# Symbol for very small variable

Is there any representation to state that a variable is close to (not equal) zero? Let me give you an example. Consider the function

$$u(x)=\alpha (e^{i\omega \delta t}-1) f(x)$$

I am interested in the function $$u(x)$$ when $$\delta t$$ is very small. For this case, it should be easy to see that

$$u(x)|_{\text{small }\delta t} \approx i \alpha \omega \delta t f(x)$$

Is there any "nice" notation to represent such an equation (without having to write small)? I thought that I could use the limit notation for that [for example, $$\lim_{\delta t \to 0} u(t)$$], but then I realized that if $$\delta t$$ goes to zero, then $$u(x)=0$$. Therefore, it is not what I need.

I found this link in the same forum, but it did not help.

• Really, what's wrong with $\delta t<<1$? Mar 23 at 15:44
• $\delta t\ll 1$ is good Mar 23 at 15:55
• I think it is okay. Nevertheless, I would need to use something like $|\delta t| << 1$ to make it clear that $\delta t$ is positive?
– jeb
Mar 23 at 16:31

I think you are asking about the first order (linear) approximation using the derivative.

You might say $$u(x)= i \alpha \omega \delta t f(x) + o(\delta t ).$$

The fact that this is an approximation for small $$\delta t$$ should be clear to your reader. If not you can say so.

• I like to "o" notation, I have to say. What would you recommend me to include in my text or even in front of the $u(x)$ term to make it clear that $\delta t$ is small? My question is based on the fact that I will need to manage this equation further, and if I include the "o" notation, they will become too overloaded. The best that I could think of was something like $u(x)|_{\delta t = \epsilon}$, such that $o(\epsilon)$ is insignificant (or can be disregarded). Thank you.
– jeb
Mar 23 at 18:41
• The insignifance to first order is implicit in the use of the $o$ notation. You can just carry the $o(\delta t)$ term along in future calculations and throw it away at the end (r sooner) as appropriate. You could add "as $\delta t \to 0$" when you first state the equation. I would not want to read an invented notation like the one you propose. Mar 23 at 19:01

The issue with taking limits is that as $$\delta t\to 0$$ both sides go to zero, but what you want to say is stronger than that.
I imagine what you actually mean by $$u(x)\approx i\alpha\omega\delta tf(x)$$ can be formalised as $$\lim_{\delta t\to 0}\frac{u(x)}{\delta t}=i\alpha\omega f(x).$$

This is simply a matter of rearranging so that the limit captures the relative error of the approximation, not just the absolute error.