I got error in left of the picture. so I did tracking. and I knew first error by using traking.
But I can't understand the reason that error is generated.
error is said that you can't divide '0'.
This is basically the same question you posted two hours ago and just marked as "being answered".
Its the same error and its NOT a division by zero!
In the sheet you posted you also can not track the error at all because of the equal sign after the definition of the function a.r(mu). If I delete that equal sign and add a a.r(mu)=.. after the definition I get the error and now it can be traced.
The reason is exactly what the error message says - you defined g.1(a,mu) as a function with two arguments, but 2 lines later you call it with just one argument.
The next error will occur when you use c(my). c is a vector defined in a table, not a function. In Prime you may access individual elements of a vector either by using the vector index (which is the preferred way) or by using the same syntax as when you call a function (which I find confusing). So c[1= and c(1)= will give you the same result. But of course the index has to be valid, so it at least has to be an integer. You use c(mu) and mu is a range of non-integer values which sure has to throw an error.
One other thing I just notice is that the first "else if" in your routine is not followed by a condition but by the assignment of a range for mu!?
I am not sure why you are working with ranges and vectors all the time, Wouldn't it be more convenient to set up your whole calculation as functions which yield single scalar values and then later call it with a vector (not a range) as argument?
P.S.: With the file you attached I don't experience the second error your screenshot shows.
