# Why is $1-Cx\approx \frac{1}{1+Cx}$?

How is this approximation derived? It is from a physics book.

$C$ is a constant and for small $Cx$ we have $$1-Cx\approx \frac{1}{1+Cx}$$

Is it also true without $C$? I.e. for small $x$, $1-x\approx \frac{1}{1+x}$?

• Notice that $(1-x)(1+x) = 1 - x^2$ and that $x^2$ is negligible. – user296602 Oct 4 '17 at 14:41
• It should have to be true without C because you can always just have $C=1$. – Tyberius Oct 4 '17 at 14:45

This is because for $|t|<1$, we have $(1-t)^{-1} = 1+t+t^2+t^3+\cdots$. Then Set $t=-Cx$.

If you are familiar with geometric series, that is $$1+x+x^2+\cdots+x^n=\dfrac{1-x^{n+1}}{1-x}$$ and when $|x|<1$ as $n\to\infty$ this expression takes the form $$1+x+x^2+\cdots=\dfrac{1}{1-x}$$ if you set $x=Ct$ then for $|Ct|<1$ we can use this series. If $Cx$ be so small then all terms $(Ct)^n$ will be negligible and we have this approximation $$1+Ct\approx\dfrac{1}{1-Ct}$$ this approximation will be better if we use three first terms, i.e. $$1+Ct+C^2t^2\approx\dfrac{1}{1-Ct}$$

It is related to the geometric series. In order to derive the formula start with

$$S_n = 1+x+x^2+...+x^n$$

multiply by $x$ to obtain $$xS_n= x + x^2 +...+x^{n+1}.$$

Subtraction will result in (note that we have a lot of chancellations):

$$(1-x)S_n=1-x^{n+1} \implies S_n = \frac{1-x^{n+1}}{1-x}=1+x+x^2+...+x^{n}$$

The series converges for $|x|<1$ to give:

$1+x+x^2+... = \frac{1}{1-x}$$Now, use$x=-Cu\$ and only use the first two terms on the left-hand side to obtain your approximation.