clc
clear all
syms x
for yr = linspace(0,1,10)
f = sin(x);
fL = [0 pi];
iL = [0 pi];
Volume = pi*int((f-yr)^2,iL(1),iL(2));
s=double(Volume)
end
h=min(s)

 Réponse acceptée

syms x
smin = inf; % initialize smin
for yr = linspace(0,1,10)
f = sin(x);
fL = [0 pi];
iL = [0 pi];
Volume = pi*int((f-yr)^2,iL(1),iL(2));
s=double(Volume);
if s<smin,
smin = s;
end
end
%h=min(s)
smin
smin = 0.9437

8 commentaires

I could not understand it why have you choosen infinity as smin? Can you please explain?
Chunru
Chunru le 28 Nov 2021
Modifié(e) : Chunru le 28 Nov 2021
You put a very large value as the initial value to ensure that the min is from the data. You can go through a simple example of s = -10, -100, 200, 1000 and examine what happens with:
smin = inf
for s = [-10, -100, 200, 1000]
if s<smin,
smin = s;
end
end
thank you so much.now i could understand it.its a good logic.once again thank you.
and how can i find the point at which it is minimum
clc
clear all
syms x
smin = inf; % initialize smin
for yr = linspace(0,1,10)
f = sin(x);
fL = [0 pi];
iL = [0 pi];
Volume = pi*int((f-yr)^2,iL(1),iL(2));
s=double(Volume);
if s<smin,
smin = s;
end
end
smin
for yr = linspace(0,1,10)
f = sin(x);
fL = [0 pi];
iL = [0 pi];
Volume = pi*int((f-yr)^2,iL(1),iL(2));
h=double(Volume)
if smin==h for yr= linspace(0,1,10);
disp(yr)
end
end
end
i have tried this one but this doesnt make any sense can you help
syms x
smin = inf; % initialize smin
locmin = nan;
for yr = linspace(0,1,10)
f = sin(x);
fL = [0 pi];
iL = [0 pi];
Volume = pi*int((f-yr)^2,iL(1),iL(2));
s=double(Volume);
if s<smin,
smin = s;
locmin = yr;
end
end
smin, yr
smin = 0.9437
yr = 1
ahh it is wrong.its taking yr=1 for max and min too.can you please check it.

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