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Express $\left|\pi - \dfrac{23}{7}\right|$ without the absolute value symbol.

I know I have to check if $\pi - \dfrac{23}{7}$ is greater than (or equal to) zero, but how can I do it analytically (without a calculator)?

I know that $\pi \gt 3=\dfrac{21}{7}$ but how to compare $\pi$ with $\dfrac{23}{7}$?

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    $\begingroup$ What facts about $\pi$ do you know, or are allowed to use, in answering this question? Because if you know that the first few digits of its decimal expansion are $3.14$, it's pretty easy. $\endgroup$
    – JonathanZ
    Jul 20, 2022 at 16:10
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    $\begingroup$ Hint: $2/7 > 2/8 = 0.25$. This is all you need, assuming you know that $\pi \approx 3.14$. $\endgroup$
    – Doug
    Jul 20, 2022 at 16:11
  • $\begingroup$ If an explanation is not required in your answer (e.g. the question is "answer only" or "multiple choice"), then it follows from the (essentially) "common knowledge fact" that $22/7$ is the best approximation for $\pi$ using one- and two-digit integers that $\pi$ has to be between $21/7$ and $23/7.$ (Why best? Common sense -- if another such approximation was better, then everyone would be learning the other approximation and not the $22/7$ approximation.) $\endgroup$ Jul 20, 2022 at 16:33
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    $\begingroup$ $\int_0^1 \frac{x^4(1-x)^4}{(1+x^2)} = \frac{22}{7} - \pi$ is the integral of a positive function, hence a positive quantity. Therefore $\frac {22}7 > \pi$. The same follows for $\frac {23}7$. I'm not sure this is what you're looking for, it is "analytically" showing what you need. $\endgroup$ Jul 20, 2022 at 16:34
  • $\begingroup$ See this question for a proof of the perimeter bound using the theory of sequences. $\endgroup$ Jul 20, 2022 at 20:07

5 Answers 5

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I will not use any approximations in this answer.

Consider the definite integral $$\int_0^1\frac{t^4(1-t)^4}{1+t^2}dt.$$ Simply expand the numerator using binomial formula and reduce the numerator in terms of the denominator. I’ll skip a few steps for the sake of brevity: $$\int_0^1\frac{t^4(1-t)^4}{1+t^2}dt=\int_0^1\left(-4t^5+t^6+t^4+\frac{4t^6}{1+t^2} \right)dt$$$$= \int_0^1\left(-4t^5+t^6+5t^4-\frac{4t^4}{1+t^2} \right)dt= \int_0^1\left(-4t^5+t^6+5t^4-4t^2+\frac{4t^2}{1+t^2} \right)dt $$$$= \int_0^1\left(-4t^5+t^6+5t^4-4t^2+4-\frac{4}{1+t^2} \right)dt$$$$=\bbox[5px, border:2px solid red]{\frac{22}{7}-\pi.}$$ This is a very nice expression containing both $\dfrac{22}{7}$ and $\pi$.

Now, note that the function $\displaystyle f(t)= \frac{t^4(1-t)^4}{1+t^2}$ is ALWAYS positive for all $t\in (0,1)$. This means that the integral $\displaystyle\int_0^1 \frac{t^4(1-t)^4}{1+t^2} dt$ is also strictly positive. Thus, we get, $$\bbox[5px, border:2px solid gold]{\frac{22}{7}-\pi>0\implies \frac{22}{7}>\pi.}$$

Hence we finally arrive at the desired conclusion: $$\pi<\frac{22}{7}<\frac{23}{7}.$$


Thus, we can write $\Bigg |\pi-\dfrac{23}{7}\Bigg|=\dfrac{23}{7}-\pi$.

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  • $\begingroup$ My mind? Blown. $\endgroup$
    – JonathanZ
    Jul 20, 2022 at 17:12
  • $\begingroup$ Thanks @JonathanZsupportsMonicaC! Actually where I am from, “Prove that $\int_0^1\frac{t^4(1-t)^4}{1+t^2}dt=\frac{22}{7}-\pi$” is a very popular question in worksheets. So all I had to do was to connect the OP’s question to the integral and write the last paragraph by myself. $\endgroup$ Jul 20, 2022 at 19:27
  • $\begingroup$ Not to detract from @insipidintegrator's fine solution (which I upvoted), but it can also be found elsewhere on Math.SE. It is a clever approach to the approximation, and a clearly non-Archimedean one. $\endgroup$
    – Brian Tung
    Jul 21, 2022 at 0:37
  • $\begingroup$ Where does this integral come from? One of the series expansions for $\pi$? It seems too simple to have been conjured from scratch just for an exercise? $\endgroup$ Jul 21, 2022 at 3:25
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    $\begingroup$ @EricSnyder I found this on mathOverflow: mathoverflow.net/questions/67384/… $\endgroup$ Jul 21, 2022 at 15:41
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Here's an idea that, like insipidintegrator's answer, avoids an approximation. Use a circumscribing polygon to upper-bound the circumference of the circle. I use a dodecagon, whose perimeter is

$$ P = 24 \tan \frac{\pi}{12} = 24(2-\sqrt3) $$

enter image description here

We want to show that this $P < 2\left(\frac{23}{7}\right) = \frac{46}{7}$.

\begin{align} 24(2-\sqrt3) < \frac{46}{7} & \Leftrightarrow 84(2-\sqrt{3}) < 23 \\ & \Leftrightarrow 168-84\sqrt{3} < 23 \\ & \Leftrightarrow 145 < 84\sqrt{3} \\ & \Leftrightarrow 21025 < 21168 \end{align}

Then, since $2\pi < P$, we have $\pi < \frac{23}{7}$. It should be pointed out that this does require you to accept that a circumscribing polygon is longer in perimeter than the circumference of the circle.

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You have already said that $\frac{21}{7}$ is 3, and that both $\pi$ and $\frac{23}{7}$ are greater than 3. But, $\frac{23}{7}$ would be $3\frac{2}{7}$, and we know that the first digit of $\frac{2}{7}$ after the decimal point is 2 (because 20 divided by 7 is 2.something). So, since $\pi$ starts with 3.1, it must be less than $\frac{23}{7}$, and thus the expression in question must be less than 0.

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It has been known for more than 22 centuries that $\pi<22/7$ (Archimedes). But if you don't know it, you can do primary school division:

\begin{array}{cccc|l} 2&3&0&0&\underline{7~~~~} \\ &2&0& &328 \\ & &6&0& \\ & & &4 \end{array}

Thus we deduce that $23/7>3.28$

Can you finish?

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We have that

$$\pi - \dfrac{23}{7} <\frac{315}{100}-\dfrac{23}{7}=\frac{2205}{700}-\dfrac{2300}{700}<0$$

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