### How large an interval with given tolerance for a Taylor polynomial?

This page treats a simple example of the first kind of question
mentioned on a previous page:
‘Given a Taylor polynomial
approximation to a function, expanded at some given point, and given a
required tolerance, *on how large an interval* around the given point
does the Taylor polynomial achieve that tolerance?’

The specific example we'll get to here is *‘For what range of
$x\ge 25$ does $5+{1\over 10}(x-25)$ approximate $\sqrt{x}$ to
within $.001$?’*

Again, with the degree-one Taylor polynomial and
corresponding remainder term, for reasonable functions $f$ we have
$$f(x)=f(x_o)+f'(x_o)(x-x_o)+{f''(c)\over 2!}(x-x_o)^2$$
for some $c$ between $x_o$ and $x$. The
**remainder** term is
$$\hbox{remainder term }= {f''(c)\over 2!}(x-x_o)^2$$
The notation $2!$ means ‘$2$-factorial’, which is just $2$, but which
we write to be ‘forward compatible’ with other things later.

*Again: no, we do not know what $c$ is, except that it is between $x_o$
and $x$*. But this is entirely reasonable, since if we really knew it
exactly then we'd be able to evaluate $f(x)$ exactly and we are
evidently presuming that this isn't possible (or we wouldn't be doing
all this!). That is, we have *limited information* about what $c$
is, which we could view as the limitation on how precisely we can know
the value $f(x)$.

To give an example of how to use this limited information, consider $f(x)=\sqrt{x}$ (yet again!). Taking $x_o=25$, we have \begin{align*} \sqrt{x}&=f(x)=f(x_o)+f'(x_o)(x-x_o)+{f''(c)\over 2!}(x-x_o)^2\\ &=\sqrt{25}+{1\over 2} { 1 \over \sqrt{25 }}(x-25)- {1\over 2!}{1\over 4}{ 1 \over (c)^{3/2 }}(x-25)^2\\ &=5+{1\over 10}(x-25)-{1\over 8}{ 1 \over c^{3/2 }}(x-25)^2 \end{align*} where all we know about $c$ is that it is between $25$ and $x$. What can we expect to get from this?

Well, we have to make a choice or two to get started: let's suppose
that $x\ge 25$ (rather than smaller). Then we can write
$$25\le c\le x$$
From this, because the three-halves-power function is *
increasing*, we have
$$25^{3/2}\le c^{3/2}\le x^{3/2}$$
Taking inverses (with positive numbers) reverses the inequalities: we
have
$$25^{-3/2}\ge c^{-3/2}\ge x^{-3/2}$$
So, *in the worst-case scenario*, the value of $c^{-3/2}$ is at
most $25^{-3/2}=1/125$.

And we can rearrange the equation:
$$\sqrt{x}-[5+{1\over 10}(x-25)]=-{1\over 8}{ 1 \over c^{3/2 }}(x-25)^2$$
Taking absolute values *in order to talk about error*, this is
$$\left|\sqrt{x}-[5+{1\over 10}(x-25)]\right|=\left|{1\over 8}{ 1 \over c^{3/2 }}(x-25)^2\right|$$
Now let's use our **estimate** $|{ 1 \over c^{3/2 }}|\le 1/125$ to write
$$\left|\sqrt{x}-[5+{1\over 10}(x-25)]\right|\le\left|{1\over 8}{1\over 125}(x-25)^2\right|$$

OK, having done this simplification, *now* we can answer questions
like *For what range of $x\ge 25$ does $5+{1\over 10}(x-25)$
approximate $\sqrt{x}$ to within $.001$?* We cannot hope to tell *exactly*, but only to give a range of values of $x$ for which we *can*
be sure *based upon our estimate*. So the question becomes: solve
the inequality
$$\left|{1\over 8}{1\over 125}(x-25)^2\right|\le .001$$
(with $x\ge 25$). Multiplying out by the denominator of $8\cdot 125$
gives (by coincidence?)
$$|x-25|^2\le 1$$
so the solution is $25\le x\le 26$.

So we can conclude that $\sqrt{x}$ is approximated to within $.001$ for all $x$ in the range $25\le x\le 26$. This is a worthwhile kind of thing to be able to find out.

#### Exercises

- For what range of values of $x$ is $x-{ x^3 \over 6 }$ within $0.01$ of $\sin x$?
- Only consider $-1\leq x\leq 1$. For what range of
values of $x$
*inside this interval*is the polynomial $1+x+x^2/2$ within $.01$ of $e^x$? - On how large an interval around $0$ is $1-x$ within $0.01$ of $1/(1+x)$?
- On how large an interval around $100$ is $10+{ x-100 \over 20 }$ within $0.01$ of $\sqrt{x}$?

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##### Calculus Refresher

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