# Why is pi here?  And why is it squared?  A geometric answer to the Basel problem

## Метаданные

- **Канал:** 3Blue1Brown
- **YouTube:** https://www.youtube.com/watch?v=d-o3eB9sfls
- **Дата:** 02.03.2018
- **Длительность:** 17:08
- **Просмотры:** 7,139,420
- **Источник:** https://ekstraktznaniy.ru/video/16253

## Описание

A most beautiful proof of the Basel problem, using light.
Help fund future projects: https://www.patreon.com/3blue1brown
An equally valuable form of support is to simply share some of the videos.
Special thanks to these supporters: http://3b1b.co/basel-thanks
This video was sponsored by Brilliant: https://brilliant.org/3b1b

Brilliant's principles list that I referenced:
https://brilliant.org/principles/

Get early access and more through Patreon:
https://www.patreon.com/3blue1brown

The content here was based on a paper by Johan Wästlund
http://www.math.chalmers.se/~wastlund/Cosmic.pdf

Check out Mathologer's video on the many cousins of the Pythagorean theorem:
https://youtu.be/p-0SOWbzUYI

On the topic of Mathologer, he also has a nice video about the Basel problem:
https://youtu.be/yPl64xi_ZZA

A simple Geogebra to play around with the Inverse Pythagorean Theorem argument shown here.
https://ggbm.at/yPExUf7b

Some of you may be concerned about the final step here where we said the 

## Транскрипт

### <Untitled Chapter 1> []

Take 1 plus 1 fourth plus 1 ninth plus 1 sixteenth and so on where you're adding the inverses of the next square number What does this sum approach as you keep adding on more and more terms? 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 pi squared divided by 6. I mean isn't that crazy? What is pi doing here? And why is it squared? We don't usually see it squared in honor of Euler whose hometown was basil This infinite sum is often referred to as the basil problem But the proof that I'd like to show you is very different from the one that Euler had I've said in a previous video that whenever you see pi show up There will be some connection to circles and there are those who like to say that pi is not fundamentally about circles and Insisting on connecting equations

### <Untitled Chapter 1> []

Take 1 plus 1 fourth plus 1 ninth plus 1 sixteenth and so on where you're adding the inverses of the next square number What does this sum approach as you keep adding on more and more terms? 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 pi squared divided by 6. I mean isn't that crazy? What is pi doing here? And why is it squared? We don't usually see it squared in honor of Euler whose hometown was basil This infinite sum is often referred to as the basil problem But the proof that I'd like to show you is very different from the one that Euler had I've said in a previous video that whenever you see pi show up There will be some connection to circles and there are those who like to say that pi is not fundamentally about circles and Insisting on connecting equations

### <Untitled Chapter 1> []

Take 1 plus 1 fourth plus 1 ninth plus 1 sixteenth and so on where you're adding the inverses of the next square number What does this sum approach as you keep adding on more and more terms? 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 pi squared divided by 6. I mean isn't that crazy? What is pi doing here? And why is it squared? We don't usually see it squared in honor of Euler whose hometown was basil This infinite sum is often referred to as the basil problem But the proof that I'd like to show you is very different from the one that Euler had I've said in a previous video that whenever you see pi show up There will be some connection to circles and there are those who like to say that pi is not fundamentally about circles and Insisting on connecting equations

### The Basel Problem [0:57]

like these ones with a geometric intuition stems from a stubborn insistence on only understanding pi in the context where we first discovered it and That's all well and good But whatever your own perspective holds as fundamental the fact is pi is very much tied to circles 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 and geometry The question is just how long and convoluted that path might be and in the case of the basil problem It's a lot shorter than you might first think and it all starts with light Here's the basic idea 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 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 and Let's just call that a brightness of one For reasons I'll explain shortly the apparent brightness of the second lighthouse is 1 fourth as much as the first and the apparent brightness of the third is 1 9th as much as the first and then 1 16th and so on and you can probably see why this is useful for the basil problem It gives us a physical representation of what's being asked Since the brightness received from the whole infinite line of lighthouses is going to be 1 plus 1 4th plus 1 9th Plus the 16th and so on So the result that we are aiming to show is that this total brightness is equal to pi squared divided by 6 times the brightness of that first lighthouse And at first that might seem useless I mean, we're just re-asking the same original question But the progress comes from a new question that this framing raises are there ways that we can rearrange these lighthouses That don't change the total brightness for the observer And if so, can you show this to be equivalent to a setup that's somehow easier to compute To start let's be clear about what we mean when we reference apparent brightness to an observer Imagine a little screen which maybe represents the retina of your eye or a digital camera sensor or something like that You could ask what proportion of the rays coming out of the source hit that screen or phrase differently What is the angle between the ray hitting the bottom of that screen and top?

### The Basel Problem [0:57]

like these ones with a geometric intuition stems from a stubborn insistence on only understanding pi in the context where we first discovered it and That's all well and good But whatever your own perspective holds as fundamental the fact is pi is very much tied to circles 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 and geometry The question is just how long and convoluted that path might be and in the case of the basil problem It's a lot shorter than you might first think and it all starts with light Here's the basic idea 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 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 and Let's just call that a brightness of one For reasons I'll explain shortly the apparent brightness of the second lighthouse is 1 fourth as much as the first and the apparent brightness of the third is 1 9th as much as the first and then 1 16th and so on and you can probably see why this is useful for the basil problem It gives us a physical representation of what's being asked Since the brightness received from the whole infinite line of lighthouses is going to be 1 plus 1 4th plus 1 9th Plus the 16th and so on So the result that we are aiming to show is that this total brightness is equal to pi squared divided by 6 times the brightness of that first lighthouse And at first that might seem useless I mean, we're just re-asking the same original question But the progress comes from a new question that this framing raises are there ways that we can rearrange these lighthouses That don't change the total brightness for the observer And if so, can you show this to be equivalent to a setup that's somehow easier to compute To start let's be clear about what we mean when we reference apparent brightness to an observer Imagine a little screen which maybe represents the retina of your eye or a digital camera sensor or something like that You could ask what proportion of the rays coming out of the source hit that screen or phrase differently What is the angle between the ray hitting the bottom of that screen and top?

### The Basel Problem [0:57]

like these ones with a geometric intuition stems from a stubborn insistence on only understanding pi in the context where we first discovered it and That's all well and good But whatever your own perspective holds as fundamental the fact is pi is very much tied to circles 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 and geometry The question is just how long and convoluted that path might be and in the case of the basil problem It's a lot shorter than you might first think and it all starts with light Here's the basic idea 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 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 and Let's just call that a brightness of one For reasons I'll explain shortly the apparent brightness of the second lighthouse is 1 fourth as much as the first and the apparent brightness of the third is 1 9th as much as the first and then 1 16th and so on and you can probably see why this is useful for the basil problem It gives us a physical representation of what's being asked Since the brightness received from the whole infinite line of lighthouses is going to be 1 plus 1 4th plus 1 9th Plus the 16th and so on So the result that we are aiming to show is that this total brightness is equal to pi squared divided by 6 times the brightness of that first lighthouse And at first that might seem useless I mean, we're just re-asking the same original question But the progress comes from a new question that this framing raises are there ways that we can rearrange these lighthouses That don't change the total brightness for the observer And if so, can you show this to be equivalent to a setup that's somehow easier to compute To start let's be clear about what we mean when we reference apparent brightness to an observer Imagine a little screen which maybe represents the retina of your eye or a digital camera sensor or something like that You could ask what proportion of the rays coming out of the source hit that screen or phrase differently What is the angle between the ray hitting the bottom of that screen and top?

### Apparent Brightness to an Observer [3:20]

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? 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 You see the first of two places this story we're thinking of screens is going to be useful is in understanding the inverse square law Which is a distinctly three-dimensional phenomenon think of all of the rays of light hitting a screen one unit away from the source as You double the distance those rays will now cover an area with twice the width and twice the height So it would take four copies of that original screen to receive the same rays at that

### Apparent Brightness to an Observer [3:20]

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? 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 You see the first of two places this story we're thinking of screens is going to be useful is in understanding the inverse square law Which is a distinctly three-dimensional phenomenon think of all of the rays of light hitting a screen one unit away from the source as You double the distance those rays will now cover an area with twice the width and twice the height So it would take four copies of that original screen to receive the same rays at that

### Apparent Brightness to an Observer [3:20]

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? 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 You see the first of two places this story we're thinking of screens is going to be useful is in understanding the inverse square law Which is a distinctly three-dimensional phenomenon think of all of the rays of light hitting a screen one unit away from the source as You double the distance those rays will now cover an area with twice the width and twice the height So it would take four copies of that original screen to receive the same rays at that

### The Inverse Square Law [4:06]

distance And so each individual one receives 1 fourth as much light This is the sense in which I mean a light would appear 1 fourth as bright two times the distance away Likewise when you're three times farther away You would need nine copies of that original screen to receive the same rays so each individual screen only receives 1 9th as much light and This pattern continues 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 and As I'm sure many of you know this inverse square law is not at all special to light 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 and Remember it's because of this inverse square law that an infinite array of evenly spaced lighthouses physically implements the Basel problem But again what we need if we're going to make any progress here is to understand how we can manipulate setups with light sources like this without changing the total brightness for the observer and The key building block is an especially nice way to transform a single lighthouse into two Think of an observer at the origin of the XY plane and a single lighthouse sitting out somewhere on that plane Now draw a line from that lighthouse to the observer and then another line perpendicular to that one at the lighthouse 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 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 And I should say by the way that the standing assumption throughout this video is that all lighthouses are equivalent They're using the same light bulb emanating the same power all of that So in other words assigning variables to things here if we call the distance from the observer to lighthouse a little a and B little B and the distance to the first lighthouse H We have the relation 1 over a squared plus 1 over B squared equals 1 over H squared This is the much less well-known Inverse Pythagorean theorem which some of you may recognize from math ologer's most recent and I'll say most excellent video on the many cousins of the Pythagorean theorem Pretty cool relation don't you think and if you're a mathematician at heart you might be asking right now how you prove it and There are some straightforward ways where you express the triangles area in two separate ways and apply the usual Pythagorean theorem 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 Imagine scaling down the whole right triangle into a tinier version and think of this miniature Hypotenuse as a screen receiving light from the first lighthouse 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? I mean the rays of light hitting one of those two legs are precisely the same as the rays that hit the hypotenuse 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 it'll be the same angle of rays 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 Why you might ask well, it's a matter of similar triangles This animation already gives you a strong hint for how it works And we've also left a link in the description to a simple GeoGebra applet for those of you who want to think this through in a slightly more interactive environment 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 The inverse Pythagorean theorem Alright buckle up now because here's where things get good We've got this inverse Pythagorean theorem, right? And that's going to let us transform a single lighthouse into two others without changing the brightness experienced by the observer With that in hand and no small amount of cleverness we can use this to build up the infinite array that we need Picture yourself at the edge of a circular lake directly opposite a lighthouse We're

### The Inverse Square Law [4:06]

distance And so each individual one receives 1 fourth as much light This is the sense in which I mean a light would appear 1 fourth as bright two times the distance away Likewise when you're three times farther away You would need nine copies of that original screen to receive the same rays so each individual screen only receives 1 9th as much light and This pattern continues 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 and As I'm sure many of you know this inverse square law is not at all special to light 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 and Remember it's because of this inverse square law that an infinite array of evenly spaced lighthouses physically implements the Basel problem But again what we need if we're going to make any progress here is to understand how we can manipulate setups with light sources like this without changing the total brightness for the observer and The key building block is an especially nice way to transform a single lighthouse into two Think of an observer at the origin of the XY plane and a single lighthouse sitting out somewhere on that plane Now draw a line from that lighthouse to the observer and then another line perpendicular to that one at the lighthouse 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 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 And I should say by the way that the standing assumption throughout this video is that all lighthouses are equivalent They're using the same light bulb emanating the same power all of that So in other words assigning variables to things here if we call the distance from the observer to lighthouse a little a and B little B and the distance to the first lighthouse H We have the relation 1 over a squared plus 1 over B squared equals 1 over H squared This is the much less well-known Inverse Pythagorean theorem which some of you may recognize from math ologer's most recent and I'll say most excellent video on the many cousins of the Pythagorean theorem Pretty cool relation don't you think and if you're a mathematician at heart you might be asking right now how you prove it and There are some straightforward ways where you express the triangles area in two separate ways and apply the usual Pythagorean theorem 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 Imagine scaling down the whole right triangle into a tinier version and think of this miniature Hypotenuse as a screen receiving light from the first lighthouse 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? I mean the rays of light hitting one of those two legs are precisely the same as the rays that hit the hypotenuse 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 it'll be the same angle of rays 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 Why you might ask well, it's a matter of similar triangles This animation already gives you a strong hint for how it works And we've also left a link in the description to a simple GeoGebra applet for those of you who want to think this through in a slightly more interactive environment 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 The inverse Pythagorean theorem Alright buckle up now because here's where things get good We've got this inverse Pythagorean theorem, right? And that's going to let us transform a single lighthouse into two others without changing the brightness experienced by the observer With that in hand and no small amount of cleverness we can use this to build up the infinite array that we need Picture yourself at the edge of a circular lake directly opposite a lighthouse We're

### The Inverse Square Law [4:06]

distance And so each individual one receives 1 fourth as much light This is the sense in which I mean a light would appear 1 fourth as bright two times the distance away Likewise when you're three times farther away You would need nine copies of that original screen to receive the same rays so each individual screen only receives 1 9th as much light and This pattern continues 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 and As I'm sure many of you know this inverse square law is not at all special to light 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 and Remember it's because of this inverse square law that an infinite array of evenly spaced lighthouses physically implements the Basel problem But again what we need if we're going to make any progress here is to understand how we can manipulate setups with light sources like this without changing the total brightness for the observer and The key building block is an especially nice way to transform a single lighthouse into two Think of an observer at the origin of the XY plane and a single lighthouse sitting out somewhere on that plane Now draw a line from that lighthouse to the observer and then another line perpendicular to that one at the lighthouse 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 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 And I should say by the way that the standing assumption throughout this video is that all lighthouses are equivalent They're using the same light bulb emanating the same power all of that So in other words assigning variables to things here if we call the distance from the observer to lighthouse a little a and B little B and the distance to the first lighthouse H We have the relation 1 over a squared plus 1 over B squared equals 1 over H squared This is the much less well-known Inverse Pythagorean theorem which some of you may recognize from math ologer's most recent and I'll say most excellent video on the many cousins of the Pythagorean theorem Pretty cool relation don't you think and if you're a mathematician at heart you might be asking right now how you prove it and There are some straightforward ways where you express the triangles area in two separate ways and apply the usual Pythagorean theorem 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 Imagine scaling down the whole right triangle into a tinier version and think of this miniature Hypotenuse as a screen receiving light from the first lighthouse 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? I mean the rays of light hitting one of those two legs are precisely the same as the rays that hit the hypotenuse 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 it'll be the same angle of rays 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 Why you might ask well, it's a matter of similar triangles This animation already gives you a strong hint for how it works And we've also left a link in the description to a simple GeoGebra applet for those of you who want to think this through in a slightly more interactive environment 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 The inverse Pythagorean theorem Alright buckle up now because here's where things get good We've got this inverse Pythagorean theorem, right? And that's going to let us transform a single lighthouse into two others without changing the brightness experienced by the observer With that in hand and no small amount of cleverness we can use this to build up the infinite array that we need Picture yourself at the edge of a circular lake directly opposite a lighthouse We're

### Inverse Pythagorean Theorem [8:59]

going to want it to be the case that the distance between you and the lighthouse Along the border of the lake is one so we'll say the lake has a circumference of two now the apparent brightness is one divided by the diameter squared and In this case the diameter is that circumference 2 divided by pi so the apparent brightness works out to be pi squared divided by 4 Now for our first transformation draw a new circle twice as big so circumference 4 and Draw a tangent line to the top of the small circle then replace the original lighthouse with two new ones where this tangent line intersects the larger circle 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 the significance of that in our diagram here is that it means the inverse Pythagorean theorem applies and the brightness from those two new lighthouses equals the brightness from the first one namely pi squared divided by 4 as The next step draw a new circle twice as big as the last with a circumference 8 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 Again, since this line is a diameter of that large circle Then the lines from those two new points to the observer are going to form a right angle 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 Good news, right? because that means we can apply the inverse Pythagorean theorem and that means that the apparent brightness from the original lighthouse is the same as the combined brightness from the two newer ones 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

### Inverse Pythagorean Theorem [8:59]

going to want it to be the case that the distance between you and the lighthouse Along the border of the lake is one so we'll say the lake has a circumference of two now the apparent brightness is one divided by the diameter squared and In this case the diameter is that circumference 2 divided by pi so the apparent brightness works out to be pi squared divided by 4 Now for our first transformation draw a new circle twice as big so circumference 4 and Draw a tangent line to the top of the small circle then replace the original lighthouse with two new ones where this tangent line intersects the larger circle 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 the significance of that in our diagram here is that it means the inverse Pythagorean theorem applies and the brightness from those two new lighthouses equals the brightness from the first one namely pi squared divided by 4 as The next step draw a new circle twice as big as the last with a circumference 8 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 Again, since this line is a diameter of that large circle Then the lines from those two new points to the observer are going to form a right angle 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 Good news, right? because that means we can apply the inverse Pythagorean theorem and that means that the apparent brightness from the original lighthouse is the same as the combined brightness from the two newer ones 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

### Inverse Pythagorean Theorem [8:59]

going to want it to be the case that the distance between you and the lighthouse Along the border of the lake is one so we'll say the lake has a circumference of two now the apparent brightness is one divided by the diameter squared and In this case the diameter is that circumference 2 divided by pi so the apparent brightness works out to be pi squared divided by 4 Now for our first transformation draw a new circle twice as big so circumference 4 and Draw a tangent line to the top of the small circle then replace the original lighthouse with two new ones where this tangent line intersects the larger circle 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 the significance of that in our diagram here is that it means the inverse Pythagorean theorem applies and the brightness from those two new lighthouses equals the brightness from the first one namely pi squared divided by 4 as The next step draw a new circle twice as big as the last with a circumference 8 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 Again, since this line is a diameter of that large circle Then the lines from those two new points to the observer are going to form a right angle 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 Good news, right? because that means we can apply the inverse Pythagorean theorem and that means that the apparent brightness from the original lighthouse is the same as the combined brightness from the two newer ones 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

### The Inverse Pythagorean Theorem [11:12]

and Even nicer these four lighthouses are all going to be evenly spaced around the lake Why? Well, the lines from those lighthouses to the center are at 90 degree angles with each other So since things are symmetric left to right that means that the distances along the circumference are 1, 2, 2, and 1 Alright, you might see where this is going, but I want to walk through this for just one more step You draw a circle twice as big so circumference of 16 now 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 and then create two new lighthouses where that line intersects with the larger circle 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 and Just as before the line from the observer to the original lighthouse is Perpendicular to the long line and those are the two facts that justify us in using the inverse Pythagorean theorem 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 going to be evenly spaced This is the final bit of geometry proofiness before the final thrust 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 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 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 What that means then is

### The Inverse Pythagorean Theorem [11:12]

and Even nicer these four lighthouses are all going to be evenly spaced around the lake Why? Well, the lines from those lighthouses to the center are at 90 degree angles with each other So since things are symmetric left to right that means that the distances along the circumference are 1, 2, 2, and 1 Alright, you might see where this is going, but I want to walk through this for just one more step You draw a circle twice as big so circumference of 16 now 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 and then create two new lighthouses where that line intersects with the larger circle 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 and Just as before the line from the observer to the original lighthouse is Perpendicular to the long line and those are the two facts that justify us in using the inverse Pythagorean theorem 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 going to be evenly spaced This is the final bit of geometry proofiness before the final thrust 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 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 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 What that means then is

### The Inverse Pythagorean Theorem [11:12]

and Even nicer these four lighthouses are all going to be evenly spaced around the lake Why? Well, the lines from those lighthouses to the center are at 90 degree angles with each other So since things are symmetric left to right that means that the distances along the circumference are 1, 2, 2, and 1 Alright, you might see where this is going, but I want to walk through this for just one more step You draw a circle twice as big so circumference of 16 now 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 and then create two new lighthouses where that line intersects with the larger circle 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 and Just as before the line from the observer to the original lighthouse is Perpendicular to the long line and those are the two facts that justify us in using the inverse Pythagorean theorem 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 going to be evenly spaced This is the final bit of geometry proofiness before the final thrust 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 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 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 What that means then is

### Inscribed Angle Theorem [13:10]

that when you draw lines from those eight new lighthouses into the center They divide the circle evenly into 45 degree angle pieces and that means the eight lighthouses are evenly spaced around the circumference with the distance of two between each one of them 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 at every step the apparent brightness to the observer remains the same pi squared over 4 and at every step the lighthouse has remained evenly spaced with a distance 2 between each one of them on the circumference 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 and Because the apparent brightness was pi squared over 4 the entire way that will also be true in this limiting case And This gives us a pretty awesome infinite series the sum of the inverse squares 1 over n squared Where n covers all of the odd integers 1 3 5 and so on but also negative 1 negative 3 negative 5 off in the leftward direction Adding all of those up is going to give us pi squared over 4 That's amazing and it's the core of what I want to show you and Just take a step back and think about how unreal this seems The sum of simple fractions that at first sight have nothing to do with geometry circles at all Apparently gives us this result that's related to pi Except now you can actually see what it has to do with geometry the number line is kind of like a limit of ever-growing circles 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 and a very loose But very fun way of speaking But wait, you might say this is not the sum that you promised us at the start of the video And well, you're right. We do have a little bit of thinking left First things first, let's just restrict the sum to only being the positive odd numbers which gets us pi squared divided by 8 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 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 one gets shifted to two gets shifted to four three gets shifted to six and so on and Because that involves doubling the distance for every lighthouse. it means that the apparent brightness would be decreased by a factor of four and That's also relatively straightforward algebra going from the sum over all the integers to the sum over the even integers Involves multiplying by 1 4th and what that means is that going from all the integers to the odd ones Would be multiplying by 3 4ths since the evens plus the odds have to give us the whole thing 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 4 thirds So taking that pi squared over 8 multiplying by 4 thirds badda boom badda bing We've got ourselves a solution to the basil problem Now this video that you just watched was primarily written and animated by one of the new three blue one brown team members Ben Hambricht an addition made possible. Thanks to you guys through patreon You

### Inscribed Angle Theorem [13:10]

that when you draw lines from those eight new lighthouses into the center They divide the circle evenly into 45 degree angle pieces and that means the eight lighthouses are evenly spaced around the circumference with the distance of two between each one of them 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 at every step the apparent brightness to the observer remains the same pi squared over 4 and at every step the lighthouse has remained evenly spaced with a distance 2 between each one of them on the circumference 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 and Because the apparent brightness was pi squared over 4 the entire way that will also be true in this limiting case And This gives us a pretty awesome infinite series the sum of the inverse squares 1 over n squared Where n covers all of the odd integers 1 3 5 and so on but also negative 1 negative 3 negative 5 off in the leftward direction Adding all of those up is going to give us pi squared over 4 That's amazing and it's the core of what I want to show you and Just take a step back and think about how unreal this seems The sum of simple fractions that at first sight have nothing to do with geometry circles at all Apparently gives us this result that's related to pi Except now you can actually see what it has to do with geometry the number line is kind of like a limit of ever-growing circles 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 and a very loose But very fun way of speaking But wait, you might say this is not the sum that you promised us at the start of the video And well, you're right. We do have a little bit of thinking left First things first, let's just restrict the sum to only being the positive odd numbers which gets us pi squared divided by 8 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 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 one gets shifted to two gets shifted to four three gets shifted to six and so on and Because that involves doubling the distance for every lighthouse. it means that the apparent brightness would be decreased by a factor of four and That's also relatively straightforward algebra going from the sum over all the integers to the sum over the even integers Involves multiplying by 1 4th and what that means is that going from all the integers to the odd ones Would be multiplying by 3 4ths since the evens plus the odds have to give us the whole thing 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 4 thirds So taking that pi squared over 8 multiplying by 4 thirds badda boom badda bing We've got ourselves a solution to the basil problem Now this video that you just watched was primarily written and animated by one of the new three blue one brown team members Ben Hambricht an addition made possible. Thanks to you guys through patreon You

### Inscribed Angle Theorem [13:10]

that when you draw lines from those eight new lighthouses into the center They divide the circle evenly into 45 degree angle pieces and that means the eight lighthouses are evenly spaced around the circumference with the distance of two between each one of them 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 at every step the apparent brightness to the observer remains the same pi squared over 4 and at every step the lighthouse has remained evenly spaced with a distance 2 between each one of them on the circumference 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 and Because the apparent brightness was pi squared over 4 the entire way that will also be true in this limiting case And This gives us a pretty awesome infinite series the sum of the inverse squares 1 over n squared Where n covers all of the odd integers 1 3 5 and so on but also negative 1 negative 3 negative 5 off in the leftward direction Adding all of those up is going to give us pi squared over 4 That's amazing and it's the core of what I want to show you and Just take a step back and think about how unreal this seems The sum of simple fractions that at first sight have nothing to do with geometry circles at all Apparently gives us this result that's related to pi Except now you can actually see what it has to do with geometry the number line is kind of like a limit of ever-growing circles 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 and a very loose But very fun way of speaking But wait, you might say this is not the sum that you promised us at the start of the video And well, you're right. We do have a little bit of thinking left First things first, let's just restrict the sum to only being the positive odd numbers which gets us pi squared divided by 8 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 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 one gets shifted to two gets shifted to four three gets shifted to six and so on and Because that involves doubling the distance for every lighthouse. it means that the apparent brightness would be decreased by a factor of four and That's also relatively straightforward algebra going from the sum over all the integers to the sum over the even integers Involves multiplying by 1 4th and what that means is that going from all the integers to the odd ones Would be multiplying by 3 4ths since the evens plus the odds have to give us the whole thing 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 4 thirds So taking that pi squared over 8 multiplying by 4 thirds badda boom badda bing We've got ourselves a solution to the basil problem Now this video that you just watched was primarily written and animated by one of the new three blue one brown team members Ben Hambricht an addition made possible. Thanks to you guys through patreon You