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I’m gonna guess that you have never had the experience of your heart rate increasing in excitement while you are imagining an infinitely large lake with lighthouses around it.
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Well, if you feel anything like I do about math, that is gonna change by the end of this video.
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Take one plus a fourth plus one-ninth plus one sixteenth and so on, where you’re adding the inverses of the next square number.
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What does this sum approach as you keep adding on more and more terms?
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Now, this is a challenge that remained unsolved for 90 years after it was initially posed, until finally it was Euler who found the answer, super surprisingly, to be 𝜋 squared divided by six.
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I mean, isn’t that crazy?
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What is 𝜋 doing here, and why is it squared?
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We don’t usually see it squared.
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In honor of Euler, whose hometown was Basel, this infinite sum is often referred to as the Basel problem.
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But the proof that I’d like to show you is very different from the one that Euler had.
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I’ve said in a previous video that whenever you see 𝜋 show up, there will be some connection to circles.
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And there are those who like to say that 𝜋 is not fundamentally about circles.
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And insisting on connecting equations like these ones with the geometric intuition stems from a stubborn insistence on only understanding 𝜋 in the context where we first discovered it.
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And that’s all well and good.
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But, whatever your own perspective holds as fundamental, the fact is, 𝜋 is very much tied to circles.
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So if you see it show up, there will be a path somewhere in the massive interconnected web of mathematics leading you back to circles in geometry.
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The question is just how long and convoluted that path might be.
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And in the case of the Basel problem, it’s a lot shorter than you might first think.
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And it all starts with light.
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Here’s the basic idea.
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Imagine standing at the origin of a positive number line and putting a little lighthouse on all of the positive integers: one, two, three, four, and so on.
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That first lighthouse has some apparent brightness from your point of view, some amount of energy that your eye is receiving from the light per unit time.
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And let’s just call that a brightness of one.
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For reasons I’ll explain shortly, the apparent brightness of the second lighthouse is one-fourth as much as the first.
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And the apparent brightness of the third is one-ninth as much as the first and then one sixteenth and so on.
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And you can probably see why this is useful for the Basel problem.
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It gives us a physical representation of what’s being asked, since the brightness received from the whole infinite line of lighthouses is gonna be one plus a fourth plus a ninth plus a sixteenth and so on.
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So the result that we are aiming to show is that this total brightness is equal to 𝜋 squared divided by six times the brightness of that first lighthouse.
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And at first that might seem useless.
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I mean, we’re just reasking the same original question.
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But the progress comes from a new question that this framing raises.
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Are there ways that we can rearrange these lighthouses that don’t change the total brightness for the observer?
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And if so, can you show this to be equivalent to a setup that’s somehow easier to compute?
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To start, let’s be clear about what we mean when we reference “apparent brightness” to an observer.
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Imagine a little screen which, maybe, represents the retina of your eye or a digital camera sensor, something like that.
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You could ask, what proportion of the rays coming out of the source hit that screen?
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Or, phrased differently, what is the angle between the ray hitting the bottom of that screen and the ray hitting the top?
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Or, rather, since we should be thinking of these lights as being in three dimensions, it might be more accurate to ask, what is the angle the light covers in both directions perpendicular to the source?
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In spherical geometry, you sometimes talk about the solid angle of a shape, which is the proportion of a sphere it covers as viewed from a given point.
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You see, the first of two places this story where thinking of screens is gonna be useful is in understanding the inverse square law, which is a distinctly three-dimensional phenomenon.
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Think of all of the rays of light hitting a screen one unit away from the source.
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As you double the distance, those rays will now cover an area with twice the width and twice the height.
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So it would take four copies of that original screen to receive the same rays at that distance.
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And so, each individual one receives one-fourth as much light.
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This is the sense in which I mean a light would appear one-fourth as bright two times the distance away.
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Likewise, when you’re three times farther away, you would need nine copies of that original screen to receive the same rays.
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So each individual screen only receives one-ninth as much light.
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And this pattern continues.
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Because the area hit by a light increases by the square of the distance, the brightness of that light decreases by the inverse square of that distance.
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And as I’m sure many of you know, this inverse square law is not at all special to light.
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It pops up whenever you have some kind of quantity that spreads out evenly from a point source, whether that’s sound or heat or a radio signal, things like that.
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And remember, it’s because of this inverse square law that an infinite array of evenly-spaced lighthouses physically implements the Basel problem.
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But again, what we need if we’re gonna make any progress here is to understand how we can manipulate set-ups with light sources like this without changing the total brightness for the observer.
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And the key building block is an especially nice way to transform a single lighthouse into two.
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Think of an observer at the origin of the 𝑥𝑦-plane and a single lighthouse sitting out somewhere on that plane.
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Now, draw a line from that lighthouse to the observer and then another line perpendicular to that one at the lighthouse.
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Now, place two lighthouses where this new line intersects the coordinate axes, which I’ll go ahead and call lighthouse 𝐴 over here on the left and lighthouse 𝐵 on the upper side.
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It turns out, and you’ll see why this is true in just a minute, the brightness that the observer experiences from that first lighthouse is equal to the combined brightness experienced from lighthouses 𝐴 and 𝐵 together.
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And I should say, by the way, that the standing assumption throughout this video is that all lighthouses are equivalent.
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They’re using the same light bulb, emanating the same power, all of that.
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So, in other words, assigning variables to things here, if we call the distance from the observer to lighthouse 𝑎 little 𝑎 and the distance from the observer to lighthouse 𝑏 little 𝑏 and the distance to the first lighthouse ℎ, we have the relation one over 𝑎 squared plus one over 𝑏 squared equals one over ℎ squared.
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This is the much less well-known inverse Pythagorean theorem which some of you may recognize from Mathologer’s most recent and I’ll say most excellent video on the many cousins of the Pythagorean theorem.
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Pretty cool relation, don’t you think?
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And if you’re a mathematician at heart, you might be asking right now how you prove it.
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And there are some straightforward ways where you express the triangles area in two separate ways and apply the usual Pythagorean theorem.
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But there is another quite pretty method that I’d like to briefly outline here that falls much more nicely into our storyline because, again, it uses intuitions of light and screens.
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Imagine scaling down the whole right triangle into a tinier version.
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And think of this miniature hypotenuse as a screen receiving light from the first lighthouse.
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If you reshape that screen to be the combination of the two legs of the miniature triangle, like this, well it still receives the same amount of light, right?
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I mean the rays of light hitting one of those two legs are precisely the same as the rays that hit the hypotenuse.
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Then the key is that the amount of light from the first lighthouse that hits this left side, the limited angle of rays that end up hitting that screen, is exactly the same as the amount of light over here coming from lighthouse 𝐴 which hits that side.
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It’ll be the same angle of rays.
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And symmetrically, the amount of light from the first house hitting the bottom portion of our screen is the same as the amount of light hitting that portion from lighthouse 𝐵.
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Why, you might ask.
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Well, it’s a matter of similar triangles.
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This animation already gives you a strong hint for how it works.
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And we’ve also left a link in the description to a simple GeoGebra applet for those of you who wanna think this through in a slightly more interactive environment.
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And in playing with that, one important fact here that you’ll be able to see is that the similar triangles only apply in the limiting case for a very tiny screen.
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All right, buckle up now cause here’s where things get good.
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We’ve got this inverse Pythagorean theorem, right?
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And that’s gonna let us transform a single lighthouse into two others without changing the brightness experienced by the observer.
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With that in hand and no small amount of cleverness, we can use this to build up the infinite array that we need.
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Picture yourself at the edge of a circular lake directly opposite a lighthouse.
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We’re gonna want it to be the case that the distance between you and the lighthouse along the border of the lake is one.
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So we’ll say the lake has a circumference of two.
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Now, the apparent brightness is one divided by the diameter squared.
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And, in this case, the diameter is that circumference, two, divided by 𝜋.
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So the apparent brightness works out to be 𝜋 squared divided by four.
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Now, for our first transformation, draw a new circle twice as big, so circumference four, and draw a tangent line to the top of the small circle.
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Then replace the original lighthouse with two new ones where this tangent line intersects the larger circle.
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An important fact from geometry that we’ll be using over and over here is that if you take the diameter of a circle and form a triangle with any point on the circle, the angle at that new point will always be 90 degrees.
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The significance of that in our diagram here is that it means the inverse Pythagorean theorem applies.
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And the brightness from those two new lighthouses equals the brightness from the first one; namely, 𝜋 squared divided by four.
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As the next step, draw a new circle twice as big as the last with a circumference eight.
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Now, for each lighthouse, take a line from that lighthouse through the top of the smaller circle, which is the center of the larger circle, and consider the two points where that intersects with the larger circle.
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Again, since this line is a diameter of that large circle, then the lines from those two new points to the observer are gonna form a right angle.
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Likewise, by looking at this right triangle here, whose hypotenuse is the diameter of the smaller circle, you can see that the line from the observer to that original lighthouse is at a right angle, with a new long line that we drew.
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Good news, right?
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Because that means we can apply the inverse Pythagorean theorem.
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And that means that the apparent brightness from the original lighthouse is the same as the combined brightness from the two newer ones.
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And of course, you can do that same thing over on the other side, drawing a line through the top of the smaller circle and getting two new lighthouses on the larger circle.
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And even nicer, these four lighthouses are all gonna be evenly spaced around the lake.
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Why?
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Well, the lines from those lighthouses to the center are at 90-degree angles with each other.
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So since things are symmetric left to right, that means that the distances along the circumference are one, two, two, two, and one.
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All right, you might see where this is going.
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But I wanna walk through this for just one more step.
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You draw a circle twice as big, so circumference of 16 now.
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And for each lighthouse, you draw a line from that lighthouse through the top of the smaller circle, which is the center of the bigger circle.
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And then, create two new lighthouses where that line intersects with the larger circle.
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Just as before, because the long line is a diameter of the big circle, those two new lighthouses make a right angle with the observer, right?
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And just as before, the line from the observer to the original lighthouse is perpendicular to the long line.
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And those are the two facts that justify us in using the inverse Pythagorean theorem.
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But what might not be as clear is that when you do this for all of the lighthouses to get eight new ones on the big lake, those eight new lighthouses are gonna be evenly spaced.
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This is the final bit of geometry proofiness before the final thrust.
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To see this, remember that if you draw lines from two adjacent lighthouses on the small lake to the center, they make a 90-degree angle.
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If, instead, you draw lines to a point anywhere on the circumference of the circle, that’s not between them, the very useful inscribed angle theorem from geometry tells us that this will be exactly half of the angle that they make with the center, in this case 45 degrees.
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But, when we position that new point at the top of the lake, these are the two lines which define the position of the new lighthouses on the larger lake.
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What that means then is that when you draw lines from those eight new lighthouses into the center, they divide the circle evenly into 45-degree angle pieces.
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And that means the eight lighthouses are evenly spaced around the circumference, with a distance of two between each one of them.
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And now, just imagine this thing playing on at every step doubling the size of each circle and transforming each lighthouse into two new ones along a line drawn through the center of the larger circle.
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At every step, the apparent brightness to the observer remains the same, 𝜋 squared over four.
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And at every step, the lighthouses remain evenly spaced with a distance two between each one of them on the circumference.
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And in the limit, what we’re getting here is a flat horizontal line with an infinite number of lighthouses evenly spaced in both directions.
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And because the apparent brightness was 𝜋 squared over four the entire way, that will, also be true in this limiting case.
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And this gives us a pretty awesome infinite series.
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The sum of the inverse squares one over 𝑛 squared, where 𝑛 covers all of the odd integers — one, three, five, and so on, but also negative one, negative three, negative five, often the leftward direction.
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Adding all of those up is gonna give us 𝜋 squared over four.
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That’s amazing!
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And it’s the core of what I wanna show you.
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And just take a step back and think about how unreal this seems.
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The sum of simple fractions that at first sight have nothing to do with geometry, nothing to do with circles at all, apparently, gives us this result that’s related to 𝜋.
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Except now, you can actually see what it has to do with geometry.
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The number line is kind of like a limit of ever-growing circles.
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And as you sum across that number line, making sure to sum all the way to infinity on either side, it’s sort of like you’re adding up along the boundary of an infinitely-large circle, in a very loose but very fun way of speaking.
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“But wait!” you might say.
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This is not the sum that you promised us at the start of the video.
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And, well, you’re right.
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We do have a little bit of thinking left.
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First things first, let’s just restrict this sum to only being the positive odd numbers, which gets us 𝜋 squared divided by eight.
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Now, the only difference between this and the sum that we’re looking for that goes over all the positive integers, odd and even, is that it’s missing the sum of the reciprocals of even numbers, what I’m coloring in red up here.
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Now, you can think of that missing series as a scaled copy of the total series that we want, where each lighthouse moves to being twice as far away from the origin.
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One gets shifted to two; two gets shifted to four; three gets shifted to six, and so on.
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And because that involves doubling the distance for every lighthouse, it means that the apparent brightness would be decreased by a factor of four.
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And that’s also relatively straightforward algebra.
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Going from the sum over all the integers to the sum over the even integers involves multiplying by one-fourth.
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And what that means is that going from all the integers to the odd ones would be multiplying by three-fourths, since the evens plus the odds have to give us the whole thing.
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So if we just flip that around, that means going from the sum over the odd numbers to the sum over all positive integers requires multiplying by four-thirds.
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So taking that 𝜋 squared over eight, multiplying by four-thirds, bada boom bada bing!
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We’ve got ourselves a solution to the Basel problem.