Math Insight

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.


  1. For what range of values of $x$ is $x-{ x^3 \over 6 }$ within $0.01$ of $\sin x$?
  2. 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 $e^x$?
  3. On how large an interval around $0$ is $1-x$ within $0.01$ of $1/(1+x)$?
  4. On how large an interval around $100$ is $10+{ x-100 \over 20 }$ within $0.01$ of $\sqrt{x}$?