# The domain of a Taylor polynomial

I recently looked into Taylor polynomials and saw that there is always a limit in the domain when using Taylor polynomials.

For instance, if I was to find Taylor polynomials for the function, $\frac{1}{x-1}$ using the Taylor polynomial function in http://geogebra.com/, it can be seen that with higher order taylor polynomials, it would be a closer approximation of the function but it would always have a limit to it, almost like an asymptote.

Why is there is a limit in the domain that can be modelled using Taylor Polynomials?

• It is not so to speak of a "limit in the domain", as any polynomial has an infinite domain. You'd better say that the approximation degrades when you step away from the evaluation point. – Yves Daoust May 9 '17 at 9:37
• Are you perhaps asking about the radius of convergence of the Taylor series? – Antonio Vargas May 9 '17 at 9:38

That is because you are trying to model something that is not a polynomial with the help of a polynomial. Assuming there is no limit at all, we would have that $\frac{1}{x-1} = T$ is a polynomial of some finite degree, but this is not the case. Hence, the taylor polynomials can only be approximations, they can never exactly give a function that is not a polynomial.

In general, let $f$ for which we want to compute the taylor polynomials at (let's make life easy) the point $0$. We assume here that everything is well-defined. Then we have

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n + R(f)$$ where $R$ is some rest that is left over after the expansion. It can then be shown that $R(f) = 0$ for many "nice" functions. In this case we can see $f(x)$ as a power series - and of course knowing more terms of this series will give better knowledge about the limit.

The reason is simple: a polynomial is unable to model a function everywhere because the asymptotic behaviors differ.

For any polynomial

$$p(x)=\sum_{k=0}^n a_kx^k,$$ the behavior for large $x$ is that of the monomial $a_nx^n$, which goes to infinity. This is absolutely unable to represent a function like $\dfrac1{x-1}$, which tends to $0$.

Now there is an even stronger reason: when the function has a singularity (this is the case with $\dfrac1{x-1}$ at $x=1$), the Taylor approximation will do its best to match the function up to the singularity, by accumulating larger and larger terms.

The price to be paid to fit the singularity is that the approximation becomes meaningless past the singularity.

Also, taking into account that $|x-x_0|^n$ grows symmetrically on both sides of $x_0$, convergence is blocked by the first singularity, say at $x_1$, causing convergence to fail also on the other side, after $2x_0-x_1$.

• Matching upto the singularity is a key feature which is not visible unless we dig deeper via complex analysis. +1 for mentioning this. – Paramanand Singh May 9 '17 at 12:26