Math Insight

Achieving desired tolerance of a Taylor polynomial on desired interval

This third question is usually the most difficult, since it requires both estimates and adjustment of number of terms in the Taylor expansion: Given a function, given a fixed point, given an interval around that fixed point, and given a required tolerance, find how many terms must be used in the Taylor expansion to approximate the function to within the required tolerance on the given interval.

For example, let's get a Taylor polynomial approximation to $e^x$ which is within $0.001$ on the interval $[-{1\over 2},+{1\over 2}]$. We use $$e^x=1+x+{x^2\over 2!}+{x^3\over 3!}+\ldots+{x^n\over n!}+ {e^c\over (n+1)!}x^{n+1}$$ for some $c$ between $0$ and $x$, and where we do not yet know what we want $n$ to be. It is very convenient here that the $n$th derivative of $e^x$ is still just $e^x$! We are wanting to choose $n$ large-enough to guarantee that $$\left|{e^c\over (n+1)!}x^{n+1}\right|\le 0.001$$ for all $x$ in that interval (without knowing anything too detailed about what the corresponding $c$'s are!).

The error term is estimated as follows, by thinking about the worst-case scenario for the sizes of the parts of that term: we know that the exponential function is increasing along the whole real line, so in any event $c$ lies in $[-{1\over 2},+{1\over 2}]$ and $$|e^c|\le e^{1/2}\le 2$$ (where we've not been too fussy about being accurate about how big the square root of $e$ is!). And for $x$ in that interval we know that $$|x^{n+1}|\le \left({1\over 2}\right)^{n+1}$$ So we are wanting to choose $n$ large-enough to guarantee that $$\left|{e^c\over (n+1)!} ({1\over 2})^{n+1}\right|\le 0.001$$ Since $$\left|{e^c\over (n+1)!} ({1\over 2})^{n+1}\right| \le {2\over (n+1)!}\left({1\over 2}\right)^{n+1}$$ we can be confident of the desired inequality if we can be sure that $${2\over (n+1)!}\left({1\over 2}\right)^{n+1}\le 0.001$$ That is, we want to ‘solve’ for $n$ in the inequality $${2\over (n+1)!}\left({1\over 2}\right)^{n+1}\le 0.001$$

There is no genuine formulaic way to ‘solve’ for $n$ to accomplish this. Rather, we just evaluate the left-hand side of the desired inequality for larger and larger values of $n$ until (hopefully!) we get something smaller than $0.001$. So, trying $n=3$, the expression is $${2\over (3+1)!}\left({1\over 2}\right)^{3+1}={1\over 12\cdot 16}$$ which is more like $0.01$ than $0.001$. So just try $n=4$: $${2\over (4+1)!}\left({1\over 2}\right)^{4+1}={1\over 60\cdot 32}\le 0.00052$$ which is better than we need.

The conclusion is that we needed to take the Taylor polynomial of degree $n=4$ to achieve the desired tolerance along the whole interval indicated. Thus, the polynomial $$1+x+{x^2\over 2}+{x^3\over 3}+{x^4\over 4}$$ approximates $e^x$ to within $0.00052$ for $x$ in the interval $[-{1\over 2},{1\over 2}]$.

Yes, such questions can easily become very difficult. And, as a reminder, there is no real or genuine claim that this kind of approach to polynomial approximation is ‘the best’.

Exercises

1. Determine how many terms are needed in order to have the corresponding Taylor polynomial approximate $e^x$ to within $0.001$ on the interval $[-1,+1]$.
2. Determine how many terms are needed in order to have the corresponding Taylor polynomial approximate $\cos x$ to within $0.001$ on the interval $[-1,+1]$.
3. Determine how many terms are needed in order to have the corresponding Taylor polynomial approximate $\cos x$ to within $0.001$ on the interval $[{ -\pi \over 2 },{ \pi \over 2 }]$.
4. Determine how many terms are needed in order to have the corresponding Taylor polynomial approximate $\cos x$ to within $0.001$ on the interval $[-0.1,+0.1]$.
5. Approximate $e^{1/2}=\sqrt{e}$ to within $.01$ by using a Taylor polynomial with remainder term, expanded at $0$. (Do NOT add up the finite sum you get!)
6. Approximate $\sqrt{101}=(101)^{1/2}$ to within $10^{-15}$ using a Taylor polynomial with remainder term. (Do NOT add up the finite sum you get! One point here is that most hand calculators do not easily give 15 decimal places. Hah!)