Roots of a function
17 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Muhammad Umar Farooq
am 30 Mär. 2016
Kommentiert: Torsten
am 31 Mär. 2016
I have two equations:
y1 = 2sinx1;
y2 = 2cos^2(x1) + 3sin(2x2+3);
here y1 = 0 while y2 = 1.
Can anyone please tell me which approach would be the best to find out the values of x1 and x2.
Thank you.
1 Kommentar
Akzeptierte Antwort
Torsten
am 31 Mär. 2016
Bearbeitet: Torsten
am 31 Mär. 2016
Try this:
function driver
y=zeros(1,2);
y(1)=0;
y(2)=1;
x0=zeros(1,2);
x=fsolve(@(x)HW1(x,y),x0)
function res = HW1(x,y)
res(1)=y(1)-2*sin(x(1));
res(2)=y(2)-(2*cos(x(1))^2 + 3*sin(2*x(2) + 3));
Best wishes
Torsten.
2 Kommentare
Torsten
am 31 Mär. 2016
Your equations don't have a unique solution. So - depending on the starting guess x0 - you'll get different solutions for x. Do you have any condition on x that could make the solution unique ?
Best wishes
Torsten.
Weitere Antworten (1)
Vlad Miloserdov
am 30 Mär. 2016
if you still need this
A=solve('0 = 2*sin(x1)','1 = 2*cos(x1)^2 + 3*sin(2*x2+3)','x1','x2');
% first ans
A.x1(1)
A.x2(1)
% second ans
A.x1(2)
A.x2(2)
5 Kommentare
Siehe auch
Kategorien
Mehr zu Function Creation 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!