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How do we specifically get to 3.14 from axioms through logic to that particular transcendental number being the case?

I'm guessing it's the sum of an infinite series. But how do we go from "there's a circle" to "this particular infinite series describes the ratio of the diameter to the circumference", other than getting out a tape measure, and noticing that the series seems to produce the same number (sans any proof). I'm looking for the proof of why the circumference/diameter = 3.14... specifically.

Jam
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Step Start
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    Archimedes did this by inscribing and circumscribing polygons and averaging their perimeter. The more sides the polygon has the more accurate an approximation it becomes. – CyclotomicField Feb 01 '20 at 03:18
  • @CyclotomicField But hey do you calculate the area of a polygon. Sure squares hexagons and maybe even $30$ gons (by figuring out $\sin|\cos {\frac {60}5}$) but how in general do you know what to perimeter of an $n$ gon is? – fleablood Feb 01 '20 at 03:51
  • @fleablood not the area, the perimeter. You do it by adding the lengths of the individual sections. He found a doubling rule so that if you had a $n$-gon approximating he could tell you the lengths of the $2n$-gons and build a better approximation. I believe he used a $32$-gon for his approximation which was improved many times by others. – CyclotomicField Feb 01 '20 at 04:01
  • Oops. "area" was a typo. My point is it's easy to say "calculate the perimeter of an $n$ gon and see it's limit as it goes to infinity" but ... we don't actually have a formula for an $n$ gon. (The doubling rule is good though. Archimedes was a pretty sharp cookie.) – fleablood Feb 01 '20 at 04:25
  • @fleablood There are formulae for the perimeter of $n$-gons, e.g., $P=2nR \sin\left(\frac{\pi}{n}\right)$ where $R$ is the polygon's radius. The virtue of Archimedes' doubling rule is that it is iterative and easier to compute. – Jam Feb 01 '20 at 04:39

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Consider the circle of radius $R$ centered at the origin. That is, the set of points such that $x^2+y^2=R^2$. The upper half of the circle is described by $y=f(x)=\sqrt{R^2-x^2}$. So if we call $\mu(R)$ the circle's perimeter, we have (from the arc-length formula) \begin{align} \mu(R)&=2\int_{-R}^R\sqrt{1+f'(x)^2}\,dx=2\int_{-R}^R\sqrt{1+\left(-\frac{x}{\sqrt{R^2-x^2}}\right)^2}\,dx\\ \ \\ &2=\int_{-R}^R\frac{R}{\sqrt{R^2-x^2}}\,dx=2\int_{-R}^R\frac1{1-(x/R)^2}\,dx\\ \ \\ &=2R\,\int_{-1}^1\frac1{\sqrt{1-x^2}}\,dx. \end{align} This in particular tells us that the perimeter is proportional to the radius. If we want to talk in terms of the diameter, we get $$ \mu(D)=D\,\int_{-1}^1\frac1{\sqrt{1-x^2}}\,dx. $$ If follows that the quotient $\mu(D)/D$ is a constant, that we name $\pi$, and is equal to $$ \pi=\int_{-1}^1\frac1{\sqrt{1-x^2}}\,dx. $$ If you didn't know anything else about $\pi$, you could now use approximations of the integral to calculate approximations of $\pi$.

For better convergence of the Riemann sums, one can easily show (via integration by parts) that $$ \int_{-1}^1\frac1{\sqrt{1-x^2}}\,dx=2\int_{-1}^1{\sqrt{1-x^2}}\,dx $$ (which in particular also shows that $\pi$ is the area of the unit disk)

Martin Argerami
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    Correct but probably meaningless to the OP. If they can understand what an integral represents, they'd have probably already known how $\pi$ is found. – Jam Feb 01 '20 at 02:52
  • So you say that OP knows about infinite series but not integrals? And I'm curious how you formally define "length" without using integrals. – Martin Argerami Feb 01 '20 at 02:59
  • In my opinion, infinite series are generally more intuitive than integrals since the notion of 'perpetual addition' is easier to explain than 'continuous addition'. Re. your question about definitions of length w/o integrals, I'd argue it can be done non-formally and intuitionistically (which is often a useful approach when explaining how theories first arose). Bear in mind that $\pi$ was not originally formally defined with any notion of measure or analysis - it relied on intuitionistic arguments like Archimedes' proof with the perimeter of polygons. – Jam Feb 01 '20 at 03:10
  • Of course, but the OP is asking "from axioms through logic", not by measuring circles with a string. – Martin Argerami Feb 01 '20 at 03:11
  • But this begs the question how to we put a quantitative value one the integral. We do we know $3.14 < integral < 3.15$ etc? – fleablood Feb 01 '20 at 03:53
  • @fleablood: Numerical Analysis is a big thing. Approximating an integral via Riemann sums is a super well-known subject. Already with the crude midpoint rule you can get decent values. – Martin Argerami Feb 01 '20 at 04:15
  • I'm not sure what you mean by "The upper half is described by". – Step Start Feb 01 '20 at 09:31
  • I meant the upper half of the circle. – Martin Argerami Feb 01 '20 at 09:53
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Using the Archimedes approach. Start with n equally spaced radii, each of length $\frac{1}{2}$. For each pair of radii form a triangle, where the third side is the chord connecting the end points of the radii. The sum of the chord lengths (which approximates the length of the circumference) will $\to \pi$, as n$\to \infty$.

herb steinberg
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  • My brain is extremely small. I can create increasingly many triangles using what you're doing, but I can't generalise the relationship between the length of all the chords and the number of triangles I have. Could someone explain how to create a formula for the length of all the chords added together as a function of the number of triangles? – Step Start Feb 01 '20 at 09:30
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This is a very abbreviated development from Rudin's "Real & Complex Analysis".

Start with the Taylor series for $\exp$. Show that $t \mapsto \exp(it)$ maps the reals onto the unit circle.

Define $\sin, \cos$ as the imaginary & real parts of $\exp$. Show $\cos'= -\sin, \sin'=\cos$. We have $\cos 0 = 1$ and with a little work can show there is a smallest positive number $t_0$ for which $\cos t_0 = 0$. Define $\pi= 2 t_0$.

copper.hat
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  • Thank you for this answer, but I can't make head nor tail of it. First I'd need to know what a Taylor series is. Then I'd need to know how these relate to sin, cos, and complex numbers, as well as apparently derivatives? This is way over my level. Can someone just explain from the level of someone who's just finished precalc how to do this and what I need to know to get it? – Step Start Feb 01 '20 at 09:39