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可视化黎曼zeta函数和解析延拓 – 译学馆
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可视化黎曼zeta函数和解析延拓

Visualizing the Riemann hypothesis and analytic continuation

黎曼ζ函数
The Riemann zeta function.
你们大部分人可能对这个现代
This is one of those objects in modern math that
数学概念有所耳闻
a lot of you might have heard of, but
但是难以理解它别担心
which can be really difficult to understand.
很快我就会解释
Don’t worry, I’ll explain that animation
你刚才看到的动画
that you just saw in a few minutes.
不少人知道这个函数 是因为求出其零点的
A lot of people know about this function because there’s a one-million-dollar prize
人能够获得一百万美元的奖励
out for anyone who can figure out
这一悬而未解的问题
when it equals 0. An open problem known as
就是”黎曼猜想”
the Riemann hypothesis. Some of you may
有人可能是从发散
have heard of it in the context of the
级数1+2+3+4
divergent sum 1 + 2 + 3 + 4…
+…听说
on and on up to

infinity.
有一种观点认为它
You see there’s a sense in which the sum
等于-1/12
equals -1/12, which seems
这看起来似乎很荒谬 甚至可以说错得离谱
nonsensical if not obviously wrong. But a
但是定义它的一种
common way to define what this
常见方法 就用到了黎曼ζ函数
equation is actually saying uses the Riemann zeta function.
不过 了解过这个函数
But as any casual Math
的数学爱好者都知道
enthusiast who started to read
它的定义涉及一个概念
into this knows its definition references this one
叫做“解析延拓”
idea called analytic continuation which
这个概念与复值函数有关 而且它既不直观又艰深难懂
has to do with complex-valued functions and this idea can be frustratingly opaque and unintuitive,
所以
so what I’d like
我想做的 是向你们展示
to do here is just show you all what
ζ函数究竟长什么样
this zeta function actually looks
并且用一种更加
like and to explain what this idea of
形象直观的方法解释
analytic continuation is in a visual and
解析延拓
more intuitive way. I’m assuming that you
我假定你了解什么是复数 并且能轻松自如地运用它
know about complex numbers and that you’re comfortable working
或许你还应该
with them, and I’m tempted to say that
了解微积分 因为解析延拓就是与
you should know calculus since analytic continuation is
导数相关的概念 但是按照我计划
all about derivatives but for the way I’m
的讲解方式 我觉得你不懂微积分也是可以的
planning to present things I think you might actually be fine without that.
接下来进入正题
So to jump right into it let’s just define
我们先来定义ζ函数 对于任意一个数
what this zeta function is for a given input where we
我们通常用变量s代表它
commonly use the variable’s’ the function
这个函数就是1/(1^s)
is 1 over one to the’s’ (which is always 1)
(也就是1) 加上1/(2^s)
+ 1 over 2 to the’s’ + 1 over 3 to
加上1/(3^s) 加上1/(4^s)
the’s’ + 1 over 4 to the’s’
这样一直下去
on and on and on
对所有自然数求和比如说
summing up over all natural numbers.
你将
So for example let’s say you plug in a
s=2代入其中
value like: s = 2
就能得到1+1
you get 1
/4+1/9+1/16+…
+ (1 over 4) + (1 over 9) + 1/16 and as
当你继续把更多的平方数的倒数加起来
you keep adding more and more reciprocals
数字恰好
of squares this just so
趋向于π^2/6
happens to approach pi squared over 6
大约是1.645
which is around 1.645 there’s a very
π在这里出现 背后有个漂亮的原因
beautiful reason for why pi shows up here
我将来可能会制作一期视频不过
and I might do a video on a later date
它也只是这个
but that’s just the tip of the iceberg
美妙函数的冰山一角
for why this function is beautiful.
你还可以代入其他的值
You can do the same thing for other inputs’s’
比如3或4
like three or four and
有时你也能得到一些有趣的结果 到目前为止
sometimes you get other interesting values and so far everything feels
一切都相当合理 你在对越来越小的数加和 而且和趋向于某个数
pretty reasonable you’re adding up smaller and smaller amounts and these sums approach some number… Great,
赞!
no
没有奇怪的东西出现 不过
craziness here! Yet if you were to read
如果你读过一些资料 你可能注意到有些人
about it you might see some people say
说ζ(-1)等于
that zeta of negative 1 equals
-1/12 但是看看这个无穷级数
-1/12 But looking at this infinite sum
这根本就说不通
that doesn’t make any sense… when you
当你取每一项的(-1)次方
raise each term to the negative 1
颠倒每一个分数
flipping each fraction you get 1
你会得到1+2+3+
+ 2 + 3 + 4
4+… 也就是所有自然数的和
on an on over all natural numbers and
显然它不趋向于任何值 更不会是-1/12
obviously that doesn’t approach anything certainly not -1/12, right? And,
对吧?挑战过黎曼猜想奖金的
as any mercenary looking into the Riemann
人也都知道
hypothesis knows this function is said
负偶数是这个函数的平凡零点
to have trivial zeros at negative even numbers
所以这意味着ζ
so for example that would mean that zeta
(-2)等于0
of negative 2 = 0, but when you plug
但是当你代入-2时
in -2 it gives you
你会得到1+4+9
1 + 4 + 9 + 16
+16+…
on and on, which again
和之前一样它明显不趋向于任何值
obviously doesn’t approach anything much
更别提0了 对吧? 我们稍后再说
less 0, right? Well we’ll get to negative
如何代入负值
values in a few minutes but
现在我们先谈谈合理的情况
for right now let’s just say the only thing that
只有当s大于1时
seems reasonable this function only makes sense when’s’ is
这个函数才有意义 因为这时级数
greater than one which is when this sum
才会收敛 目前为止
converges so far it’s simply not defined
它还没有为其他值定义可以说
for other values.
波恩哈德·黎曼
Now with that said Bernhard Riemann was
能够称得上是“复分析之父”
somewhat of a father to complex
这个理论所研究的函数
analysis which is the study of functions that
变量和取值都是复数
have complex numbers as inputs and outputs.
他所考虑的
So rather than just thinking about how
不是如何将实轴上
this sum takes a number’s’
的某个数s变换到实轴上
on the real number line to another number on the
的另一个数 而是着眼于代入
real number line his main focus was on understanding what
一个复数s会有什么结果
happens when you plug in a complex value for’s’,
比如说
so for example maybe instead of
你代入的数可能不是2
plugging in 2, you would plug in 2 + i
而是2+i 如果你从未见过
now if you’ve never seen the idea of
一个数的虚数次方
raising a number to the power of a complex value you
可能一开始你会觉得陌生
can feel kind of strange at first because it no longer
因为它不再与多次相乘有关 但是数学家发现
has anything to do with repeated multiplication but mathematicians found
了一种很好很自然的方法
that there is a very nice
将指数的定义
and very natural way to extend the definition of
从熟悉的实数域拓展到
exponents beyond their familiar territory of real numbers and into the
复数域中
realm of complex values. It’s not super
就这期视频的内容来说 理解复指数并不关键
crucial to understand complex exponents for where I’m going
但我还是觉得
with this video but I think it’ll still be
总结一下要点还是好的
nice if we just summarize the gist of it here
它的基本思想在于
the basic idea is that when you write
当你取一个数(如1/2)的复数次方时
something like one half to the power
你把它分成
of a complex number you split it up as
1/2的实部次方乘以1
one-half to the real part times one-half
/2的虚部次方
to the pure imaginary part we’re good on
我们熟悉1/2的实部次方 这部分没有问题
one half to the real part there’s no issues there but what about raising
但是取一个数的纯虚数次方呢?
something to a pure imaginary number?
它的结果就是复
Well the result is going to be some
平面单位圆上的某个复数
complex number on the unit circle in the
当你让
complex plane as you let that pure
指数沿着虚轴上下移动
imaginary input walk up and down the
结果就
imaginary line the resulting output
会沿着单位圆移动
walks around that unit circle
如果1/2作为底数
For a base like one half the output
结果会以较慢的速度在单位圆上移动
walks around the unit circle somewhat slowly but
但是如果底数
for a base that’s farther
离1很远 比如1/9
away from one like 1/9 then as you let
当指数沿着
this input walk
虚轴上下移动时
up and down the imaginary axis the corresponding output
结果便会以更快的速度在单位圆上移动
is going to walk around the unit circle more quickly.
如果你从未见过这个
If you’ve never seen this
想知道究竟
and you’re wondering what
为什么会这样 可以去看看视频简介/评论中的优质资源链接
on earth this happens I’ve left a few links to good resources in the description for here
这里我就直接使用结果 而不再解释原因
i’m just going to move forward with the what without the why.
重点在于 当你计算(1/
The main takeaway is that when you raise something
2)^(2+i)时
like 1/2 to the power of 2 + i which
也就是(1/2)^2乘以(1/
is one-half squared times one-half to
2)^i (1/2)^i那
the i that one-half to the i part is
部分就在单位圆上
going to be on the unit circle meaning
意味着它的模是1
it has an absolute value of one. So when
所以(1/2)^2与它相乘时
you multiply it it doesn’t change the size
它并不改变这个数的模
of the number it just takes that
只是将1/4旋转了一下
one fourth and rotates at somewhere.
所以当你将2+i代入
So if you were to plug in 2 + i to
ζ函数中
the zeta function one way to think about
一种思考方法是 首先取每一项的平方
what it does is to start off with all of

the terms raised to the power of 2 which
可以看作
you can think
是把长度为平方数倒数的
of is piecing together lines whose length of the reciprocals of
线段拼凑起来
squares of numbers which like I said
正如我之前所说 它收敛到π^2/6
before converges to pi² over six
然后当指数从2
then when you change that input from two
变成2+i时 每条线段都经过
up 2 + i each of these lines gets
了一定的旋转
rotated by some amount but importantly
但重点在于这些线段的长度不发生变化
the lengths of those lines won’t change
所以和仍旧收敛
so the sum still converges it just does
它只是以螺旋的方式收敛到复平面
so in a spiral to some specific point on
上的某个点
the complex plane. Here let me show what
现在让我展示一下s变化时会发生什么
it looks like when I vary the input is represented
复平面上的黄点代表s
with this yellow dot on the
而螺旋线的终点
complex plane where this spiral sum is
代表这个和
always going to be showing the
的收敛值 也就是ζ(s)
converging value for zeta of s
这也就是说 用这个无穷级数定义
what this means is that zeta(s) defined
的ζ(s)是一个完全
as this infinite sum is a perfectly
合理的复函数
reasonable complex function as long as
只要代入值的实部大于1
the real part of the input is greater
也就是s落在复平面
than one meaning the input’s’ sits somewhere
的右半部分即可
on this right half of the complex plane again this is
这是因为s的实部
because it’s the real part
决定每个数的模
of s that determines the size of each
而虚部只决定
number while the imaginary part just
它的旋转
dictate some rotation.
我现在
So now what I want to
想将这个函数可视化
do is visualize this function it
给定复平面右半部分
takes in inputs on the right half
的一个数 函数就会在复
of the complex plane and spits out outputs
平面上输出另一个数
somewhere else in the complex plane a
理解复函数的一个好方法是通过变换
super nice way to understand complex functions is to visualize them
来将其可视化
as transformations meaning you look at
意思是说你考虑函数所有能代入的值
every possible input to the function and
然后让它们移动到对应的像
just let it move over to the corresponding output…
我们先花点
for example let’s
时间来可视化一个比ζ(s)更
take a moment and try to visualize something a little bit easier than the
简单的函数 比如f(s)=s^2
zeta function: say f(s) = s²
代入2会得到4
When you plug in s = 2 you get 4
所以2代表
so we’ll end up moving that
的点最终会移动到4的位置
point at two over to the point at four
代入-1则会得到1
when you plug in -1 you get 1 so
所以-1代表
the point over here
的点会移动到1的位置
at negative 1 is going to end up moving over to the point at 1.
当你代入i
When you plug in
按照定义 它的平方是-1
i by definition its square is -1
所以它会移动到-1的位置
so it’s going to move over here
因为整个平面
to negative 1 now I’m going to add on a
要开始移动了 所以我在这里添加一个
more colorful grid and this is just because things are about
彩色网格 这样做是为了更好
to start moving and it’s kind of nice to have
区分移动前后的网格线
something to distinguish grid lines during that movement.
这里
From here I’ll tell the
我让计算机将这个平面上的
computer to move every single point on
每一个点
this grid over to its corresponding
在f(s)=s^2的作用下移动到像的位置上
output under the function f(s) = s²
这就是最终结果
Here’s what it looks like
这里信息量有点大
I can be a lot to take in so I
所以我重新播放一遍
‘ll go ahead and play it again and this time
这一次关注其中一个标记点
focus on one of the marked points and
注意它是如何移动到其平方
notice how it moves over to the point
数代表的位置上的
corresponding to its square. It can be a
同时观察所有点移动
little complicated to see all
或许会有些混乱
of the points moving all at once but the reward
但是好处在于它给出了丰富的图像 让我们了解复
is that this gives us a very rich picture for what
函数究竟在做什么 而且这一切都在二维空间中发生!
the complex function is actually doing and it all happens in
回到ζ函数
just two dimensions… so back to the zeta
我们有一个无穷级数
function we have this infinite sum which
它是复数s的函数
is a function of some complex number s and we feel good and happy
并且我们能顺利代入实部大于1
about plugging in values of s whose real part
的复数s
is greater than one and getting
然后根据螺旋线的收敛得到有
some meaningful output via the converging
意义的结果 为了将这个函数可视化
spiral cell so to visualize this
我取出复平面
function i’m going to take the portion of the grid
的右半部分
sitting on the right side of the complex plane here where
也就是实部大于1的部分
the real part of numbers is greater than one and
让计算机将每一个点
I’m gon na tell the computer to move
移动到它对应的位置实际上
each point of this grid to the appropriate output
在1附近
it actually helps if I add a few
添加些额外的网格线有助于演示
more grid lines around the number one
因为这部分会被向外拉伸很多很好
since that region gets stretched out by quite a bit
首先我们得好好
alright so first of all let’s just
欣赏这究竟有多么漂亮好家伙!
appreciate how beautiful that is
如果这都无法
I mean damn that doesn’t make you want
激起你学习复函数的兴趣 那你真的有些无情不过
to learn more about complex functions you have no heart.
变换后的网格非常想要继续
But also this transformed grid is just begging to be
向左延伸 比如说
extended a little bit for example let’s
我们来关注这样两条直线 它们由虚部为i或-i的全体复数组成
highlight these lines here which represent all of the complex numbers
它们
with imaginary part i or -i after the
在变换后成为两段美妙的弧线
transformation these lines make such lovely arcs before they just abruptly
却在一处戛然而止 难道说
stopped don’t you want to just you
你就不想补全这两段弧吗? 事实上你可以想象
know continue those arcs in fact you can imagine how
存在着一个改良版的函数
some altered version of the function
它的定义域延伸
with the definition that extends into
至左半平面
this left half of the plane might be
这样也许能够补全这
able to complete this picture
幅美妙的图景 这正是研究复函数
with something that’s quite pretty well this is exactly what mathematicians
的数学家所做的
working with complex functions do! They
他们将函数延拓至原
continue the function beyond the
定义域之外
original domain where was defined now as
只要我们代入实部小于
soon as we branch over into inputs where
1的复数
the real part is less than 1 this
我们原先讨论过的无穷
infinite sum that we originally used to
级数和就不再有意义
define the function doesn’t make sense
你会得到形如1
anymore you’ll get nonsense like adding
+2+3
1 + 2 + 3 + 4
+4+…的荒谬结果
on a non up to infinity
但是
But just looking at this transformed
看看右半平面(无穷级数
version of the right half
收敛的区域)变换后的结果
of the plane where the some does make sense it’s just
它在乞求我们扩大
begging us to extend the set
定义域的范围
of points that were considering as inputs even if
即便经过某种方式延拓后的函数
that means defining the extended function
不一定用到那个无穷级数
in some way that doesn’t necessarily use that sum of course that
这自然留下了一个问题
leaves us with the question how would you define that function
我们究竟如何在余下的平面内定义这个函数?
on the rest of the plane?
你可能会觉得
You might think that you could extend it
存在无数种方法进行延拓
any number of ways maybe you define an
比如说你可以定义一种延拓
extension that makes it so the point at say…
它将s=-1映射为
s = -1 moves over to
-1/12 但你也可以对它
-1/12 but maybe you squiggle on
进行扭曲 使得s落在其他位置上
some extension that makes it land on any other value I
只要你放开头脑
mean as soon as you open yourself up to the idea
大胆想象
of defining the function differently for values outside that
不根据无穷级数来定义收敛域之外
domain of convergence that is not based
的函数值
on this infinite sum the world is your
大千世界便尽在掌握 你就有无限种方式来进行延拓
oyster and you can have any number of extensions right?
对吗?
Well not exactly I mean
并不尽然当然 你可以给小孩一支笔
yes you can give any child a marker and
让他随意延长那些弧线
have them extend these lines any
但如果加上一个
which way but if you add on the restriction
限制条件:
that this new extended function has to
延拓后的复函数处处可导
have a derivative everywhere it locks us
那就迫使我们选择唯一的延拓方式了
into one and only one possible extension
好啦好啦
I know I know…
我知道我曾经说过 看这期视频不需要知道导数
I said that you wouldn’t need to know about derivatives
即便你学过微积分
for this video and even if you do know calculus maybe
你也可能还没学到复函数的导数是什么意思
you have yet to learn how to interpret derivatives for complex
幸运的是
functions but luckily for us there is a
你可以记住一种非常好
very nice geometric intuition that you
的几何直觉 每当我说“处处可导”时
can keep in mind for when I say a phrase
你就可以这么想
like has a derivative everywhere here to
为了说明这句话的意思
show you what I mean let’s look back at
我们回头看看之前的例子:f(s)=s^2
that f(s) = s² example
我们还是把这个函数
again we think of this function as a
看作变换 它将复平面上
transformation moving every point s of
的每一个点s移动到s^2
the complex plane over to the point s²
那些懂微积分的人
for those of you who know
知道这个函数能
calculus you know that you can take the derivative
在任何地方求导
of this function at any given input but there
但是作为变换 它有一条有趣的性质事实上
‘s an interesting property of that transformation that
这条性质和“处处
turns out to be related and
可导”几乎等价
almost equivalent to that fact if you look at
如果任取原空间中两条
any two lines in the input space that
相交的直线
intersect at some angle and consider
然后考虑变换之后的像 它们的交角并
what they turn into after the transformation they will still intersect
不会发生改变
each other at that same angle.
直线可能会有所弯曲 但是这不要紧
The lines might get curved and that’s okay but the
重要的是
important part is that the angle
它们的交角并不改变
at which they intersect remains unchanged
取任何一对直线
and this is true for any pair of lines
这一点都成立
that you choose!
所以当我说一个
So when I say a function has a
函数“处处可导”时 我希望你心里
derivative everywhere I want you to think about
想的是这个保角特性
this angle preserving property that anytime two lines
即任意一对直线的交角在变换
intersect the angle between them remains
之后保持不变
unchanged after the transformation at a
一眼瞥去
glance this is easiest to appreciate by
你会注意到网格线在变换
noticing how all of the curves
后仍然彼此正交
that the gridlines turn into still intersect each
这是最容易理解保角性的
other at right angles.
处处可导的复函数被称为是“解析函数”
Complex functions that have a derivative everywhere are called analytic
所以你可以
so you can think of this
把“解析的”理解为
term analytic as meaning angle
“保角的” 我得承认这并不完全准确
preserving admittedly i’m lying to a
但也几乎无差
little here but only a little bit a
我稍微提醒一下
slight caveat for those
那些追根究底的观众
of you who want the full details is that inputs where
在那些导数为0的点处
the derivative of a function is 0 instead
交角并非保持不变
of angle being preserved they
而是变成了某个整数倍
get multiplied by some integer, but those points are
但这些点只是少数中的少数
by far the minority and for almost all
几乎所有情况下 解析函数都保持
inputs to an analytic function angles
交角不变 因此我说“解析”时
are preserved so when I say analytic you think angle
你完全可以将它直观地理解为“保角”
preserving I think that’s a fine intuition to have
现在请你仔细体会
now if you think about it for a moment
我非常希望你能明白
and this is the point that i really want
这是一个
you to appreciate this is a very
非常强的约束条件
restrictive property the angle between
任意一对直线的交角必须
any pair of intersecting lines has to
保持不变 不过
remain unchanged and yet pretty much any
那些能叫出名字的函数几乎全都是解析的
function out there that has a
黎曼
name turns out to be analytic the field of
协助建立了复分析
complex analysis which Riemann helped to
的现代形式
establish in its modern form is almost
而这门理论几乎完全是关于利用
entirely about leveraging the properties
解析函数的性质
of analytic functions to understand
来理解数学的其他领域及自然科学中出现的结果和规律
results in patterns and other fields of math and science.
在右半平面上以无穷级数定义
The zeta function defined by this infinite sum on the
的ζ函数是一个解析函数
right half of the plane is an analytic
注意一下
function notice how all of these curves
网格线在变换
that the gridlines turn
后仍然彼此正交
into still intersect each other at right angles
复函数的一个出人意料
so the surprising fact about complex
的性质在于
functions is that if you want to extend
如果你想把一个解析函数延拓至原
an analytic function beyond the domain
定义域之外
where was originally defined for example
比如将ζ函数延拓至
extending this zeta function into the
左半平面
left half of the plane then if you
并且你要求延拓后的函数
require that the new extended function
仍然解析
still be analytic that is that it still
也就是说它处处保角 那么你只能按照唯一的方式来
preserves angles everywhere it forces you into only one possible
进行延拓(如果它存在的话)
extension if one exists at all
这就像是一幅无限而
it’s kind of like an infinite continuous
连续的拼图
jigsaw puzzle for this requirement of
保角的条件把你限制在唯一
preserving angles walks you into one and
的延拓方式上
only one choice for how to extend it
以唯一一种保持解析性
this process of extending an analytic
的方式延拓解析函数的过程
function in the only way possible that
或许你已经猜到了
still analytic is called as you may have
就叫做“解析延拓”
guessed”analytic continuation” so that’s
以上就是完整黎曼ζ函数的定义过程
how the full Riemann’s zeta function is defined
对于右半平面
for values of s on the right
也就是实部大于1的s
half of the plane where the real part
直接把它们代入这个和
is greater than one just plug them into
看它收敛到何处
this sum and see where it converges and
这种收敛可能看上去像是某种螺旋
that convergence might look like some kind of spiral since raising each
因为每一项取复数次幂有着
of these terms to a complex power has the
旋转的效果
effect of rotating each one then for the
对于余下的平面
rest of the plane we know
我们知道存在唯一一种
that there exists one and only one way to extend
延拓方式
this definition so
使得函数仍然解析
that the function will still be analytic that is so that
也就是说它仍然处处保角
it still preserves angles at every single point
所以我们就说
so we just say that by
不管这个延拓的结果是什么
definition the zeta function on the left
左半平面的ζ
half of the plane is whatever that
函数就是它了
extension happens to be and that’s a
而且这是一个有效的定义 因为只有一种可能
valid definition because there’s only one possible
的解析延拓 注意
analytic continuation notice that’s a very implicit definition
这个定义很不直接 这只是说使用“拼图”
it just says use the solution of this
得到的结果 通过繁杂抽象的推导 我们知道它必须存在
jigsaw puzzle which through more abstract derivation we know must exist
但这并没有指明我们如何求解它
but it doesn’t specify exactly how to
数学家基本上理解
solve it mathematicians have a pretty good grasp
了这个延拓的形式
on what this extension looks like but
但是某些重要部分
some important parts
仍旧是个谜 实际上是个价值一百万美元的谜
of that remain a mystery a million-dollar mystery in fact let’s
我们来花点时间讨论黎曼猜想
actually take a moment and talk
这个价值百万的问题 这个函数的零点
about the Riemann hypothesis the million-dollar problem the places where this function equals
原来是相当重要的
zero turn out to be quite important that
也就是说哪些点在变换之后
is which points get mapped onto the origin
与原点重合
after the transformation one thing we
关于这个延拓 我们知道所有
know about this extension is
的负偶数被映射到0
that the negative even numbers get map to 0 these
这些点通常被称为“平凡零点”
are commonly called the trivial zeros
这个命名方式源于数学家
the name here stems from a long-standing
的一个悠久的传统 他们能很好地理解的东西就是“平凡的”
tradition of mathematicians to call things trivial when they understand
即便这个事实乍一看来
quite well even when it’s a fact that is
一点都不显然
not at all obvious from the outset we
我们也知道剩下
also know that the rest
的零点落在这条竖直的
of the points that get map to 0 sit somewhere in this
带状区域中 它被称为“临界带”
vertical strip called the critical strip
而这些非平凡零点
and the specific placement of those
的具体分布
non-trivial zeros encodes a surprising
蕴含着有关素数的海量信息
information about prime numbers it’s
为何这个函数包含这么多关于素数的信息?
actually pretty interesting why this function carry so much
说起来非常有趣 我想我之后会
information about primes and I definitely think i’ll
制作一期有关的视频
make a video about that later on but right
不过这期视频已经很长了
now things are long enough so
我就先不解释了
I’ll leave it unexplained Riemann hypothesized that
黎曼猜想道:所有非平凡零点全部落在临界
all of these non-trivial zeros sit right
带的正中央
in the middle of the strip on the line
即实部为1/2的数所构成的直线上
of numbers s who’s real part is
它被称为
one-half this is called the critical
“临界线” 如果这是真的
line if that’s true it gives us a
它能让我们深刻理解素数
remarkably tight grasp on the pattern of
分布的规律 同样也包含由此
prime numbers as well as many other
衍生出的其他众多数学规律
patterns in math stem from this now so
到目前为止我在展示ζ函数时
far when I shown what the zeta function looks
都只展示了
like I’ve only shown what it does
屏幕上的这一部分
to the portion of the grid on the screen
这可能低估了它的复杂程度
and that kind of under sells its complexity
当我标注出临界线
so if I were to highlight
然后应用变换
this critical line and apply the
它似乎根本
transformation it might not seem to
不通过原点 但是
cross the origin at all however use with
变换后的临界线延伸
the transformed version of more and more
之后是这样的
of that line looks like
注意这条线
notice how its passing through the
多次穿过了原点
number zero many many times if you can
如果你能证明
prove that all
所有非平凡零点落在这条线上
of the non-trivial zeros sit somewhere on this line the clay math
克雷数学研究所就会奖励你一百万美元
Institute gives you 1 million dollars and you’d
同时你也将证明了成百上千的现代
also be proving hundreds if not thousands of modern
数学结论 因为它们都是在假定黎曼
math results that have already been shown contingent
猜想成立的前提下得到的
on this hypothesis being true
我们还知道这个延拓后的
another thing we know about this extended function is
函数将-1映射到-1/12
that map’s the point -1 over to negative -1/12
如果你把-1代入一开始的和里面
and if you plug this into the original
这似乎
sum it looks
是在说1+2+
like we’re saying 1 + 2 + 3 + 4
3+4+…等于-
on and on up to infinity equals
1/12 我们要是仍然
-1/12 now they might seem
把它称为一个和式 似乎有些狡猾
disingenuous to still call this is a sum since the
因为ζ函数在左半平面内的定义
definition of the zeta function on the left half of
并不是直接从“和”而来的
the plane is not defined directly from this sum
而是来源于这个
instead it comes from analytically
和在原定义域外
continuing this own beyond the domain
的解析延拓
where it converges that is solving the
也就是完成这个从右半平面开始的拼图
jigsaw puzzle that began on the right half
即便如此
of the plane that said you have to
你得承认这个
admit that the uniqueness
解析延拓有唯一性
of this analytic continuation the fact that the
这个拼图只有一种解法
jigsaw puzzle has only one solution is
这强烈暗示着 延拓得到的值与原始和之间有着
very suggestive of some intrinsic connection between these extended values
某种内在联系 最后还有一个
and the original sum the last animation
非常漂亮的动画
and this is actually pretty cool i’m going to
我会向你们展示ζ函数的导函数是什么样的
show you guys what the derivative of the zeta function
不过在此之前
looks like but before that it matters to me to
我要让你知道是谁让这期视频成为可能 这对我来说很重要首先
let you guys know who’s making these videos possible
有着和你一样的观众直接在Patreon上资助我
first and foremost there’s the viewers like you supporting directly on patreon
这期视频也有audible
and this particular video was also
.com的赞助 audible
supported in part by audible.com which
.com提供有声书与其他有声读物 我拥有Audible已经
provides audio books and other audio materials actually use audible almost
有段时间了 几乎每天都使用它
every day and I have for a while now if
不论你是乘车上下班 慢跑 做饭
you don’t already have listening to literature as one of
还是做别的事情 如果你还没有养成“听书”
your habits whether that’s why you’re commuting or jogging
这个习惯
or cooking or whatever it can be a real
它能大大改变你的生活 我最近听到的一本
life-changer one particularly good book
特别棒的书是乔希·维茨
that i recently listen to is the art of
金的《学习之道》
learning by Josh Waitzkin which I got
这本书是在我哥哥的强烈推荐下入手的
due to a very emphatic recommendation of my brother
我想你们应该会相当喜欢它
and it’s one that I think you guys
作者乔希
would like a lot Josh Waitzkin the
·维茨金在他的童年时期一直
author was a national chess champion
是全美象棋冠军
throughout his childhood and later
后来他开始学习太极拳
in life he took up the martial art tai-chi-chuan and
在短短几年内就成为了国际冠军所以说
became a world champion in just a few years so
这个人明白学习的秘诀
the man knows what it takes to learn and I
我发现他所说的学习
found a lot of what he says about
象棋与太极的方法
chest and taichi translates pretty
都能很好地转化到数学的学习上
meaningfully to learning math as well
特别是他提到的
especially what he says about starting
“从残局入手” 访问audible
with the endgame you can listen to the
.com/3blue1brown
art of learning or any other audio book
你就可以免费听《学习之道》或其他有声书
for free by visiting audible.com/3blue1brown
你可以通过这个链接
going to that URL gives you
免费试用一个月 而且也让Audible
a one month free trial and it lets
知道你是我介绍来的
audible know that you came from me which
这能鼓励他们赞助更多像这样的视频
encourages them to support future videos like this
我使用这个产品
one again it’s a product i’ve
有一段时间了
used for a while now and it’s
我也非常乐意推荐它
one that i am more than happy to recommend alright
这就是结尾的动画——ζ函数
here’s that final animation what the
的导函数是什么样的
derivative of the zeta function looks like

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视频来源

https://www.youtube.com/watch?v=sD0NjbwqlYw

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