Doubt about the derivative of a Taylor series

[Hak]

Homework Statement

While studying the calculation of the Lagrangian for a relativistic free particle, I came across this equation below
$$L((v')^2) = L(v^2) + \frac{\partial L(v^2)}{\partial (v^2)}\big ((v'^2) - v^2) \big)$$, with $$L'(v^2) = \frac{\partial L(v^2)}{\partial (v^2)}$$,

obtained by using the Taylor expansion for $L$ at the point $(v')^2$.

Relevant Equations

/

My doubt arises over the definition of L'(v^2). If we are using $x= v'^2$, shouldn't the derivative be made with respect to that very term? In essence, shouldn't it be: L'(v^2) = \frac{\partial L(v^2)}{\partial (v'^2)}? In the article I read, L'(v^2) = \frac{\partial L(v^2)}{\partial (v^2)} is assumed. Could you explain to me why the latter definition is right and mine is wrong? Thank you very much.

 

[anuttarasammyak]

The quantity you assumed is zero because v^2 and v’^2 are independent variables.

 

[Hak]

The quantity you assumed is zero because v^2 and v’^2 are independent variables.

Thank you for your answer, but I did not understand. Could you explain it to me in more detail? Thank you very much.

 

[anuttarasammyak]

L(v^2) is not function of v’^2 thus its derivative with v’^2 is zero.

 

[Hak]

L(v^2) is not function of v’^2 thus its derivative with v’^2 is zero.

OK, thanks, maybe I got it. You said, though, that $v'^2$ and $v^2$ are independent variables. Actually, $v'$ is a function of $v$. How do you explain it, then? Thanks again.

 

[PeroK]

My doubt arises over the definition of L'(v^2). If we are using $x= v'^2$, shouldn't the derivative be made with respect to that very term? In essence, shouldn't it be: L'(v^2) = \frac{\partial L(v^2)}{\partial (v'^2)}? In the article I read, L'(v^2) = \frac{\partial L(v^2)}{\partial (v^2)} is assumed. Could you explain to me why the latter definition is right and mine is wrong? Thank you very much.

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 $L$".

 

[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 $L$".

OK, thank you very much for the detailed answer. I am unclear, however, about one point: does your digression mean to imply that it makes no difference whether I put, in this specific case, $$\frac{\partial L}{\partial (v^2)}$$ or $$\frac{\partial L}{\partial (v'^2)}$$? If, on the other hand, I have misunderstood, why is $$\frac{\partial L}{\partial (v^2)}$$ correct and $$\frac{\partial L}{\partial (v'^2)}$$ not? Thank you very much.

 

[PeroK]

I am unclear, however, about one point ...

I suspected you might be!

 

[Hak]

I suspected you might be!

You suspected right! Could you, however, dispel this doubt of mine, if you can? Thank you very much.

 

[PeroK]

You suspected right! Could you, however, dispel this doubt of mine, if you can? Thank you very much.

I can't read your equations.

 

[Hak]

I can't read your equations.

I edited my message. Sorry.

 

[PeroK]

OK, thank you very much for the detailed answer. I am unclear, however, about one point: does your digression mean to imply that it makes no difference whether I put, in this specific case, $$\frac{\partial L}{\partial (v^2)}$$ or $$\frac{\partial L}{\partial (v'^2)}$$? If, on the other hand, I have misunderstood, why is $$\frac{\partial L}{\partial (v^2)}$$ correct and $$\frac{\partial L}{\partial (v'^2)}$$ not? Thank you very much.

Neither is more correct than the other. That's why mathematicians prefer the derivative notation $f', f''$ etc. for Taylor series, whenever possible.

Read my Insight if you would like to underatand this issue fully. That's why I wrote it.

 

[Hak]

Neither is more correct than the other. That's why mathematicians prefer the derivative notation $f', f''$ etc. for Taylor series, whenever possible.

Read my Insight if you would like to underatand this issue fully. That's why I wrote it.

Thank you very much. I will read your article, it sounds very interesting!

 

[anuttarasammyak]

Actually, v′ is a function of v. How do you explain it, then? Thanks again.

Usually Taylor-Mclauring expansion form is free from how much deviation $\triangle x$ from x is. Maybe I am misunderstanding your setting. What is v'(v), v' as a function of v, in your problem ?

[EDIT] When we know it we may expect to calculate your qunatity by
\frac{\partial L(v^2)}{\partial v'^2}=\frac{\partial L(v^2)}{\partial v^2}\frac{\partial v^2}{\partial v'^2}

 

[Hak]

Usually Taylor-Mclauring expansion form is free from how much deviation $\triangle x$ from x is. Maybe I am misunderstanding your setting. What is v'(v), v' as a function of v, in your problem ?

Yes.

 

[Hak]

[EDIT] When we know it we may expect to calculate your qunatity by
\frac{\partial L(v^2)}{\partial v'^2}=\frac{\partial L(v^2)}{\partial v^2}\frac{\partial v^2}{\partial v'^2}

What does it mean? Does this answer disagree with that of @PeroK?

 

[anuttarasammyak]

It is familiar derivative chain rule https://en.wikipedia.org/wiki/Chain_rule .

 

[PeroK]

Usually Taylor-Mclauring expansion form is free from how much deviation $\triangle x$ from x is. Maybe I am misunderstanding your setting. What is v'(v), v' as a function of v, in your problem ?

[EDIT] When we know it we may expect to calculate your qunatity by
\frac{\partial L(v^2)}{\partial v'^2}=\frac{\partial L(v^2)}{\partial v^2}\frac{\partial v^2}{\partial v'^2}

That's not what's meant in this context.

 

[PeroK]

What does it mean? Does this answer disagree with that of @PeroK?

As explained above, that's not what's meant in this context.

 

[Hak]

It is familiar derivative chain rule.

Ok, thanks.

 

[Hak]

As explained above, that's not what's meant in this context.

I already understood that. Thank you.

 

[PeroK]

As explained above, that's not what's meant in this context.

Just to emphasis the point. We have a function $L$ with a Taylor series expansion:
$$L(x) = L(a) + L'(a)(x-a) + \frac 1 2 L''(a)(x - a)^2 + \dots$$Now, if you want to replace the derivative $L'$ with the differential form, what do you use: $a$ or $x$? I think that it looks better with $x$, as derivative with respect to $a$ looks odd.
$$L(x) = L(a) + \frac{dL}{dx}(a)(x-a) + \frac 1 2 \frac{d^2L}{dx^2}(a)(x - a)^2 + \dots$$
We can let $a = v^2$ and $x = v'^2$:
$$L(v'^2) = L(v^2) + L'(v^2)(v'^2-v^2) + \frac 1 2 L''(v^2)(v'^2 - v^2)^2 + \dots$$And we have the same issue: do we want to write $L' \equiv \frac{dL}{d(v^2)}$ or $L' \equiv \frac{dL}{d(v'^2)}$? In this context, the chain rule isn't relevant, as the question is purely notational.

 

[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 think there is an abuse of notation here. The partial derivative should be denoted $L_{v^2}$, not $L' (v^2)$, don't you think? I was the one who first posed the question in this way, so it is my fault for falling into error.

 

[PeroK]

I think there is an abuse of notation here. The partial derivative should be denoted $L_{v^2}$, not $L' (v^2)$, don't you think?

It's not really a partial derivative, as $L$ is really a function of one variable in this case. The problem is that for partial derivatives, there is no argument-free notation for the derivative. See my Insight.

I was the one who first posed the question in this way, so it is my fault for falling into error.

There isn't a totally satisfactory solution, as we are stuck with the notation. The notation is standard. It's not an error, but a deficiency in the differential notation.

 

[Hak]

The problem is that for partial derivatives, there is no argument-free notation for the derivative.

I read the Insight, but I could not understand this point. I have some difficulty understanding the type of notation, perhaps because there are different schools of thought? Could you clarify this for me? Thank you very much.

 

[PeroK]

I read the Insight, but I could not understand this point. I have some difficulty understanding the type of notation, perhaps because there are different schools of thought? Could you clarify this for me? Thank you very much.

I can't say any more than in is the Insight. It's all in there.

 

[Hak]

Just to emphasis the point. We have a function $L$ with a Taylor series expansion:
$$L(x) = L(a) + L'(a)(x-a) + \frac 1 2 L''(a)(x - a)^2 + \dots$$Now, if you want to replace the derivative $L'$ with the differential form, what do you use: $a$ or $x$? I think that it looks better with $x$, as derivative with respect to $a$ looks odd.

Re-reading this statement of yours, I have my doubts. I understand that you state that the notation is uncertain, but you say that it is better to express the derivative with $x$ and not with $a$, because the latter case is odd. Since in this particular case $x = v'^2$ and $a = v^2$, you are saying that it is better to put $v'^2$ in the denominator, right? Doesn't this contradict the article you referred me to, where it is considered less strange to enter $v^2$ in the denominator, not $v'^2$? I don't understand, I'm confused. I would be grateful if you would clarify this point. Thank you very much for everything.

 

[anuttarasammyak]

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 ?

 

[PeroK]

Re-reading this statement of yours, I have my doubts. I understand that you state that the notation is uncertain, but you say that it is better to express the derivative with $x$ and not with $a$, because the latter case is odd. Since in this particular case $x = v'^2$ and $a = v^2$, you are saying that it is better to put $v'^2$ in the denominator, right? Doesn't this contradict the article you referred me to, where it is considered less strange to enter $v^2$ in the denominator, not $v'^2$? I don't understand, I'm confused. I would be grateful if you would clarify this point. Thank you very much for everything.

I said it doesn't matter. You could use $a$ or $x$ or $v^2$ or $v'^2$. Or, something else. It's only notation.

 

[Hak]

I said it doesn't matter. You could use $a$ or $x$ or $v^2$ or $v'^2$. Or, something else. It's only notation.

OK, thanks.

 

[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!