What is "identical to" operator?

[find_the_fun]

What does mean "identically equal to" in the context of differential equations? In class the prof wrote \(\displaystyle \mu_x \equiv 0\). I asked what it meant and he said "it means identical to". Can someone elaborate, for example what purpose does it surve? If it just means a function always has that value, why not use the regular equal sign (=)? For example, isn't it perfectly valid to write \(\displaystyle f(x)=5\)?

And out of curisousity, we were told \(\displaystyle \mu_x\) is Newton's way of writing partial derivatives, is that correct or did someone else come up with it?

 

[magneto1]

A function $f$ is said to be "identically 0", notationally $f\equiv 0$
or $f(x) \equiv 0$ when the function evaluates to $0$ over the domain. If you write $f(x) = 0$, you will need to add another qualifier, e.g. for all $x \in D$ to mean the same thing. Otherwise, $f(x) = 0$ just tells me that $f$ evaluates to $0$ at $x$ but there may be some $x' \neq x$ where $f(x') \neq 0$.

 

[find_the_fun]

But isn't x variable meaning its value can vary to anything, so how can two variables be different? I'm serious, I was never properly taught the difference between a variable and a constant. I could understand f(A) not equal f(B) where A not equal B if A and B are constants, but if x and x' are variables then I don't see how it's true f(x) not equal f(x')

 

[magneto1]

Variables need not vary to the same values. Two variables $x,y$ can take on different value. If $f: x \mapsto x^3$, clearly, if $x \neq y$, $f(x) \neq f(y)$.

 

[find_the_fun]

Ok well there's still some things I don't get. You say if you use = instead of \(\displaystyle \equiv\) then you have to specify the domain. For example \(\displaystyle f(x) = 1\) for all x in domain. But why don't you have to do this in other cases? For example just saying \(\displaystyle f(x)=x^2\) is unambigious, a reader wouldn't assume the writer forgot to add anoter case. But by your argument one could say there is x' such that \(\displaystyle f(x') \neq x'^2\)

Also would it be wrong to write \(\displaystyle f(x)=0x\) instead of using \(\displaystyle \equiv\)?

 

[find_the_fun]

A function $f$ is said to be "identically 0", notationally $f\equiv 0$
or $f(x) \equiv 0$ when the function evaluates to $0$ over the domain. If you write $f(x) = 0$, you will need to add another qualifier, e.g. for all $x \in D$ to mean the same thing. Otherwise, $f(x) = 0$ just tells me that $f$ evaluates to $0$ at $x$ but there may be some $x' \neq x$ where $f(x') \neq 0$.

Can't you say the same thing about other functions? If someone says \(\displaystyle f(x)=x^2\) couldn't you say that's ambigious as for \(\displaystyle x'\) f may not be \(\displaystyle x^2\)? What's so special about a constant?

 

[magneto1]

When one says $f(x) = x^2$, it is not ambiguous on what the function does -- as the exact mapping is provided, it is ambiguous on the domain of the function. If I am to ask you is $f: x \mapsto x^2$ is invertible, you cannot give me a yes or no answer without assuming the domain or specifying one.

Imagine the writing, "If polynomial $p$ has such and such properties, then $p(x) = 0$". What exactly does it mean? Do they mean there is some $x$ such that $p(x) = 0$ but not others? or do they mean all $p(x) = 0$ for all $x$, in which case, we can simply say $p$ is identically 0, or $p \equiv 0$.

The importance of this is not the notational difference between $=$ vs. $\equiv$. While there are convention, and you may disagree with said convention, the importance is that when you utilize variables, they should be qualified in some ways.