# Expansion of 1/(1+quantity)

1. Jul 4, 2011

### yayscience

I'm watching some lectures on electromagnetism from NPTEL on Youtube.
Lecture 3 on Coulomb's Law, at 29minutes:

The professor expands this:
$$\frac{1}{1+\frac{2dsin(\theta)}{r_1}}$$
into:
$$1-\frac{2dsin(\theta)}{r_1}$$

I'm not seeing this expansion. Could someone provide a hint (or a good resource for similar types of problems? Anything would be helpful.)

Thanks all!

2. Jul 4, 2011

### IronHamster

You can use taylor polynomials, if you're familiar with those.

$$\frac{1}{1+x} = f(x) = f(0) + f'(0) x + \frac{f''(0)x^2}{2!} + . . .$$

Notice that the nth derivative evaluated at 0 = (-1)^n. So

$$\frac{1}{1+x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + . . .$$

Now let x = 2dsin(ɵ)/r.

$$\frac{1}{1+\frac{2dsin\theta}{r_1}} = 1 - \frac{2dsin\theta}{r_1} + \frac{(\frac{2dsin\theta}{r_1})^2}{2!} - . . .$$

For small d or large r, the expansion the professor gives is a good approximation because the squared terms are very small.

If you haven't experienced Taylor Polynomials before, say so and I'll explain what I just did.

3. Jul 4, 2011

### gb7nash

What's the restriction on d? r1? Theta?

I'm assuming that d and r1 are positive integers and theta is any angle. Algebraically speaking, they're not equivalent. Choosing d = 1, r1 = 1, and theta = pi/2, you get:

1/(1+2sin(pi/2)) = 1/3

and

1 - 2sin(pi/2) = -1

It looks like he just pulled that out of thin air, but what I think he's doing is using an approximation using the fact that d << r1.

4. Jul 5, 2011

### Redbelly98

Staff Emeritus
The expression (1+x)n is well approximated by 1+nx -- provided that |x| is much much smaller than 1, i.e. |x|«1.

For the example at hand -- n=-1, x=2d(sinθ)/r1 -- this means the approximation holds if
d sinθ « r1

Loosely speaking, this means either d or sinθ must be small, or r1 must be large.

5. Jul 5, 2011

### gb7nash

Couldn't θ be arbitary? sinθ is bounded between -1 and 1, so all you should really need is that d << r1.