lsq

薄板样条，一种很好的拟合曲线的方法。是最近比较常用的拟合方法。(thin plate spline is so nice fitting method. Features: (1) support arbitrary control points (2) allow regularization (3) estimate the warp parameters from corresponding pairs )

第一个把专家青睐的数学模型技术（NNM 算法模型）引入词汇记忆软件中，并取得了巨大的成功。NNM 算法模型与经典的“艾滨豪斯遗忘曲线”真正的融合于一体，使得该著名的理论在计算机与数学模型下第一次得到了淋漓尽致的发挥。(First a mathematical model of the popular technology expert (NNM algorithm model) the introduction of vocabulary learning software, and has achieved great success. NNM algorithm model and the classic " Ai Bin House forgetting curve" true integration in one, making the well-known theory in computer and mathematical models has been most vividly the first time under play.)

无功功率补偿SVG，已经调试好了，可以用的(Reactive power compensation SVG, has been debugged, can be used.)

面向对象的有限元程序设计。与传统的有限元程序相比，面向对象有限元程序更易于编写、更易于维护和扩充，程序代码的可重用成分更大。(object oriented finite element programming)

划分网格一个开源的库，有限元的算法，很好用(Meshing an open source library = for finite element)

基于互相关分析的时延估计算法；基于互相关分析的时延估计算法(cross-correlation analysis for time delay calculation)

适用于ABAQUS的自定义材料子程序，适用于平面单元的随动强化J2模型。(Applies the custom material subroutine in ABAQUS, kinematic hardening J2 model applies to plane unit.)

ADAM （Adaptive Moment Estimation）是另外一种自适应学习率算法，它结合动量梯度 下降法，在不同参数方向上采用不同学习率，保留前几次迭代的梯度，能够很好 的适应于稀疏数据。(ADAM (Adaptive Moment Estimation) is another adaptive learning rate algorithm, which combines momentum gradient. The descent method, which uses different learning rates in different parameter directions and retains the gradients of previous iterations, is very good. It is suitable for sparse data.)

说明：  简支梁-车辆模型，用于ansys建模和车桥耦合分析的初级阶段(Simply Supported Beam-Vehicle Model)

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.)