Piecewise function handle
in [Matlab]
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From: Greig on 31 Mar 2010 23:31 I would like to define a piecewise function handle but I'm not too sure how to do it. Basically something like: fn=@(x) if(x<0) 1 elseif (x>1) 0 else (1/n)(1-x)^((1/n)-1) end where n is a constant. Cheers.
From: Gene on 1 Apr 2010 12:58 "Greig " <greig(a)abc.com> wrote in message <hp141p$cvu$1(a)fred.mathworks.com>... > I would like to define a piecewise function handle but I'm not too sure how to do it. > Basically something like: > > fn=@(x) if(x<0) 1 > elseif (x>1) 0 > else (1/n)(1-x)^((1/n)-1) > end > > where n is a constant. > Cheers. Greig: One way to to this is to use 0-1 (logicals) fn = @(x) 1*(x<=0) ... + (1/n)*(1-x).^(1/n-1).*(x > 0).*(x <= 1) ... + 0 * (x > 1); Of course, the last line can be omitted, as can the 1* in the first line. BTW your function is not well-defined at x = 1. You may want to move the x = 1 condition emc
From: Greig on 1 Apr 2010 21:24
"Gene" <ecliff(a)vt.edu> wrote in message <hp2jau$eat$1(a)fred.mathworks.com>... > "Greig " <greig(a)abc.com> wrote in message <hp141p$cvu$1(a)fred.mathworks.com>... > > I would like to define a piecewise function handle but I'm not too sure how to do it. > > Basically something like: > > > > fn=@(x) if(x<0) 1 > > elseif (x>1) 0 > > else (1/n)(1-x)^((1/n)-1) > > end > > > > where n is a constant. > > Cheers. > > Greig: > One way to to this is to use 0-1 (logicals) > > fn = @(x) 1*(x<=0) ... > + (1/n)*(1-x).^(1/n-1).*(x > 0).*(x <= 1) ... > + 0 * (x > 1); > > Of course, the last line can be omitted, as can the 1* in the first line. BTW your function is not well-defined at x = 1. You may want to move the x = 1 condition > > emc > Thank for that. You're right about the limit x=1, it's the reason I'm defining it as a piecewise function, I forgot the "=>". Cheers |