# 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!)