# Approximating division

1. Mar 10, 2014

### d.arbitman

What is the name and where can I find the derivation for the following approximation?

(X-(ΔX/2)) / (X+ (ΔX/2)) ≈ 1- ΔX/X

Assuming ΔX << X, and X = (X1 + X2) / 2 and ΔX = X1 - X2

Thanks fellas!

EDIT: Revised the expression.

Last edited: Mar 11, 2014
2. Mar 11, 2014

### maajdl

(X-(ΔX/X)) / (X+ (ΔX/X)) ≈ 1- 2ΔX/X²

in addition, if the X1 and X2 are physical measurements, like distances,
then (X-(ΔX/X)) and (X+(ΔX/X)) are invalid expressions because X and ΔX/X do not have the same dimensions.

3. Mar 11, 2014

### d.arbitman

You're correct. I revised the expression to make sense.

4. Mar 11, 2014

### d.arbitman

Solved.
There's a series expansion and the two terms I listed are just the first two terms of the series.

5. Mar 11, 2014

### Staff: Mentor

You can do this by using long division (which is another technique to produce an infinite series). Since Δx is small in comparison to x, (Δx)2 will be miniscule, so you can discard all terms in (Δx)2 or higher powers.

6. Mar 11, 2014

### d.arbitman

Yes, the higher order terms are minuscule and it's okay to discard them for hand analysis. How would I go about performing long division to achieve an infinite series? I've never done it.

Thanks.

7. Mar 11, 2014

### Staff: Mentor

8. Mar 11, 2014

### d.arbitman

9. Mar 11, 2014

### Staff: Mentor

When I worked on this yesterday, I noticed that the stated approximation wasn't working out to what you had, so I abandoned my efforts.

It's hard to lay out long division on a computer, but I'll do the best I can

Code (Text):

____________
x + 1/2 Δx ) x - 1/2 Δx

1. Divide x in the dividend (the numerator) by x in the divisor (the denominator). The partial quotient is 1.
Code (Text):

1
____________
x + 1/2 Δx ) x - 1/2 Δx

2. Multiply the partial quotient (1) times the divisor, and put the answer beneath the dividend
Code (Text):

1
____________
x + 1/2 Δx ) x - 1/2 Δx
x + 1/2 Δx
________________

3. Subtract. You should get -Δx

4. Now divide -Δx by x to get -Δx/x.
5. Continue this process until you get tired of doing it.
$$1 + \frac{-Δx}{x} + \text{other terms}$$
The "other terms" are those in (Δx)2 and higher powers, which as I mentioned, can be discarded.

BTW, except in probability, variables are almost always written in lower case. IOW, x rather than X. In probability, so-called random variables are usually written in upper case.

10. Mar 11, 2014

### d.arbitman

Thank you, I mean it. I just copied the variables from one of my engineering textbooks which omits the derivation.

I didn't even know that in mathematics that there's a capitalization convention. I'm used to lower case for small signals and upper case for DC (+AC).

11. Mar 11, 2014

### AlephZero

Alternatively, remember that $1/(1 - a)$ is the sum of the geometric series $1 + a + a^2 + a^3 + \dots$ and multiply the series by $(1+a)$. (And then let $a = -x/2$).

You didn't say what context this came from, but it is an example of a Padé approximation, which is a technique used to create numerical solutions of differential equations etc. Specifically, $(1-\frac x 2)/(1+ \frac x 2)$ can be a nice numerical approximation to $e^{-x}$, for all positive values of $x$, not just for "small" values of $x$ like the first few terms of a Taylor series.