fdiscont

acer — Sun, 02 Oct 2011 08:50:15 Z

It looks like `fdiscont` has more success than symbolic `discont` here.

Try these,

plot(F(x),x=-1..27,y=0..0.5,gridlines=true,discont=true,usefdiscont);

plot(F(x),x=-1..27,y=0..0.5,gridlines=true,discont=[usefdiscont]);

The following also gets it right. I suspect that's because `discont` needs an expression (like your unevaluated F(x) as the first argument to `plot`). So, since `discont` doesn't accept an operator, like just F instead of F(x), then plot has to fall back to using `fdiscont`.

plot(F,-1..27,0..0.5,gridlines=true,discont=true);

The first pair were about 2-3 times faster than the last one.

It might look more legible (to some people), but creating an operator like `F` only so that it can ever get used as a function call like F(x) is something that often leads to confusion because it muddles unnecessarily with the distinction between operator and expression. You didn't encounter it here (maybe because the `Statistics` author went gone to extra pains to anticipate it) but this confusion often results in an error message, when the operator is prematurely evaluated at symbolic `x` while it is only prepared to handle numeric values for parameter x.

Another possibility that works is,

plot(CDF(X,x),x=-1..27,y=0..0.5,gridlines=true,discont=[usefdiscont]);

where once again, the piecewise symbolic result from CDF(X,x) may not get best result from symbolic `discont`.

acer

MaplePrimes - answers and comments on Question, Why doesn't discont=true work in this case?

fdiscont