Whats the value of B ??

2 Ansichten (letzte 30 Tage)
mr mo
mr mo am 10 Jun. 2016
Kommentiert: Walter Roberson am 10 Jun. 2016
Hello every one.
I wanna find the value of B, that make the variable Z=0.8 or very close to 0.8. How can I do that in Matlab ?
here is the sample of my code .
n=10;
C=[0.2 0.55 0.7 0.8 0.99 0.1 0.24 0.33 0.44 0.74];
P=exp(-B*C);
P=P/sum(P);
z = sum (P(1 : n/2));
With best regards.

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 10 Jun. 2016
syms B
n = 10;
C = [0.2 0.55 0.7 0.8 0.99 0.1 0.24 0.33 0.44 0.74];
P = exp(-B.*C);
P2 = P./sum(P);
z = sum(P2(1 : n/2));
Bval = vpasolve(z == 0.8,B);
However, this will give only one of the 89 solutions. It happens to be the only real-valued solution so that might be good enough for you.
  2 Kommentare
mr mo
mr mo am 10 Jun. 2016
Thanks for your help
But I want to use the variable B in the Next other lines of my code, and here the variable B becomes symbolic, How can I use B in other lines ?? with best regards
Walter Roberson
Walter Roberson am 10 Jun. 2016
If you construct the other lines symbolically in B then use
subs(expression, B, Bval)
If the other lines expect B numerically, then use
syms Bs
n = 10;
C = [0.2 0.55 0.7 0.8 0.99 0.1 0.24 0.33 0.44 0.74];
P = exp(-Bs.*C);
P2 = P./sum(P);
z = sum(P2(1 : n/2));
B = double( vpasolve(z == 0.8, Bs) );

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Torsten
Torsten am 10 Jun. 2016
function main
x0=1;
sol=fzero(@fun,x0)
function y=fun(x)
n=10;
C=[0.2 0.55 0.7 0.8 0.99 0.1 0.24 0.33 0.44 0.74];
for k=1:length(x)
xk=x(k);
P=exp(-xk*C);
P=P/sum(P);
y(k)=0.8-sum(P(1:n/2));
end
Best wishes
Torsten.

Kategorien

Mehr zu Sequence and Numeric Feature Data Workflows finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by