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yuyi2009

铜虫 (小有名气)

[求助] matlab程序问题求解答

function[c,err,yc]=bisect(f,a,b,delta)
% Input  - f is the function input as a string 'f'
%        - a and b are the left and right endpoints
%        - delta is the tolerance
% Output - c is the zero
%        - yc = f(c)
%        - err is the error estimate for c
ya = feval(f,a);
yb = feval(f,b);
if ya * yb > 0, end
max1 = 1+ round((log(b-a)-log(delta))/log(2));
for k =1:max1
    c =(a+b)/2;
    yc = feval(f,c);
    if yc==0;
        a=c;
        b=c;
    elseif yb * yc > 0
        b=c;
        yb=yc;
    else
        a=c;
        ya=yc;
    end
    if b - a < delta, break, end
end
c=(a+b)/2;
err=abs(b-a);
yc=feval(f,c);     % 求函数在c处的函数值

for T=200:10:300
   K = exp(-9.3932+3204.71/T+0.835931*log(T)...
       +2.35267*10^-3*T-1.8736*10^-6*T^2+5.1606*10^-10*T^3)
   f=inline('(4*K-1)*x^2-8*K*x+4*K');
   bisect(f,0,1,10^-5)
end
主程序调用函数出错
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Bruce_lee

铜虫 (著名写手)

【答案】应助回帖

★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★
感谢参与,应助指数 +1
yuyi2009: 金币+50, ★★★★★最佳答案, 再次感谢 2013-11-24 23:47:56
function problem_solve
%--------------------------------------------
disp('----------------分割线-----------------');
%--------------------------------------------
function[c,err,yc]=bisect(f,a,b,delta)
% Input  - f is the function input as a string 'f'
%        - a and b are the left and right endpoints
%        - delta is the tolerance
% Output - c is the zero
%        - yc = f(c)
%        - err is the error estimate for c
ya = feval(f,a);
yb = feval(f,b);
if ya * yb > 0, end
max1 = 1+ round((log(b-a)-log(delta))/log(2));
for k =1:max1
    c =(a+b)/2;
    yc = feval(f,c);
    if yc==0;
        a=c;
        b=c;
    elseif yb * yc > 0
        b=c;
        yb=yc;
    else
        a=c;
        ya=yc;
    end
    if b - a < delta, break, end
end
c=(a+b)/2;
err=abs(b-a);
yc=feval(f,c);     % 求函数在c处的函数值
end
%---------------------------------------------
function f=subfun(x,K)
f=(4*K-1).*x.^2-8*K.*x+4*K;
end
%---------------------------------------------
for T=200:10:300
   K = exp(-9.3932+3204.71/T+0.835931*log(T)...
       +2.35267*10^-3*T-1.8736*10^-6*T^2+5.1606*10^-10*T^3)
   [c,err,yc]=bisect(@(x)subfun(x,K),0,1,10^-5)
   disp('----------------分割线-----------------');
end
end


楼主,希望能帮助到你。
努力就有收获!
7楼2013-11-24 23:37:40
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zhg19

金虫 (小有名气)

【答案】应助回帖

感谢参与,应助指数 +1
LZ详细描述一下问题。。。
2楼2013-11-24 17:21:08
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yuyi2009

铜虫 (小有名气)

引用回帖:
2楼: Originally posted by zhg19 at 2013-11-24 17:21:08
LZ详细描述一下问题。。。

给了一个与温度相关的平衡方程,求200-300每隔10的平衡常数,以及平衡转化率。。。要求用二分法求解~
3楼2013-11-24 17:46:06
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zhg19

金虫 (小有名气)

引用回帖:
3楼: Originally posted by yuyi2009 at 2013-11-24 17:46:06
给了一个与温度相关的平衡方程,求200-300每隔10的平衡常数,以及平衡转化率。。。要求用二分法求解~...

主要是程序出现的问题是什么呢
4楼2013-11-24 22:09:26
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