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It is $\pi$-day and the internet is full of stories about $\pi$. One story mentions that "an approximation -- $22/7$ -- is used in many calculations."

I have never actually used $22/7$ as an approximation for $\pi$ in a calculation ... or at least I don't have a memory of doing that.

My questions are

  1. Is it true that today $22/7$ is an approximation that is used in many calculations? Are there, for example, any computers that use this approximation? Is $22/7$ in general (in the USA) often used in handcalculations that involve $\pi$?

  2. I could imagine that before the widespread use of calculators, people would actually use $22/7$ as an approximation. Are there any historical examples of this?

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  • $\begingroup$ See e.g. the second line of this page (Wikipedia), and references [2,3,4,5] within: "Some Egyptologists have claimed that the ancient Egyptians used an approximation of π as 22/7 from as early as the Old Kingdom. This claim has met with skepticism." $\endgroup$ – Clement C. Mar 14 '16 at 15:11
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    $\begingroup$ $22/7$ is not accurate enough for many modern purposes. But a very long time ago (read: millenia ago) it was frequently used as an approximation for $\pi$ $\endgroup$ – Justin Benfield Mar 14 '16 at 15:11
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    $\begingroup$ @BigbearZzz To me that sounds like teaching children the concept of $\pi$ completely wrong from the very beginning. $\endgroup$ – molarmass Mar 14 '16 at 15:24
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    $\begingroup$ @molarmass That's sad, isn't it? Go and ask, says, a high school student in Thailand, a smart one while we are at it, what is $\pi$. Chances are that he'll say that it's 3.14159265 or, in a sad case, $\frac{22}7$. 95% won't even know that it's the ratio of circumference over diameter. $\endgroup$ – BigbearZzz Mar 14 '16 at 15:27
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    $\begingroup$ The second question is a better fit for the History of Science and Mathematics site. Why don't you ask there? $\endgroup$ – Rory Daulton Mar 14 '16 at 16:37
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For the first part of your question about whether computers use this approximation, I think the answer is generally "no." Most modern languages contain some sensible numerical constant in a library that has $\pi$ to machine precision. The preference should be to use that. Before that, like old-style Fortran, the "correct" way was to make your own constant using $\pi = 4 \tan^{-1} 1$. (Of course that doesn't mean that no one ever hand-coded 3.14 or 22/7 into their program, but there were always better, more accepted solutions.)

For the second part, there's a list of historical approximations at Wikipedia, for example. It lists an old one going back to Archimedes:

In the 3rd century BCE, Archimedes proved the sharp inequalities 223⁄71 < π < 22⁄7, by means of regular 96-gons (accuracies of 2·10−4 and 4·10−4, respectively).

https://en.wikipedia.org/wiki/Approximations_of_%CF%80#Early_history

If true that shows it used as a bound.

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    $\begingroup$ Why the down-vote? $\endgroup$ – Brick Mar 14 '16 at 15:30
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    $\begingroup$ I observed that the amount of downvotes flying around in this thread is phenomenal. While I understand that the question may not be fully appropriate for this site, it still baffles me that people would throw downvotes everywhere just for that. $\endgroup$ – BigbearZzz Mar 15 '16 at 13:55
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The two better known rational approximations of $\pi$ are $\dfrac{22}7$ ($3$ correct figures) and $\dfrac{355}{113}$ ($7$ correct figures).

Personally I don't use them as I consider that they bring no real benefit over the straight decimal representations $3.14$ and $3.141593$.


These approximations did not arise by accident: they are the most efficient for a given magnitude of the denominator and can be retrieved from the theory continuous fractions.

Repeatedly that the inverse of the fractional part, and you get a sequence of integer parts

$$3,7,15,1,292\dots$$ so that

$$\pi=3+\frac1{7+\frac1{15+\frac1{1+\frac1{292+\cdots}}}}.$$

The first approximations can be drawn from this:

$$3,\frac{22}7,\frac{333}{106},\frac{355}{113},\frac{103993}{33102}\cdots$$

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You're right that 22/7 isn't used in actual calculations today. It's still the first of an interesting sequence of remarkably good approximations, in the following sense.

To approximate $\pi$ using tenths, just look at the beginning of the decimal expansion $\pi \approx 3.1$. You can be sure that you are within $1/20$ of the correct value. With $\pi \approx 22/7$ you can be sure that you are within $1/14 = 0.071428\ldots$, but in fact you're much closer than that: $\pi- 22/7 = 0.003069\ldots$. So $7$ is a very good denominator to use when looking for rational fractions near $\pi$. The next really good approximations are $333/106$ and $355/113$.

These good approximations come from the continued fraction for $\pi$ - see http://mathworld.wolfram.com/PiContinuedFraction.html .

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One mathematical use comes from the continued fraction expansion $$3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{1 + \frac{1}{292 + \frac{1}{1 + \ldots}}}}} $$

The "convergents" of this expansion are the approximations that you get by truncating the continued fraction. The first convergent is $3$, the next is $3 + \frac{1}{7} = \frac{22}{7}$, the next is $3 + \frac{1}{7 + \frac{1}{15}} = \frac{333}{106}$, and so on.

The mathematical importance of the continued fraction convergents $\frac{m}{n}$ of any real number $t$ is that they are very good approximations in a very precise sense: they are the only fractions such that $$\bigl|t - \frac{m}{n}\bigr| < \frac{1}{c \, n^2} $$ where $c = \sqrt{5}$ (thanks to @Ian for the constant which I could not dig out of my brain at the right moment).

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  • $\begingroup$ $c=\sqrt{5}$ (the upper bound is attained when $t=\frac{1+\sqrt{5}}{2}$). $\endgroup$ – Ian Mar 14 '16 at 15:36
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    $\begingroup$ Continued Fraction approximations are guaranteed to have $\left|r-\frac pq\right|\le\frac1{q^2}$ and if $\left|r-\frac pq\right|\le\frac1{2q^2}$, then $\frac pq$ is guaranteed to be a continued fraction approximation for $r$. $\endgroup$ – robjohn Mar 14 '16 at 15:42
  • $\begingroup$ @robjohn Interesting addition, but the first part is not optimal, as the Hurwitz theorem referenced above shows. The second part is quite interesting, though: only the continued fractions can be below $\frac{1}{2q^2}$ but they are actually below $\frac{1}{\sqrt{5} q^2}$. $\endgroup$ – Ian Mar 14 '16 at 16:17
  • $\begingroup$ @Ian: Consider $106^2\left|\pi-\frac{333}{106}\right|=0.9350557349$ Something is amiss. $\endgroup$ – robjohn Mar 14 '16 at 16:24
  • $\begingroup$ @robjohn I seem to have misunderstood the Hurwitz theorem: it actually says that the $t$ in question must be irrational, and it only says that a subsequence of the continued fraction convergents satisfies that inequality. So for instance it holds for $355/113$ with $\pi$ but not for $333/106$. $\endgroup$ – Ian Mar 14 '16 at 16:33
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No, of course computers don't use $\frac{22}{7}$ as an approximation for $\pi$.

But people often do, when they need a quick back-of-the-envelope calculation of something, don't want to pull out a calculator/computer, and only need a few significant digits of precision anyway.

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    $\begingroup$ Do people really use it? When I still took physics classes either I needed to guesstimate an order of magnitude and then $\pi = 3$, or I need a nontrivial amount of significant places and I used a calculator. And in mathematics $\pi = \pi$ of course. I cannot recall ever using $\pi \approx 22/7$. $\endgroup$ – Najib Idrissi Mar 14 '16 at 15:14
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    $\begingroup$ I disapprove those downvotes. $\endgroup$ – Yves Daoust Mar 14 '16 at 15:14
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    $\begingroup$ Why does this answer get so many downvotes? $\endgroup$ – BigbearZzz Mar 14 '16 at 15:15
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    $\begingroup$ @BigbearZzz: All part of my secret plan to earn the Peer Pressure badge, perhaps? $\endgroup$ – hmakholm left over Monica Mar 14 '16 at 15:15
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    $\begingroup$ I guess the downvotes come from the fact that people expect some sort of source for the assertion. It's not a math question where you can prove without doubt that your answer is correct just by writing down a proof, it's a sociology/history/i-don't-know question... A bare assertion is useless. $\endgroup$ – Najib Idrissi Mar 14 '16 at 15:16

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