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高数重要极限证明原创中英文对照版
在德化等地区,都构建了全面的区域性战略布局,加强发展的系统性、市场前瞻性、产品创新能力,以专注、极致的服务理念,为客户提供成都网站建设、成都网站设计 网站设计制作按需设计网站,公司网站建设,企业网站建设,高端网站设计,全网营销推广,成都外贸网站制作,德化网站建设费用合理。重要极限
Important Limit
作者 赵天宇
Author:Panda Zhao
我今天想在这里证明高等数学中的一个重要极限:
Today I want to prove animportant limit of higher mathematics by myself:
想要证明上述极限,我们先要去证明一个数列极限:
If we want to give evidence ofthe limit, first of all, there are a limit of a series of numbers according toa certain rule we need to certify:
想要证明这个极限,我首先要介绍一个定理和一个法则:
Before we begin to prove thelimit, there are one theorem and one rule that are the key point we need to introduce:
1. 牛顿二项式定理(Binomialtheorem)
定理的定义为:
Definition of Binomial theorem:
其中 ,称为二项式系数,又有 的记法。
Among the formula: we define the as binomialcoefficient, it can be remembered to.
牛顿二项式定理(Binomial theorem)验证和推理过程:
The process of the ratiocination of Binomialtheorem:
采用数学归纳法
We consider to use the mathematical inductionto solve this problem.
当n = 1时(While n = 1:),
;
假设二项展开式在n=m时成立。
We can make a hypothesis that the binomial expansionequation is true when n = m.
设n=m+1,则:So if we suppose that n equal mplus one, we will CONTINUE to deduce:
具体步骤解释如下:
The specific step of interpretation :
第三行:将a、b乘入;
The 3rd line: a and b are multiplied into the binomial expansion equation.;
第四行:取出k=0的项;
The 4th line: take out of theitem which includes the k = 0 in the binomial expansion equation.;
第五行:设j=k-1;
The 5th line: making a hypothesisthat is j = k-1;
第六行:取出k=m+1项;
The 6th line: What we need totake out of the item including k=m+1 in the binomial expansion equation.
第七行:两项合并;
The 7th line: Combining the twobinomial expansion equation.
第八行:套用帕斯卡法则;
The 8th line: At this line weneed to use the Pascal’s Rule to combine the binomial expansion equation whichare
.;
接下来介绍一下帕斯卡法则(Pascal’s Rule)。
So at this moment, we should get someknowledge about what the Pascal’s Rule is. Let’s see something about it:
帕斯卡法则(Pascal’s Rule):组合数学中的二项式系数恒等式,对于正整数n、k(k<=n)有:
Pascal’s Rule: a binomial coefficientidentical equation of combinatorial mathematics. For the positive integer n andk (k<=n), there is a conclusion:
通常也可以写成:
There is also commonly written:
代数证明:
Algebraic proof:
重写左边:
We can rewrite the left combinatorial item:
通分;reductionof fractions to a common.
合并多项式;combining the polynomial.
证明完成;The Pascal’s Rule has been proved.
接下来只要要证明是单调增加并且有界的,那么就可以得到它存在极限,我们通常称它的极限为e。
So what is our next step? The progression ofnumbers according to a certain rule of should be proved that it is a monotonicincrease sequence and has a limitation. If we can do these things, we will drawa conclusion that the sequence has an limitation which we generally call e.
类似的,我们可以得到:
We can analogously get the:
可见, 和相比,除了前两个1相等之外,后面的项都要小,并且多一个值大于0的项目,因此:
Thus it can be seen, comparing with , all of the items of the are lower thanthese items in except the 1stand the 2rd one are equaling. In addition it has an item whose value is biggerthan zero that is in the . So we can get a point :
所以数列是单调递增的得证,接下来证明其有界性:
Because of the point, we can prove thesequence is an monotonic increase sequence, so we remain only one thing shouldbe proved that is the sequence’s limitation. So let’s get it :
可见{ }是有界的,所以根据数列极限存在准则可得:
Thus it can be seen , the sequence of has a limitation , as we know, we can draw aconclusion by the means of the rule of limitation of sequence exiting: