Update transcript.txt (#276)

added words missing and deleted words that were not said in the video

2018/wallis-product/english/transcript.txt

@@ -1,10 +1,10 @@
 Alright, I think you're going to like this. 
 I want to show you a beautiful result that reveals a surprising connection between a simple series of fractions and the geometry of circles. 
 But unlike some other results like this that you may have seen before, this one involves multiplying things instead of adding them up. 
-The video you're about to watch is particularly exciting for us at 3Blue1Brown, because it came about a little differently from most of the videos we've done. 
+Now, the video you're about to watch is particularly exciting for us at 3Blue1Brown, because it came about a little differently from most of the videos that we've done. 
 If you step back and think about it, the value of any kind of math presentation comes from a combination of the underlying math and all of the choices that go into how to communicate it. 
-And for almost all of the content on this channel, the underlying math is something that's well known in the field, either based on general theory or some particular paper, and my hope is for the novelty to come from the communication half. 
-And with this video, the result we're discussing, a very famous infinite product for pi known as the Wallace product, is indeed well known math. 
+And for almost all of the content on this channel, the underlying math is something that's well known in the field, it’s either based on general theory or some particular paper, and my hope is for the novelty to come from the communication half. 
+And with this video, the result we're discussing, a very famous infinite product for pi, known as the Wallace product, is indeed well known math. 
 However, what we'll be presenting is, to our knowledge, a more original proof of this result. 
 For context, after watching our video on the Basel problem, Sweeter, the new 3b1b member who some of you may remember from the video about color and winding numbers, well, he spent some time thinking about the approach taken in that video, as well as thinking about the connection between the Basel problem and the Wallace product, and he's tumbled into a new proof of the relationship between the Wallace product and pi. 
 I mean, I'll leave open the possibility that an argument of this style is hidden somewhere in the literature beyond what our searching pulled up, but I can at least say that it was found independently, and that if it does exist out there, it has done a fantastic job hiding itself from the public view. 
@@ -25,11 +25,11 @@ Back then, the quantity we cared about involved looking at the distance between
 This is why we had the whole narrative with lighthouses in the first place, since the inverse square law gave a really nice physical interpretation to this quantity, it was the total amount of light received by that observer. 
 But despite that nice physical interpretation, there's nothing magical about adding inverse square distances, that just happened to be what was useful for that particular problem. 
 Now to tackle our new problem, of 2 over 1 times 2 over 3 times 4 over 3 times 4 over 5 and so on, we're going to do something similar but different in the details. 
-Instead of using the inverse square distances, just look at the distances themselves directly, and instead of adding them up, we'll be multiplying them, giving a quantity I'll be referring to as the distance product for the observer. 
-And even though this distance product no longer has a nice physical analogy, I still want to illustrate it with lighthouses and an observer, because, well, it's pretty, and also more fun than abstract geometric points. 
-For this proof of the Wallace product, we're going to need two key facts about this distance product, two little lemmas. 
+Instead of using the inverse square distances, just look at the distances themselves directly, and instead of adding them up, we'll be multiplying them, giving a quantity I'll be referring to as the distance product for the observer, that will be important. 
+And even though this distance product no longer has a nice physical analogy, I still kinda want to illustrate it with lighthouses and an observer, because, well, I don't know, it's pretty, and also it’s just more fun than just abstract geometric points. 
+Now, for this proof of the Wallace product, we're going to need two key facts about this distance product, two little lemmas. 
 First, if the observer is positioned halfway between two lighthouses on the circle, this distance product, the thing that you get by multiplying together the lengths of all these lines, works out to be exactly 2, no matter how many lighthouses there are. 
-And second, if you remove one of those lighthouses and put the observer in its place, this distance product from all the remaining lighthouses equals the number of lighthouses you started with. 
+And second, if you remove one of those lighthouses and put the observer in its place, this distance product from all the remaining lighthouses happens to equal the number of lighthouses you started with. 
 Again, no matter how many lighthouses there are. 
 And if those two facts seem crazy, I agree! 
 I mean, it's not even obvious that the distance product here should work out to be an integer in either case. 
@@ -43,7 +43,7 @@ Similarly, cubing this number is going to triple the angle it makes with the hor
 So, for example, on screen right now there are seven evenly spaced points around the unit circle, which I'll call l0, l1, l2, and so on, and they're rotated in such a way that l0 is sitting at the number 1 on that right hand side. 
 So because the angle that each one of these makes with the horizontal is an integer multiple of 1 seventh of a turn, raising any one of these numbers to the 7th power rotates you around to landing on the number 1. 
 In other words, these are all solutions to the polynomial equation x to the 7th minus 1 equals 0. 
-But on the other hand, we could construct a polynomial that has these numbers as roots a totally different way, by taking x minus l0 times x minus l1 up to x minus l6, I mean you plug in any one of these numbers and that product will have to equal 0. 
+But on the other hand, we could construct a polynomial that has these numbers as roots a totally different way, by taking x minus l0 times x minus l1, on and on and on, up to x minus l6, I mean you plug in any one of these numbers and that product will have to equal 0. 
 And because these two degree-7 polynomials have the same seven distinct roots and the same leading term, it's just x to the 7th in both cases, they are in fact one and the same. 
 Now take a moment to appreciate just what a marvelous fact that is. 
 This right hand side looks like it would be an absolute nightmare to expand. 
@@ -77,26 +77,26 @@ The distance 0 lighthouses ends up annihilating all other factors.
 But suppose we just got rid of that one troublesome lighthouse, and considered only the contributions from all the other ones. 
 What would that distance product work out to be? 
 Well, now instead of considering the polynomial observer to the n-1, which has a root at all of these n roots of unity, we're looking at the polynomial observer to the n-1 divided by observer-1, which has a root at all of the roots of unity except for the number 1 itself. 
-And a little algebra shows that this fraction is the same thing as 1 plus observer plus observer squared, up to observer to the n-1. 
+And a little algebra shows that this fraction is the same thing as 1 plus observer plus observer squared, on and on and on, up to observer to the n-1. 
 And so if you plug in observer equals 1, since that's the number he's sitting on, what do you get? 
 All of the terms here become 1, so it works out to be n, which means the total distance product for this setup equals the number of original lighthouses. 
-This does depend on the total number of lighthouses, but only in a very simple way. 
+Now this does depend on the total number of lighthouses, but only in a very simple way. 
 Think about this, this is incredible, the total distance product that an observer sitting at one of the lighthouses receives from all other lighthouses is precisely n, where n is the total number of lighthouses, including the ignored one. 
 That is our second key fact. 
 And by the way, proving geometric facts with complex polynomials like this is pretty standard in math, and if you went up to your local mathematician and showed him or her these two facts, or other facts like these, they'd recognize both that these facts are true, and how to prove them using the methods we just showed. 
 And now, so can you! 
 So next, with both these facts in our back pocket, let's see how to use them to understand the product we're interested in, and how it relates to pi. 
 Take this setup, with n lighthouses evenly spaced around a unit circle, and imagine two separate observers, what I'll call the keeper and the sailor. 
 Put the keeper directly on one of the lighthouses, and put the sailor halfway between that point and the next lighthouse. 
-The idea here will be to look at the distance product for the keeper divided by the distance product for the sailor, and then we're going to compute this ratio in two separate ways.
+The idea here will be to look at the distance product for the keeper divided by the distance product for the sailor, and then we are going to compute this ratio in two separate ways. 
 From the first key fact, we know that the total distance product for the sailor is 2. 
 And the distance product for the keeper? 
 Well, it's 0, since he's standing right on top of 1. 
 But if we got rid of that lighthouse, then by our second key fact, the remaining distance product for that keeper is n. 
 And of course, by getting rid of that lighthouse, we've also gotten rid of its contribution to the sailor's distance product, so that denominator now has to be divided by the distance between the two observers. 
 And simplifying this just a little bit, it means that the ratio between the keeper's distance product and the sailor's is n times the distance between the two observers, all divided by 2. 
-But we could also compute this ratio in a different way, by considering each lighthouse individually. 
-For each lighthouse, think about its contribution to the keeper's distance product, meaning its distance to the keeper, divided by its contribution to the sailor's distance product, its distance to the sailor. 
+But, we could also compute this ratio in a different way: by considering each lighthouse individually. 
+For each lighthouse, think about its contribution to the keeper's distance product, meaning just its distance to the keeper, divided by its contribution to the sailor's distance product, its distance to the sailor. 
 And when we multiply all of these factors up over each lighthouse, we have to get the same ratio in the end, n times the distance between the observers, all divided by 2. 
 Now that might seem like a super messy calculation, but as n gets larger, this actually gets simpler for any particular lighthouse. 
 For example, think about the first lighthouse after the keeper, in the sense of counter-clockwise from him. 
@@ -112,11 +112,11 @@ Combining this over all of the lighthouses, we get the product 2 over 1 times 2
 This is the product that we're interested in studying, and in this context, each one of those terms reflects what the contribution for a particular lighthouse is as n approaches infinity. 
 And when I say contribution, I mean the contribution to this ratio of the keeper's distance product to the sailor's distance product, which we know at every step has to equal n times the distance between the observers divided by 2. 
 So what does that value approach as n approaches infinity? 
-The distance between the observers is half of 1 over n of a full turn around the circle, and since this is a unit circle, its total circumference is 2 pi, so the distance between the observers approaches pi divided by n, and therefore n times this distance divided by 2 approaches pi divided by 2. 
+Well, the distance between the observers is half of 1 over n of a full turn around the circle, and since this is a unit circle, its total circumference is 2 pi, so the distance between the observers approaches pi divided by n, and therefore n times this distance divided by 2 approaches pi divided by 2. 
 So there you have it! 
 Our product, 2 over 1 times 2 over 3 times 4 over 3 times 4 over 5, on and on and on, must approach pi divided by 2. 
 This is a truly marvelous result, and it's known as the Wallace product, named after 17th century mathematician John Wallace, who first discovered this fact in a way more convoluted way. 
-And also, a little bit of trivia, this is the same guy who discovered, or well, invented, the infinity symbol. 
+And also, a little bit of trivia, this is the same guy who discovered, or well, invented, the infinity symbol.
 And, actually, if you look back at this argument, we've pulled a little bit of sleight of hand in the informality here, which the particularly mathematically sophisticated among you might have caught. 
 What we have here is a whole bunch of factors which we knew multiplied together to get n times the distance between the observers divided by 2, and then we looked at the limit of each factor individually as n went to infinity, and concluded that the product of all of those limiting terms had to equal whatever the limit of n times the distance between the observers divided by 2 is. 
 But what that assumes is that the product of limits is equal to the limit of products, even when there's infinitely many factors. 
@@ -141,13 +141,13 @@ And indeed, we'll find that if you intermix these two halves differently, for ex
 It's only when you specifically combine them in this one-for-one manner that you get a product that converges to pi halves. 
 This is something that falls out of the way that dominated convergence justifies us in commuting limits the way we did, and again, for more details, see the supplemental post. 
 Still, those are just technicalities. 
-The conceptual gist for what's going on here is exactly what we just showed. 
+The conceptual gist for what's going on here is exactly what we just showed.
 And in fact, after doing all that work, it would be a shame not to take a quick moment to talk about one more neat result that falls out of this argument. 
 Arguably, this is the coolest part of the whole proof. 
 You see, we can generalize this whole discussion. 
 Think back to when we discovered our first key fact, where we saw that you could not only consider placing the sailor precisely halfway between lighthouses, but any fraction, f, of the way between adjacent lighthouses. 
 In that more general setting, the distance product for the sailor wasn't necessarily 2, but it was chord of f, where f is that fraction of the way between lighthouses. 
-If we go through the same reasoning that we just did with the sailor at this location instead and change nothing else, what we'll find is that the ratio of the keeper's distance product to the sailor's distance product is now n times the distance between them divided by chord of f, which approaches f times 2 pi divided by chord of f as n gets larger. 
+And if we go through the same reasoning that we just did with the sailor at this location instead and change nothing else, what we'll find is that the ratio of the keeper's distance product to the sailor's distance product is now n times the distance between them divided by chord of f, which approaches f times 2 pi divided by chord of f as n gets larger. 
 And in the same way as before, you could alternatively calculate this by considering the contributions from each individual lighthouse. 
 If you take the time to work this out, the kth lighthouse after the keeper will contribute a factor of k divided by k-f to this ratio. 
 And all the lighthouses before the keeper, they contribute the same thing, but you're just plugging in negative values for k. 
@@ -156,8 +156,8 @@ Put another way, since chord of f is 2 times the sine of f pi, this product is t
 Now rewriting this a little bit more, what you get is a pretty interesting fact. 
 Sine of f times pi is equal to f pi times this really big product, the product of 1 minus f over k over all non-zero integers k. 
 So what we found is a way to express sine of x as an infinite product, which is really cool if you think about it. 
-So not only does this proof give us the Wallace product, which is incredible in its own right, it also generalizes to give us the product formula for the sine. 
+So not only does this proof give us the Wallace product, which is incredible in its own right? It also generalizes to give us the product formula for the sine. 
 And what's neat about that is that it connects to how Euler originally solved the Basel problem, the sum that we saw in the previous video. 
 He was looking at this very infinite product for sine. 
 I mean, connecting these formulas for pi to circles is one thing, but connecting them to each other is another thing entirely. 
-And once again, if you want more details on all of this, check out the supplementary blog post. Thank you.
+And once again, if you want more details on all of this, check out the supplementary blog post.