but how do we know this convergence is to the actual number this function should give if we can't check by another means?"

For a series like $\sin(x)$, we can get the value expected using trigonometry for comparison to the series expansion and show they match, and use this to compare the accuracy of $n$ terms of the expansion. For a function where you cannot simply compare the actual value to the series resultant because the series is the only way to get a result, how do you show that the series converges to the actual function value and not another value? How do you establish how good the approximation is for the first n terms of the series? Especially for functions with huge radii of convergence, why should the students expect derivative information taken around a single number to give accurate values extremely far from that number? How do you give students an intuition for the number of terms needed to get an accurate approximation vs distance from the expansion point?

Answer 1

Taylor series (or Maclaurin Series) are the only way to get values for some functions

This is not true in the example you give. There is at least one other way to get the value of erf, which is to do numerical integration of the integral you wrote down as a definition.

BTW, you don't need to say "Taylor series (or Maclaurin Series)," because a Maclaurin series is a Taylor series.

how do we know this convergence is to the actual number this function should give if we can't check by another means?

Well, if the only information you have about this function is its Taylor series, then you can't determine whether the Taylor series converges to the correct value (at a point inside its radius of convergence) -- because you have no other information about its correct value.

I'm sure there are functions we can define such that nobody on earth can prove any nontrivial facts about exact values of the function. For example, let $F_n$ be the nth Fibonacci number, and define the function

$$f(x)=\sum \frac{x^n}{F_n!}.$$

This function is analytic everywhere on the real line. I don't know, maybe someone can prove something about some exact value of this function other than the trivial fact that $f(0)=1$, but since I made this example up essentially at random, it seems unlikely.

But very few real-world examples are like this. In most cases, we have some reason why we're interested in this function, which implies other things about it. E.g., some books do define $e^x$ in terms of its Taylor series, but then they prove things like $(e^x)'=e^x$ and $e^{x+y}=e^xe^y$ based on that definition. This gives you a body of facts that can all be correlated with one another.

Especially for functions with huge radii of convergence, why should the students expect derivative information taken around a single number to give accurate values extremely far from that number?

In general, they should not expect this. Analytic functions are in some sense just a infinitesimal subset of the set of all functions. (WP gives the following more rigorous statement of this fact: "And in fact the set of functions with a convergent Taylor series is a meager set in the Fréchet space of smooth functions.") But many of the important functions we use a lot in math are ones that have nice properties, and the nice properties are the reason we study those functions to study. Once such nice property is if a function is analytic.

There are certainly techniques for proving whether certain Taylor series converge to certain values, but they may not be appropriate to teach in a second-semester freshman calc course. For example, if I'm remembering my long-ago complex analysis correctly, then the function $1/(x^2+1)$ is going to have a Taylor series about $x=0$ that converges to the correct value throughout its radius of convergence of 1, and this is because it's formed by the composition of functions that are analytic except at $x=\pm i$.

One can certainly say things about the error incurred by truncating a Taylor series, e.g., putting bounds on this error. But I don't think this has much to do with your question, since functions like $\exp(-1/x^2)$ would have small bounds on the truncation error, but the error relative to the desired value is large.

Answer 2

If the power series $\sum_{j=0}^\infty a_j z^j$ converges to some function $f(z)$, then the Maclaurin series of $f(z)$ is $\sum_{j=0}^\infty a_j z^j$.

But the converse need not hold.

It could happen that the Maclaurin series of a function $f(z)$ is $\sum_{j=0}^\infty a_j z^j$, but $\sum_{j=0}^\infty a_j z^j$ converges to some function other than $f(z)$. Calculus textbooks often use the example $$ f(z) = \begin{cases} \exp\left(\frac{-1}{z^2}\right),\qquad &z>0\\ 0,\qquad &z \le 0\end{cases}. $$

Answer 3

Functions whose Taylor series converge to the original function are called analytic. How do you know if a function is analytic? For teaching basic calculus it probably suffices to know that $\exp,\cos,\sin$ are analytic and that analytic functions are closed under sums, products, division, composition, inversion, derivation and anti-derivation. In particular all elementary functions are analytic at every point in the interior of their domain of definition and the function from you example is analytic. https://en.wikipedia.org/wiki/Analytic_function

Answer 4

I agree with your assertion this issue is not dealt with in many introductory calculus texts for a variety of reasons. There are two rather different concepts of convergence at play:

• Convergence of the Taylor series centered at $x_o$ let's say $\displaystyle \sum_{n=0}^{\infty} a_n(x-x_o)^n$. The Interval of Convergence (IOC) is the set of all $x$ for which numerical series $\displaystyle \sum_{n=0}^{\infty} a_n(x-x_o)^n$ converges. For a given $x$ and set of coefficients we have numerous tools to decide the IOC. Ratio or Root Tests cover much ground here.

• Convergence of the Taylor series to the function for all $x$ near $x_o$; let $\displaystyle T_k(x) = \sum_{n=0}^{k} a_n(x-x_o)^n$ does $T_k \rightarrow f$ as $k \rightarrow \infty$ ? Actually, what sort of convergence is this ? In truth, this is not the sort of convergence we actually give careful discussion of in Calculus II (in the typical US curriculum). This is convergence of a sequence of functions. The work around for Calculus II is Taylor's Theorem with remainder. Essentially, it says $$ f(x) = T_k(x)+R_k(x) $$ where $R_k(x) = \frac{f^{(k+1)}(c)}{(k+1)!}(x-x_o)^{k+1}$ for some $c$ between $x$ and $x_o$. The Mean Value Theorem is the basic example of this result and the proof of the general result can follow a very similar path. So, we get to trade the original question of convergence for the easier task of somehow arguing $R_k(x) \rightarrow 0$ as $k \rightarrow \infty$ for appropriate $x$. It is a pain. It is certainly harder than the series convergence and divergence analysis which already kills the given audience. So, many books try to downplay this topic. Personally, I try to prove cosine or sine is analytic because I have typically built sine and cosine from radian measure and limits etc... all without power series and I just need $-1 \leq \cos \theta, \sin \theta \leq 1$. I think the argument for the exponential function is not too bad. But generally, unless I was teaching an honors Calculus II it is something I touch on then walk away from saying we will just assume the function in question is analytic. If the students are shown the example given by Gerald Edgar then they can appreciate the distinction in concepts of convergence. If you show them how to prove analyticity then perhaps they can appreciate why you don't ask them to show it usually. We have many other more pressing issues at this point in calculus: ability to apply convergence tests logically, ability to find Taylor series via non-rediculous methods, mastery of geometric series techniques...

Convergence of a sequence of functions is usually dealt with in a later analysis course where we consider pointwise convergence as well as uniform convergence. Anyway, the problem for Calculus II is we cannot even compare $f$ to its Taylor series unless we have a definition of $f$ which allows us to make the comparison.

One rather slippery solution here:

• If $f$ is defined by a given power series then it necessarily converges to its Taylor expansion.

The above may seem tautological, but if you use power series to formulate the definition of a function then it really is just that simple. However, more generally, to even hope to answer if $f$ is analytic we need some formulation or information about $f$ which allows us to compare $f$ to its Taylor series.

Finally, if you want to see how to bound the error in a given Taylor series for an analytic function then I have a few examples worked out and visualized in my notes. See Pages 204-208 or so

Often we can use the alternating series estimation theorem to get around the sort of arguments I face in the notes. For example, $$ \sin(1) = 1- \frac{1}{6}+ \frac{1}{5!}+ \cdots $$ is an alternating series hence $\sin(1) = 5/6$ to within an error of $1/5!$. Many problems allow this simple run-around.

I wonder, have you thought about showing the derivative of a power series is in fact the derivative of the function as well ? This is actually quite technical and I skipped it for years without a student ever noticing... or at least without noticing and saying something. Salas Hille and Etgen actually does have arguments for this. But, most schools choose lighter fare for their course. After all, retention, enrollment and worse yet what if engineering starts their own course (gulp).