Any formula that polynomial approximation is not possible (e.g. Taylor expansion)?

Question

Polynomial approximation such as Taylor expansion can deal with many popular functions (sin function, cosine function, etc).

I'm curious what kind of functions cannot be approximated by polynomial approximations?

Answer 1

The Stone–Weierstrass theorem says that every continuous function defined on a compact interval of the real line can be uniformly approximated as closely as desired by a polynomial function.

Answer 2

A great example of a function that doesn't play well with Taylor series is $\ln(x)$, or $\log(x)$ if you're not American. The derivative of $\ln(x)$ is $\frac{1}{x}$, so it has to be centered around some point other than zero. Recalling the formula for the Taylor series, we have

$$f(x)=\displaystyle\sum_{n=0}^{\infty}\frac{d^nf}{dx^n}(a)\frac{(x-a)^n}{n!}.$$

From this, we can let $a=1$ and we have

$$\ln(x)=\ln(1)+\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}(n-1)!\frac{(x-1)^n}{n!}=-\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{(x-1)^n}{n}$$

It's easy to see that this sum does not converge for all $x\geq2$. We can center it around different points to get better approximations near those points, but due to the sum's lack of convergence, my statement above is actually false, because it assumes that the sum is convergent. A more true statement would be

$$\ln(x)=-\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{(x-1)^n}{n}\ni 0<x\leq2.$$

This also shows up for the function $x^p$, where $p$ is some non-integer real power. Allow me to demonstrate: using the Taylor Series formual mentioned earlier coupled with the power rule, we have:

$$x^p=\displaystyle\sum_{n=0}^{\infty}\displaystyle\prod_{k=0}^{n-1}(p-k)\frac{(x-1)^n}{n!}.$$

This also fails to converge for sufficiently large values of $x$. Again, we can center it around some point other than $1$, but it ultimately won't converge for all values of $x$.Therefore, this statement also needs some cleaning up before it's true. We can just be lazy here and say

$$x^p\approx\displaystyle\sum_{n=0}^{\infty}\displaystyle\prod_{k=0}^{n-1}(p-k)\frac{(x-1)^n}{n!},$$

because finding the boundaries of convergence for this can be quite difficult. Let me know if you need any further clarification; I hope this helped!

Answer 3

As mentioned, thanks to the Stone-Weierstrass theorem you can uniformly approximate any continuous function on a closed interval of the real line by a polynomial function.

Concerning Taylor series problems occur when your function is non-analytic. A classical example is: $$f(x)=\begin{cases}e^{-1/x^2}&x\neq0\\0&x=0\end{cases}$$ this function is $\mathcal{C}^\infty(\mathbb{R})$ and all its derivatives $f^{(n)}(x)$ are null for $x=0$, hence its Taylor expansion around $x=0$ is the zero function. This is obviously not a "good" approximation of $f$

See Non-analytic smooth function (wikipedia) for further details.