I am trying to better understand the purpose of the "Radius of Convergence".
For example, consider a Taylor Series of a function initially expanded over a certain point - we know how to calculate the Approximation Error of this Taylor Series at any other point on this function.
When it comes to Power Series, it is my understanding that the Radius of Convergence describes a "bound" over which the Power Series will converge - outside of this "bound", the Power Series is said to diverge.
- Thus, in the case of Taylor Series - does the Radius of Convergence describe a "region" over which the Taylor Approximation has 0 error when approximating the original function?
Is my interpretation correct?
Thanks!
When dealing with power series (which Taylor series are), there is no error. It is simply a function $f$ defined by a series on the form
$$f(x)=\sum_{n=0}^\infty a_nx^n.$$
Now of course this does not always necessarily converge for any choices of the sequence $\{a_n\}_{n\in\mathbb{N}}$ and value of $x$. Indeed, given some such sequence, the convergence depends on $x$. This is why you introduce the idea of a radius of convergence. If $R>0$ is a radius of convergence of the power series $f(x)$, it means that $f(x)$ converges for all $x\in\mathbb{C}$ such that $\lvert x\rvert <R$. You can actually prove that if $f(x)$ converges, then $\lvert x\rvert$ is a radius of convergence of $f(x)$. Notice, however, that there is no mention of an error term. Now you could still somehow relate the convergence of a Taylor series to its error term, and indeed a Taylor series will converge if the error term of its Taylor polynomial vanishes.
Consider the $N$'th order Taylor expansion of the exponential function. We have that
$$e^x=\sum_{n=0}^N\frac{x^n}{n!}+R_N(x),$$
where there error term $R_N$ can be bounded by
$$\lvert R_N(x)\rvert \leq \frac{\lvert x\rvert^{N+1}}{(N+1)!}e^{\lvert x\rvert}$$
using the Lagrange remainder theorem. Now it can easily be seen that for any fixed $x\in\mathbb{R}$,
$$\lim_{N\to\infty}R_n(x)=0.$$
We thus have that for any $x\in\mathbb{R}$,
$$e^x=\lim_{N\to\infty}(e^x-R_N(x))=\lim_{N\to\infty}\sum_{n=0}^N\frac{x^n}{n!}=\sum_{n=0}^\infty\frac{x^n}{n!}.$$
Thus, as the left hand side converges (indeed it is just $e^x$), we have that the right hand side converges, and se we can see how we get that the Taylor series of the exponential function converges precisely because its error term vanishes.