Differentiatial equation

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Discussion Overview

The discussion revolves around the interpretation and manipulation of differential equations, specifically focusing on the validity of expressions involving derivatives and functions. Participants explore the implications of applying the product rule and the meaning of derivative notation in various contexts.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether the expression $$ ∅*(\frac{d}{dξ})=∅*(\frac{d1}{dξ}) $$ holds true and explores the implications of multiplying a function by a derivative.
  • Another participant argues that the derivative symbol without a function next to it does not constitute a meaningful quantity, asserting that $$ ∅*(\frac{d}{dξ})\neq \frac{d∅}{dξ} $$.
  • A later reply reiterates the previous point about the derivative symbol, emphasizing that it serves as an instruction to differentiate whatever follows it.
  • One participant expresses confusion about transitioning between two equations, suggesting that they can only do so if $$ ∅*(\frac{d}{dξ}) = \frac{d∅}{dξ} $$ is accepted.
  • Several participants question the standard notation used in the equations, suggesting that the placement of terms may not align with common conventions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity of the expressions involving derivatives. There are competing views on the interpretation of the derivative notation and its application in the context of the equations discussed.

Contextual Notes

There are limitations regarding the assumptions made about the functions involved and the definitions of the terms used in the equations. The discussion does not resolve these ambiguities.

TimeRip496
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Does $$ ∅*(\frac{d}{dξ})=∅*(\frac{d1}{dξ}) $$?

If is true,
Does multiplying a function and a derivative equals to the derivative of that function? For e.g. $$ ∅*(\frac{d}{dξ})=\frac{d∅}{dξ} $$ where ∅ is a function of ξ

But isn't it supposed to be like this(based on the product rule), $$ ∅*(\frac{d}{dξ}) = ∅*(\frac{d1}{dξ}) = \frac{d}{dξ}*∅-1*\frac{d∅}{dξ} $$ ?

What if ∅ is a constant or is not a function of ξ?
 
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TimeRip496 said:
Does $$ ∅*(\frac{d}{dξ})=∅*(\frac{d1}{dξ}) $$?
Obviously no. Derivative symbol with nothing next to the right of it constitutes no meaningful quantities, no numerical value can be associated with it (if the variable is given a number), it's just an instruction to differentiate whatever stands on the right. If you put something to the right of a derivative (like you did in the RHS of that equation), you have given a numerical value to the entire expression.
Therefore
$$
∅*(\frac{d}{dξ})\neq \frac{d∅}{dξ}
$$
 
blue_leaf77 said:
Obviously no. Derivative symbol with nothing next to the right of it constitutes no meaningful quantities, no numerical value can be associated with it (if the variable is given a number), it's just an instruction to differentiate whatever stands on the right. If you put something to the right of a derivative (like you did in the RHS of that equation), you have given a numerical value to the entire expression.
Therefore
$$
∅*(\frac{d}{dξ})\neq \frac{d∅}{dξ}
$$
Thanks!
upload_2016-1-31_19-20-24.png

upload_2016-1-31_19-20-46.png

But then how do I get from equation (12) to equation (13)? The only way I can do it is when
$$
∅*(\frac{d}{dξ}) = \frac{d∅}{dξ}.
$$
 
Where did you get source from? Is it the same source as the one with harmonic oscillator in another thread of yours?
 
blue_leaf77 said:
Where did you get source from? Is it the same source as the one with harmonic oscillator in another thread of yours?
Yes.
Source: http://vixra.org/pdf/1307.0007v1.pdf
 
I believe that's not the common and standard way to write the derivative of a function; in equation (12), ##\phi_0## should be on the right of the bracketed terms.
 
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blue_leaf77 said:
I believe that's not the common and standard way to write the derivative of a function; in equation (12), ##\phi_0## should be on the right of the bracketed terms.
Ok thanks again for your help!
 

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