Doubt about the derivative of a Taylor series

  • #1
Hak
709
56
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 [tex]L'(v^2) = \frac{\partial L(v^2)}{\partial (v^2)}[/tex],

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 [tex]L'(v^2)[/tex]. 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: [tex]L'(v^2) = \frac{\partial L(v^2)}{\partial (v'^2)}[/tex]? In the article I read, [tex]L'(v^2) = \frac{\partial L(v^2)}{\partial (v^2)}[/tex] is assumed. Could you explain to me why the latter definition is right and mine is wrong? Thank you very much.
 
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  • #2
The quantity you assumed is zero because v^2 and v’^2 are independent variables.
 
  • #3
anuttarasammyak said:
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.
 
  • #4
L(v^2) is not function of v’^2 thus its derivative with v’^2 is zero.
 
  • #5
anuttarasammyak said:
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.
 
  • #6
Hak said:
My doubt arises over the definition of [tex]L'(v^2)[/tex]. 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: [tex]L'(v^2) = \frac{\partial L(v^2)}{\partial (v'^2)}[/tex]? In the article I read, [tex]L'(v^2) = \frac{\partial L(v^2)}{\partial (v^2)}[/tex] 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##".
 
  • #7
PeroK said:
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.
 
  • #8
Hak said:
I am unclear, however, about one point ...
I suspected you might be!
 
  • #9
PeroK said:
I suspected you might be!
You suspected right! Could you, however, dispel this doubt of mine, if you can? Thank you very much.
 
  • #10
Hak said:
You suspected right! Could you, however, dispel this doubt of mine, if you can? Thank you very much.
I can't read your equations.
 
  • #11
PeroK said:
I can't read your equations.
I edited my message. Sorry.
 
  • #12
Hak said:
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.
 
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  • #13
PeroK said:
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!
 
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  • #14
Hak said:
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
[tex]\frac{\partial L(v^2)}{\partial v'^2}=\frac{\partial L(v^2)}{\partial v^2}\frac{\partial v^2}{\partial v'^2}[/tex]
 
Last edited:
  • #15
anuttarasammyak said:
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.
 
  • #16
anuttarasammyak said:
[EDIT] When we know it we may expect to calculate your qunatity by
[tex]\frac{\partial L(v^2)}{\partial v'^2}=\frac{\partial L(v^2)}{\partial v^2}\frac{\partial v^2}{\partial v'^2}[/tex]
What does it mean? Does this answer disagree with that of @PeroK?
 
  • #18
anuttarasammyak said:
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
[tex]\frac{\partial L(v^2)}{\partial v'^2}=\frac{\partial L(v^2)}{\partial v^2}\frac{\partial v^2}{\partial v'^2}[/tex]
That's not what's meant in this context.
 
  • #19
Hak said:
What does it mean? Does this answer disagree with that of @PeroK?
As explained above, that's not what's meant in this context.
 
  • #20
anuttarasammyak said:
It is familiar derivative chain rule.
Ok, thanks.
 
  • #21
PeroK said:
As explained above, that's not what's meant in this context.
I already understood that. Thank you.
 
  • #22
PeroK said:
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.
 
  • #23
PeroK said:
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.
 
  • #24
Hak said:
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.
Hak said:
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.
 
  • #25
PeroK said:
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.
 
  • #26
Hak said:
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.
 
  • #27
PeroK said:
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.
 
  • #28
I am afraid I am cofusing the notation. Let me say ##v^2=x## for simlicity in writing

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

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

[tex]\frac{\partial L(x')}{\partial x}=\frac{\partial L(x)}{\partial x'}=0[/tex]

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

Way 2
[tex]\frac{\partial L(x)}{\partial x}|_{x=x_0}=\frac{\partial L(x)}{\partial x}(x_0)[/tex]
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 ?
 
Last edited:
  • #29
Hak said:
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.
 
  • #30
PeroK said:
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.
 
  • #31
anuttarasammyak said:
I am afraid I am cofusing the notation. Let me say ##v^2=x## for simlicity in writing

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

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

[tex]\frac{\partial L(x')}{\partial x}=\frac{\partial L(x)}{\partial x'}=0[/tex]

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

Way 2
[tex]\frac{\partial L(x)}{\partial x}|_{x=x_0}=\frac{\partial L(x)}{\partial x}(x_0)[/tex]
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.
 
  • #32
PeroK said:
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.
 
  • #33
Hak said:
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.
 
  • #34
PeroK said:
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'?
 
  • #35
Hak said:
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!
 

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