Quotient rule and binomial theorem

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Discussion Overview

The discussion centers around the potential relationship between the quotient rule in calculus and the binomial theorem. Participants explore whether a similar formulation exists for the quotient rule as seen with the product rule and its connection to the binomial theorem.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant suggests that if the product rule can be related to the binomial theorem, then a similar relationship might exist for the quotient rule.
  • Another participant provides a link that may contain relevant information, but does not elaborate on its content.
  • A different participant reiterates the initial idea about relating the product rule to the binomial theorem but expresses skepticism about the possibility of doing the same for the quotient rule.
  • Another participant proposes rewriting the quotient as a product of a function and the inverse of another function, suggesting the application of the product rule to this formulation.

Areas of Agreement / Disagreement

There is no consensus on whether the quotient rule can be related to the binomial theorem. Some participants express uncertainty or skepticism regarding this possibility.

Contextual Notes

The discussion does not clarify the assumptions or definitions that might be necessary to establish a relationship between the quotient rule and the binomial theorem.

Jhenrique
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If it's possible to relate the product rule with the binomial theorem, so:

(x+y)^2=1x^2y^0+2x^1y^1+1x^0y^2
D^2(fg)=1f^{(2)}g^{(0)}+2f^{(1)}g^{(1)}+1f^{(0)}g^{(2)}

So, is it possible to relate the quotient rule with the binomial theorem too?
 
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Jhenrique said:
If it's possible to relate the product rule with the binomial theorem, so:

(x+y)^2=1x^2y^0+2x^1y^1+1x^0y^2
D^2(fg)=1f^{(2)}g^{(0)}+2f^{(1)}g^{(1)}+1f^{(0)}g^{(2)}

So, is it possible to relate the quotient rule with the binomial theorem too?
I don't think so.
 
Write f/g as fg^{-1} and apply the product rule to that.
 

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