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

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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!