帮忙翻译一段(数学研究类的)，拒绝机译！（有大加分！！）

1. 帮忙翻译一段(数学研究类的)，拒绝机译！（有大加分！！）

结论
常微分方程是一个强大的工具，它被广泛的用于金融和经济领域。我们先介绍常微分方程的基本概念以及常用的求解方法（经典方法和拉普拉斯变换）。常微分方程的应用在确定性和随机性系统中都有展现。作为确定性系统应用的例子，我们考察一个简单的GDP模型和公司的投资模型，对于随机性系统，我们考察资本结构管理模型，计算使得公司权益最大化的最优默认触发级别。

fairly easy~

2. 帮忙翻译一段(数学研究类的)，拒绝机译！（有大加分！！）

微分方程的理论已经成为金融经济学研究中的一个基本工具,尤其是在计算机普遍应用的今天.

一个常微分方程仅含一个未知函数,且该函数与它的导函数仅有一个自变量,但是非常微分方程,例如偏微分方程,含有一个拥有超过一个自变量的未知函数.

在这一章,我们将要引入常微分方程以及它在金融经济学研究中应用.

有许多方法可以解决一个常微分方程.

在这些方法中,最常用的方法是对于线形微分方程的经典解决技巧和拉普拉斯变换.

拉普拉斯变换在解决积分微分方程的过程中是强有力的.

它原自工程学当中的应用.

作为一个有效的工具,它同样在金融经济学中有广泛的利用(详情及出处 参见Churchill (1972) 和 Schiff (1999)) .

3. 请高手翻译几句话，如果翻译的准确度，还有加分。

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4. 请翻译大神帮忙翻译一段文献，不要翻译工具的。翻译通顺就好！！分数可以追加的！！

Our last example is also a geometric one, but here the “divisions”, i. e. the boxes, are not whole parts of the given geometric figure, indeed, they are points in it. 
我们最后的一个例子仍然是关于几何的，但是这里用的是“分割”，例如，这些方框，不是这个几何图形的全貌，事实上，这几个方框只是这个几何图形中的几个点。
The difficulty is realizing that it is these precise points that will help (or having a hunch of what those points are by making a few drawings). 
理解这里的困难之处在于这几个精确的点是否有用（或者说可以通过做一些辅助来预测到这几个点是干什么用的）。
We are given a square and 9 lines cutting through it. Each line cuts the square into two quadrilaterals, and the ration of the areas of those quadrilaterals is 2:3.
我们有一个有9条线穿过的正方形，每条线都将这个正方形切成两个面积比为2:3的四边形。
Prove that at least three of these 9 lines pass through the same point. 
试证明这9条线里面至少有3条相交于同一点。
Since each line divides the square in two quadrilaterals, each line cuts the square at two non consecutive edges.
因为每条线都将这个正方形分成两个四边形，每条线都切割在了正方形的两条不相交的边上。
The figures obtained are trapezoids with two right angles, and their area is equal to their base (here the side of the square) multiplied by the height of their “midline”, the line perpendicular to their two parallel edges and at equal distance from those two edges. 
最后得出来的图形是一些有着两个直角的梯形，并且这些梯形的面积都等于它们的底边（这里指正方形的边）乘以它们高的“中线”，这条中线垂直于梯形的两条平行的边上，并且这条中线到两边的距离相等。
Since for any of those trapezoids their base is the same, the ration of their midlines must be 2:3. 
因为所有梯形的底边都相同，那么他们的中线和中线的比必然是2:3。
But there are only two possible midlines for the trapezoids, so we get four different points, call them midpoints, two on each midline, where the midline of two trapezoids intersects with one of the 9 given lines.
但是这些梯形有两条可行的中线，所以我们取4个点，叫它们中点，每条中线上都有两个中点，并且每两个梯形的中线和已知的9条线相交。
Every one of those lines must pass through one of these midpoints. 
每条线都必须通过一个中点。
We have 4 points and 9 lines, so there is at least one point through which pass three lines. 
现在我们有4个点和9条线，所以这里至少有3条线相交于一点。
We have shown that, given 9 lines, each cutting a square into two trapezoids whose areas are in the ratio 2:3, at least three of the lines pass through the same point. 
我们已经知道，已知的9条线，每条线将正方形切割成两个面积比为2:3的梯形，至少有3条线会相交于一点。
We now move on to more difficult problems, which generally necessitate a little more twist, or work, before applying the pigeonhole principle
我们现在再看一看更难的问题，在应用鸽巢原理之前需要一点更多的思考和工作。 

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本人是英语系大三学生，长时间不接触数学，有不恰当的地方还请见谅！：）

5. 请翻译大神帮忙翻译一段文献，不要翻译工具的。翻译通顺就好！！分数可以追加的！！

Of course, this result is well-known now, but in the 1950’s this topic was only beginning to surface.当然，这个结果现在众所周知，但在1950年这个话题仅仅是表面的。
We color the edges of the complete graph with 6 vertices (and so 15 edges) with two colors.  我们用6个顶点（15边）两种颜色来为这个图表着色。
Show we can always find three points such that the edges joining them are of the same color.所以我们总是能够找到三个点它们的边是用相同颜色连接。
The proof is quite simple. Take any vertex, name it a. It has 5 edges, so by the pigeonhole principle at least three of those are of the same color. 
样张很简单，任意一个顶点，称作a，它有五条边，根据鸽棚规则它们中的三条都有同样颜色的边
Say ab, ac and ad are of the same color, blue (without loss of generality). Then if either bc, bd or cd are blue, we are done.
比如ab，ac和ad是同样的颜色——蓝色（在一般情况下）。那么bc，bd和cd也是蓝色，我们就完成了。
But if none is, then bcd is a red triangle. Either way, we have found three points in the complete graph of three vertices such that the three edges joining them are of the same color.Remarks. 
但如果没有一个是这样，那么bcd是一个红色的三角形。任何一种方式，我们发现三个顶点的整个图表的三点是用相同颜色标识连在一起。
A very similar problem in graph theory, also easily solved using the pigeonhole principle, is the following : suppose there are n people at a party, then show at least two of them have the same number of acquaintances, where a person is not acquainted with himself or herself. 
一个和图表理论非常相近的问题，用鸽篷理论也能轻易解决。如下列所述，假如宴会上有n个人，找出至少两个有同等数量熟人的，其中有一个和他或者她不认识。
This is easy, for the maximum number of acquaintances for one person is n-1. For everyone to have a different number of friends would mean someone at the party has no friends at all, a contradiction to the fact that someone is friends with everyone (let alone that this person would not be at the party if he or she did not know anyone)! 这非常简单，因为每个熟人的最大数是n-1.宴会里有不同数量的朋友的这个人意味着没有朋友。与事实矛盾的是一些人是每个人的朋友（如果一个人不认识任何人他会孤立）



Also, consider the complete graph of six vertices,assume all edges have different length. 同时，考虑六个至高点的图表，假设所有的边长度不同。


Some edges form triangles between themselves.Prove there is at least one edge that is both the shortest edge of a triangle and the longest edge of another triangle.来自三角形的一些表证明至少有一条边是另一个三角形最短或者最长的边。

 To prove this, let us start by coloring blue all edges that are the shortest edge of at least one triangle, and coloring the rest, if any, red. 
为证明这个，让我们至少为一个三角形的所有边着成蓝色，其他任何的边着成红色。

Then, we know there is at least one triangle such that its edges are all blue (we cannot have a triangle all red since its shortest edge should be blue). Take the longest edge of that triangle; clearly it is blue because it is the shortest edge of some other triangle.
那么，我们知道至少有一个三角形的所有边都是蓝色（我们不可能有个所有边是红色的三角形因为最短的边必须是蓝色），找出那个三角形的最长的边；很明显是蓝色因为它是其它三角形最短的边。


终于翻译完了，可能有不尽人意的地方，望采纳！谢谢！