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

[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.

 

[member 428835]

NVM, I solved it! Assume $\Sigma \ll 1$ and neglect H.O.T. Thanks!

 

[Charles Link]

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).

 

[member 428835]

Beat me to it, sorry. Didn't see your post till after my second one. Thanks!

 

[jedishrfu]

Everytime i see these approximations its always a Taylor expansion of some sort.

 

[fresh_42]

NVM, I solved it! Assume $\Sigma \ll 1$ and neglect H.O.T. Thanks!

If I made no mistake, then you need $\dfrac{\Sigma}{a} \approx 0.$