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Best answer by wayne

Prior to your problem with z; x and y are predefine vectors, but z was always input as a scalar in the functions.

Up to the point of your problem, the variable z was never assigned a value, out side of an argument list.

In the last equation cf does not have an argument list (x,y,z), which I think you want. At least you need cf(z)=

Since the vectors x and y were defined prior to the function definition, they do not actually have to be in the argument list, unless you want to use the functions again with different x and y vectors.

1 reply

wayne12-AmethystAnswer
12-Amethyst
September 11, 2011

Prior to your problem with z; x and y are predefine vectors, but z was always input as a scalar in the functions.

Up to the point of your problem, the variable z was never assigned a value, out side of an argument list.

In the last equation cf does not have an argument list (x,y,z), which I think you want. At least you need cf(z)=

Since the vectors x and y were defined prior to the function definition, they do not actually have to be in the argument list, unless you want to use the functions again with different x and y vectors.

1-Visitor
September 11, 2011

Thanks: something stupid, as I thought. cf should be cf(x,y,z).