# Meaning of interval of convergence when approximating functions

Let's say I have a Taylor series approximation, $p(x)$, of a function $f(x)$ at $a$:

$$p(x)=\sum_{n=0}^\infty{\frac{f^{(n)}(a)}{n!}(x-a)^n}$$

And that this Taylor series has a radius of convergence of $r$.

Does the radius of convergence mean, that $\forall x\in(a-r,a+r)$, $p(x)$ approximates $f(x)$ perfectly (assuming that there are infinite terms in the series)? So that as long as $x$ is in the interval of convergence, $p(x)=f(x)$? Also, is there some other meaning to the radius/interval of convergence (in this context)?

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No, convergence does not mean that the series is exact there. For example, the function which is $0$ for $x = 0$ and $e^{-1/x^2}$ for $x \neq 0$ (which is infinitely differentiable) has, when expanded at $x=0$, a Taylor series which has infinite radius of convergence and yet which doesn't approximate the function very well at all.

Sort of conversely, the standard Taylor series expansion for $\ln(1+x)$ diverges once $x > 2$, but we may center our Taylor series differently and it will converge (and equal $\ln(1+x)$). This property is called being analytic.

But if the series converges and the remainder estimates go to $0$ in the limit (as you add more terms), then the Taylor series is equal to the original function.

I wrote a blog post for some of my calculus students about Taylor series before. In section $3$ I introduce a different function whose Taylor series doesn't approximate the function itself, and discuss it a bit.

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Thanks, that blog post was very interesting. So seeing that the interval of convergence doesn't tell anything about how well it approximates the function, what is it useful for? Is it just a meaningless interval that tells when the series sums to a finite value? – Aapeli Jun 16 '14 at 8:22
It is certainly meaningful to know where the series converges. – Philip Hoskins Jun 16 '14 at 9:21

The interval of convergence tells you for which values of $x$ the series converges. However, it doesn't necessarily tell you that it converges to $f\left(x\right) .$ The Taylor series of a function about some point can converge to a different function. Consider

$$f(x) = \begin{cases} e^{-1/x^2}&\text{if } x\not=0\\ 0&\text{if } x=0 \end{cases}.$$

Since the function and each of its derivatives is $0$ at $x=0$, the Taylor series expansion about $a=0$ is zero, but the function isn't.

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