- #1

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## Main Question or Discussion Point

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##?

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- Thread starter TheCanadian
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- #1

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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##?

- #2

fresh_42

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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

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Got it, thank you. Although why exactly did you specify ##\rho = 1##?

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##?

- #4

fresh_42

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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.Got it, thank you. Although why exactly did you specify ##\rho = 1##?

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

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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.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.

- #6

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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 ##?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.

- #7

fresh_42

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

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

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I got the expression:

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

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?

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

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

fresh_42

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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

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I agree with most of what you've said, but wouldn't:

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.

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

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

fresh_42

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No. I simply changed the variable names we all are used to: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

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

##\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'tsince L'Hôpital's rule requires both parts to yield an indefinite value?

Edit: You are right, I forgot the ##-f(x_0)## part in ##\frac{df}{dx}##. Shame on me . But this mistake shows, that we have to use L'Hôpital and not just the definition of ##\partial##.

- #12

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Haha that makes more sense. Also, just to clarify, although there is no general constraint on ## a ##, when evaluatingNo. 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'trequireany 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 . But this mistake shows, that we have to use L'Hôpital and not just the definition of ##\partial##.

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

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

fresh_42

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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?)

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