L'Hospital's rule for Laplacian

  • #1
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Main Question or Discussion Point

Screen Shot 2016-07-25 at 11.19.35 AM.png


In the above expression for the Laplacian, how exactly does the author apply l'Hospital's rule? And is this transformation only valid for ## \rho = 0##?
 

Answers and Replies

  • #2
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To start with: I'm not sure.
But ##ρ=0## looks like the point where the expression isn't defined. How about differentiating according to the product rule, application of L'Hôpital on the first summand ##\frac{1}{ρ} \frac{\partial e}{\partial ρ}## to get the second derivative at ##ρ=1##?
 
  • #3
367
13
To start with: I'm not sure.
But ##ρ=0## looks like the point where the expression isn't defined. How about differentiating according to the product rule, application of L'Hôpital on the first summand ##\frac{1}{ρ} \frac{\partial e}{\partial ρ}## to get the second derivative at ##ρ=1##?
Got it, thank you. Although why exactly did you specify ##\rho = 1##?
 
  • #4
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Got it, thank you. Although why exactly did you specify ##\rho = 1##?
I had no better idea to get rid of ##\frac{1}{\rho}## and get the ##2## of the solution. Otherwise it would have been ##(1+\frac{1}{\rho})## instead.
But as I said, I'm not sure and it was a bit of a quick and dirty calculation. You've possibly done it better.
 
  • #5
367
13
I had no better idea to get rid of ##\frac{1}{\rho}## and get the ##2## of the solution. Otherwise it would have been ##(1+\frac{1}{\rho})## instead.
But as I said, I'm not sure and it was a bit of a quick and dirty calculation. You've possibly done it better.
It's as you said: use product rule, but then only apply L'Hôpital's rule to ##\frac {1}{\rho} \frac {\partial e}{\partial \rho}##. This gives the extra term needed to get a coefficient of 2.
 
  • #6
367
13
I had no better idea to get rid of ##\frac{1}{\rho}## and get the ##2## of the solution. Otherwise it would have been ##(1+\frac{1}{\rho})## instead.
But as I said, I'm not sure and it was a bit of a quick and dirty calculation. You've possibly done it better.
Just to add to my previous post: since L'Hôpital's rule is only applicable at ## \rho = 0## to derive their expression, wouldn't it only be valid for ## \rho = 0## and no other values of ## \rho ##?
 
  • #7
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Without going through the list of conditions I used
$$\lim_{\rho \rightarrow a} \frac{f(\rho)}{g(\rho)} = \lim_{\rho \rightarrow a} \frac{f'(\rho)}{g'(\rho)}$$
and simply substituted ##\frac{\partial e}{\partial \rho}## by ##\frac{\partial^2 e}{\partial \rho^2}## which left me with a factor ##\frac{1}{\rho}## that I adjusted by letting ##\rho \rightarrow a = 1##. (I told you it was quick and dirty.) How did you get rid of this factor?
 
  • #8
367
13
Without going through the list of conditions I used
$$\lim_{\rho \rightarrow a} \frac{f(\rho)}{g(\rho)} = \lim_{\rho \rightarrow a} \frac{f'(\rho)}{g'(\rho)}$$
and simply substituted ##\frac{\partial e}{\partial \rho}## by ##\frac{\partial^2 e}{\partial \rho^2}## which left me with a factor ##\frac{1}{\rho}## that I adjusted by letting ##\rho \rightarrow a = 1##. (I told you it was quick and dirty.) How did you get rid of this factor?
I got the expression:

$$\frac {1}{\rho} ( \frac {\partial e}{\partial \rho} + \rho \frac {\partial ^2 e}{\partial \rho ^ 2})$$

This gives me:

$$\frac {1}{\rho} \frac {\partial e}{\partial \rho} + \frac {\partial ^2 e}{\partial \rho ^ 2} $$

Now this is where I could be wrong, but I applied L'Hôpital's rule to only the first term here for when ## \rho = 0## which gives me

$$ \lim_{\rho \rightarrow 0} \frac {\partial^2 e}{\partial \rho ^2} + \frac {\partial ^2 e}{\partial \rho ^ 2} $$

Thus it only seems to match their result when ##\rho \rightarrow 0##.
 
  • #9
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But this isn't the rule de L'Hôpital. It looks like the definition of differentials. If we take ##f(\rho)= \frac{\partial e}{\partial \rho}## you took ##\lim_{\rho \rightarrow 0} \frac{f(\rho)}{\rho} = f'(\rho)## but wouldn't it be ##\lim_{\rho \rightarrow 0} \frac{f(\rho_0 + \rho)}{\rho} = f'(\rho_0)##?
Thus you are also left with a choice, at which point ##\rho_0## you want to evaluate the differentials. Omitting the index ##0## only disguises this fact by implicitly assuming ##f(\rho_0 + \rho) = f(\rho + \rho) = f(2\rho) = f(\rho)## or as you said ##\rho_0 = 0##
L'Hôpital gives a general case with factor ##(1+\frac{1}{\rho})##. What makes more sense in the context? ##\rho_0 = 0## or ##\rho_0 = 1##?
I think it is in any case, how ever we may turn it, a bit of a sloppy calculation.
 
  • #10
367
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But this isn't the rule de L'Hôpital. It looks like the definition of differentials. If we take ##f(\rho)= \frac{\partial e}{\partial \rho}## you took ##\lim_{\rho \rightarrow 0} \frac{f(\rho)}{\rho} = f'(\rho)## but wouldn't it be ##\lim_{\rho \rightarrow 0} \frac{f(\rho_0 + \rho)}{\rho} = f'(\rho_0)##?
Thus you are also left with a choice, at which point ##\rho_0## you want to evaluate the differentials. Omitting the index ##0## only disguises this fact by implicitly assuming ##f(\rho_0 + \rho) = f(\rho + \rho) = f(2\rho) = f(\rho)## or as you said ##\rho_0 = 0##
L'Hôpital gives a general case with factor ##(1+\frac{1}{\rho})##. What makes more sense in the context? ##\rho_0 = 0## or ##\rho_0 = 1##?
I think it is in any case, how ever we may turn it, a bit of a sloppy calculation.
I agree with most of what you've said, but wouldn't:

$$\lim_{\rho \rightarrow 0} \frac{f(\rho_0 + \rho)}{\rho} = f'(\rho_0)$$

only be a valid expression if in this case

$$ \lim_{\rho \rightarrow 0} f(\rho_0 + \rho) = 0 $$

since L'Hôpital's rule requires both parts to yield an indefinite value?
 
  • #11
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I agree with most of what you've said, but wouldn't:

$$\lim_{\rho \rightarrow 0} \frac{f(\rho_0 + \rho)}{\rho} = f'(\rho_0)$$

only be a valid expression if in this case

$$ \lim_{\rho \rightarrow 0} f(\rho_0 + \rho) = 0 $$
No. I simply changed the variable names we all are used to:

##\lim_{\rho \rightarrow 0} \frac{f(\rho_0 + \rho)}{\rho} = \lim_{h \rightarrow 0} \frac{f(x_0+ h)}{h} = f'(x_0)##

since L'Hôpital's rule requires both parts to yield an indefinite value?
No. L'Hôpital's rule can be formulated as I did above in #7 (plus a bunch of conditions that have to hold). It doesn't require any infinity (beside those implied by taking differentials of course). It simply holds for ##a \in \{± \infty \}## as well.

Edit: You are right, I forgot the ##-f(x_0)## part in ##\frac{df}{dx}##. Shame on me :sorry:. But this mistake shows, that we have to use L'Hôpital and not just the definition of ##\partial##.
 
  • #12
367
13
No. I simply changed the variable names we all are used to:

##\lim_{\rho \rightarrow 0} \frac{f(\rho_0 + \rho)}{\rho} = \lim_{h \rightarrow 0} \frac{f(x_0+ h)}{h} = f'(x_0)##


No. L'Hôpital's rule can be formulated as I did above in #7 (plus a bunch of conditions that have to hold). It doesn't require any infinity (beside those implied by taking differentials of course). It simply holds for ##a \in \{± \infty \}## as well.

Edit: You are right, I forgot the ##-f(x_0)## part in ##\frac{df}{dx}##. Shame on me :sorry:. But this mistake shows, that we have to use L'Hôpital and not just the definition of ##\partial##.
Haha that makes more sense. Also, just to clarify, although there is no general constraint on ## a ##, when evaluating

$$\lim_{\rho \rightarrow a} \frac{f(\rho)}{g(\rho)} = \lim_{\rho \rightarrow a} \frac{f'(\rho)}{g'(\rho)}$$

these are only equivalent when

$$ \lim_{\rho \rightarrow a} \frac{f(\rho)}{g(\rho)} = \frac {\pm \infty}{\pm \infty}$$ or

$$ \lim_{\rho \rightarrow a} \frac{f(\rho)}{g(\rho)} = \frac {0}{0}$$

correct?
 
  • #13
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Essentially. (##g'(x) \neq 0## for ##x \neq a## in a neighborhood of ##a## must also hold.)
It probably is sufficient to require the latter, for one may switch nominator and denominator.
(It's been quite a while since I ... what was I about to say?)
 
  • #14
vanhees71
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Just a remark. The above Laplacian is the radial part in cylinder coordinates, and one must be utmost careful with points on the axis ##\rho=0##, because there the coordinates are singular. If in doubt with an expression, it's better to translate it into Cartesian coordinates and analyse the situation there!
 

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