### Determining tolerance/error in Taylor polynomials.

This section treats a simple example of the second kind of question
mentioned above:
‘Given a Taylor polynomial approximation to a function,
expanded at some given point, and given an interval around that
given point, *within what tolerance* does the Taylor polynomial
approximate the function on that interval?’

Let's look at the approximation
$1-{x^2\over 2}+{x^4\over 4!}$ to $f(x)=\cos^x$ on the interval
$[-{1\over 2}, {1\over 2}]$. We might ask *‘Within what
tolerance does this polynomial approximate $\cos x$ on that
interval?’*

To answer this, we first recall that the error term we have after
those first (oh-so-familiar) terms of the expansion of cosine is
$${-\sin c\over 5!}x^5$$
For $x$ in the indicated interval, we want to know the *worst-case
scenario* for the size of this thing. A sloppy but good and simple
*estimate* on $\sin c$ is that $|\sin c|\le 1$, regardless of
what $c$ is. This is a very happy kind of estimate because it's not so
bad and because it doesn't depend at all upon $x$. And the biggest
that $x^5$ can be is $({1\over 2})^5\approx 0.03$. Then the *error
is estimated as*
$$|{-\sin c\over 5!}x^5|\le {1\over 2^5\cdot 5!}\le 0.0003$$
This is not so bad at all!

We could have been a little clever here, taking advantage of
the fact that a lot of the terms in the Taylor expansion of cosine at
$0$ are already zero. In particular, we could *choose* to view the
original polynomial $1-{x^2\over 2}+{x^4\over 4!}$ as *including* the *fifth-degree* term of the Taylor expansion as
well, which simply happens to be zero, so is invisible. Thus, instead
of using the remainder term with the ‘5’ in it, we are actually
entitled to use the remainder term with a ‘6’. This typically will
give a better outcome.

That is, instead of the remainder we had must above, we would have an
error term
$${-\cos c\over 6!}x^6$$
Again, in the *worst-case scenario* $|-\cos c|\le 1$. And still
$|x|\le {1\over 2}$, so we have the *error estimate*
$$|{-\cos c\over 6!}x^6|\le {1\over 2^6\cdot 6!}\le 0.000022$$
This is less than a tenth as much as in the first version.

But what happened here? Are there two different answers to
the question of how well that polynomial approximates the cosine
function on that interval? Of course not. Rather, there were two *approaches* taken by us to *estimate* how well it approximates
cosine. In fact, we still do not know the *exact* error!

The point is that the second estimate (being a little wiser) is *closer*
to the truth than the first. The first estimate is *true*, but is
a *weaker* assertion than we are able to make if we try a little
harder.

This already illustrates the point that ‘in real life’ there is often no single ‘right’ or ‘best’ estimate of an error, in the sense that the estimates that we can obtain by practical procedures may not be perfect, but represent a trade-off between time, effort, cost, and other priorities.

#### Exercises

- How well (meaning ‘within what tolerance’) does $1-x^2/2+x^4/24-x^6/720$ approximate $\cos x$ on the interval $[-0.1,0.1]$?
- How well (meaning ‘within what tolerance’) does $1-x^2/2+x^4/24-x^6/720$ approximate $\cos x$ on the interval $[-1,1]$?
- How well (meaning ‘within what tolerance’) does $1-x^2/2+x^4/24-x^6/720$ approximate $\cos x$ on the interval $[{ -\pi \over 2 },{ \pi \over 2 }]$?

#### Thread navigation

##### Calculus Refresher

#### Similar pages

- Taylor polynomials: formulas
- Classic examples of Taylor polynomials
- Computational tricks regarding Taylor polynomials
- Prototypes: More serious questions about Taylor polynomials
- How large an interval with given tolerance for a Taylor polynomial?
- Achieving desired tolerance of a Taylor polynomial on desired interval
- Integrating Taylor polynomials: first example
- Integrating the error term of a Taylor polynomial: example
- The idea of the derivative of a function
- Derivatives of polynomials
- More similar pages