# Digression on squaring

pbuk
Gold Member
We need an umpire perhaps.
For which bit? You could take Mathworld as an authoritative external reference.

Ok. So y = sin(x) is a function but
No, that relation is not a complete description of a function; in order to make it complete we need to specify the domain of the function, which is the set of values that contains exactly all values that ## x ## is allowed to take, and the range of the function, which is a set of values that contains at least every corresponding value of ## y ##.

X=sin-1(y) is not.
It might be: if we specify the domain ## y \in [0, 1] ## and the range ## x \in [0, \frac{\pi}2] ## then we have a function but there is no such function with the range ## x \in [0, \pi] ##.

Thete will be ‘functions’ which are functions in a limited range. Difficult but my insistence on some sort of qualification needing to be added to any function is justified.
I'm not sure what this bit means.

Anything as sweeping as the one-to-one condition has to be qualified with limits.
Note that a function can also be many-to-one (but not one-to-many or many-to-many).

Do people ever actually do that?
Yes, mathematicians often do - but only when it matters. Like most scientists mathematicians often omit a lot of detail and assumed prior knowledge, if we didn't then mathematical texts would be unreadable. OK, even more unreadable.

I probably ought to say that there is also something called a 'many valued function', of which ## y = \sin^{-1}x, x \in \mathbb R, y \in [-1, 1] ## is an example but this is not a special kind of function, it is a completely different thing ('many-valued' is not an adjective here, rather the whole expression is a compound noun).

Perhaps the vocabulary has changed in 50 years.
Perhaps - it is only 40 years since I learned this, almost to the day. I wouldn't be surprised though if the single-valued criterion goes back to the 19th century.

pbuk
Gold Member
My only reason for carrying on with this is to deal with this:
...
"By definition" is too strong.
Try searching for some definitions, for example here and here and here.

The one to one stipulation seems pointless.
Well it is a one-or-many to one stipulation, but I am afraid that regardless of whether it seems pointless to you or not it is a stipulation that everyone else has agreed on.

etotheipi, PeroK and weirdoguy
Svein
etotheipi
Gold Member
For which bit? You could take Mathworld as an authoritative external reference.

No, that relation is not a complete description of a function; in order to make it complete we need to specify the domain of the function, which is the set of values that contains exactly all values that ## x ## is allowed to take, and the range of the function, which is a set of values that contains at least every corresponding value of ## y ##.

It might be: if we specify the domain ## y \in [0, 1] ## and the range ## x \in [0, \frac{\pi}2] ## then we have a function but there is no such function with the range ## x \in [0, \pi] ##.

I'm not sure what this bit means.

Note that a function can also be many-to-one (but not one-to-many or many-to-many).

Yes, mathematicians often do - but only when it matters. Like most scientists mathematicians often omit a lot of detail and assumed prior knowledge, if we didn't then mathematical texts would be unreadable. OK, even more unreadable.

I probably ought to say that there is also something called a 'many valued function', of which ## y = \sin^{-1}x, x \in \mathbb R, y \in [-1, 1] ## is an example but this is not a special kind of function, it is a completely different thing ('many-valued' is not an adjective here, rather the whole expression is a compound noun).

Perhaps - it is only 40 years since I learned this, almost to the day. I wouldn't be surprised though if the single-valued criterion goes back to the 19th century.
You are probably right. I never considered, until now, that my Analysis course was actually more elementary than they implied.
But it's true that (back to the original point) that relying on the modulus of a value could be fraught (whether or not you use the word "function"). Calculators happily give you answers for inverse trig functions so perhaps they should all flash a warning when those buttons are pressed.

A.T.
"By definition" is too strong.
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.

Mark44
Mentor
@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.
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.
Afaics, it's important to qualify statements about the 'meaning' of Root x.
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}##.
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, 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.)
Perhaps the vocabulary has changed in 50 years. (Which is possible.)
No, it hasn't. From Calculus and Analytic Geometry, by Abraham Schwartz, 2nd Ed., published in 1967: (emphasis added by me)
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 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##).
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.
"By definition" is too strong. You are stating that the default actually exists.
No, "by definition" is not too strong. One definition of the absolute value is ##|x| = \sqrt{x^2}##, for all real numbers x.

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.

Klystron, weirdoguy, etotheipi and 1 other person
Gold Member
I do take all of your points and I clearly have mistaken how the word "function" is strictly defined. So the term is mi-used a lot? Perhaps and it wouldn't be the only commonly mis-used term.

Mark44
Mentor
I do take all of your points and I clearly have mistaken how the word "function" is strictly defined. So the term is mi-used a lot? Perhaps and it wouldn't be the only commonly mis-used term.
Yes, it's misused a lot. A test that can be used to determine whether a graph represents a function is the vertical line test. If a vertical line is swept along the horizontal axis and never intersects more than one point, the graph is that of a function. This would show that the graph of a circle isn't that of a function, nor is the graph of the other conic sections, including, say, ##y^2 = x##.
Also, we have had many threads over the years where people are confused about the meaning of ##\sqrt x##, thinking that it represents two values.

sophiecentaur