It broke my head before realising there is a typo here in "A wave function is any function such that f(x,t)=f(x−vt)"!

[MaxMath]

I was reading this about the subject of wave functions when having a systematic review of Maxwell's equations. It broke my head before realising there is a typo here in "A wave function is any function such that [imath]f(x,t)=f(x−vt)[/imath]". It really annoyed me. It should be [imath]y(x,t)=f(x−vt)[/imath]!

 

[mario99]

When you study partial differential equations, you will realize that the general solution to the wave equation is:

[imath]\displaystyle u(x,t) = \frac{1}{2}[f(x - ct) + f(x + ct)] + \frac{1}{2c}\int_{x - ct}^{x + ct} g(s) \ ds[/imath]

Which can also be written in short like this:

[imath]\displaystyle u(x,t) = F(x - ct) + G(x + ct)[/imath]

Describing two waves traveling in opposite directions. The [imath]F[/imath] is the function of the wave traveling to the right while [imath]\displaystyle G[/imath] is the function of the wave traveling to the left. When writing the wave function like this:

[imath]\displaystyle y(x,t) = f(x - vt)[/imath]

The function is describing a single wave traveling to the right. Therefore, it's not wrong to use the same variable again when you're only describing one wave! But it's better to use different variables for a better illustration.

I prefer the two different variables notation.

 

[MaxMath]

The function is describing a single wave traveling to the right. Therefore, it's not wrong to use the same variable again when you're only describing one wave! But it's better to use different variables for a better illustration.

I prefer the two different variables notation.

Thank you. It sounds like what I think is a typo may not necessarily be one; it's one of the legitimate ways of saying that (though probably not the best one).

This is interesting. The reason is that I find it hard to get my head around this expression [imath]f(x,t) = f(x - vt)[/imath], based on my (non mathematics-purist) understanding of function or the notion of function (single or multivariable).

It took me some time to finally find a way to properly (hopefully) explain this.

The notion [imath]f(x,t)[/imath], for a function, means an expression with two variables as the arguments of the function [imath]f[/imath], which allows us to 'plug' any values, or symbols, such as [imath]something[/imath] and [imath]something_{else}[/imath], into the actual formula of [imath]f[/imath], such as [imath]\sqrt{x - vt}[/imath] ([imath]v[/imath] as constant), into the places of [imath]x[/imath] and [imath]t[/imath], respectively. So we get the evaluation of the function [imath]f[/imath], [imath]\sqrt{something-v\cdot something_{else}}[/imath], when its two arguments are so assigned, i.e. [imath]f(something,something_{else})[/imath].

Since the symbols for the variables (or arguments) are arbitrary, to avoid unhelpful distraction from the form, let's make a convention on the use of variables -- always use x, y, z for the first, second and third variables (if there are) for any function. (Anything else in an expression of a function is not a variable.)

So the proper expression is [imath]f(x,y)=\sqrt{x - v y}[/imath]. What [imath]f[/imath] stands for is the rest of the expression (right hand side) with [imath]x[/imath] and [imath]y[/imath] taken out, i.e. [imath]\sqrt{\triangle - v \square}[/imath] ([imath]\triangle[/imath] and [imath]\square[/imath] as placeholders for the variables).

With the notion of [imath]\displaystyle f(x,t) = f(x - vt)[/imath], now of course I understand what it means by [imath]f[/imath] on the right hand side is [imath]f(u)[/imath] (or [imath]f(x)[/imath], if in the 'proper' form following our convention) with [imath]x - v t[/imath] (or [imath]x - v y[/imath]) as the value for its single variable. The problem however, is that what [imath]f(x)[/imath] stands for, for the same example, will now be [imath]\sqrt{\triangle}[/imath]. This is obviously different from [imath]\sqrt{\triangle - v \square}[/imath], in line with my intuition (how can a two-variable function all of the sudden be expressed as a same single-variable function?!).

By the way, can you explain the reason for "Therefore" (I now highlighted with bold typeface)?

 

[MaxMath]

So I think the accurate, and responsible, way is to say ---

A wave function is any function [imath]f(x,t)[/imath] that is in the form of [imath]g(x−vt)[/imath].

 

[mario99]

By the way, can you explain the reason for "Therefore" (I now highlighted with bold typeface)?

Of course, I can. "Therefore" means that the variable [imath]y[/imath] always represents the wave [imath]f(x-vt)[/imath] so let's just give it the same initial name [imath]f[/imath].

There is nothing better than an example to clear up some thoughts. Let's take an arbitrary function [imath]f(x) = \sqrt{2x}[/imath]. Let's apply the definition of a wave function using the ugly notation:

[imath]f(x,t) = f(x - vt) = \sqrt{2x - 2vt}[/imath]

It's not wrong to use the same variable again (maybe for saving some ink), but because we did that we have to be very careful with this sloppy notation. Of course [imath]f(x)[/imath] and [imath]f(x,t)[/imath] are completely different functions except when [imath]t = 0[/imath], and when we say the function [imath]f[/imath], the reader will be confused which function we mean. Therefore, we will always have to mention the function [imath]f[/imath] along with its parameter (parameters), so that the reader will know exactly which function we are refering to.

When you study the wave equation in partial differential equations, you will see this initial condition a lot:
[imath]u(x,0) = f(x)[/imath]

So I think the accurate, and responsible, way is to say ---

A wave function is any function [imath]f(x,t)[/imath] that is in the form of [imath]g(x−vt)[/imath].

I agree with you in almost everything. I don't agree with this notation [imath]f(x,t) = f(x - vt)[/imath] (as the best notation), I just accept it as valid because I have seen it in my different places. To avoid confusion (like what happened to you), it's better for the authors to introduce new variables whenever possible.

Note: I did not mention every point you have written in your post because I agree with what you have said.

 

[MaxMath]

Of course, I can. "Therefore" means that the variable [imath]y[/imath] always represents the wave [imath]f(x-vt)[/imath] so let's just give it the same initial name [imath]f[/imath].

There is nothing better than an example to clear up some thoughts. Let's take an arbitrary function [imath]f(x) = \sqrt{2x}[/imath]. Let's apply the definition of a wave function using the ugly notation:

[imath]f(x,t) = f(x - vt) = \sqrt{2x - 2vt}[/imath]

I don't think we have anything in disagreement. Thank you for your thoughts.

This reminds me, again, that it's often very easy to be fooled by the form of mathematics, and it makes me very nervous with such sloppy practice. Because in even one line of mathematics expression, I can hardly tell the very nuanced problem in it. If this goes on and reaches a certain size of an argument, there is virtually no chance of me knowing if I have made a grave mistake in my derivations leading to what I thought was a big breakthrough!

For this exact reason, I believe 1) it's very important to have the right notion for concepts and theories of mathematics (calculus, specifically the notation of [imath]dy/dx[/imath], might be an example of the bad ones -- or one might say exactly the opposite?!); and 2) we should stick with a high standard when it comes to the choice of notation.

Part of the reason why I'm so fussy about the notion (and the concept) of function is due to this post, where I ended up struggling to get a crystal clear idea of what exactly function is in solving a practical problem.

For example, can we say that this (probably the simplest function)
[imath]f = 0[/imath]
is a two-variable function
[imath]f(x, y) = x - y[/imath]
where
[imath]y = x[/imath]?

Keen to hear the thoughts of people conversant with pure/advanced mathematics, so hopefully to get the essence out of it.

 

[mario99]

Part of the reason why I'm so fussy about the notion (and the concept) of function is due to this post, where I ended up struggling to get a crystal clear idea of what exactly function is in solving a practical problem.

I thought that we have given you the good reasons why your notation was invalid in that post. And eventually you were convinced and agreed! Am I wrong?

For example, can we say that this (probably the simplest function)
[imath]f = 0[/imath]
is a two-variable function
[imath]f(x, y) = x - y[/imath]
where
[imath]y = x[/imath]?

Keen to hear the thoughts of people conversant with pure/advanced mathematics, so hopefully to get the essence out of it.

Yes, I would here say [imath]y = x[/imath]. But it's very rare to see them write [imath]f = 0[/imath]. They will write [imath]f(x,y) = 0[/imath] or they will introduce new variable [imath]y = f(x,y)[/imath] and then they will write [imath]y = 0[/imath].

 

[MaxMath]

I thought that we have given you the good reasons why your notation was invalid in that post. And eventually you were convinced and agreed! Am I wrong?

Just release you were there – yeah, I think I agreed. But I only kind-of understood, or even I did, it's not to a degree that I hope it was.

 

[mario99]

Just release you were there – yeah, I think I agreed. But I only kind-of understood, or even I did, it's not to a degree that I hope it was.

Yes, I was there. And I understand why you compare this thread with that thread.

In there, we had [imath]v(s)[/imath] and [imath]v(t)[/imath], and when the integral was written as [imath]\int v \ dt[/imath], it was confusing to know if this is [imath]v(s)[/imath] or [imath]v(t)[/imath]. But later we knew how to deal with this confusion and the problem was resolved.

The funny thing, in there, I made a mistake, and you have corrected me. But then you used my mistake, and I had a hard time to show you it was a mistake while you were the one who first showed me it was a mistake.

 

[MaxMath]

The funny thing, in there, I made a mistake, and you have corrected me. But then you used my mistake, and I had a hard time to show you it was a mistake while you were the one who first showed me it was a mistake.

Yes, very funny, embarrassingly funny!