Digression on squaring

  • #1
sophiecentaur
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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 (vaveraget). Starting from 0, the average velocity will be v/2 so the distance travelled will be vt2/2.

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

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  • #2
anuttarasammyak
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I could ask you why not just use the magnitude of a value instead of using the square of the value.
[tex]|v|=\sqrt{v^2}[/tex] Same thing :wink:
[tex]f(|\mathbf{v}|)=f(\sqrt {\mathbf{v}^2})=g(\mathbf{v}^2)[/tex]
 
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  • #3
sophiecentaur
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[tex]|v|=\sqrt{v^2}[/tex] Same thing :wink:
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.
 
  • #4
etotheipi
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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.
 
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  • #5
jbriggs444
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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.
 
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  • #6
etotheipi
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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{}##.
 
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  • #8
jbriggs444
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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.
 
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  • #9
sophiecentaur
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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.
 
  • #10
etotheipi
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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.
 
  • #11
PeroK
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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
$$|a| = \sqrt{a^2}$$
 
  • #12
sophiecentaur
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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.
 
  • #13
etotheipi
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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!
 
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  • #14
sophiecentaur
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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.
 
  • #15
etotheipi
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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.
1600535792275.png


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.
 
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  • #16
PeroK
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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".
 
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  • #17
sophiecentaur
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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.
 
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  • #19
sophiecentaur
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@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 realise that the writer is discussing a limited region of uses of the word.
 
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  • #20
sophiecentaur
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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.)
 
  • #21
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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.
 
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  • #22
sophiecentaur
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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.
 
  • #23
etotheipi
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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]##.
 
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  • #24
sophiecentaur
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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.
 
  • #25
etotheipi
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@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!
 
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