# Approximating division

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:

maajdl
Gold Member
(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.

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

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

Mark44
Mentor
Solved.
There's a series expansion and the two terms I listed are just the first two terms of the series.
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.

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.

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.

Mark44
Mentor

How do I handle the fractional parts? or do I split the numerator into two?

Mark44
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:
            ____________
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:
                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:
                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.

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:
            ____________
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:
                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:
                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.

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).

AlephZero