MSE_calculation

产生 50 个标准正态分布的随机数, 经验累计分布函数图.(Random number generated 50 standard normal distribution, noting their distribution characteristics, and to draw experience cumulative distribution function diagram.)

利用时域有限差分求解CFD问题，其中差分格式为LUSGS，是求解流体问题的理论基础(Solved using finite-difference time-domain CFD problems, in which the differential form LUSGS )

本程序是贝叶斯网络结构学习的K2算法程序，可获取离散变量的贝叶斯网络(This procedure is K2 Bayesian network structure learning algorithm program, available discrete variables Bayesian Network)

说明：  fortran 学习软件非常有用啊，请管理员加我。

求解非线性不动点方程组x=G(x)的非线性塞德尔迭代方法，初值x0(For solving nonlinear fixed point equation x = G (x) the nonlinear Seidel iterative method, initial value x0)

最新的关于贝叶斯聚焦算法IEEE文章。该算法通过利用信号DOA概率密度函数实现加权的阵列波前差值法，能实现宽带相干信号的DOA估计，并能减小阵列聚焦误差，提高宽带阵列频域波束形成鲁棒性。(In this paper, we present and study a Bayesian focusing transformation (BFT) for coherent wideband array processing, which takes into account the uncertainty of the direction of arrivals (DOAs). The Bayesian focusing method minimizes the mean-square error of the transformation over the probability densityfunctions(pdfs)oftheDOAs,thusachievingimprovedfocusing accuracy over the entire bandwidth)

《基于L 矩估计的GEV-GPD分布拟合及其频数分配和分位数计算》 L 矩估计各分布参数(包括GUMBEL分布,GEV分布,GPD分布) ，其中包括子程序(1)计算频数分配(PINSHU),(2)数组排序(BSORT),(3)L矩估计GUMBEL,GEV及GPD参数,(4)检验拟合效果TEST,TEST1.( L moment GEV-GPD calculation, which can be used to calculate distributed parameters of L moment, etc. )

分别用梯形法和矩形法求定积分，附加应用实例(, Respectively, with the trapezoidal method and the rectangle method to find definite integrals, additional application examples)

说明：  最小二乘法拟合曲线的一个小程序，采用C++程序，(Least-squares fitting curve C++ program)

说明：  利用GA遗传算法解决欺骗函数最优问题，具体问题描述如下，如有问题请与我联系(The deceptive functions are a family of functions in which there exists low-order building blocks that do not combine to form the higher-order building blocks. Here, a deceptive problem that consists of 25 copies of the order-4 fully deceptive function DF2 is constructed for this paper. DF2 can be described as follows: f(0000)=28 f(0001)=26 f(0010)=24 f(0011)=18 f(0100)=22 f(0101)=6 f(0110)=14 f(0111)=0 f(1000)=20 f(1001)=12 f(1010)=10 f(1011)=2 f(1100)=8 f(1101)=4 f(1110)=6 f(1111)=30 This problem has a maximal function value of 750.)