第1个回答 2012-08-26
n = 10; % 输入 n
i = 1:n;
x = rand(1,n) % 输入 x
x1 = sum((i-1).*x)/sum(x);
f = @(N0) sum(1./(N0-i+1)) - n/(N0-x1);
x0 = 10; % Make a starting guess at the solution
[N0,fval] = fsolve(f,x0) % Call solver追问
i = 1:n;
x = rand(1,n) % 输入 x
x1 = sum((i-1).*x)/sum(x);
f = @(N0) sum(1./(N0-i+1)) - n/(N0-x1);
x0 = 10; % Make a starting guess at the solution
[N0,fval] = fsolve(f,x0) % Call solver追问
sorry,我对matlab不是很熟
请问最后三个步骤是什么意思呢？
f = @(N0) sum(1./(N0-i+1)) - n/(N0-x1);
x0 = 10; % Make a starting guess at the solution
[N0,fval] = fsolve(f,x0) % Call solver
% 定义一个变量为N0的函数，即你的方程
f = @(N0) sum(1./(N0-i+1)) - n/(N0-x1);
x0 = 10; % 设定N0的初值
% 解方程 N0为解，fval为与N0对应的f值
[N0,fval] = fsolve(f,x0) % Call solver
你可查阅与fsolve相关的资料
相似回答
