# Thread: Simple Substituting and Rearranging

1. ## Simple Substituting and Rearranging

Hello,
may someone be so kind to explain how to arrive, step by step, from equation 23 to 28?

Most of all I would like to understand the approximation with delta: if I substitute eq26 in 25 I get a different result (e.g. delta^3 terms).

See the attached image.

Thank you very much.

PS
eq24 may be taken as it is, I mean, phi is simply "(A D Cs / x') - (c D Cs/2)"

2. ## Re: Simple Substituting and Rearranging

I've uploaded a bigger version of the image, split in two figures.

3. ## Re: Simple Substituting and Rearranging

I find it a bit hard to see how to go from (23) to (24) without seeing (17)...

4. ## Re: Simple Substituting and Rearranging

eq 23 and 24 are not linked, eq24 does not need to be explained, phi is simply

5. ## Re: Simple Substituting and Rearranging

I can update the problem since there is some progress:

I have the following equation

$q=aq^3+b$ [eq1]

eq1 is approximated with $q=a^{-0.5} + \delta$ [eq2]

$\delta=b/3$

Where does this approximation come from and why is $\delta=b/3$?

Thank you very much.

6. ## Re: Simple Substituting and Rearranging

I've found the solution !!

Premise the result was not [tex]\delta=b/3[\tex] but [tex]\delta=-b/2[\tex]

we have two equations, basically:

$aq^3-q+b=0$ and $q=a^{-0.5}+\delta$ with a small δ

Writing those two equations in a system, we have that the δ that satisfy both equations is a δ that satisfy this 3rd order equation:

$a\delta^3+3a^{-0.5}\delta^2+2\delta+b=0$

ignoring third order and second order terms in δ we have simply

$2\delta+b=0$ so the result $\delta=-b/2$