Cubic Algebraic Solution: Understanding the Approximation in Equations 4 and 5

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Discussion Overview

The discussion revolves around the approximation found in equations 4 and 5 related to cubic algebraic solutions. Participants explore the mathematical reasoning behind the approximation and its implications, particularly focusing on series expansions and assumptions regarding small quantities.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant seeks clarification on how the approximation $$a_0 \approx a\left( 1- \frac{1}{a^2}\Sigma \right)$$ is derived from the equation $$a^3=a_0^3+3 a_0 \Sigma$$.
  • Another participant suggests that assuming $$\Sigma \ll 1$$ and neglecting higher-order terms (H.O.T.) leads to the solution.
  • A correction is made regarding the absence of a factor of 2 in the denominator of the equation, which alters the approximation to $$a_o^3=a^3(1-3 (a_o/a^3) \sum ) \approx a^3(1-\frac{3}{a^2} \sum )$$.
  • One participant mentions that Taylor expansions are commonly used for these types of approximations.
  • Another participant reiterates the assumption that $$\Sigma \ll 1$$ is crucial for the approximation to hold.

Areas of Agreement / Disagreement

There is no consensus on a single method of deriving the approximation, as participants present different perspectives and corrections. The discussion includes both agreement on the use of Taylor expansions and differing views on the specific details of the approximation.

Contextual Notes

Participants reference assumptions about the smallness of $$\Sigma$$ and the neglect of higher-order terms, but the implications of these assumptions are not fully resolved. The discussion also highlights the importance of precise mathematical notation in conveying the correct relationships.

member 428835
Hi PF!

Here equations 4 and 5 imply $$a^3=a_0^3+3 a_0 \Sigma \implies a_0 \approx a\left( 1- \frac{1}{a^2}\Sigma \right).$$ Can someone explain how this approximation is found?

Edit: I place in calculus thread because there must be some series expansion going on or a neglecting of terms.
 
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NVM, I solved it! Assume ##\Sigma \ll 1## and neglect H.O.T. Thanks!
 
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There is no ##2 ## in the denominator. It should read ##a^3=a_o^3+3 a_o \sum ##. This will make ##a_o^3=a^3(1-3 (a_o/a^3) \sum ) \approx a^3(1-\frac{3}{a^2} \sum ) ##. ## \\ ## Meanwhile ## (1+\Delta)^{1/3} \approx 1+\Delta/3 ##, (Taylor expansion).
 
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Beat me to it, sorry. Didn't see your post till after my second one. Thanks!
 
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Everytime i see these approximations its always a Taylor expansion of some sort.
 
joshmccraney said:
NVM, I solved it! Assume ##\Sigma \ll 1## and neglect H.O.T. Thanks!
If I made no mistake, then you need ##\dfrac{\Sigma}{a} \approx 0.##
 

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