1
$\begingroup$

Why is it possible to do the following approximation?

$$\frac{\mathrm{d} x}{x+\mathrm{d} x} \approx \frac{\mathrm{d}x}{x} $$

Why does that $\mathrm{d} x$ in the denominator "count" less than the one in the numerator?

$\endgroup$
  • 1
    $\begingroup$ What is ${\rm d}x$ here? Usually it's used to denote a very small quantity (${\rm d}x \ll x$). Then the approximation makes sense as $x+{\rm d}x \approx x$. $\endgroup$ – Winther Sep 14 '16 at 14:42
  • $\begingroup$ If $\mathrm dx$ is infinitesimal and $x$ is not $\approx 0$, we have $x+\mathrm dx\approx x\not\approx 0$ $\endgroup$ – Hagen von Eitzen Sep 14 '16 at 14:42
5
$\begingroup$

The $dx$ in the denominator is dwarfed by the $x$. The difference between $\frac{0.000001}{1.000001}$ and $\frac{0.000001}{1}$ is miniscule.

$\endgroup$
3
$\begingroup$

It's $\displaystyle \frac{dx}{x+dx}=\sum\limits_{k=0}^\infty (-1)^k (\frac{dx}{x})^{k+1}$.

Because of $\displaystyle (\frac{dx}{x})^{k+1}<<\frac{dx}{x}$ for $k\in\mathbb{N}$ you can approximate $\displaystyle \frac{dx}{x+dx}\approx \frac{dx}{x}$.

$\endgroup$
1
$\begingroup$

This is only valid as long as $x\not\approx 0$.

We have $x+\mathrm dx\approx x$ because $\mathrm dx\approx 0$ and addition is continuous. Then provided $x\not\approx 0$, the claim follows by continutity of division.

$\endgroup$
1
$\begingroup$

You are allowed to approx when terms are added as the difference between the results will be negligible however you cannot do the same with multiplication or division $10000+0.0001=10000.0001$ which is quite close however $10000*0.0001$ is definitely nowhere near $10000 $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.