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I'm trying to find a proof that the following definitions of $\pi$ are equivalent:

  1. $\pi$ is the smallest positive real number such that $e^{\pi i}=-1$.

  2. $\pi$ is half the circumference of a circle with radius $1$, namely $$\pi :=\int_{-1}^{1}\frac{1}{\sqrt{1-x^2}}dx.$$

  3. $\pi$ is the area of a circle of radius $1$, namely

$$\pi :=2\int_{-1}^{1}\sqrt{1-x^2}dx.$$

The worst part is that I remember seeing a proof that the first definition was equivalent to one of the other two, yet I don't remember the proof nor where I saw it, and I now can't seem to find it. I would appreciate any help.

Blue
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Sam
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2 Answers2

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Here's a calculus proof that 2 is equivalent to 3.

Integration by parts gives \begin{align} \int_{-1}^1\sqrt{1-x^2}\,dx &=\left[x\sqrt{1-x^2}\right]_{x=-1}^1+\int_{-1}^1\frac{x^2}{\sqrt{1-x^2}}\,dx\\ &=\int_{-1}^1\frac{1-(1-x^2)}{\sqrt{1-x^2}}\,dx\\ &=\int_{-1}^1\frac{dx}{\sqrt{1-x^2}}-\int_{-1}^1\sqrt{1-x^2}\,dx \end{align} and so $$2\int_{-1}^1\sqrt{1-x^2}\,dx=\int_{-1}^1\frac{dx}{\sqrt{1-x^2}}.$$

Angina Seng
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Consider the exponential function $f(z)=e^z=\sum \frac{z^k}{k!}$ which is entire and is the unique such that satisfies $f'=f, f(0)=1$ (just look at the Taylor series conditions implied by $f'=f$) so in particular we get the fundamental property that $f(z+w)=f(z)f(w)$ (both satisfy the same $g'=g, g(0)=f(w)$ so if $f(w)=0, g(z)=f(z+w)=0$ so $f=0$ contradiction, hence $f(z+w)/f(w)$ satisfies the differential equation and intial condtion of $f(z)$ so it is $f(z)$!)

But since the Taylor coefficients are real, $f(\bar z)=\bar f(z)$ so $e^{-ix}= \overline {e^{ix}}, x \in \mathbb R$ and in particular $|e^{ix}|^2=e^{ix}\overline {e^{ix}}=e^{ix}e^{-ix}=e^0=1$ so $f$ maps $i\mathbb R$ into the unit circle $S$ also as a group homomorphism; from here it is immediate that $f(i\mathbb R)=S$ (the circle doesn't have non-trivial connected subgroups for example) and actually $f$ restricted there is periodic since by compacity of $S$ the map cannot be injective and onto so there are $f(ix)=f(iy), x \ne y$ or $f(ia)=1, a=x-y \ne 0$ and choosing the smaller positive such and calling it $2b$ we get $f(2ib)=1, f^2(ib)=1$ so $f(ib)=-1$ as by the group homomorphism property any $f(ia)=1, a \ne 0$ gives a period of $f$ and $b <2b$ so it cannot be such!

We claim that $b$ is the area of the unit circle and half the circumference of same, so giving the equivalence between the 3 definitions (as for example if $c$ is the area say, we know that the $b$ above which is defined independently of $c$ is also the area so $b=c$ etc)

Since by defintion of $2b$ we get a parametrization of the unit circle by $f(it), 0 \le t \le 2b$ as $f$ is then injective on $(0,2b)$, we can apply the usual length formula ($L=\int_0^{2b}|f'(it)|dt$) noting that $|f'(it)|=1$ so getting that the circumference of the unit circle is $2b$

Similarly we easily see that $rf(e^{it})$ is a bijective map from $[\epsilon, 1]\times [0,2b)$ into the annulus with outer circle $S$ and inner circle of radius $\epsilon$ and with Jacobian $r$, so by the usual change of variable formula from $2$d calculus we get that the area of the annulus is $b(1-\epsilon^2)$ so letting $\epsilon \to 0$ we get that the area of the unit circle is $b$ and we are done!

Note that the formulas 2,3 in the OP are obtained by using different parameterizations of the unit circle and disc respectively, so if one wants to get a direct integral equivalence, just apply the change of variable formulas with respect to the parametrizations given by $f(it)=(u(t),v(t))$ and $rf(it)$ respectively

Conrad
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  • What is the formal argument as to why $f(i\mathbb{R}) = S$? I do not see we can define such group homomorphism on $S$ without first having proved precisely that $f(i\mathbb{R}) = S$. – Sam Mar 04 '24 at 15:32
  • not sure I understand the question - we prove that $e^{z+w}=e^ze^w$ using the ODE; we prove that $e^{\bar z}=\overline{ e^{z}}$ using the realness of the Taylor coefficients; but $\overline { ix}=-ix$ when $x$ is real so $e^{-ix}=e^{(\overline{ix})}=\overline {e^{ix}}$ hence $1=e^{ix}e^{-ix}$ by the first property above (as $ix+(-ix)=0$ and $e^0=1$) meand that $e^{ix}\overline {e^{ix}}=1$ hence $|e^{ix}|=1$ – Conrad Mar 04 '24 at 15:45
  • That I undnerstand, but that seems only to show that $f(i\mathbb{R})\subseteq S$. – Sam Mar 04 '24 at 15:48
  • $f(i\mathbb{R})=C$ is a connected (nontrivial) subgroup of $S$ and $S$ has no such but itself which is easy to see as connectdness means the image is an arc with nonzero measure, but a high enough power of the arc (eg the arc $C$ is $\theta \le 2\pi$ in arclength then $C^{[2\pi/\theta]+1})$ has length greater than $2\pi$ so covers the whole $S$ why by the group property it is contained in $f(i\mathbb{R})$ – Conrad Mar 04 '24 at 15:53
  • How are you defining a group structure on $S$? – Sam Mar 04 '24 at 15:54
  • complex multiplication (and again you do not need a definition of $\pi$ or anything - just note that the circle is a compact Jordan curve, has an arclength $P$ and its nontrivial connected subsets are arcs of positive such and squaring doubles the arclength just by integration for example, assuming that arclength of arc is less than half so the square is injective) – Conrad Mar 04 '24 at 15:55