牛顿如何彻底改变了圆周率的计算方式 veritasium 2021-03-16

圆周率的直观理解

本视频将探讨我们过去计算圆周率(Pi: 一个圆的周长与其直径之比的数学常数,约等于3.14159)的荒谬方式。

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This video is about the ridiculous way we used to calculate Pi.

在两千年的时间里,最成功的方法是极其缓慢和乏味的,但后来艾萨克·牛顿(Isaac Newton: 英国物理学家、数学家、天文学家和自然哲学家)的出现改变了局面。

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For 2000 years the most successful method was painstakingly slow and tedious, but then Isaac Newton came along and changed the game.

你可以说他“速通”了圆周率的计算,我将向你展示他是如何做到的。

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You could say he speed-ran Pi and I'm gonna show you how he did it.

但首先,我们用披萨来理解圆周率。

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But first Pi with pizzas.

切下披萨的饼边,然后将其放在相同的披萨上。

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Cut the crust off of pizza and lay it across identical pizzas.

你会发现它能横跨三个多一点的披萨,这就是圆周率。

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And you'll find that it goes across three and a bit pizzas, this is Pi.

一个圆的周长(circumference)大约是其直径(diameter)的3.14倍,但圆周率也与圆的面积(area)有关。

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The circumference of a circle is roughly 3.14 times its diameter but Pi is also related to a circles area,

面积就是πr²

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area's just Pi R squared.

但为什么是πr²呢?

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But why is it Pi R squared?

将一个披萨切成非常薄的薄片,然后将这些薄片排成一个矩形(rectangle)。

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Well cut a pizza into really thin slices and then form these slices into a rectangle.

现在,这个矩形的面积就是长乘以宽。

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Now the area of this rectangle is just length times width.

矩形的长度是周长的一半,因为一半的饼边在一边,另一半在另一边,所以长度是πr

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The length of the rectangle is half the circumference because there's half the crust on one side and half on the other, so the length is Pi R.

而宽度就是一片披萨的长度,也就是原始圆的半径(radius)。

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And then the width is just the length of a piece of pizza which is the radius of the original circle.

所以面积是πr乘以r,即面积是πr²

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So area is Pi R times R, area is Pi R squared.

因此,一个单位圆(unit circle: 半径为1的圆)的面积就是π,请记住这一点,因为它稍后会派上用场。

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So the area of a unit circle then is just Pi, keep that in mind because it'll come in handy later.

古老而耗时的方法:多边形逼近

那么,我们过去计算圆周率的荒谬方法是什么呢?

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So what was the ridiculous way we used to calculate Pi?

嗯,这是最显而易见的方法。

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Well, it's the most obvious way.

很容易证明圆周率必须介于三和四之间。

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It's easy to show that Pi must be between three and four,

取一个圆,并在其中画一个边长为一的六边形(hexagon)。

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take a circle and draw a hexagon inside it, with sides of length one.

一个正六边形可以分成六个等边三角形(equilateral triangles)。

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A regular hexagon can be divided into six equal lateral triangles.

所以圆的直径是二。

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So the diameter of the circle is two.

现在六边形的周长是六,而圆的周长必须大于此,所以圆周率必须大于六除以二。

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Now the perimeter of the hexagon is six and the circumference of the circle must be larger than this, so Pi must be greater than six over two.

所以圆周率大于三。

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So Pi is greater than three.

现在在圆的外面画一个正方形(square)。

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Now draw a square around the circle,

正方形的周长是八,这比圆的周长要大,所以圆周率必须小于八除以二。

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the perimeter of the square is eight which is bigger than the circles circumference, so Pi must be less than eight over two.

所以圆周率小于四。

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So Pi is less than four.

这实际上已经为人所知数千年了。

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This was actually known for a thousands of years.

然后在公元前250年,阿基米德(Archimedes: 古希腊数学家、物理学家、工程师、发明家和天文学家)改进了这种方法。

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And then in 250 BC, Archimedes improved on the method.

Alex: 首先他从六边形开始,就像你做的那样,然后他将六边形平分,得到一个十二边形(dodecagon)。

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So first he starts with the hexagon, just like you did and then he bisects the hexagon to dodecagon.

所以那是一个有12条边的正多边形。

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So that's a 12 sided, regular 12 sided shape.

他计算其周长,该周长与直径的比率将小于圆周率。

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And he calculates its perimeter, the ratio of that perimeter to the diameter will be less than Pi.

他对一个外接(circumscribed)的十二边形也做了同样的事情,并找到了圆周率的上限。

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He does the same thing for a circumscribed 12-gon and finds an upper bound for Pi.

计算现在变得更加复杂,因为他必须提取平方根(square roots)以及平方根的平方根,并将所有这些转化为分数。

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The calculations now become a lot more tricky because he has to extract square roots and square roots of square roots and turn all these into fractions,

但他算出了十二边形,然后是二十四边形,四十八边形,等到他算到九十六边形时,他差不多就受够了,但最终他将圆周率限定在3.1408到3.1429之间。

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but he works out the 12-gon, then the 24-gon, 48-gon and by the time he gets to the 96-gon he sort of had enough, but he gets, in the end he gets Pi to between 3.1408 and 3.1429.

所以对于两千多年前来说,这还不错。

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So for over 2000 years ago, that's not too bad.

Derek: 是的,这似乎是你在圆周率计算中所需的所有精度。

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Yeah, that seems like all the precision you'd need in Pi.

Alex: 对,所以这远远超出了任何实际目的所需的精度。

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Right, so this goes way beyond precision for any practical purpose.

这现在是展示你实力的表现,展示你拥有多么强大的数学能力,能够以非常高的精度计算像圆周率这样的常数。

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This is now a matter of flexing your muscles. This is showing off just how much mathematical power you have, that you can work out a constant like Pi to very high precision.

传统方法的局限性

在接下来的两千年里,每个人都以这种方式继续将多边形(polygons)不断地平分到令人眩晕的程度,圆周率的计算流传于中国、印度、波斯和阿拉伯数学家之间,每个人都按照阿基米德的思路为这些界限做出了贡献。

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So for the next 2000 years, this is how everyone carried on bisecting polygons to dizzying heights as Pi passed through Chinese, Indian, Persian and Arab mathematicians, each contributed to these bounds along our committee's line.

在16世纪末,法国人弗朗索瓦·韦特(Francois Viete)比阿基米德多倍增了十几次,计算了一个有393,216条边的多边形的周长。

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And in the late 16th century, Frenchman Francois Viete doubled a dozen more times than Archimedes, computing the perimeter of a polygon with 393,216 sides

然而在17世纪初,他被荷兰的鲁道夫·范·科伊伦(Ludolph van Ceulen)超越。

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only to be out done at the turn of the 17th century by the Dutch Ludolph van Ceulen.

他为此付出了25年的努力,以高精度计算了一个拥有2的62次方条边的多边形的周长。

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He spent 25 years on the effort computing to high accuracy the perimeter of a polygon with two to the 62 sides.

那是四百六十一亿亿六千八百六十一万八千四百二十七万三千八百七十九万零四条边。

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That is four quintillion, 611 quadrillion, 686 trillion, 18 billion, 427 million, 387,904 sides.

所有这些辛勤工作的回报是什么?

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What was the reward for all of that hard work?

仅仅是圆周率的35位正确小数。

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Just 35, correct decimal, places of Pi.

他将这些数字刻在了自己的墓碑上。

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He had these digits inscribed on his tombstone,

20年后,他的记录被克里斯托弗·格林伯格(Christoph Grienberger)超越,他得到了38位正确小数。

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20 years later, his record was surpassed by Christoph Grienberger who got 38, correct decimal places.

Derek: 但他是最后一个这样做的人。

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But he was the last to do it like this

Alex: 差不多吧。

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Pretty much.

是的,因为此后不久,艾萨克·牛顿爵士(Sir Isaac Newton)登场了。

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Yeah, because shortly thereafter we get Sir Isaac Newton on the scene.

一旦牛顿引入了他的方法,就再也没有人会去平分n边形了。

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And once Newton introduces his method nobody is bisecting n-gons ever again.

牛顿的突破:二项式定理

那一年是1666年,牛顿当时只有23岁。

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The year was 1666 and Newton was just 23 years old.

他因为腺鼠疫(bubonic plague)的爆发而在家隔离。

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He was quarantining at home due to an outbreak of bubonic plague.

牛顿当时正在研究像(1+x)²这样简单的表达式。

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Newton was playing around with simple expressions like one plus X, all squared.

你可以把它乘开得到1+2x+x²。

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You can multiply it out and get one plus two X plus X squared.

或者(1+x)³呢?

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Or what about one plus X all cubed?

同样,你可以把所有项乘开,得到1+3x+3x²+x³。

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Well, again, you can multiply out all the terms and get one plus three X plus three X squared plus X cubed.

你也可以对(1+x)⁴或(1+x)⁵等做同样的事情。

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And you could do the same for one plus X to the four or one plus X to the five and so on.

但牛顿知道有一种模式可以让他跳过所有繁琐的算术,直接得到答案。

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But Newton knew there was a pattern that allowed him to skip all the tedious arithmetic and go straight to the answer.

如果你看这些方程中的数字,即x和x²等的系数(coefficients),它们实际上就是帕斯卡三角形(Pascal's Triangle: 一个由二项式系数排列成的三角形数表)中的数字。

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If you look at the numbers in these equations the coefficients on X and X squared and so on, well, they're actually just the numbers in Pascal's triangle.

(1+x)的幂次对应着三角形的行。

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The power that one plus X raised to corresponds to the row of the triangle

帕斯卡三角形非常容易构建,它从古希腊人、印度人、中国人和波斯人那里就已经为人所知,许多不同的文化都发现了它。

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And Pascal's triangle is really easy to make, it's something that's been known from ancient Greeks in Indians and Chinese Persians, a lot of different cultures discovered this.

你所要做的就是,每当你有一行时,你只需将两个相邻的数字相加,就能得到下一行的值。

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All you do is whenever you have a row you just add the two neighbors and that gives you the value of the row below it.

所以这是一个非常快速简单的方法,你可以在一秒钟内计算出(1+x)¹⁰的系数,而无需坐在那里进行所有的代数运算。

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So that's a really quick easy thing you can compute the coefficients for one plus X to the 10 in a second instead of sitting there doing all the algebra.

Alex: 当我开始看那些古老文献时,让我着迷的是,即使我不懂那些语言,我也不知道那些数字系统,但很明显,他们都在写同样的东西,而今天在西方世界,我们称之为帕斯卡三角形。

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The thing that fascinated me when I started looking at those old documents was how even like, I don't speak those languages, I don't know those numbers systems and yet it is obvious, it is clear as day that they're all writing down the same thing which today in the Western world, we call Pascal's triangle.

Derek: 这就是数学的美妙之处。

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That's the beauty of Mathematics.

它超越文化,超越时间,超越人类。

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It transcends culture, it transcends time, it transcends humanity.

在我们消失很久之后,它仍将存在,古代文明,外星文明也会知道帕斯卡三角形。

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It's gonna be around well after we're gone and ancient civilizations, alien civilizations we'll know Pascal's triangle.

随着时间的推移,人们总结出了帕斯卡三角形中数字的通用公式。

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Over time, people worked out a general formula for the numbers in Pascal's triangle.

所以你可以计算任何一行中的数字,而无需计算它之前的所有行。

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So you can calculate the numbers in any row without having to calculate all the rows before it,

对于任何表达式(1+x)ⁿ,它等于1加上n乘以x,加上n乘以(n-1)乘以x²除以二阶乘(two factorial: 2! = 2 × 1 = 2),加上n乘以(n-1)乘以(n-2)乘以x³除以三阶乘(three factorial: 3! = 3 × 2 × 1 = 6),以此类推。

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for any expression one plus X to the N it is equal to one plus N times X plus N times N minus one X squared on two factorial plus N times N minus one times and N minus two times X cubed on three factorial and so on.

这就是二项式定理(binomial theorem: 一个用于展开形如 (x+y)^n 的代数表达式的公式)。

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And that's the binomial theorem.

之所以称之为“二项式”,是因为只有两项,1和x,而“定理”是指这是一个可以严格证明的定理,这个公式正是你在帕斯卡三角形中看到的系数。

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So binomial, because there's only two terms, one in X by is two, there's two normals and a theorem is that this is a theorem that you can rigorously prove that this formula is exactly what you'll see as the coefficients in Pascal's triangle.

Alex: 所以所有这些在牛顿时代就已经为人所知了。

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So all of this was known in Newton's day already.

Derek: 是的,没错,每个人都知道这个。

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Yeah, exactly, everybody knew this.

每个人都看到了这个公式,但没有人想到用它来做牛顿所做的事情,那就是“打破”这个公式。

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Everybody saw this formula and yet nobody thought to do with it the thing that Newton did with it which is to break the formula.

标准的二项式定理坚持你只能在n是正整数时应用它,这很合理。

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The standard binomial theorem insist that you apply it only when N is a positive integer, which makes sense.

这整个事情都是关于计算(1+x)自身相乘一定次数,但牛顿说,管它呢,直接应用这个定理。

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This whole thing is about working out one plus X times itself a certain number of times, but Newton says, screw that just apply the theorem.

数学是关于发现模式,然后扩展它们,并试图找出它们在哪里失效。

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Math is about finding patterns and then extending them and trying to find out where they break.

所以他尝试了(1+x)⁻¹。

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So he tries one plus X to the negative one.

也就是1/(1+x)。

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So that's one over one plus X.

如果我只是盲目地将n等于负一代入公式的右侧,会发生什么?

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What happens if I just blindly plug in N equals negative one for the right-hand side of the formula?

你得到的是各项交替出现:加一减一,加一减一,如此永远下去。

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And what you get is the terms alternate back and forth. Plus one minus one, plus one minus one, and so on forever.

所以那是1-x,下一项将是+x²,再下一项将是-x³,然后是+x⁴,-x⁵。

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So that's one minus X, the next term will be a plus X squared, the next one will be a minus X cubed plus X to the fourth minus X to the fifth.

所以那只是一个交替级数(alternating series),系数带有正负号。

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So that just alternating series with plus and minus signs as the coefficient.

Derek: 所以它变成了一个无穷级数(infinite series: 无限个项相加的数学表达式)。

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So it becomes an infinite series.

Alex: 是的,没错。

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Yeah, that's right.

如果你不使用正整数,二项式定理,也就是牛顿的二项式定理,会给你一个无限的和。

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If you, don't a positive integer the binomial theorem, Newton's binomial theorem will give you an infinite sum.

Derek: 但你如何理解这一点?

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But how do you understand that?

对于所有正整数,它只是一个有限的项集,而现在我们得到了一个无限的项集。

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Like for all positive integers it was just a finite set of terms and now we've got an infinite set of terms.

Alex: 是的,所以发生的情况是,如果你有一个正整数,你记得那个公式,系数看起来像n乘以(n-1)乘以(n-2)等等,当你达到n减n时,如果n是一个正整数,你最终会达到那里,n减n就是零。

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Yeah, so what happens is if you have a positive integer you remember that formula, the coefficient looks like N times N minus one times N minus two and so on, when you get to N minus N, if N is a positive integer, you will eventually get there and N minus 10 is zero.

所以那个系数以及它之后的所有系数都为零,这就是为什么它只是一个有限的和,它是一个有限的三角形。

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So that coefficient and all the coefficients after it are all zero and that's why it's just a finite sum, it's a finite triangle.

但一旦你用正整数超出了三角形的范围,你永远不会达到n减n,因为n不是一个正整数,所以你得到了这个无穷级数。

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But once you get outside of the triangle with positive integers, you never hit N minus N because N is not a positive integer, so you get this infinite series.

Derek: 所以我认为最大的问题是,这真的有效吗?

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So I think the big question is, does this actually work?

牛顿的无穷级数真的能给出1/(1+x)的值吗?

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Does Newton's infinite series actually give you the value of one over one plus X?

Alex: 对,它可能毫无意义。

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Right, and it might be nonsense.

当你这样做时,有很多数学公式可能会完全失效。

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There's lots of math formulas that could break completely when you do this.

我们制定规则是有原因的,但我们也应该始终知道规则在多大程度上可能继续有效。

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There's, we have rules for a reason but we should always know the extent to which the rules have a chance of working farther.

如果你把整个级数乘以(1+x),然后把所有项乘开,你会看到除了最前面的1之外,所有项都抵消了。

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If you take that whole series and you multiply it by one plus X and you multiply all that out you'll see all the terms cancel, except that leading one.

所以那个大级数乘以(1+x)等于1。

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And so that big series times one plus X is one.

换句话说,那个大级数就是1/(1+x),这就是牛顿向自己证明在不适用之处应用公式是合理的。

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In other words, that big series is one over one plus X, that's how Newton justified to himself that it makes sense to apply the formula where it shouldn't be applicable.

所以牛顿确信二项式定理即使对于负数n值也有效,这意味着帕斯卡三角形在零行之上还有更多内容。

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So Newton is convinced the binomial theorem works even for negative values of N, which means there's more to Pascal's triangle above the zeroeth you could add a zero and a one that add to make that first one.

然后那一行将继续是负一、正一、负一、正一,一直延伸到无穷。

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And then that row would continue minus one, plus one, minus one, plus one, all the way out to infinity.

在标准三角形之外,隐含的值处处为零。

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And outside the standard triangle the implied value everywhere is zero.

这与此相符。

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And this fits with that.

交替的正负一在它们下面的每一行中相加都等于零。

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The alternating plus and minus ones add to make zero everywhere in the row beneath them.

你可以用二项式定理或仅仅通过观察哪些数字会相加得到下面的数字来扩展所有负整数的模式。

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And you can extend the pattern for all negative integers either using the binomial theorem or just looking at what numbers would add together to make the numbers underneath.

这里有一个惊人的发现。

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And here's something amazing.

如果你暂时忽略负号,这些数字与主三角形中的数字完全相同,排列模式也一样。

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If you ignore the negative signs for a minute these are the exact same numbers arranged in the same pattern as in the main triangle.

整个东西只是被旋转了。

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The whole thing has just been rotated on its side.

但牛顿并没有止步于整数,接下来他尝试了分数幂,比如(1+x)的½次方。

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But Newton doesn't stop with the integers, next he tries fractional powers like one plus X to the half.

那么,(1+x)的½次方意味着什么呢?

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So now what does it mean, you take one plus X to the one half.

嗯,这和√(1+x)是一回事。

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Well, that's the same thing as square root of one plus X.

他想知道这是否也有相同的展开式。

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And he wants to understand does that have the same expansion.

将n等于½代入二项式定理,他得到了一个无穷级数。

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Putting N equals a half into the binomial theorem he gets an infinite series.

Derek: 这让我觉得我们实际上可以进入帕斯卡三角形,把它放大,并在我们熟悉的行之间添加分数。

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That makes me think that we could actually go into Pascal's triangle blow it up and add fractions in between the rows that we're familiar with.

Alex: 没错,甚至有一个帕斯卡三角形的连续统(continuum of Pascal's triangles),在零和一之间有一个连续的数字,你可以将其作为幂次代入。

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Exactly, there's even a continuum of Pascal's triangles, between zero and one there's this a continuum of numbers that you could put in for powers.

Derek: 你可以把每个分数,比如二分之一、四分之一、三分之一,都看作存在于它自己的平面中,在每个平面中,成对的数字相加就构成了它们下面的数字。

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And you can think of each fraction like a half, a quarter, a third as existing in its own plane where in each plane pairs of numbers add to make the number beneath them.

而且不再需要是正整数了。

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And doesn't have to be a positive integer anymore.

Alex: 它不需要是正整数,不需要是负整数,甚至不需要是整数。

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It doesn't have to be a positive integer, it doesn't have to be a negative integer it doesn't have to be an integer.

所以现在我们让n等于½,他算出了这个东西,然后他可以做各种各样的事情。

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So now we're gonna take N to be a half and he works this thing out and then he could do all kinds of things.

例如,他可以非常快速高效地计算出√3,因为我们可以把3写成4-1。

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For example, he could work out the square root of three very quickly and efficiently 'cause the square root of three we can write three is four minus one.

如果我们提取出4,那么我们得到√4,也就是2乘以√(1-¼)。

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And if we pull out a four, then we get a squire root four which is just two times the square root of one minus a quarter.

如果你在这个级数中将x代入-¼,你将得到一个非常快速收敛的级数展开式,它将迅速给你高精度的√3。

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If you put in minus a quarter for X in this series, you'll get a very rapidly converging series expansion that will quickly give you square root of three to high accuracy.

微积分与圆周率的计算

现在,牛顿特别关注n等于½的情况,因为单位圆(unit circle)的方程是x²+y²=1。

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Now, Newton is particularly interested in N equals a half because the equation for a unit circle is X squared plus Y squared equals one.

如果你解出y,那么圆的上半部分就等于√(1-x²)。

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And if you solve for Y, well the top part of the circle is equal to one minus X squared to the half.

这基本上是他一直在研究的相同表达式,他只需将x替换为-x²,这会增加一些负号并使每项中x的幂次翻倍。

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This is basically the same expression he's been looking at, he just has to replace X by minus X squared, which adds in some minus signs and doubles the power of X on each term.

但他现在得到了一个圆的方程,其中每一项都只是一个无理数(irrational number)乘以x的某个幂次。

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But now he's got an equation for a circle where each term is just irrational number times X raised to some power.

现在我们有两种不同的方式来表示同一事物。

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Now we have two different ways of representing the same thing.

每当你遇到这样的情况,奇迹即将发生,烟花即将绽放。

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And whenever you have something like that magic is about to happen, fireworks is about to go off.

Derek: 但他如何用这个来计算圆周率呢?

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But how does he use this to calculate Pi?

Alex: 幸运的是,他刚刚发明了微积分(calculus: 研究变化率和累积的数学分支),或者他称之为流数术(theory of Fluxions: 牛顿对微积分的称呼)。

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Well, luckily for us, he had just invented calculus or what he called the theory of Fluxions.

他意识到,如果他在x从零到一的范围内对那条曲线进行积分(integrate),他将得到曲线下的面积,也就是一个四分之一圆(quarter circle)。

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He realizes that if you integrate under that curve as X goes from zero to one, you're getting the area under the curve, which is a quarter circle.

他知道单位圆的面积正好是πr²,但r是1,所以面积是π。

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And he knows that the area of a unit circle is exactly Pi R squared except R is one, so the area is Pi.

而我们只想要四分之一,所以面积是π/4。

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And we want just a quarter, so the area is Pi over four.

另一方面,他有这个漂亮的级数,他知道如何对x的某个幂次进行积分。

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On the other side, he has this nice series and he knows how to integrate X to some power.

你只需将x的每个幂次增加一,然后除以新的幂次。

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You just increase each power of X by one and divide by the new power.

现在你得到了一个无穷项的级数,其中只涉及简单的分数运算。

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And now you have an infinite series of terms which just involve simple arithmetic with fractions.

你代入x等于一,就可以计算出任意高精度(arbitrarily high precision)的圆周率。

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You put an X equals one and you can calculate Pi to an arbitrarily high precision.

牛顿的最终优化与影响

但牛顿更进一步,增加了最后一个调整。

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But Newton goes even further adding one final tweak.

一篇不好的数学论文没有新思想,只是推动每个人都知道但没人费心去做的事情。

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A not good math paper has zero ideas, it's just pushing through things that everybody already knows but nobody bothered to do.

然后有一些好的数学论文,它们有一个真正令人震惊的新思想。

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Then there are good math papers that have one new idea that's really shockingly.

牛顿此时已经有了第四个新思想,他即将提出第五个新思想。

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New Newton's on new idea number four at this point and he's about to have new idea number five.

第五个新思想是,他不再从零到一积分,而是只从零到二分之一积分。

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And new number five is instead of integrating from zero to one he's gonna integrate just from zero to a half.

当你有一个无穷级数时,你希望各项的大小尽可能快地减小。

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When you have an infinite series you want the terms to decrease in size, as fast as possible.

这样你就不必计算那么多项就能得到一个相当好的答案。

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That way you don't have to calculate as many of them to get a pretty good answer.

牛顿看到,如果他不是从零到一积分,而是从零到二分之一积分,那么当他将二分之一代入x时,每一项的大小将额外缩小一个x²的因子,在本例中是四分之一。

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And Newton sees if he integrates not from zero to one, but from zero to a half, then when he subs in a half for X, each term will shrink in size by an additional factor of X squared, which in this case is a quarter.

但如果你只积分到二分之一,你计算的曲线下面积是什么?

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But if you only integrate to a half, what is the area under the curve that you're computing?

嗯,它是圆的这一部分,你可以将其分解为一个30度的圆扇形(sector),其面积为π/12,加上一个底边为二分之一、高为√3/2的直角三角形(right triangle)。

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Well, it is this part of a circle, which you can break into a 30 degree sector of the circle, which has an area of Pi on 12 plus a right triangle with a base of a half and a height of root three on two.

所以那个积分应该得到这个表达式。

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So that integral should come out to this expression.

重新排列圆周率,你得到以下结果。

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And rearranging for Pi, you get the following.

现在,如果你只评估前五项,你得到圆周率等于3.14161,这只差十万分之二。

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Now, if you evaluate only the first five terms you get Pi equals 3.14161, that's off by just two parts in a 100,000.

为了达到范·科伊伦那个有四百六十一亿亿条边的多边形的计算能力,你只需要计算牛顿级数的50项。

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And to match the computational power of Van Ceulen four quintillion sided polygon, you would only need to compute 50 terms in Newton series.

以前需要数年才能完成的工作,现在只需要几天。

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What, before it took years now would take only days.

所以再也没有人会通过平分多边形来寻找圆周率了。

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So no one was bisecting polygons to find Pi ever again.

你为什么要那样做呢?

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Why would you, yeah you do all that work and somebody comes along and beats you in a second.

这有点像,一旦有人建造了起重机,而另一个人还在爬梯子往房子上放砖头,那样就不是你现在建造房子的方式了,我们有了新技术,你是不是疯了,我们要建一个100层的房子,我们要建一个5层的房子,它会倒塌。

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It's sort of like a, once someone builds a crane and then somebody else is still climbing up on a ladder to put a brick on a house, like that's just not how you build houses anymore, we have new technology, are you out of your mind, we're gonna build a 100 story house, we're gonna build a five story thing that's gonna fall over.

你在纽约市就能看到。

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You see it in New York city.

你看到,当技术出现时,确实如此。

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You see, literally we're technology came along.

一排排的五层建筑,突然间这里有一栋20层的,那里有一栋30层的,还有一栋90层的。

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There's rows and rows of five story buildings and all of a sudden here's a 20 story and here's the 30 story, and here's a 90 story.

所以这一切都取决于谁拥有技术。

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So it's all about who has the technology.

对我来说,这是一个关于显而易见的方法并非总是最好的方法的故事,而且尝试玩弄模式并将其推向超出预期范围的地方通常是个好主意。

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For me, this is a story about how the obvious way of doing things is not always the best way and that it's often a good idea to play around with patterns and push them beyond the bounds where you expect them to work.

因为一点点洞察力和数学知识就能走得很远。

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Because a little bit of insight and mathematics can go a very long way.

赞助商信息:Brilliant

这段视频由Brilliant赞助,这是一个提供互动课程和测验的网站,让你深入学习本视频中展示的主题,如微积分、神经网络(neural networks: 模拟人脑结构和功能的计算模型)、Python编程(programming in python: 一种高级通用编程语言),他们都能满足你的需求。

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Hey, this video was sponsored by Brilliant a website with interactive courses and quizzes that let you dive deep into the topics like the ones I've shown in this video, calculus, neural networks, programming and python, they've got you covered.

有时我会被问到为什么我的视频不深入细节或解决数值问题,答案是,我认为视频不是学习这些技能的最佳方式。

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Now, I sometimes get asked why I don't get into the nitty-gritty detail or solve numerical problems in my videos and the answer is because I don't think a video is the best way to learn those skills.

最好的方式是像你在Brilliant上那样,亲自参与解决问题。

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The best way is to engage yourself in problem solving like you can on Brilliant.

我喜欢他们通过支架式教学法(scaffold: 一种教学策略,通过提供支持来帮助学习者逐步掌握新技能和知识)引导你学习一个主题的方式,在学习过程中建立你的理解和信心。

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I love the way they scaffold you through a topic, building your understanding and your confidence as you go.

作为一名拥有科学教育博士学位(PhD in science education)的人,我可以说这是唯一真正的学习方式,你必须感到有点不适才能获得理解。

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And as someone with a PhD in science education, I can say this is the only real way to learn, you have to be a little uncomfortable to gain understanding.

当我做这些测验时,我发现我的大脑真的在工作。

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When I do these quizzes, I find that my brain is really working.

所以我可以保证,无论你处于哪个教育水平,Brilliant都会有适合你的内容。

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So I can guarantee that whatever level of education you're at, Brilliant we'll have something for you.

它是观看有趣的教育视频的完美补充。

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It's the perfect compliment to watching fun educational videos.

对于本频道的观众,Brilliant为前314名(或π乘以100)注册用户提供年度订阅20%的折扣,只需访问brilliant.org/veritasium。

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And for viewers of this channel, Brilliant are offering 20% off an annual subscription to the first 314 or Pi 100 people to sign up, just go to brilliant.org/.veritasium

我会在描述中附上这个链接。

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I will put that link down in the description.

所以我要感谢Brilliant对Veritasium的支持,也要感谢你的观看。

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So I wanna thank Brilliant for supporting Veritasium and I wanna thank you for watching.

📌 文中提及的人物和组织

人物: Isaac Newton, Derek Muller

公司/组织: Brilliant

媒体/书籍: infinite-series

关键字: history pi-calculation science