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I've been trying to prove that $\pi > 3$ by using the following definition:

$$\pi = 2\int_{-1}^1{ {\sqrt{1-t^2}}}\, dt$$

Which comes from finding what the area of the unit circle is. (This path can be found in Spivak's Calculus, in case someone wants to read about this topic)

I've done it already using sums and geometry, but I'm having a really bad time trying to find a good starting point, let alone the entire path for this proof.

Any help would be greatly appreciated.

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  • $\begingroup$ Actually $$\pi = 2\int_{-1}^1\sqrt{1-t^2}\,dt<2\int_{-1}^1{ \frac{1}{\sqrt{1-t^2}}}$$ $\endgroup$ – Yiorgos S. Smyrlis Mar 30 '18 at 6:26
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Let $f(t) = \sqrt{1-t^2}$. Since $f$ is even $$ 2 \int_{-1}^{1} f(t)\,dt = 4 \int_{0}^{1} f(t)\,dt $$

Bound $f$ from below with easily integrable functions, like straight lines. For example, in this case, the concatenation of the line segments connecting in order the points $(0, f(0))$, $\left(\frac{1}{2}, f(\frac{1}{2})\right)$, $\left(\frac{1}{2} + \frac{1}{4}, f(\frac{1}{2} + \frac{1}{4})\right)$, $\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8}, f(\frac{1}{2} + \frac{1}{4} + \frac{1}{8})\right)$ and $(1, f(1))$ is sufficient.

approximation by lines below

Let $L(t)$ the concatenation of these lines. Show that $3 < 4 \int_{0}^{1} L(t)\,dt$ and $L(t) \le f(t)$ for $t \in [0,1]$. Then conclude by the integral monotonicity that $$ 3 < 4 \int_{0}^{1} L(t)\,dt \le 4 \int_{0}^{1} f(t)\,dt $$

Actually you can partition the interval the way you prefer and with enough line segments you will be able to conclude the same.

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The definition $\pi=2\int_{-1}^{1}\sqrt{1-x^2}\,dx$ is actually equivalent to $\pi=\Gamma\left(\frac{1}{2}\right)^2$, or to $\pi=2\arcsin(1)$, or to $\pi=6\arcsin\frac{1}{2}$, or to $$ \pi = \sum_{n\geq 0}\frac{3\binom{2n}{n}}{16^n(2n+1)}=\color{red}{3}+\frac{1}{8}+\frac{9}{640}+\frac{15}{7168}+\ldots $$ hence $\pi>3$ or even $\pi>\frac{25}{8}$ is just a consequence of the structure of the Maclaurin series of $\frac{1}{\sqrt{1-x^2}}$ and $\arcsin(x)$. This essentially is Newton's historical approach to the computation of $\pi$, rapidly superseded by Machin's identity $\pi=16\arctan\frac{1}{5}-4\arctan\frac{1}{239}$. Since $\pi=4\int_{0}^{1}\frac{dt}{1+t^2}$ and

$$ \int_{0}^{1}\frac{t^4(1-t)^4}{1+t^2}\,dt = \frac{22}{7}-4\int_{0}^{1}\frac{dt}{1+t^2} $$ we also have $\pi<3+\frac{1}{7}$ (this is the Archimedean approximation).
An Archimedean-like geometric approach is the following: a circle with radius $1$ can be decomposed as the union of an octagon with side length $\sqrt{2-\sqrt{2}}$ and eight circular segments. Such segments can be approximated by parabolic segments, whose area is simply $\frac{2}{3}\text{base}\cdot\text{height}$. The parabolic segments are slightly smaller than the corresponding circle segments, hence the following construction

enter image description here

leads to the lower bound $$\pi > \frac{16}{3}\sqrt{2-\sqrt{2}}-\frac{2}{3}\sqrt{2}= 3.13914757\ldots$$ whose accuracy is comparable with the actual Archimedean approximation. In such context $\pi>3$ simply follows from the fact that the area of a regular dodecagon inscribed in the unit circle is $3$.
The parabolic method applied to the regular dodecagon leads to the nice bound $$ \pi > 4\sqrt{6}-4\sqrt{2}-1 = 3.1411\ldots $$ which also explains the proximity between $\pi$ and $\sqrt{2}+\sqrt{3}$.
Back to sneaky tricks involving hypergeometric series, $$ \sum_{n\geq 1}\frac{1}{n^2(n+1)^2}=\frac{\pi^2-9}{3},\qquad \sum_{n\geq 1}\frac{1}{n^3(n+1)^3}=10-\pi^2 $$ clearly prove that $3<\pi<\sqrt{10}$, or that $\sqrt{78}<2\pi\sqrt{2}<\sqrt{79}$.

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Hint:

It can be shown rigorously that

$$\sqrt{1-t^2}\ge 1-t^2+\frac{t^4(1-t^2)}{0.65}$$

and the integral of the polynomial is $\frac{412}{273}>\frac32$.

enter image description here

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    $\begingroup$ Interesting, what was your approach to come up with said polynomial? $\endgroup$ – Marko A. Rodríguez Mar 30 '18 at 18:06
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    $\begingroup$ @MarkoA.Rodríguez: empirical. I looked at the shape of $\sqrt{1-t^2}-(1-t^2)$ and found a polynomial that fitted. [$1-t^2$ alone is not enough, it lead to a ratio $4/3$, so you need to find extra terms.] $\endgroup$ – Yves Daoust Mar 30 '18 at 18:33
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In the OP $\pi$ is defined as the area of the unit circle, and it is thus larger than the area of any inscribed, in the unit circle, canonical polygon.

We shall show that the area $A_{12}$ of the canonical 12-gon, inscribed in the unit circle is EXACTLY 3, and hence $\pi>3$.

Indeed, $$ A_{12}=12\times T_{12} $$ where $T_{12}$ is the area of the isosceles triangle with sides $1$, $1$ and $2\sin (\pi/12)$. In this triangle: $$ \mathrm{Height}=\cos(\pi/12)=\frac{\sqrt{6}+\sqrt{2}}{4},\quad \mathrm{Basis}=2\sin(\pi/12)=2\cdot\frac{\sqrt{6}-\sqrt{2}}{2} $$

and hence $T_{12}=\frac{1}{4}$ and $A_{12}=3$, and consequently, $\pi>3$.

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    $\begingroup$ Or, the circumference $2 \pi$ of the unit circle is greater than the perimeter $6$ of the inscribed regular hexagon, but that's using a different definition of $\pi$ than the one that the OP specifically stated. $\endgroup$ – dxiv Mar 30 '18 at 6:40
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    $\begingroup$ @dxiv This proof can be rescued if you calculate the area of the triangle joining the vertices $O=(0,0),\ A=(1,0),\ B=(\sqrt{3}/2,1/2)$ using $OA$ as base rather than using $AB$ as base. The other triangles can be easily shown to be SSS-congruent to this one. $\endgroup$ – user1551 Mar 30 '18 at 10:05
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I presume the integral is: $$\pi = 2\int_{-1}^1{ \frac{1}{\sqrt{1-t^2}}}dt$$ Hint: use the standard integral: $$2\int_{-1}^1{ \frac{1}{\sqrt{1-t^2}}}dt=2 \bigg[\arcsin t \bigg]_{-1}^{1}= 2\bigg[ \ arcsin (1) - \arcsin(-1)\ \bigg]>3$$

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    $\begingroup$ I just noticed I had incorrectly written the integral, sorry for the inconvenience. $\endgroup$ – Marko A. Rodríguez Mar 30 '18 at 6:28
  • $\begingroup$ I will provide a second answer with your correct integral $\endgroup$ – Joe Goldiamond Mar 30 '18 at 6:35
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    $\begingroup$ Assuming you know $\arcsin 1 =\frac{\pi}{2}$, you simply get $\pi = \pi$, I don't understand how you compare it to $3$. $\endgroup$ – Alvin Lepik Mar 30 '18 at 7:18
  • $\begingroup$ $\pi <4<2 \int_{-1}^1\frac{1}{\sqrt{1-t^2}}dt$ $\endgroup$ – Yiorgos S. Smyrlis Mar 30 '18 at 7:42
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$$\pi = 2\int_{-1}^1{ {\sqrt{1-t^2}}dt}$$ Use the hint:: $t=\cos u$ to obtain: $$\pi = 2\int{ {\sqrt{(1-\cos^2u}}) du} = -2\int \sin (u)\sin (u)du= -2\int \sin^2(u) du$$ Can you continue from here?

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  • $\begingroup$ Definitely want to try it myself from this point onwards, but could I ask for another small hint on how to make a path for the other side of the inequality? $\endgroup$ – Marko A. Rodríguez Mar 30 '18 at 6:49
  • $\begingroup$ You missed $dt = -\sin u du$ $\endgroup$ – GoodDeeds Mar 30 '18 at 6:57
  • $\begingroup$ @GoodDeeds: Thanks, rushed a little! $\endgroup$ – Joe Goldiamond Mar 30 '18 at 7:04
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    $\begingroup$ This doesn't answer the question. It seems like you're showing that the given integral evaluates to $\pi$, but that is being assumed as a definition. This doesn't show that $\pi > 3$; to do that you would need to bound the integrand somehow. $\endgroup$ – User8128 Mar 30 '18 at 7:13
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See Trapezoidal rule, e.g. at Wikipedia.

You can show the function to be integrated is concave, so any approximation of its graph with a polygonal chain gives you a lower bound of the integral. Try to split the integration interval into some three, four pieces and see what happens.

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  • $\begingroup$ I doubt whether this is in the spirit of the question $\endgroup$ – King Tut Mar 30 '18 at 7:41
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$$\pi = 2\int_{-1}^1{ {\sqrt{1-t^2}}}dt$$ is the area of the unit disk.We can consider the area of regular dodecagon inside the unit circle which is very easy to compute.

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    $\begingroup$ Unfortunately, the area of the hexagon inscribed in the unit circle is $\frac{3\sqrt{3}}{2}<3$. The area of the dodecagon inscribed in the unit circle is 3. $\endgroup$ – Yiorgos S. Smyrlis Mar 30 '18 at 7:40
  • $\begingroup$ thank you @ Yiorgos S. Smyrlis $\endgroup$ – Riemann Mar 30 '18 at 7:43
  • $\begingroup$ thank you @ Yiorgos S. Smyrlis $\endgroup$ – Riemann Mar 30 '18 at 7:43

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