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hhspectrum
Hilbert Haung spectrum
- 2012-07-19 00:52:09下载
- 积分:1
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fit
基于Levenberg-Marquardt的曲线/面拟合(C library for Levenberg-Marquardt least-squares minimization and curve fitting)
- 2014-02-14 04:20:28下载
- 积分:1
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newton
数值分析的牛顿法,是用C编写的,希望大家互相参考!!1(Pleiades值tub cavity Xu Yang Lai Hey, Did using C met写cavity, submerged mode希望deceive gaze intently EC banana!! 1)
- 2008-05-14 13:25:28下载
- 积分:1
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wave_propagation
西安电子科技大学吴振森教授的波传播与电磁散射课件,里面有很多fortran计算的代码,此课程为博士课程,供专业人士学习.(The courseware of the Xidian University Professor Wu Zhensen wave propagation and electromagnetic scattering, which have many fortran code, this course is Dr. courses for professionals to learn.)
- 2012-09-15 11:17:28下载
- 积分:1
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simplex_step1
单纯形法找初始可行解,上课留的作业,希望对别人有点用吧(Simplex method to find an initial feasible solution , hope it is useful for others.)
- 2014-12-15 15:24:33下载
- 积分:1
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zxecf
一个集交互式和命令式于一体的程序,采用最小二乘法进行曲线的拟合(A set of interactive and command-style in one program, using least-squares curve fitting method)
- 2009-12-15 10:49:31下载
- 积分:1
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standard
通过标准方法实现结构加速度的线性积分问题(Implementation structure by standard methods of linear integral of acceleration)
- 2020-11-27 14:29:30下载
- 积分:1
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Python for Data Analysis, 2nd Edition
Python for Data Analysis
DATA WRANGLING WITH PANDAS,
NUMPY, AND IPYTHON
- 2018-09-15 23:03:37下载
- 积分:1
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cavity
数值模拟方腔内Rayleigh-Benard对流。采用交错网格投影法,能量方程采用松耦合方法。(Numerical Simulation of square cavity Rayleigh-Benard convection. Using staggered grid projection method, the energy equation using loosely coupled approach.)
- 2020-07-17 10:38:48下载
- 积分:1
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GA-DF2
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.)
- 2020-05-10 09:50:49下载
- 积分:1