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How to plot this function?

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Michaël
Michaël 2011 年 3 月 14 日
Hello,
Could you please help me plot this function with Matlab ?
<http://img638.imageshack.us/img638/3205/fonction.png >
I'm a beginner. Your help will be valuable to me.
Many thanks

採用された回答

Matt Fig
Matt Fig 2011 年 3 月 15 日
If you want to do it manually, try this:
f = @(z) quadv(@(x) (1/(3*sqrt(2*pi)))*exp(-(x-30).^2/18),z,100);
Z = 0:.25:50;
G = zeros(1,length(Z));
for ii = 1:length(Z)
G(ii) = f(Z(ii));
end
plot(Z,G)
xlabel('z')
ylabel('(1/\surd(2*pi))\inte^{-(x-30)^2/18}')

その他の回答 (4 件)

Matt Tearle
Matt Tearle 2011 年 3 月 14 日
Take a look at erfc. A simple change of coordinates should get you there.
Then plotting can be done by
x = linspace(xmin,xmax,npoints);
y = constant*erfc(x);
plot(x,y)
  2 件のコメント
Michaël
Michaël 2011 年 3 月 14 日
Thank you for your answer.
But can't I put my function directly (without going through a pre-determined function)?
Thanks
Matt Tearle
Matt Tearle 2011 年 3 月 15 日
Why would you want to? This function is defined via an improper integral, which means you'll have to do an infinite-domain integral for each value of z. This will be slow and will introduce all sorts of numerical errors. On the other hand, the error function is a well-known function -- its implementation in MATLAB is stable and tested. Why reinvent the wheel?
But if you really want to, you can use something like quad to do the integral.

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Michaël
Michaël 2011 年 3 月 14 日
Thank you for your answer.
But can't I put my function directly (without going through a pre-determined function)?
Thanks

Michaël
Michaël 2011 年 3 月 14 日
How can't we put : (-1+x.)^2 ?

Michaël
Michaël 2011 年 3 月 15 日
Thank you very much !!!

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