Taylor expansion

1. Dec 24, 2004

quasar987

$$V = 2\pi \sigma(\sqrt{R^2+a^2}-R)$$

Show that for large R,

$$V \approx \frac{\pi a^2 \sigma}{R}$$

I figured if I could develop the MacLaurin serie with respect to an expression in R such that when R is very large, this expression is near zero, then the first 1 or 2 terms should be a fairly good aproximation. But I can't find such an expression.

2. Dec 24, 2004

learningphysics

The given expression approaches zero for large R.

I get the given answer, if you find the Taylor series with respect to a. The first derivative term is 0, and the second derivative term is the answer. I think for a rigorous answer, you'll have to show that the condition for convergence is satisfied...

3. Dec 24, 2004

quasar987

I did a few similar exercices since my original post and now I clearly see how to do this one:

$$V = 2\pi \sigma(\sqrt{R^2+a^2}-R) = 2\pi \sigma R(\sqrt{1+a^2/R^2}-1)$$

The first two terms of the binomial epansion of $(1+a^2/R^2)^{1/2}$ give the answer.

4. Dec 24, 2004

quasar987

It's funny how your way works too though... it means there's no difference in how V "behaves" when it's R that's very large from when it's a that's very near 0. Is this pure coincidence? In which kind of function will this sort of symetry happen ?

5. Dec 24, 2004

learningphysics

Not coincidence... I believe it's the same Taylor expansion. If each of us continued our expansion we'd get the same terms.

6. Dec 25, 2004

arildno

The proper expansion parameter is a/R; hence, it doesn't matter whether a is small or R big. In both cases, a/R is small.
This is readily seen by rewriting V as:

$$V=\frac{2\pi\sigma{a}^{2}}{R(1+\sqrt{1+(\frac{a}{R})^{2}})}=2\pi\sigma{a}\frac{\epsilon}{\sqrt{1+\epsilon^{2}}},\epsilon=\frac{a}{R}$$

Last edited: Dec 25, 2004
7. Dec 25, 2004

....I see !