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3.如何用MATLAB写这个公式的源码代码?
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5.MATLAB编程语如何还原为数学公式
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n = size(b, 1);
c = zeros(n);
for i = 1:n
for j = 1:n
c(i,j) = sum(b(i,:) - b(j,:));
end
end
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syms x y z s t
f1=x+y-z+s+t;
f2=x^2-y^2+2*z;
f3=x*y+s*t;
f4=z-s*t;
f5=x^2+y^2-2*t;
[x,y,z,s,t]=solve(f1,f2,f3,f4,f5);
解å¾ï¼
x =
0
1/2*2^(1/2)
1/2*2^(1/2)
-1/2*2^(1/2)
-1/2*2^(1/2)
y =
0
1+(-2-2*2^(1/2))^(1/2)
1-(-2-2*2^(1/2))^(1/2)
1+(-2+2*2^(1/2))^(1/2)
1-(-2+2*2^(1/2))^(1/2)
z =
0
-(2^(1/2)-1)*(1+(-2-2*2^(1/2))^(1/2))/(1/2*2^(1/2)*(1+(-2-2*2^(1/2))^(1/2))-1/2*2^(1/2)-1-(-2-2*2^(1/2))^(1/2))
-(2^(1/2)-1)*(1-(-2-2*2^(1/2))^(1/2))/(1/2*2^(1/2)*(1-(-2-2*2^(1/2))^(1/2))-1/2*2^(1/2)-1+(-2-2*2^(1/2))^(1/2))
-(-1-2^(1/2))*(1+(-2+2*2^(1/2))^(1/2))/(-1/2*2^(1/2)*(1+(-2+2*2^(1/2))^(1/2))+1/2*2^(1/2)-1-(-2+2*2^(1/2))^(1/2))
-(-1-2^(1/2))*(1-(-2+2*2^(1/2))^(1/2))/(-1/2*2^(1/2)*(1-(-2+2*2^(1/2))^(1/2))+1/2*2^(1/2)-1+(-2+2*2^(1/2))^(1/2))
s =
0
1/(1/2*2^(1/2)*(1+(-2-2*2^(1/2))^(1/2))-1/2*2^(1/2)-1-(-2-2*2^(1/2))^(1/2))*(1+(-2-2*2^(1/2))^(1/2))
1/(1/2*2^(1/2)*(1-(-2-2*2^(1/2))^(1/2))-1/2*2^(1/2)-1+(-2-2*2^(1/2))^(1/2))*(1-(-2-2*2^(1/2))^(1/2))
1/(-1/2*2^(1/2)*(1+(-2+2*2^(1/2))^(1/2))+1/2*2^(1/2)-1-(-2+2*2^(1/2))^(1/2))*(1+(-2+2*2^(1/2))^(1/2))
1/(-1/2*2^(1/2)*(1-(-2+2*2^(1/2))^(1/2))+1/2*2^(1/2)-1+(-2+2*2^(1/2))^(1/2))*(1-(-2+2*2^(1/2))^(1/2))
t =
0
1/2*2^(1/2)*(1+(-2-2*2^(1/2))^(1/2))
1/2*2^(1/2)*(1-(-2-2*2^(1/2))^(1/2))
-1/2*2^(1/2)*(1+(-2+2*2^(1/2))^(1/2))
-1/2*2^(1/2)*(1-(-2+2*2^(1/2))^(1/2))
如何用MATLAB写这个公式的代码?
如何用MATLAB写求解微分方程组的代码?这个代码可以按下列几个方面来写:
第一个,根据题主提供的公式微分方程组,创建其自定义函数,源码线上商城平台源码怎么用即内容为
dydx=ode_fun (t,公式z)
x=z(1);y=z(2);
dydx1=r1*x*(1-x/K1).*x/(x+e)-q*x*y/(a+x)-m1*x;
dydx2=r2*y*(1-y/K2)+e*q*x*y/(a+x)-m2*y;
第二个,确定时间t的源码源码下载baidow范围,如
tspan=[0,公式dos软件源码1]
第三个,确定x、源码y的公式边界条件,如x(0)=0,源码y(0)=0
第四个,使用ode函数,公式求解其方程的源码数值解,即
[t,公式z]=ode(@ode_fun,tspan,[0 0])
ode函数是采用四、五阶龙格库塔法求解微分方程(组)。源码rocketmq源码分析
第五个,公式使用plot函数,源码小猪 2017 源码绘制x-t和y-t曲线图,绘制x(t) 与 y(t) 相平面图,即
plot(t,z,'b') %绘制x-t和y-t曲线图
plot(z(:,1),z(:,2),'r-') %制x(t) 与 y(t) 相平面图
完善上述代码,运行可以得到其数值解以及图形。
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å¦ææ»ç¨æ·çæ°æ®å·²ç¥ï¼å个ç¨æ·çï¼çå·²ç¥ï¼å³å¯ä»¥å°SFL1,rFL1,SFL2,rFL2,çåæåéå½¢å¼ï¼ä¾å¦å¯ä»¥åæå¦ä¸å½¢å¼ï¼Sr1=SFL1.*rFL1;
Sr2=SFL2.*rFL2;
PFL=sum(Sr1(1:NFC))/sum(Sr2(1:NSC));
MATLAB编程语如何还原为数学公式
A=0; B=0; C=0;
for j=1:n
A=A+(cos(seta(j)))^2/D(j)^2*sigma(j)^2;
B=B+sin(seta(j))*cos(seta(j))/D(j)^2*sigma(j)^2;
C=C+(sin(seta(j)))^2/D(j)^2*sigma(j)^2;
end
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A=magic(3);
b=3*ones(3,3);
c=cross(A,b)
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c =
-3 -
- -
- -3
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é¢ä¸»ç»åºçæ°ç»æ¹ç¨ï¼å¯ä»¥éè¿å¾ªç¯ï¼ç¨vpasolveæ±è§£ãæ±è§£æ¹æ³å¦ä¸ï¼syms lambda
for Lq=1:;
lambda0(Lq,:)=vpasolve(Lq==lambda^2*(.^2+)/2./(1-lambda*.));
end
lambda=lambda0