资源列表
111
- Fortran 赋值语句-Fortran
mentekaluo
- 蒙特卡罗算法求 Pi . 随机算法. 概率算法-Monte Carlo method to caculat Pi
LSM
- 递推最小二乘法估计的实现方法，很简单，有需要的就下吧-Recursive least squares estimation method is very simple, there is a need for it on the next
GaBin
- 遗传算法解决复杂背包问题，用Java编写。 该背包拥有3个属性，500个东西，50个包。 求解包与包之间物品重量差最小，并且同一个背包的物品属性有特殊要求。 该程序容易修改。-Genetic Algorithm on Bin Packing Proglem. 500 items, 3 parameters of each item. 50 containers, target: All 50 containers has the smallest weight differe
fp_tree
- 数据挖掘的树结构下C加加的源代码，在商业上很有应用价值。-Data Mining the tree structure of simple C source code, useful in commercial applications.
FFT.ZIP
- The latest official release of FFTW is version 3.2.1, available from our download page see the release notes for what s new. (Subscribe to the fftw-announce mailing list to get release announcements.) This release has better threading support, f
Problem1
- solve the Vanderpol equation using Runge-Kutta-Gill method
spline3
- 三次样条插值（简称Spline插值）是通过一系列形值点的一条光滑曲线，数学上通过求解三弯矩方程组得出曲线函数组的过程。-Cubic spline interpolation (hereinafter referred to Spline interpolation) is through a series of point-shaped curve of a smooth, mathematically by solving the three-moment equations derived
ROMBERG
- 在变步长求积过程中的三个加速公式，将粗糙的积分近似值迅速加工成精度较高的积分近似值的求积方法为龙为个求积算法。-Variable step size in order to plot the course of the three accelerated formula rough approximation of the points quickly processed into a high precision quadrature integral approximation method
nt_cotes
- 牛顿－科茨求积公式 是常微分方程的数值解法-Newton- Cotes quadrature formula is the numerical solution of ordinary differential equations
shuzhifenxikechengsheji
- 考虑在一个固定区间上用插值逼近一个函数。显然，Lagrange插值中使用的节点越多，插值多项式的次数就越高。我们自然关心插值多项式增加时，Ln(x)是否也更加靠近被逼近的函数。龙格(Runge)给出的一个例子是极著名并富有启发性-Consider a fixed-interval interpolation using a function approximation. Obviously, Lagrange interpolation nodes are used the more the n
nd
- 本程序利用牛顿插值法对函数表，求出各阶均差值并通过输入x值得到函数值。例如（课本 ） 给出 的函数表 X 0.40 0.55 0.65 0.80 0.90 y 0.41075 0.57815 0.69675 0.88811 1.02652 均差求至4阶，并由此计算 的近似值 运行环境:C-Free -This procedure for the use of Newton' s interpolation function table, find the diffe