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Using If for syms

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Hossein Alishahi
Hossein Alishahi 2021 年 3 月 6 日
コメント済み: Hossein Alishahi 2021 年 3 月 6 日
Hi,
I hope you are keeping safe.
I have a urgent question in matlab. Could you please give me sime advice in this regard. I wrote following code where
"function F = fun(x,y,alpha1,alpha2,beta1,beta2,lambda)
F= log((1+alpha1*x./(1+alpha2*y))) -lambda*log((1+beta1*x./(1+beta2*y)));
end"
Main Code
Z=fun(x,P-x,alpha1,alpha2,beta1,beta2,lambda);
z=solve(Z==0);
for i=1:length(z)
z(i,1)=piecewise(z(i,1)<=0,0)
y(i)=P-z(i,1)
S(i)=fun(z(i,1),y(i),alpha1,alpha2,beta1,beta2,lambda)
end
I want to z(i,1)=0 if z(i,1)<=0 or complex and if z(i,1)>0, claculate S(i) but there is a problem with piecewise() which cannot be solved for me.
Could you please help me to write it. P, alpha1, alpha2, beta1, beta2, and lambda are desired positive valuse.
Many Thanks in advance.

回答 (1 件)

Walter Roberson
Walter Roberson 2021 年 3 月 6 日
subs(piecewise(x<3,0),x,5)
ans = NaN
That is, if no condition is matched by the piecewise then the output is undefined, NaN. Not 0 or empty.
piecewise(imag(z(i))~=0 || z(i)<0, 0, S(i))
  6 件のコメント
Walter Roberson
Walter Roberson 2021 年 3 月 6 日
I will need the values for the constants to test with
Hossein Alishahi
Hossein Alishahi 2021 年 3 月 6 日
Thank you for your consideration.
I give you two case which meet all states
1.P=3, alpha1=1, alpha2=0.2, beta1=1, beta2=2, and lambda=2.
2.P=3, alpha1=1, alpha2=2, beta1=3, beta2=4, and lambda=5
Thanks in advance

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