Simple Substituting and Rearranging

[Caccioppoli]

Hello,
may someone be so kind to explain why the following equation

\(\displaystyle q=aq^3+b \) [eq#1]

can be approximated with \(\displaystyle q=a^{-0.5} + \delta \) [eq#2]

with \(\displaystyle \delta=b/3 \)

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

Thank you very much.

 

[DrPhil]

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)"

to get eq.(25), it should say "put x'=1/q into eq.(23)." Both sides are multiplied by q^3, and then divided by the coefficient of the q-term on the right.

delta is defined to be a small change. Any approximations involving delta are first-order - d^2 or d^3 terms are ignored. Thus

\(\displaystyle \displaystyle \left( q_0 + \delta \right)^3 \approx q_0^3 + 3\ q_0^2\ \delta \)

I hope that is enough to get you through!

 

[Caccioppoli]

to get eq.(25), it should say "put x'=1/q into eq.(23)." Both sides are multiplied by q^3, and then divided by the coefficient of the q-term on the right.

delta is defined to be a small change. Any approximations involving delta are first-order - d^2 or d^3 terms are ignored. Thus

\(\displaystyle \displaystyle \left( q_0 + \delta \right)^3 \approx q_0^3 + 3\ q_0^2\ \delta \)

I hope that is enough to get you through!

Thank you for your Help DrPhil! But, also if I ignore \(\displaystyle \delta^2\) or \(\displaystyle \delta^3\) terms I still can't arrive to the result.

PS

I've updated the problem since I have done some progress.

I need to explain why \(\displaystyle \delta=b/3\)

 

[DrPhil]

Hello,
may someone be so kind to explain why the following equation

\(\displaystyle q=aq^3+b \) [eq#1]

can be approximated with \(\displaystyle q=a^{-0.5} + \delta \) [eq#2]

with \(\displaystyle \delta = b/3 \)

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

Thank you very much.

I don't have a copy of the original problem - it is generally a bad idea to edit away the fundamental post of the thread.

Eq. 1 would be much simpler if b=0. The zero-order approximation is

\(\displaystyle q_0 = a\ q_0^3, \; \; \; \; q_0 = a^{-1/2} \)

The next order of approximation is

\(\displaystyle q = q_0 + \delta \)

The assumption is that \(\displaystyle \delta\) will turn out to be small compared to q_0, so that the approximation of q^3 is

\(\displaystyle q^3 \approx q_0^3 + 3 \delta q_0^2 \)

Then eq.#1 becomes

\(\displaystyle q_0 + \delta = a \left[ q_0^3 + 3 \delta q_0^2 \right] + b \)

The q_0 on the left cancels aq_0^3 on the right, so we get

\(\displaystyle \delta = a [3 \delta q_0^2] + b \)

And since q_0^2 = 1/a,

\(\displaystyle \delta = 3 \delta + b \)

The solution I get is \(\displaystyle \delta = - b/2\), so either I made a mistake or something is mistranslated from the original equation to this point. Another approach to try would be to use Newton's method to find the first-order approximation.

 

[Caccioppoli]

You're right! The solution is \(\displaystyle \delta=-b/2\), I mistranlsated the original paper

THANKS!