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1)  Zero mean independent
零均值独立
2)  independent threshold
独立阈值
1.
The principles of several wavelet restraining the noise methods including the Mallat forced restraining the noise,acquiescent threshold restraining the noise meuhod,the independent threshold restraining the noise method and the restraining the noise based on the Wavelet analysis are described,and this paper compares these methods in the grain insect image processing.
对Mallat算法强制去噪、默认阈值去噪、独立阈值去噪三种方法及基于小波去噪原理进行了阐述,并比较了这几种小波去噪方法在处理粮虫图像去噪的优缺点。
3)  independent value
独立价值
1.
Right lease, the property will have the right to request an independent value of its functions.
物权请求权优先于债权请求权,行使物权请求权比行使债权请求权更有利于保护物权人,有关物权请求权与债权请求权的关系的这两个流行观点值得质疑,这从另一个角度说明了物权请求权的独立价值功能。
2.
Controversy on whether the adaptability scale of the principle of credibility can become the basic principle of the law of civil lawsuits has long existed,and the key of which lies in whether the principle of credibility is of independent value and significance,and in whether it is suitable for the existence of the law of civil lawsuit.
关于诚实信用原则的适用范围能否成为民事诉讼法的基本原则,理论上的争论早已存在,其中关键在于诚实信用原则是否具备独立价值和意义,以及适合民事诉讼法的土壤能否生存。
3.
The independent value refers to the value on itself which stands alone from its substantial consequences.
刑事诉讼具备工具价值与独立价值两大类型价值。
4)  non zero mean value
非零均值
1.
This article subjected to the third section of Longevity Study of Mast under Natural Wind Loads,Mainly studies the dynamic reliability analysis of mast structure under the action of non zero mean value random loads.
本文系桅杆在自然风中生命链研究的第三部分 ,主要研究桅杆在非零均值随机激励作用下的动力可靠性。
5)  zero-mean-value
零均值
1.
A computation method of mechanical part size with reliability is pres- ented in this paper according to the fatigue theory when the part is suffering from zero-mean-value random fatigue load.
本文根据现有的疲劳理论,提出了进行零均值随机变幅载荷作用下零件疲劳可靠度尺寸的计算方法。
6)  Zero mean normalization
零均值化
补充资料:均值不等式

几个重要不等式(一)

一、平均值不等式

设a1,a2,…, an是n个正实数,则,当且仅当a1=a2=…=an时取等号

1.二维平均值不等式的变形

(1)对实数a,b有a2+b2³2ab          (2)对正实数a,b有

(3)对b>0,有,   (4)对ab2>0有,

(5)对实数a,b有a(a-b)³b(a-b)                (6)对a>0,有

(7) 对a>0,有                   (8)对实数a,b有a2³2ab-b2

(9) 对实数a,b及l¹0,有

二、例题选讲

例1.证明柯西不等式

证明:法一、若或命题显然成立,对¹0且¹0,取

代入(9)得有

两边平方得

法二、,即二次式不等式恒成立

则判别式

例2.已知a>0,b>0,c>0,abc=1,试证明:

(1)

(2)

证明:(1)左=[]

=

³

(2)由知

同理:

相加得:左³

例3.求证:

证明:法一、取,有

a1(a1-b)³b(a1-b), a2(a2-b)³b(a2-b),…, an(an-b)³b(an-b)

相加得(a12+ a22+…+ an2)-( a1+ a2+…+ an)b³b[(a1+ a2+…+ an)-nb]³0

所以

法二、由柯西不等式得: (a1+ a2+…+ an)2=((a1×1+ a2×1+…+ an×1)2£(a12+ a22+…+ an2)(12+12+…+12)

=(a12+ a22+…+ an2)n,

所以原不等式成立

例4.已知a1, a2,…,an是正实数,且a1+ a2+…+ an<1,证明:

证明:设1-(a1+ a2+…+ an)=an+1>0,

则原不等式即nn+1a1a2…an+1£(1-a1)(1-a2)…(1-an)

1-a1=a2+a3+…+an+1³n

1-a2=a1+a3+…+an+1³n

…………………………………………

1-an+1=a1+a1+…+an³n

相乘得(1-a1)(1-a2)…(1-an)³nn+1

例5.对于正整数n,求证:

证明:法一、

>

法二、左=

=

例6.已知a1,a2,a3,…,an为正数,且,求证:

(1)

(2)

证明:(1)

相乘左边³=(n2+1)n

证明(2)

左边= -n+2(

= -n+2×[(2-a1)+(2-a2)+…+(2-an)](

³ -n+2×n

说明:补充资料仅用于学习参考,请勿用于其它任何用途。
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