final_robotSearch-1.0_2learn-it

说明：  从别的网上下来的，是一个简单又经典的分子动力学模拟程序，很适合初级使用。(for molecular simulation)

给定样本点，采用高斯核密度估计，求出概率密度分布函数。(It is good to use this method to evaluate pdf)

详细介绍了有限元法基础知识，对有限元入门者很有帮助(Details the basics of the finite element method, finite element beginners helpful)

国外最近的一篇变步长nlms仿真，该文件有文中的几种仿真(Recent foreign nlms a variable step simulation, there is the text of the document of several simulation)

说明：  自己编的，实现l-bfgs解无约束优化问题(Own, and the realization of l-bfgs Unconstrained optimization problems)

自己编写的期望最大化（EM）算法的MATLAB实现，里面有较为运行方法和程序说明，对新手有很好的帮助( I have written the expectation-maximization (EM) algorithm in MATLAB, which has run more methods and procedures described, there is a good help for the novice)

说明：  模拟电场中的等势线，利用计算机数学软件MATLAB模拟仿真真空中自由电荷，电偶极子，带电细棒，尖端导体等附近静电场电场线及等势线分布图型(Equipotential line in simulated electric field)

This source file contains example PitchCntrl(), UserHSSBr(), ! UserVSCont(), and UserYawCont() routines that are used to interface ! FAST with a master controller implemented as a dynamic-link-library ! (DLL) in the style of Garrad Hassan s Bladed wind turbine software ! package. All four routines call routine BladedDLLInterface(), which ! contains a call to the Bladed-style DLL that evaluates as DISCON(). ! Routine BladedDLLInterface() USEs a MODULE named ! BladedDLLParameters(), which stores values of PARAMETER constants ! used in the interface.

电力芯片RN8302计算高次谐波时的FFT变换算法，最高可实现51次谐波计算。(FFT transform algorithm for power chip RN8302 to calculate the high harmonics, the highest can achieve the 51 harmonic calculation.)

假设有一个能装入总体积为T的背包和n件体积分别为w1 , w2 , … , wn 的物品，能否从n件物品中挑选若干件恰好装满背包，即使w1 +w2 + … + wn=T，要求找出所有满足上述条件的解。例如：当T=10，各件物品的体积{1，8，4，3，5，2}时，可找到下列4组解：（1，4，3，2），（1，4，5），（8，2），（3，5，2）。 [需求分析](Suppose there are a load of the backpack of the total volume of T and n volumes were w1, w2, ... wn items, whether selected from n items a number of exactly filled backpack, even if w1+w2+ ...+ wn = T, find out the solution of all the above conditions are met. For example: when T = 10, the volume of items {1,8,4,3,5,2}, can be found in the following four solutions: (1,4,3,2), (1,4,5 ), (8,2), (3,5,2). [Needs Analysis])