Taylor's inequality giving a large accuracy range

Question

I tried using a basic example that I could graph on desmos, and I'm looking at $f(x)=1/(1-x)$ and seeing how accurate the first-degree Maclaurin series is for when $1/3 \leq x \leq 1/2$. Unless I've done it wrong, it seems to tell me that the accuracy of the Taylor polynomial on this interval is within $2$.

From the graph, the actual maximum difference is $0.5$, but I suppose the inequality gives me a guarantee that I'm within $2$ (even though I could be more accurate in reality?)

Answer 1

What you get from Taylor's formula is that $$ \frac{1}{1-x} - (x+1) = \frac{f''(\xi)}{2}\cdot x^2, \quad \xi \in (0,x), \quad x \in (0,1/2). $$

So, as you say, the maximum error for $x \in(1/3, 1/2)$ is bounded by $\frac{16}{2} \cdot (1/2)^2 = 2$. This is an upper bound for the error, which can be much larger than the actual error. On the other hand, you can also get a lower bound for the error, by substituting $f''(\xi)$ by the smallest value it can take, i.e., $\frac{2}{2}\cdot (1/2)^2 = \frac 14$.

Without any a priori knowledge on the localtion of $\xi$, we can only say that the error will be somewhere between $\frac 14$ and 2.

Answer 2

The first lesson to learn is that if you are going to ignore $[0,\frac13]$ then maybe you should consider expanding the series about some other point closer to your interval of interest.

Anyway, you can improve the naive estimate with more work. For example, start with the integral form remainder (since $f$ is (real-)analytic we might as well use it instead of Schlömilch remainder) $$ R_k(x)=\int_0^x \frac{f^{(k+1)}(t)}{k!}(x-t)^k\,\mathrm{d}t $$ (if you actually do the integration it gives you back what you expect, i.e., $f(x)-f_k(x)$). You need a different strategy to estimate the integral instead of the rough "bound $\lvert f^{(k+1)}(t)\rvert$ by its sup", e.g. splitting the interval $(0,x)$ into a small piece around $0$ where both $(x-t)$ and $f''(t)$ is large, and a larger piece where the integrand is smaller. Of course, if you say evaluate the integral $\int_0^c$ exactly and $\int_c^x$ using naive estimate, then you are effectively doing the same procedure as before but using the Taylor series around $x=c$, which you may or may not want to do.