I have a variable, x, and I want to know whether it is pointing to a function or not. I had hoped I could do something like: >>> isinstance (x, function) But that gives me: Traceback (most

2015年1月2日 · For example f may be from A -> B and g is from C -> D and if the Range of g is not a subset of Domain of f the composition won't be a function. So essentially we have to …

2 Is there a way to see if a relation is a function without having to do a "vertical line test" (where you draw a vertical line on the graph and if there line touches two points then it's not a …

You shouldn't have equality with f and the sine function is f is the pdf. Might also be good to point out that from 0 to 2 pi the sine is positive over half the cycle and negative over the other.

Thanks--so from this and the above, it is clear that I have asked about something that is not part of c++ (but apparently is possible in Java, so there are probably a couple of people before me …

A function is always defined otherwise it is not a function. What you need in the example, is to prove that there exists a function $f$ satisfying some given property, and that such a function …

Whether or not a function is supposed to be called with new should be discernable from its documentation. When I create a function, how do I make it NOT a constructor? To create a …

Make sure to remove the parenthesis on your function and only use the function name in the if-condition or the function will be invoked instead of giving you true or false.

For something like this, I would just make sure the function is one-to-one / injective. This is because if a function is injective, then it has an inverse as well.

1 If you want to do something with the output of your function in the case it returns something, first pass it to a variable and then check if the type of that variable is "undefined".