필터 지우기
필터 지우기

Find function's minimum for specific constant values

조회 수: 1 (최근 30일)
Valeria
Valeria 2018년 6월 4일
댓글: Walter Roberson 2018년 6월 5일
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?

답변 (2개)

Walter Roberson
Walter Roberson 2018년 6월 4일
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.
  댓글 수: 2
Walter Roberson
Walter Roberson 2018년 6월 4일
Note that if there are multiple minima then this will not find them all to check which is the global minimum.
Walter Roberson
Walter Roberson 2018년 6월 5일
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

댓글을 달려면 로그인하십시오.


Sid Parida
Sid Parida 2018년 6월 4일
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);
  댓글 수: 3
Valeria
Valeria 2018년 6월 5일
Matlab doesn't recognize its own function, although it is in its help file... Whattt
Walter Roberson
Walter Roberson 2018년 6월 5일
Which function is not recognized?

댓글을 달려면 로그인하십시오.

태그

Community Treasure Hunt

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

Start Hunting!

Translated by