Doubt about the derivative of a Taylor series

[Hak]

I am afraid I am cofusing the notation. Let me say $v^2=x$ for simlicity in writing

Way 1
\frac{\partial L(x)}{\partial x}=\frac{\partial L(x)}{\partial x}(x)
is a function of variable x and replacing x wih x'

\frac{\partial L(x')}{\partial x'}=\frac{\partial L(x')}{\partial x'}(x')
is a function of variable x'.　　Say variable x and variable x' are independent

\frac{\partial L(x')}{\partial x}=\frac{\partial L(x)}{\partial x'}=0

Say variable x is a function of x' and vice versa
\frac{\partial L(x)}{\partial x'}=\frac{\partial L(x)}{\partial x}\frac{\partial x}{\partial x'}
\frac{\partial L(x')}{\partial x}=\frac{\partial L(x')}{\partial x'}\frac{\partial x'}{\partial x}

Way 2
\frac{\partial L(x)}{\partial x}|_{x=x_0}=\frac{\partial L(x)}{\partial x}(x_0)
is a vaule of the partial differential at $x=x_0$ where I wrote x_0 instead of x' in ordet to mention cleary that it is a value not a variable.

Which way ( or another one ) do you use dashed one in your OP ?

Sorry, I cannot understand what you are talking about. Maybe @PeroK? Thanks anyway.

 

[Hak]

This is a subtle point and is covered in full in my Insight on the chain rule:

https://www.physicsforums.com/insights/demystifying-chain-rule-calculus/

The specific answer in this case is that $L'$ is a well-defined function. But, if you want to use the $\frac{dL}d$ or $\frac{\partial L}{\partial}$ notation, then you must specifiy a variable in the denominator. It's a sort of dummy variable and is there only because you can't omit it. You need to put something in the denominator.

In this case $$\frac{\partial L}{\partial (v^2)} \equiv \frac{\partial L}{\partial (v'^2)} \equiv L'$$They all just mean "the derivative of the function $L$".

I don't quite understand what "well-defined function" means. Could you please explain it to me? What is the difference between a "well-defined function" and a "function"? I have found conflicting opinions on the net... Thank you very much.

 

[PeroK]

I don't quite understand what "well-defined function" means. Could you please explain it to me? What is the difference between a "well-defined function" and a "function"?

Technically nothing. Well-defined is used just for emphasis. A derivative (of a function) is a function.

 

[Hak]

Technically nothing. Well-defined is used just for emphasis. A derivative (of a function) is a function.

OK, thank you very much. But why, in this case, did you want to emphasise that $L'$ is a 'well-defined function'?

 

[PeroK]

OK, thank you very much. But why, in this case, did you want to emphasise that $L'$ is a 'well-defined function'?

I said $L'$ was a well-defined function. In many physics textbooks, informal notation leads to functions not being well-defined - usually in the sense that the same symbol is used for two different functions. That's covered in my Insight!

 

[Hak]

I said $L'$ was a well-defined function. In many physics textbooks, informal notation leads to functions not being well-defined - usually in the sense that the same symbol is used for two different functions. That's covered in my Insight!

I asked you precisely because it is a passage in your Insight that I did not understand. As far as I know, you cannot use prime notation without specifying the variable that is changing. Therefore, you cannot write $L'$ if you are referring to a partial derivative, but must write, in this case, $L' (v^2)$. This is what I know. Does this not conflict with the definition of a well-defined function? This I did not understand. Could you explain it? Thank you very much.

 

[PeroK]

I asked you precisely because it is a passage in your Insight that I did not understand. As far as I know, you cannot use prime notation without specifying the variable that is changing.

Yes you can. If $f$ is a function, then $f'$ is its derivative. There is no need to write $f(x)$. Technically, $f(x)$ is a function value (not a function). Although, we'd all struggle if we had to obey that at all times. So, we use informal notation to communicate more effectively.

Therefore, you cannot write $L'$ if you are referring to a partial derivative, but must write, in this case, $L' (v^2)$.

No. You need to specify the argument to which the partial derivative applies. It's common, in fact, with partial derivatives to omit the point at which the function is being evaluated. That's also in the Insight.

This is what I know.

It's wrong, so not such a good thing to know.

Does this not conflict with the definition of a well-defined function?

No.

 

[Hak]

No. You need to specify the argument to which the partial derivative applies. It's common, in fact, with partial derivatives to omit the point at which the function is being evaluated. That's also in the Insight.

I know, in fact I watched it. I wasn't referring to that. I was referring to partial derivatives when the function is of a single variable (hence, not really partial derivatives, as in this precise case). It is not possible, as you say, to write the partial derivative without specifying the changing variable, i.e. the argument to which it is to be applied. I confused the argument with the evaluation point. Perhaps, the evaluation point is in the numerator and the variable that is changing in the denominator, no? If I use prime notation ($L'$), how can I tell which variable is changing? That was my point, there was a lot of confusion on my part, sorry. In your Insight, it is written that it is possible to use the prime notation without referring to the variable, which must instead be written in the partial derivative. However, I was referring to the school of thought according to which it is better to indicate the Lagrangian with the subscript, rather than with the prime notation. I simply meant that the notation is limiting, you cannot write the partial derivative with a strict prime notation, you have to sacrifice some things. I don't see a solution other than that. Right?

 

[vanhees71]

Hm, it's a somewhat difficult point of what's the right notation.

My favorite form of Taylor's theorem goes as follows:\
$$f(x+a)=f(x)+f'(x) a + \frac{1}{2} f''(x) a^2+\cdots=f(x) + a \mathrm{d}_x f(x) + \frac{a^2}{2} \mathrm{d}_x^2 f(x) + \cdots = \sum_{j=0}^{\infty} \frac{a^j}{j!} \mathrm{d}_x^j f(x),$$
and this can be written in the convenient operator equation
$$f(x+a)=\exp(a \mathrm{d}_x) f(x).$$
It's clear that $\mathrm{d}_x=\frac{\mathrm{d}}{\mathrm{d} x}$ operates on functions of the indpendent variable, $x$!

 

[Hak]

Hm, it's a somewhat difficult point of what's the right notation.

My favorite form of Taylor's theorem goes as follows:\
$$f(x+a)=f(x)+f'(x) a + \frac{1}{2} f''(x) a^2+\cdots=f(x) + a \mathrm{d}_x f(x) + \frac{a^2}{2} \mathrm{d}_x^2 f(x) + \cdots = \sum_{j=0}^{\infty} \frac{a^j}{j!} \mathrm{d}_x^j f(x),$$
and this can be written in the convenient operator equation
$$f(x+a)=\exp(a \mathrm{d}_x) f(x).$$
It's clear that $\mathrm{d}_x=\frac{\mathrm{d}}{\mathrm{d} x}$ operates on functions of the indpendent variable, $x$!

Thank you. I did not understand, however, whether what you said is in agreement or contradiction with what @PeroK says. At first glance, it seems like the first option, that is, a different way of saying the same thing.

 

[PeroK]

Hm, it's a somewhat difficult point of what's the right notation.

My favorite form of Taylor's theorem goes as follows:\
$$f(x+a)=f(x)+f'(x) a + \frac{1}{2} f''(x) a^2+\cdots=f(x) + a \mathrm{d}_x f(x) + \frac{a^2}{2} \mathrm{d}_x^2 f(x) + \cdots = \sum_{j=0}^{\infty} \frac{a^j}{j!} \mathrm{d}_x^j f(x),$$
and this can be written in the convenient operator equation
$$f(x+a)=\exp(a \mathrm{d}_x) f(x).$$
It's clear that $\mathrm{d}_x=\frac{\mathrm{d}}{\mathrm{d} x}$ operates on functions of the indpendent variable, $x$!

It's important to note, especially from the point of view of functional analysis, that a function is defined independently of any variable. And that a function of a single variable has a unique derivative. And the differential operator in this case is unique. So, although certain notations require a variable such as $x$, it's still a dummy variable.

 

[PeroK]

... whereas, when we invoke a "change of variables", we end up with two different functions denoted by the same letter. In this case, the variables are no longer dummy variables, as they determine which function is intended. And, as there are two different functions, there are two different derivatives. The variable again indicates which function is intended.

 

[vanhees71]

This latter notation is, unfortunately, very common in physics. E.g., in some textbooks they label a function and its Fourier transform with the same symbol, and only the name of the variable distinguishes them. That's very bad notation and leads to a lot of confusion though!