By - Swastika Deb
In this article, we are going to prove a simple looking problem that is an infinite series which goes like,
Isn’t it crazy? How did Pi pop in the picture? We are going to prove this problem using simple high school geometry tools and it is very different than the proof Euler gave.
Imagine you are standing on the origin of a number line, with lighthouses on each point that is a unit distance apart. The brightness on each point reduces by the square of that number at each step, why is that so? Let’s check out.
As the distance doubles, the length and the breadth also doubles making the area of the screen 4 times the area of the small screen, hence the brightness reduces to 1/4th. This process continues and is called the “inverse square law”. You might have heard of this if you have studied gravitation or electric field that evenly spreads out in all directions.
Good with this? Lets dive deeper.
Consider a triangle ABC,we have that
So if we place three light sources on A,D and C, the brightness of A and C combined equals D. You can prove this on your own if you want to. (Hint: Equate areas)
Cool so far? Let’s see how we can correlate now.
Now we build a set up.
Brightness on EG from D is same as brightness on CG from B. Similarly brightness on EF from D is same as brightness from A to CF.
This is an application of similar triangles which can be proved. Also I’ll link a geogebra diagram at the end so you can slide it and understand. Now we consider a circle with circumference 2, so the area becomes π^2 /4.
Now construct another circle with circumference 4 as shown in the figure.
Now the brightness on the centre of the big circle equals the brightness of the two extreme points as observed from the bottom point that is π^2 /4.
Now we draw the diameter of the big circle.
And place two light sources on the two points as shown. The brightness on point A is combined of the two extreme points placed on the diameter.
Now we make another diameter as done previously and get the same thing (Note that the diameter in this case is passing through point B).
The light sources are equally spaced due to the symmetry of the figure. Now we take another circle of circumference 8 and placing the light sources on the diametrically opposite points.
Ending up with something like this. Now if we keep doing this till infinity, we get a straight line with light sources being evenly spread (due to symmetry present in the circle).
There is an infinite number of light sources, equally spaced in both directions.
And because the apparent brightness was π /4, 2 that we observed taking circumference 4,8,16. So it will also be true in this limiting case where the circle has an infinite radius.
This gives us an awesome infinite series!
This gives us π^2 /4.(Why the reciprocals of odd 2 squared and not even? Try to figure this out seeing at what distance the light sources are placed on the circle).
Phew, wanna walk a little more? We are very close!
The sum of reciprocals of positive odd squared will be π /8, just dividing it by 2.
We somehow need to bring in the reciprocals of even squared. Well, if we think carefully it’s just the scaled copy of the sum we want.
Consider the sum as
Yesssssssssssssss! We finally get equals π^2 /6!
That was indeed an amazing proof, the way we used simple geometry tools and involved pi in this crazy little problem was thrilling.
Thank you so much for the patience and reading this article.
Geogebra link- https://www.geogebra.org/m/yPExUf7b
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