I have to admit, I am not familiar with the use of $\mathscr{O}\left[\lambda^{n}\right]$ notation. Apparently it doesn't mean what I thought.

In John L. Friedman's lecture notes on Lie derivatives, forms, densities, and integration

He writes:

A vector field $\mathbf{w}$ is Lie-derived by $\mathbf{v}$ if, for small $\lambda$, $\lambda\mathbf{w}$ is dragged along by the fluid flow. To make this precise, we are requiring that the equation [eq. (3)]


be satisfied to $\mathscr{O}\left(\lambda\right)$.

At first I thought I knew exactly what that meant. Then I tried to put it into more rigorous terms and realized that I don't really know what it means in this context. Since equation 3 is linear in $\lambda$, I expect it to be accurate to $\mathscr{O\left(\lambda^{2}\right)}$.

What does it mean to say equation 3 is "satisfied to $\mathscr{O}\left(\lambda\right)$", or to use $\mathscr{O}\left(\lambda^{2}\right)$ in equation 4



I know it means "don't worry about error terms. They will vanish when the limit is taken." But that's not very satisfying. How is $\mathscr{O}\left(\lambda^{2}\right)$ stated in terms of a Taylor polynomial with a remainder?

Edit to add:

If $$\mathbf{r}\left(t\right)+\lambda\mathbf{w}\left(\mathbf{r}\left(t\right)\right)=\overline{\mathbf{r}}\left(t\right),$$

satisfied to $\mathscr{O}\left(\lambda\right),$ means


then $\mathscr{O}\left(\lambda\right)=\mathbf{k}\lambda,$ where $\mathbf{k}\ne\vec{0}$ is a constant. So,


Which is pretty clearly not what is intended.

If it means $$\overline{\mathbf{r}}\left(t\right)-\mathbf{r}\left(t\right)=\mathscr{O}\left(\lambda\right),$$

then $\mathscr{O}\left(\lambda\right)=\mathbf{w}\left(\mathbf{r}\left(t\right)\right).$ Which makes sense. But that implies


That is what has me confused. Are equation 3 satisfied to $\mathscr{O}\left(\lambda\right)$ and the last statement above equivalent?

  • $\begingroup$ WP. Can you use this mainstream language to focus the question? $\endgroup$ – Cosmas Zachos Jul 31 '18 at 0:24
  • $\begingroup$ Note that $\mathcal O(\lambda^2)\subset \mathcal O(\lambda)$. $\endgroup$ – AccidentalFourierTransform Jul 31 '18 at 0:27
  • $\begingroup$ @CosmasZachos Trying to relate that discussion to the topic at hand seems futile. $\endgroup$ – Steven Thomas Hatton Jul 31 '18 at 1:21
  • 3
    $\begingroup$ I've seen that phrase used quite often in physics (and use it myself!). It means what I believe you think it means - that the equation is correct if one is only interested in terms of order $\lambda$ but may not be correct if we start including terms of order $\lambda^2$. If $A=B$ to ${\cal O}(\lambda)$ then $A=B+{\cal O}(\lambda^2)$. $\endgroup$ – Prahar Jul 31 '18 at 2:20
  • $\begingroup$ Re your last sentence of edit: of course they are equivalent. It means they are satisfied at order ${\cal O}(\lambda)$ so you neglect ${\cal O}(\lambda^2)$ terms. That's what the WP example on the exponential specifies for you, no? In any case, you are hung up on language : in your edit, the second alternative is meant, incontrovertibly. $\endgroup$ – Cosmas Zachos Jul 31 '18 at 13:28

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