Find function's minimum for specific constant values
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I have a function mew which is a derivative of another function, fun:
syms v a b
fun = -a*v + (2*b*v^(2/3))/(exp(0.5) - 2*(v^(1/3))*log(exp(0.5)*v^(1/3)));
mew=diff(fun,v)
I must find mew's minimum for a=0, b=1.3
How?
回答（2 个）
Walter Roberson
2018-6-4
G = subs(mew, [a, b], [0, 1.3]);
sol = solve(G);
mewmin = subs(G, v, sol)
crosscheck = subs(G, v, sol+1)
The reason for doing the crosscheck is that solve() is only able to find a single numeric root, and without further checking you do not know whether it will be a maxima or a minima. The fact that the crosscheck comes out with a larger value (less negative) than the min gives evidence that we are indeed working with a minima not a maxima. But you should really look at sign( subs(diff(mew), v, sol ) to be sure.
2 个评论
Walter Roberson
2018-6-4
Note that if there are multiple minima then this will not find them all to check which is the global minimum.
Walter Roberson
2018-6-5
MATLAB finds sol as
6396480385065521911547770910773962672562070583/(1427247692705959881058285969449495136382746624*lambertw(0, (3445831591435597800619981388529*exp(-1))/1267650600228229401496703205376)^3)
However, if you change the exp(0.5) to exp(sym(0.5)) then the more accurate solution is
exp(3/2)/LambertW(1)^3
Sid Parida
2018-6-4
Try this:
You can play wit the lb, and ub to get the global minimum:
syms v a b
fun = -a*v + (2*b*v^(2/3))/(exp(0.5) - 2*(v^(1/3))*log(exp(0.5)*v^(1/3)));
mew = diff(fun,v);
subbed_ab = matlabFunction(subs(mew, {a, b}, {0.0, 1.3}));
lower_bound = -100000;
upper_bound = 100000;
min_val= fminbnd(subbed_ab, lower_bound, upper_bound);
3 个评论
Walter Roberson
2018-6-4
note that fminbnd gets stuck in local minima so if there are multiple minima you could have a problem.
Valeria
2018-6-5
Walter Roberson
2018-6-5
Which function is not recognized?
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2018-6-5
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