The function $e^x$ has a Maclaurin series $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}.$$
It also has various Taylor series expansions centered at $x$-values other than zero, for example $x=3$: $$e^x = \sum_{n=0}^{\infty} \frac{e^3}{n!} (x-3)^n.$$
If we approximate $e^{1/2}$ using the first series above, our approximation converges to the exact value very quickly (|error| < 0.001 after 4 terms). If we use the second series, it converges more slowly (|error| < 0.001 after 13 terms).
I suspect this happens because $x=\frac{1}{2}$ is closer to zero than it is to 3. Do Taylor series always converge more quickly near the center of convergence? If not, is there a general result that tells us when this happens?
Best Answer
Taylor series is: $$f(x)=f(x_0)+(x-x_0)f^{'}(x_0)+\frac{(x-x_0)^2}{2!}f^{''}(x_0)+....+\frac{(x-x_0)^n}{n!}f^{(n)}(x_0)+R_n$$
where $R_n$ is Lagrange remainder,
$$R_n=\frac{(x-x_0)^{n+1}}{(n+1)!}f^{(n)}(x^*)$$
( here $x^*$ is value between $x_0$ and $x$ )
So, the rate of convergence is depending upon $(x-x_0)$ with respect to $x$, hence this difference will decide the convergence.
For example,
To calculate $e^\frac{1}{2}$, series: $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}.$ is better and
To caculate $e^\frac{5}{2}$, series: $e^x = \sum_{n=0}^{\infty} \frac{e^3}{n!} (x-3)^n.$ is better
(See differences between $x_0$ and $x$ to understand the above statement)
Refer: Taylor Series