- #1
TheCanadian
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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##?
fresh_42 said: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##?
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.TheCanadian said:Got it, thank you. Although why exactly did you specify ##\rho = 1##?
fresh_42 said: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.
fresh_42 said: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.
fresh_42 said: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?
fresh_42 said: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.
No. I simply changed the variable names we all are used to:TheCanadian said: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. 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.since L'Hôpital's rule requires both parts to yield an indefinite value?
fresh_42 said: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 . But this mistake shows, that we have to use L'Hôpital and not just the definition of ##\partial##.
L'Hospital's rule for Laplacian is a mathematical theorem that helps us evaluate the limit of a function as it approaches a point where the function is not defined, such as infinity or zero.
L'Hospital's rule for Laplacian is commonly used in calculus and differential equations to simplify complex expressions and solve problems involving limits. It is also used in various fields of science, such as physics, chemistry, and engineering, to analyze and model systems that involve rates of change.
The formula for L'Hospital's rule for Laplacian states that the limit of a quotient of two functions, f(x) and g(x), as x approaches a point c where both functions are undefined, is equal to the limit of the quotient of the derivatives of f(x) and g(x) as x approaches c. This can be written as: lim [f(x)/g(x)] = lim [f'(x)/g'(x)] as x→c.
No, L'Hospital's rule for Laplacian can only be applied to functions that are differentiable and have a well-defined derivative at the point of interest. It is important to check the conditions of the theorem before applying it to a specific function.
Yes, L'Hospital's rule for Laplacian may not always provide the correct answer. It should be used with caution and other methods should also be considered to verify the result. Additionally, it is important to note that the limit may not exist in some cases, even after applying L'Hospital's rule.