How to write a basic loop

[EDIT: Thu May 26 02:39:10 UTC 2011 - Reformat - MKF]
z = 0.05;
y = 1/20;
f = @(x) (10.^(-15).*x.^(3).*exp((x.^2).^(-1)/4).*y.^(-1)).^(1/5)-z;
fzero(f,1);
This is my given command line and I need to write a loop in order to solve for a range of z and y values? How would I go about this? I have tried just giving a y value such as:
for z = 0.05:.06,
f = @(x) (10.^(-15).*x.^(3).*exp((x.^2).^(-1)/4).*.05.^(-1)).^(1/5)-z;
fzero(f,1)
and letting it try and run a simple loop for z but it doesn't seem to work. It seems easy but its giving me a lot of trouble. Thanks ahead of time for any help.

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Matt Fig
Matt Fig el 26 de Mayo de 2011

2 votos

Your question is confusing because you say you want to solve for a range of x and y values but you are looping over z values. Could you show the values you actually want to use?
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EDIT After some clarifying comments.
Here is an example with some range for y and z that I made up.
opts.Display = 'off';
y = 0:.01:1; % Some y values.
z = 0:.01:.14; % Some z values.
H = nan(length(y),length(z));
for ii = 1:length(y)
for jj = 1:length(z)
f = @(x) (10.^(-15).*x.^(3).*...
exp((x.^2).^(-1)/4).*y(ii).^(-1)).^(1/5)-z(jj);
try
H(ii,jj) = fzero(f,1,opts);
catch
end
end
end
H % Display the results. Row is y, column is z

3 comentarios

Matt Fig
Matt Fig el 26 de Mayo de 2011
O.k., but still,... what values of y and z?
William Duhe
William Duhe el 26 de Mayo de 2011
O.k I was a little jumbled up in my initial writing.... I need to solve for x, with varying z and y values. Such as the original equation I posted:
z = 0.05;
y = 1/20;
f = @(x) (10.^(-15).*x.^(3).*exp((x.^2).^(-1)/4).*y.^(-1)).^(1/5)-z;
fzero(f,1);
I want to run a loop for that with different values of z and/or y. I hope this makes sense, I'm still kind of a rookie to Matlab and its rough for me =)
William Duhe
William Duhe el 26 de Mayo de 2011
This is awesome, I have a meeting tomorrow to figure out a little bit more about what range of values I'm going to be running, I'll let ya know if I have any other questions thanks!

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