This might be a more appropriate question for math.stackexchange, but it's about a problem I'm considering giving my students, so here it goes.

One of the later exercises in Section 7.4 of James Stewart's Calculus textbook (this section is on partial fraction decomposition) states the following:

The rational number $\frac{22}7$ has been used as an approximation for the number $\pi$ since the time of Archimedes. Show that $$ \int_0^1 \frac{x^4(x-1)^4}{x^2+1}\,dx = \frac{22}7-\pi. $$

Now, this isn't a particularly difficult calculation; a fairly straightforward polynomial long division gives you that this integrand is $x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac{4}{x^2+1}$, and that's pretty easy to integrate. But the later exercises in Stewart are usually about some specific insight or deeper conceptual illustration. And mentioning the $\frac{22}7$ approximation for $\pi$ is pretty specific. But what insight is this supposed to give students? Is there a more clever way to solve this problem that uses geometry or numerical approximation in some specific way? Or is this really just a simple rational function integration problem with a little extra dressing? To put it bluntly, what's the point of this particular problem?

  • $\begingroup$ I do not think such exercises have any point in a country students have readily access to computers. $\endgroup$ – ablmf Sep 7 '20 at 17:19
  • 2
    $\begingroup$ I disagree, the specifics don’t, but the technique is. $\endgroup$ – razivo Sep 7 '20 at 18:24
  • 5
    $\begingroup$ See Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$? (asked by Noam D. Elkies, by the way) for a lot of analysis and references for this result. Also of interest is Integral proofs that $355/113 > \pi$ by Stephen Kenneth [Australian Mathematical Society Gazette 32 #4 (September 2005), pp. 263-266]. $\endgroup$ – Dave L Renfro Sep 7 '20 at 19:04
  • 3
    $\begingroup$ As for the purpose of this, I'm virtually certain that, in the case of Stewart's book, this is simply offered as a "stand alone" curious result that students would find interesting. One might think there would be no need to mention the relevance of $22/7,$ but perhaps this fraction is not as ubiquitously known to non-math students as it used to be (even with non-college bound students!) before calculators became commonplace in schools (beginning after the mid 1970s). $\endgroup$ – Dave L Renfro Sep 7 '20 at 19:15
  • $\begingroup$ @DaveLRenfro I hadn't heard about the generalizations to $(x-x^2)^{2n}$ described in this article, that's very interesting that you can get better approximations for larger values of $n$. I guess I just didn't see it as a fascinating stand-alone result. (Not nearly as spectacular as, for example, $\frac \pi 2 = \prod_{n=1}^{\infty} \frac{4n^2}{(2n-1)(2n+1)}$, imo.) But maybe some of my students would disagree... $\endgroup$ – D Ford Sep 7 '20 at 19:59

This is a classic example of transforming an approximation problem from the real line to the polynomial one.
Let’s image we found a really good approximation, $p(x)\approx\frac{4}{x^2+1}$.
$$\int^1_0 p(x)-\frac{4}{x^2+1} = p-\pi$$ Means that $p$ is a really good approximation of $\pi$.
The natural conclusion is to select a few points from $\frac{4}{x^2+1}$ in the range and use the polynomial gotten as $p(x)$ to get a $\pi$ approximation.
In conclusion, if $p(x)$ is a good approximation of $\frac{4}{x^2+1}$ then $\int^1_0 p(x)$ is a good approximation of $\pi$.
Note: $p(x)$ need only be a good approximation in $[0,1]$.

  • 1
    $\begingroup$ Okay... so basically you're saying $p(x) = x^6 - 4x^5 + 5x^4 - 4x^2 + 4$ is a pretty good approximation for $\frac{4}{x^2+1}$ in the $L^1([0,1])$-topology? But how would a student know to attempt to approximate $\frac{4}{x^2+1}$ at all? And even if they did know they should do that, the most sensible way to construct $p(x)$ is with a Taylor series, which is a ways off. Besides, if this problem is about function approximation, why express the integrand as a single rational expression at all? $\endgroup$ – D Ford Sep 7 '20 at 18:57
  • $\begingroup$ Well, first, almost all students will know to construct a polynomial passing through some points, they learn it in high school algebra, this could also be a nice thing to bring up again when you do learn Taylor series. I don’t know why it was chosen to express it as a singular rational function. $\endgroup$ – razivo Sep 7 '20 at 19:22
  • 4
    $\begingroup$ The point is that the integrand is positive, so 22/7 >π. $\endgroup$ – Steven Gubkin Sep 8 '20 at 0:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.