Why do we use squared quantities in equations and formulae?

[sophiecentaur]

That is because not direction of velocity but magnitude of velocity matters. v² has no information of direction. Same for c².

This sort of thing can read like magic unless you're familiar with Maths and there's more to it than just the sign. I could ask you why not just use the magnitude of a value instead of using the square of the value.
Squares often come into equations and formulae when there are two quantities multiplied together and one quantity is also due to two multiplied quantities.
So, velocity times time is distance (vt) and velocity is acceleration times the time it's applied (at). This means the distance travelled, after a time t will be the average velocity times time (v_averaget). Starting from 0, the average velocity will be v/2 so the distance traveled will be vt²/2.

Squared quantities often come out of the area of a graph.

 

[anuttarasammyak]

I could ask you why not just use the magnitude of a value instead of using the square of the value.

$$|v|=\sqrt{v^2}$$ Same thing
$$f(|\mathbf{v}|)=f(\sqrt {\mathbf{v}^2})=g(\mathbf{v}^2)$$

 

[sophiecentaur]

$$|v|=\sqrt{v^2}$$ Same thing

Not in all equations of that type and you missed out the +/- bit. I think the square is better because there is better continuity around zero. Root x / root y has two possible outcomes which is not good when trying to take things further in algebra.

 

[etotheipi]

Not in all equations of that type and you missed out the +/- bit. I think the square is better because there is better continuity around zero. Root x / root y has two possible outcomes which is not good when trying to take things further in algebra.

There is no plus-minus, $|a| = \sqrt{a^2}$ is definition.

 

[jbriggs444]

There is no plus-minus, $|a| = \sqrt{a^2}$ is definition.

But it's not the definition of $|a|$. It's the definition of $\sqrt{a^2}$. There is nothing in taking a square root that demands that the result be positive or negative. We have decided by fiat that the $\sqrt{}$ notation denotes the positive square root.

 

[etotheipi]

But it's not the definition of $|a|$. It's the definition of $\sqrt{a^2}$. There is nothing in taking a square root that demands that the result be positive or negative. We have decided by fiat that the $\sqrt{}$ notation denotes the positive square root.

Yes, I meant definition of $\sqrt{}$.

 

[PeroK]

But it's not the definition of $|a|$. It's the definition of $\sqrt{a^2}$. There is nothing in taking a square root that demands that the result be positive or negative. We have decided by fiat that the $\sqrt{}$ notation denotes the positive square root.

It is an alternative definition of $|a|$:

https://en.wikipedia.org/wiki/Absolute_value#Definition_and_properties

 

[jbriggs444]

It is an alternative definition of $|a|$:

https://en.wikipedia.org/wiki/Absolute_value#Definition_and_properties

That's a property, not a definition.

Edit: I take it back. It is a definition:

|x| denotes the unique non-negative number that, when squared, yields $x^2$.

Although, if you have a definition of "non-negative" in hand and are working in a field, it seems simpler to just do the definition by cases.

 

[sophiecentaur]

That's a property, not a definition.

Edit: I take it back. It is a definition:

|x| denotes the unique non-negative number that, when squared, yields $x^2$.

Although, if you have a definition of "non-negative" in hand and are working in a field, it seems simpler to just do the definition by cases.

I must be missing something - unless this discussion is about a limited set of circumstances where the positive root is the only one to consider. But, in general - even is a simple case of ballistics - there can often be two correct answers (two possible elevations for a gun) to obtain a given range.

Afaics, it's important to qualify statements about the 'meaning' of Root x.

 

[etotheipi]

I must be missing something - unless this discussion is about a limited set of circumstances where the positive root is the only one to consider. But, in general - even is a simple case of ballistics - there can often be two correct answers (two possible elevations for a gun) to obtain a given range.

Afaics, it's important to qualify statements about the 'meaning' of Root x.

If you want to solve an equation $x^2 = 4$, then$$\begin{align*}

x^2 &= 4 \\

\sqrt{x^2} &= \sqrt{4} \\

|x| &= 2 \implies x = \pm 2

\end{align*}
$$The point is that the square root of a number $a^2$ is by definition positive, i.e. $\sqrt{a^2} = |a|$. You need this constraint in order for $f(x) = \sqrt{x}$ to be a function (i.e. it's a one-to-one map). That's not to say that $a$ isn't potentially negative.

 

[PeroK]

I must be missing something - unless this discussion is about a limited set of circumstances where the positive root is the only one to consider. But, in general - even is a simple case of ballistics - there can often be two correct answers (two possible elevations for a gun) to obtain a given range.

Afaics, it's important to qualify statements about the 'meaning' of Root x.

Nevertheless, if $a \in \mathbb R$, then
$$|a| = \sqrt{a^2}$$

 

[sophiecentaur]

in order for f(x)=x to be a function

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

You are making assumptions here and you really should be more explicit - if you want avoid confusing people.

 

[etotheipi]

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

You are making assumptions here and you really should be more explicit - if you want avoid confusing people.

No, I do mean just function. In order for the map $f(x) = \sqrt{x}$ to be a function, each $x$ in the domain is mapped to a unique element of the codomain.

Nothing I say has anything to do with differentiability, or continuity. It's just the definition of what a function is!

 

[sophiecentaur]

I did a course on Mathematical Analysis many years ago and it was frequently necessary to specify the nature of any function that was discussed. What would you call a set of symbols that involve discontinuities, if not a Function?

Are you suggesting that a function can only be single valued or that it can, perhaps only operate 'one way'? That's a very limiting sort of definition. OK if that's the way you want it, you still need to make that clear every time, imo.

I know that there are lots of subsets of use of Maths that are self consistent and give perfectly good answers but the field of Maths surely has to include general cases.

 

[etotheipi]

Are you suggesting that a function can only be single valued or that it can, perhaps only operate 'one way'? That's a very limiting sort of definition. OK if that's the way you want it, you still need to make that clear every time, imo.

What would you call a set of symbols that involve discontinuities, if not a Function?

What do discontinuities have to do with anything? A function can be continuous, or it can be discontinuous.

 

[PeroK]

Are you suggesting that a function can only be single valued

A function is, by definition, single-valued. If dealing with multiple values (like complex roots and logs etc.), you would need to specify "multi-valued".

 

[sophiecentaur]

A function is, by definition, single-valued.

So a conic section like a circle is not 'a function'? I think you need to re-think this.

 

[PeroK]

So a conic section like a circle is not 'a function'? I think you need to re-think this.

A circle is not the graph of a function. Technically it is the graph of a relation, of which a function is a special case.

https://en.wikipedia.org/wiki/Function_(mathematics)

 

[sophiecentaur]

@etotheipi I think I would go along with my University course, rather than one person's view that we can find on Wiki. It reads 'convincingly until you realize that the writer is discussing a limited region of uses of the word.

 

[sophiecentaur]

A circle is not the graph of a function. Technically it is the graph of a relation, of which a function is a special case.

https://en.wikipedia.org/wiki/Function_(mathematics)

I guess, with that, we have to part company. If y²+x² = 1 is not y, as a function of x or vice versa then we are definitely on a different wavelength. Perhaps the vocabulary has changed in 50 years. (Which is possible.)

 

[weirdoguy]

rather than one person's view that we can find on Wiki.

It's not "one person's view", it's every textbook/teacher/lecturer's that I have encountered in my whole life point of view. There is something called implicit function: https://en.wikipedia.org/wiki/Implicit_function that you probably have in mind. But still, the very definition of a function requires single-valuedness. If you want mulitple values you have something called relations that are much more general than functions.

 

[sophiecentaur]

Ok. So y = sin(x) is a function but
X=sin-1(y) is not. 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.
Anything as sweeping as the one-to-one condition has to be qualified with limits. Do people ever actually do that?
We need an umpire perhaps.

 

[etotheipi]

Yes, you do have to be careful when trying to find inverses of functions that aren't one-to-one. In those cases, you have to restrict the domains so that they are one-to-one in that interval.

For $f(x) = \sin{x}$, the range of $f^{-1}(x) = \arcsin{x}$ will be $\mathcal{R} = [-\frac{\pi}{2}, \frac{\pi}{2} ]$. This is analogous in concept to the square root example, because for instance $a = \sin{(\arcsin{(a)} + 2n \pi)} = \sin{( - \arcsin{(a)} + [2m+1] \pi)}$, with $a \in [-1, 1]$.

 

[sophiecentaur]

My only reason for carrying on with this is to deal with this:

A function is, by definition, single-valued. If dealing with multiple values (like complex roots and logs etc.), you would need to specify "multi-valued".

"By definition" is too strong. You are stating that the default actually exists. A function is a function. If it is relevant that there are certain ranges in which it applies then it would be stated but I would say that the limits are qualifiers of and parts of the function.

Yes, you do have to be careful when trying to find inverses of functions that aren't one-to-one. In those cases, you have to restrict the domains so that they are one-to-one in that interval.

Yes, I agree except why distinguish between a function and its inverse? It's the same relationship that's being described and only a re-arrangement of the components. Why exclude all the familiar functions from being functions just because the direction through the operation is different? The one to one stipulation seems pointless.

 

[etotheipi]

@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.

To give an example, the reason why you can 'do the same thing to both sides' when doing algebra is formalised by the so-called substitution property of equality. That is, if $a = b$, then $f(a) = f(b)$. Evidently this rests on functions being single-valued.

There is no wiggle room here!

 

[pbuk]

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]

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.

 

[Svein]

That is the problem with real numbers - there are traps lying around! The complex numbers have solved this kind of dilemmas by introducing Riemann surfaces (https://en.wikipedia.org/wiki/Riemann_surface).

 

[sophiecentaur]

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]

@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.

 

[sophiecentaur]

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]

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]

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.

Ahh - and the 'quadratic solving formula' actually includes the term +/-√(4ac) to take care of that. It was hiding from me in plain sight!