Should the max function and the fminbnd function produce the same result? Also, is there something wrong with my MATLAB syntax?
2 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Ruten9
el 23 de Oct. de 2015
Comentada: Ruten9
el 23 de Oct. de 2015
this is my code guys. my height.m file is: function [h] = height(t)
h = @(t) ((-9.8).*(2.^(-1))).*(t.^2) + 125.*t + 500;
%h(h<0) = 0; %I DON'T UNDERSTAND WHY THIS DOESN'T WORK. if h(t) <0 h=0 end end
My actual program is: %Part A clear all; clc; t = 0:.01:30; h = height(t); %the h(h<0) is acting weird
%Part C (maxima)---I swapped the order for the plotting %max function [xpsudeomax,ymax] = max(h(t)); xrealmax = t(ymax); maxfunction = [xrealmax,ymax] %THIS IS LESS ACCURATE THAN THE BELOW VERSION-WHY?
%fminbnd function [xmax2,ymaxpsudeo]= fminbnd(@(t) (-1.*h(t)),0,30); yrealmax = -1.*ymaxpsudeo; format long g fminbndmax = [xmax2,yrealmax] %Is fminbnd and max function suppposed to give the %exact same answer? %THIS IS THE ACCURATE ONE-WHY?
0 comentarios
Respuesta aceptada
John D'Errico
el 23 de Oct. de 2015
NO. max is not an optimization tool. It just finds the maximum element of an array. VERY different.
Anyway, fminbnd is a MINIMIZER.
Finally, as you have written it, h is a function handle. You cannot then do a vectorized array operation on it, like this:
h(h<0) = 0;
Más respuestas (0)
Ver también
Categorías
Más información sobre Solver Outputs and Iterative Display en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!