finding variable of a given equation
2 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Irfan
am 19 Okt. 2024
Kommentiert: Sam Chak
am 20 Okt. 2024
I want to get the value of b given the equation, where AonAstar = 16 and gamma = 1.22
AonAstar = ((b)^(1/gamma)*(1-(b)^((gamma-1)/gamma))^(1/2))/(((gamma-1)/2)^(1/2)*(2/(gamma+1))^((gamma+1)/(2*(gamma-1))));
0 Kommentare
Akzeptierte Antwort
Walter Roberson
am 19 Okt. 2024
Bearbeitet: Walter Roberson
am 19 Okt. 2024
Q = @(v) sym(v);
AonAstar = Q(16);
gamma = Q(1.22);
syms b
eqn = AonAstar == ((b)^(1/gamma)*(1-(b)^((gamma-1)/gamma))^(1/2))/(((gamma-1)/2)^(1/2)*(2/(gamma+1))^((gamma+1)/(2*(gamma-1))));
fplot(lhs(eqn) - rhs(eqn), [-2 2])
b_solution = vpasolve(eqn, b)
disp(char(b_solution))
We can see from the graph that there are no solutions over the reals.
Weitere Antworten (1)
John D'Errico
am 19 Okt. 2024
gamma is a really bad choice of variable names, since gamma is itself a very useful function in MATLAB. Try to avoid doing these things, because one day soon you will need to use gamma yourself as a function. I've used g instead, to replace gamma.
AonAstar = 16;
g = 1.22;
syms b
I'll subtract the right hand side from AonAstar, to create a problem where we will look for a zero.
eqn = AonAstar - (b^(1/g)*sqrt(1-b^((g-1)/g)))/((sqrt(g-1)/2)*(2/(g+1))^((g+1)/(2*(g-1))))
fplot(eqn,[0,1])
For a solution to exist, this curve must cross zero, and must do so in the interval [0,1], since it is only real valued in that interval.
As you can see, it never even comes close to zero, being always positive.
5 Kommentare
John D'Errico
am 20 Okt. 2024
It is a habit we need to train our minds to avoid if we are to use MATLAB successfully. And yes, it is far too easy to name variables alpha, beta, gamma, and yes, I fail at it at times myself. My tendency is to use contractions, like alph, bet, gam, or I might deliberately mispell the greek letter, perhaps as gammuh. Or I'll capitalize the name, using Gamma. At least all of these variations in names remind me of the relation of my code to my thinking.
Siehe auch
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!