number of zeros of a holomorphic function

One basic problem in mathematics is to solve equations. Giving a polynomial non-trivial of complex coefficients on one variable, we know that it has as many zeros as its degree. The simplest holomorphic function which has infinite zeros is perhaps the sinus and cosinus functions.

Of course, from complex analysis we know that a holomorphic function on the whole plane can not have infinite zeros on a bounded region. So we can talk about the number of zeros of a holomorphic function inside a circle of certain radius.

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《小城之春》电影

我看的《小城之春》是费穆拍摄的于1948年9月上映的电影。

电影中演员很少,只有五个人:年轻少妇周玉纹(韦伟饰),她的生病的丈夫戴礼言(石羽饰),戴礼言的妹妹戴秀(张鸿眉饰),佣人老黄(崔超明饰)和戴礼言的朋友章志忱(李玮饰)。

戴礼言和周玉纹结婚八年了,但是戴礼言生了六年的病,并且从前富裕的家境由于战争的破坏而败落,戴礼言对此忧心忡忡,整天唉声叹气,脾气变得非常古怪,夫妻关系也逐渐恶化。周玉纹对此也是不知所措,每天和丈夫说不了几句话,只能趁着买菜的时间到村边的城墙上走一走散散心。一天,学医的章志忱求学归来,来到戴家。章志忱和周玉纹从小一起长大,并且是旧日的情人,只是由于章志忱没有找媒人向周家提亲而错过了这段佳缘。现在的章志忱和戴礼言形成了鲜明的对比:一个英俊潇洒,生气勃勃,一个萎靡不振,心情沮丧。章志忱的到来给戴家带来了不少欢乐。特别是戴秀非常喜欢章志忱。但是,周玉纹见到往日的情人,心理七上八下:想和章志忱亲近但又碍于妇道。最后,戴礼言明白了事情的原委,服药自杀未遂。章志忱深知自己做得也不对,主动告别戴家。

这个电影虽然只有五个人,但是讲述的故事非常完整。现在看着当时的黑白片,别有一番韵味。

电影取名为‘小城之春’。个人觉得有两个方面的含义。一个是表面上的,就是章志忱的到来确实给戴家添加了不少欢乐,特别是戴礼言和章志忱初次见面的时候,平时寡言少语、愁眉苦脸的戴礼言见到章志忱竟然眉开眼笑,兴奋异常。其次是戴秀一直围在章志忱身边唱歌,央求章志忱给她讲故事,等等。当然,对周玉纹也可以算是一种欢乐。另一方面,章志忱给戴家带来了希望。这也要分两点说。一个是,章志忱一直强调戴礼言的病并无大碍,是可以治愈的。这让戴家人看到了希望。还有一个,就是他让戴礼言重拾了活下去的勇气,修补了戴礼言和周玉纹的夫妻感情。这一点可以通过电影的最后几分钟看出来。戴礼言自杀未遂(是被章志忱救活的)后,章志忱就辞别,戴秀和老黄送他到火车站。他们一路有说有笑。同时,周玉纹站在城墙上目送他们。而戴礼言竟破天荒的也出来了,来到周玉纹的旁边,一起远望。他们一起看的不仅仅是人,更是生活的勇气。

如何让一个绝望的人重获生活的勇气?这大概是这部电影要解决的一个问题。

建筑的外部空间

最近陆续读了两遍芦原义信的《外部空间的设计》。

外部空间,顾名思义,就是相对于内部空间而言的。内部空间,都是人为形成的。比如,一间房子的内部,就是一个内部空间。房子的外部,就是外部空间。但是,并不是这整个外部空间都对我们有用,通常,我们只关心房子周围的空间。所以,外部空间,准确的说应该是围绕着内部空间的那部分空间。

外部空间也可以加入人为的元素。还是以房子为例。我们盖了一间房子,在房子的周围,我们可以什么也不布置,这时这个外部空间就是任由自然处理的空间,对房子的作用不好说,我们一般称这样的外部空间是消极的。我们也可以在房子的周围的一片空地围上栅栏或者篱笆,在房子和栅栏之间种花养小动物等等。这时这个外部空间就有了人为的痕迹,我们一般称之为积极的。

外部空间还有很多其他的例子。比如说,在野外野炊,临时搭起一个小棚子。这个小棚子所覆盖的空间就是外部空间(这时内部空间缩小为零)。城市的广场,也是一种外部空间(比如巴黎的Place Saint Quebec,Place Saint Germain,意大利的Pizza del Campo,等等)。

外部空间在建筑学中的作用是不可小觑的。外部空间设计得好,能够增加内部空间的吸引力。否则,就是增加内部空间的孤立感。人们都喜欢和周围事物有关系都建筑,这就是为什么要人为干预外部空间。

外部空间都一个重要标准就是视觉感受。比如说有一排楼房,如何处理楼房之间的距离?过于狭窄,会产生拥挤感,过于分开,会产生孤立感。通常而言,楼房之间的距离应该是楼房高度的一到二倍之间(当然这也仅仅是一个经验数值,真正进行设计的时候还是要建筑师自己拿主意,这里提到这个说法是要说明建筑师进行设计的时候要考虑到这一点)。这也大概说明了为什么高楼林立的城市会给人一种疲劳的感觉。

另一个重要标准就是外部空间和内部空间的搭配情况。这个问题更加困难,因为这在很大程度上取决于内部空间是如何设计的。一栋住宅,它的外部空间可能仅仅是一圈篱笆围成的地面。而一幢图书馆楼,或者大学校园,它们的外部空间都是不一样的。

外部空间有时候能和内部空间互换,也就是说,将内部空间的顶盖放到外部空间上,得到的新的建筑群落仍是和谐的。这时很有意思的一点(意大利的Pizza del Campo就是一个很好的例子)。这让人联想到一幅画中内容和背景的关系。有时候,一幅画的哪部分是内容,哪部分是背景是不能区分的(比如Escher的一些作品)。从另一个角度看,内容是不能离开背景而存在的。内容需要放到背景中去解读。换言之,背景也是内容的一部分。

Jacobi sums and Beta functions

(0)In this post we will talk a little about the similarity between these two terms appearing in the title. It is, in fact, what I learned in a course on Galois theory in Ecole Polytechnique.

(1)In number theory one common problem is to determine the number of solutions to some equation.

For example, suppose that p is a prime number, and thus \mathbb{Z}/p\mathbb{Z}=\mathbb{F}_p is a finite field.The simplest case of these problems is a linear equation in indeterminant x:ax=b(a,b\in \mathbb{F}_p). Of course to determine the number of its solutions is a very simple problem.

Then what about this quadratic one: x^2=b(b\in \mathbb{F}_p)(we suppose that p\neq 2)? This, again, is not difficult. We just have to utilize the method of quadratic residue: if b\neq 0 is a quadratic residue modulo p, then this equation has two solutions, while otherwise it has no solution(for b=0, there is always only one solution). Using the Legendre symbol, we can say that the number of solutions to this quadratic equation is just 1+(\frac{b}{p}).

(2)There are three directions to generalize what we have said, generalize the polynomial in the equation either  to higher degrees, either to more variables, or generalize the equation to a larger field.

Let’s do the first kind of generalization. So the problem becomes now: determine the number of solutions to the equations x^n=b(n>1) in \mathbb{F}_p.There is a little problem here: whether n|(p-1)? If n\not|(p-1), then consider the map f:\mathbb{F}_p\longrightarrow \mathbb{F}_p, a\mapsto a^n and its restriction f' to (\mathbb{F}_p)^*. It’s easy to see that f' is a group homomorphism, and observing that n doesn’t divide the cardinal of the group (\mathbb{F}_p)^*,i.e. p-1, we get that f' is just a bijection, and thus so is f. So the problem in this case(n\not |(p-1)) is not very interesting. Therefore in the rest we want to consider the more interesting case, n|(p-1).(In case that n>p-1, we can apply Fermat’s little theorem to the case n\leq p-1) . In order to get a formula similar to the above one, we need to introduce the concept of characters of an abelian group. Basically, a character \xi of an abelian group G is just a group homomorphism \xi:G\longrightarrow \mathbb{C}^*. In our case, we define a character \xion the finite field \mathbb{F}_p to be a character on the group (\mathbb{F}_p)^* and extend to the whole field by setting \xi(0)=0. Furthermore we can define some kind of composition between two characters \xi,\phi over a same group: the point-wise multiplication, (\xi\circ\phi)(a)=(\xi(a))(\phi(a)). It’s easy to verify that the resulting mapping is again a character. In fact, in noting that (\mathbb{F}_p)^* is a cyclic group, we verify easily that the set of all the characters over \mathbb{F}_p forms a cyclic group of order p-1(We denote this set as C_p,and the neutral element as $e$, we can regard C_p as the dual group of (\mathbb{F}_p)^*=\mathbb{Z}/(p-1)\mathbb{Z}). Now we can state the corresponding formula for the number NS(x^n=b) of solutions to the equation x^n=b:

NS(x^n=b)=\sum_{\xi\in C_p,\xi^n=e}\xi(b)

We can prove this assertion by simply writing down all the \xi explicitly.

In all these formulas, we see the same form: a sum of some characters satisfying some conditions. What does this remind us? In fact, in some way, this reminds us of generating functions: if we have a set of numbers \{a_n\} indexed by positive integers, then the most natural way is to give this set a generating function f(x)=\sum_n a_n x^n, sometimes under some conditions for convenience we write x=e^{2i\pi\theta}. What we will do next is similar: we want to give C_p a generating function. Noting that there are just p-1 elements in this set, so this generating function is a finite sum. We can try this one g_{\xi}(a)=\sum_{b\in\mathbb{F}_p}\xi(b)e^{2i\pi ab/p}(a\in\mathbb{F}_p)(it is indeed well defined). It is easy to find the correspondence between this generating function and the generic one.

This generating function has a special name, it is called the Gauss sum of \xi.

It is worth pointing out that the Gauss sum has some similarity with the \Gamma function: \Gamma(s)=\int_0^{\infty}t^se^{-t}\frac{dt}{t}. Just comparing term by term: integral v.s sum, t^s v.s \xi(b), e^{-t} v.s e^{2i\pi ab/p}.

(3)Now we want to deal with the second kind of generalization: adding more variables in the equation.

The simplest case is this one: find the number of solutions to the equation x^2+y^2=b(b\neq 0). Now we make use of a little trick to find out the number NS(x^2+y^2=b). In fact we can write NS(x^2+y^2=b)=\sum_{a\in\mathbb{F}_p}NS(x^2=a)NS(y^2=b-a)=\sum_a(1+(\frac{a}{p}))(1+(\frac{b-a}{p})). Expanding out, and noting that \sum_{a\in\mathbb{F}_p}(\frac{a}{p})=0(the number of non zero quadratic residues equals that of non-quadratic residues), we obtain NS(x^2+y^2=b)=p+\sum_{a,a\neq 0}(\frac{a(b-a)}{p}). To evaluate the second term, we write \sum_{a\neq 0}(\frac{a(b-a)}{p})=\sum_{a\neq 0}(\frac{b/a-1}{p})=\sum_{a\neq 0}(\frac{a-1}{p})=-(\frac{-1}{p})=-(-1)^{(p-1)/2}.

So here we see another form of sum, that is \sum_a \xi(a)\psi(b-a). In other words, if we define J(\xi, \psi)=\sum_{a\in\mathbb{F}_p}\xi(a)\psi(1-a), then after some similar calculations, we have that NS(x^n+y^n=1)=\sum_{\xi^n=e,\psi^n=e}\sum_a\xi(a)\psi(1-a)=\sum_{\xi^n=e,\psi^n=e}J(\xi,\psi). This definition gives the Jacobi sum J(\xi,\psi).

(4)So after a lengthy introduction, we define the Jacobi sum, and understand that Jacobi sum can be used to calculate the number of solutions to some kind of equations.

Now since the title of this post includes another name, ‘Bessel functions’, next we want to formally talk a little bit about the similarity between Jacobi sum and Bessel functions.

The Beta function is defined to be B(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}dt. Compare this function with the Jacobi sum, their similarity is very clear.

Furthermore, we know that there is some relation between the Beta function and the \Gamma function, that is

B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.

In fact, we also have such relation between the Jacobi sum and the Gauss sum, that is

J(\xi, \psi)=\frac{g_{\xi}(1)g_{\psi}(1)}{g_{\xi\circ\psi}(1)}.