I can't believe the hint that is provided; can someone explain how it is true?

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Homework Help Overview

The discussion revolves around a mathematical identity involving the second derivative of a function A with respect to r, specifically examining the equality between two expressions. Participants are trying to understand the validity of a hint provided in a problem set related to differential equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to manipulate the given equation to show both sides are equal, but some express confusion about the steps taken. There are questions about the legality of treating certain operations as multiplication and the application of the chain rule.

Discussion Status

Some participants have identified misunderstandings in the manipulation of the derivatives and have begun to clarify the steps involved. There is acknowledgment of the need to properly apply differentiation rules, particularly regarding the chain rule. Multiple interpretations of the problem are being explored, and guidance has been offered to help clarify the confusion.

Contextual Notes

It is assumed that A is a function of r and is twice differentiable. Participants are working within the constraints of a homework problem, which may impose specific requirements on the approach to the solution.

Tsunoyukami
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I'm working on a question for a problem set that has the following hint:

For any function A:

[itex]\frac{d^{2}A}{dr^{2}} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

I don't understand how this is true; this is how I try to show that both sides are equal but it doesn't work out...any help would be appreciated.

[itex]\frac{d^{2}A}{dr^{2}} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

[itex]\frac{d}{dr}\frac{dA}{dr} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

[itex](\frac{d}{dr} + \frac{2}{r}) \frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

[itex]\frac{d}{dr} + \frac{2}{r} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2})[/itex]

[itex]\frac{d}{dr} + \frac{2}{r} = \frac{d}{dr}[/itex]

[itex]\frac{2}{r} = \frac{d}{dr} - \frac{d}{dr}[/itex]

[itex]\frac{2}{r} = 0[/itex]

This is only true if r [itex]\rightarrow[/itex] [itex]\infty[/itex]. I think I must be doing something wrong - can anyone please explain what I'm doing wrong and show me how the hint is a true statement? Thanks!
 
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You're treating everything as if it's multiplication, that's not legal.
 
Last edited by a moderator:
Tsunoyukami said:
I'm working on a question for a problem set that has the following hint:

For any function A:

[itex]\frac{d^{2}A}{dr^{2}} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

I don't understand how this is true; this is how I try to show that both sides are equal but it doesn't work out...any help would be appreciated.

[itex]\frac{d^{2}A}{dr^{2}} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

[itex]\frac{d}{dr}\frac{dA}{dr} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

[itex](\frac{d}{dr} + \frac{2}{r}) \frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

[itex]\frac{d}{dr} + \frac{2}{r} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2})[/itex]

[itex]\frac{d}{dr} + \frac{2}{r} = \frac{d}{dr}[/itex]
The step above makes no sense. d/dr(r2) does NOT mean d/dr * r2, so you can't divide by r2 to end up with d/dr.

There are steps above that that also don't make sense, for essentially the same reason.
Tsunoyukami said:
[itex]\frac{2}{r} = \frac{d}{dr} - \frac{d}{dr}[/itex]

[itex]\frac{2}{r} = 0[/itex]

This is only true if r [itex]\rightarrow[/itex] [itex]\infty[/itex]. I think I must be doing something wrong - can anyone please explain what I'm doing wrong and show me how the hint is a true statement? Thanks!

The assumption here is that A = A(r); i.e., A is a function of r. Also, it is assumed that A is twice differentiable.

Start on the right side and calculate (1/r2) * d/dr(r2 *dA/dr).
 
Mindscrape said:
You're treating everything as if it's multiplication, that's not legal.

[tex]\frac{1}{r^2}\frac{d}{dr}(r^2 \frac{dA}{dr}) =\frac{1}{r^2}( 2r\frac{dA}{dr}+r^2\frac{d^2A}{dr^2})[/tex]

Oh man, I can't believe I missed that - thanks so much it makes perfect sense now! It slipped my mind that the [itex]\frac{d}{dr}[/itex] would act on the [itex]r^{2}[/itex] and that I would have to use the chain rule. Thanks a lot!

Mark44 said:
The step above makes no sense. d/dr(r2) does NOT mean d/dr * r2, so you can't divide by r2 to end up with d/dr.

There are steps above that that also don't make sense, for essentially the same reason.The assumption here is that A = A(r); i.e., A is a function of r. Also, it is assumed that A is twice differentiable.

Start on the right side and calculate (1/r2) * d/dr(r2 *dA/dr).

I see my mistake, thank you. I knew I was doing something fishy but I wasn't certain what it was. Thank you again for the prompt replies!
 

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