# Basel Problem - Area of $\frac 16$ of Circle with Radius $\sqrt{\pi}$.

There are several proofs to the solution of the well-known Basel Problem, i.e. $$\sum_{n=1}^\infty \frac 1{n^2}=\frac {\pi^2}6$$

Is is possible to create a geometrical interpretation of this identity in the form of the area of $\frac 16$ of a circle with radius $\sqrt{\pi}$?

• I havent actually watched this yet, but 3blue1brown usually has great geometric insights youtube.com/watch?v=d-o3eB9sfls – Elliot G Jun 3 '18 at 17:13
• Suppose there were such an interpretation. Then the right hand side has units of area, which means that the terms on the left side do as well. But then each of the $n$s on the left need to represent something of units of reciprocal length, which may be difficult/unnatural. (I'm not saying it's impossible, but this is an issue you;d have to deal with.) – Mark S. Jun 3 '18 at 17:17
• @ElliotG The 3Blue1Brown video has great insights but it does not take the area as OP has thought of here. Still it is a good video I'd recommend for understanding the Basel problem – The Integrator Jun 3 '18 at 17:18
• It is more likely to have a geometric interpretation of $\zeta(2)$ as the volume of a torus. By creative telescoping it is simple to prove that $\zeta(2)$ is related to $\arcsin^2\left(\frac{1}{2}\right)$. – Jack D'Aurizio Jun 3 '18 at 17:30
• You mean radius $\pi$, right? If the radius is $\sqrt{\pi}$, then the area is $\pi/6$... – Hans Lundmark Jun 3 '18 at 20:31

(For the best expierience- please use a compass)

I. An approximation for $$\sqrt{\pi}$$ : II. How to split a circle into 6 equal part:

EDIT: Epilog and history:

I have start w/ some famous square roots (without a compass): On this way we can win by hand roots like $$\sqrt{3}$$; $$\sqrt{10}$$ since $$=\sqrt{ 3^2+1^2}$$

Note: $$\sqrt{3}<\sqrt{\pi}$$ and $$\pi<\sqrt{10}$$

or the golden ratio- we need add to $$1$$ with the compass the $$\sqrt{5}$$ (or vice versa) and split in the middle:

• What does this have to do with the Basel Problem? – Joe Sep 20 '18 at 19:54
• @Joe : Topic "Area 1/6 of a circle with Radius $\sqrt{\pi}$" - I have show a sample approximation how to win a lenght of $\sqrt{\pi}$- may be as ~$\sqrt{3}$ ( but is required first $\sqrt{2}$) and much better. – Krzysztof Myśliwiec Sep 20 '18 at 20:02
• @Joe I post a solution with the golden ratio – Krzysztof Myśliwiec Sep 24 '18 at 18:22

TOPIC: Area of $$\frac{1}{6}$$ of Circle with Radius $$\sqrt{\pi}$$

Last time I show how to get an easy approximation for radius via Pythagorean theorem: $$(\sqrt{\pi})^2\geq1.7^2+0.5^2$$ and have give a suggestion to do different- let me show for $$\pi\leq\frac{22}{7}$$ (by using intercept theorem for $$\frac{\sqrt{7}}{7}$$) and fully geometrical solving this time:

Topic: Radius $$\sqrt{\pi}$$

Today (and this gonna be my last post- unless I get requests) with the golden ratio:

$$\phi=\frac{1+\sqrt{5}}{2}$$ ; Property : $$\phi^2=\phi+1$$

I found this pythagorean relation: $$(\sqrt{\pi})^2\leq\phi^2+(\frac{\phi}{4})^2+0.6^2$$ Accuracy:  \begin{align} \frac{17}{16}\phi^2+0.36\approx3.141661\\ \pi\approx3.141593 \\ { For-comparison:} \frac{22}{7}\approx3.14\color{red}{2857} \\ \end{align}