Mark44
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Strongly agree. In the example posted by @sophiecentaur, ##x^2 + y^2 = 1##, y is most definitely NOT a function of x. The same is true for the other conic sections. For the circle equation, solving for y yields ##y = \pm \sqrt{1 - x^2}##, so for each x, with ##x \ne \pm 1##, there are two y values.etotheipi said:@sophiecentaur I really have a hard time understanding what you're trying to say. The fact that functions are single valued (by definition) is a critical property.
No. Although a positive real number has two square roots, the symbol ##\sqrt x## is taken by convention to mean the principal, or positive square root. Why else do you think there's a +/- symbol in the Quadratic Formula; i.e., the solutions to ##ax^2 + bx + c = 0## are given by ##x = \frac{b^2 \pm \sqrt{b^2 - 4ac}}{2a}##.sophiecentaur said:Afaics, it's important to qualify statements about the 'meaning' of Root x.
No, he (@etotheipi) meant plain old functions. BTW, ##x^2 = y## is continuous and has derivatives of all orders. ##\frac{dy}{dx} = 2x##. OTOH, if you meant ##y^2 = x \Leftrightarrow y = \pm \sqrt x##, then y is not a function of x. (Each positive x value is paired with two y values.)sophiecentaur said:I think you mean a continuously differentiable function. There are plenty of functions with first derivatives that aren't continuous. Triangular waves, cycloids and x2 = y
No, it hasn't. From Calculus and Analytic Geometry, by Abraham Schwartz, 2nd Ed., published in 1967: (emphasis added by me)sophiecentaur said:Perhaps the vocabulary has changed in 50 years. (Which is possible.)
The relationship here is either a one-to-one relationship (e.g. y = f(x) = 2x) or many-to-one (e.g., y = g(x) = sin(x). By definition, a function cannot be one-to-many (e.g. ##y^2 = x## or equivalently ##y = \pm \sqrt x##).Function; domain; range. We are given a set of numbers, which we shall call the domain D, and instructions for associating a number y with each number x of D. The set of all numbers y associated with numbers x of D shall be called the range R. The correspondence thus created between the sets D and R shall be called a function.
The First Edition appeared in 1960, which is 60 years ago. I'm positive the definition of a function didn't change between the two editions.
No, "by definition" is not too strong. One definition of the absolute value is ##|x| = \sqrt{x^2}##, for all real numbers x.sophiecentaur said:"By definition" is too strong. You are stating that the default actually exists.
A.T. said:How can "by definition" be too strong? It just a convention about what a word means in a certain context, that is most widely accepted.