# Taylor approximation vs setting to zero

I was recently solving a problem where I had to find the velocity of a particle. The correct result is the following: \begin{equation}v^2 = v_0^2 + \frac{k\delta^2}{m}\left(1-e^{-\frac{\Delta^2}{\delta^2}}\right)\end{equation}

where $k,\delta > 0$, $v_0$ the initial velocity and $\Delta$ is the distance from the origin.

Now, what I was supposed to do was to find out, how $v$ looks like when $\Delta << \delta$.

I thought when $\Delta << \delta$, I could say that the exponent is approximately equal to 0 so therefore: \begin{equation}v^2 \approx v_0^2\end{equation}

However, the correct solution is to use $e^x \approx 1+x$ for small x.

I can see why this is valid, however what I don't understand is, how to decide when to use taylor expansions for small values vs when to say that something is effectively so small, that I can set it 0 or ignore it.

• Well, they should have been much clearer in their statement of the problem. They should have asked for the linearized approximation with respect to $\Delta^2$. – Chet Miller Dec 13 '17 at 13:22
• This would depend on the precise wording of the question, but if they're asking for what $v$ "looks like" as a function of $\Delta$, then that usually carries the implication that the quantity of interest is a functional dependence on $\Delta$, rather than just a single point. The precise language of the problem is likely to have similar flags that indicate that a Taylor expansion (which is the only way to get a functional dependence out) is in order, but it's hard to confirm without seeing it. – Emilio Pisanty Dec 13 '17 at 13:45
• You've provided the zero'th-order approximation (zero'th power of Δ), but the implied question was about a non-trivial approximation. In this case, it turns out the first-order approximation is still trivial; the second-order approximation is intended. – MSalters Dec 13 '17 at 14:44

I would say you should always do the Taylor expansion and then if it justifiable you can set the $O(\frac{\Delta}{\delta})$ terms to 0.
Applying this to your equation, you get after Taylor expanding: $$v^2 = v_0^2 + \frac{k\Delta^2}{m}$$
Now I only know that $\frac{\Delta}{\delta}\ll1$, and $k,\delta > 0$, I don't know the relative sizes of $k$, $m$, $\Delta$. Without this information you can't say if $v_0 \gg\sqrt{\frac{k}{m}}\Delta$ to justify $v\approx v_0$.
For example, if $v_0=\delta=1$, $\Delta=m=10^{-3}$ and $k=10^3$ then $\frac{\Delta}{\delta}\ll1$ as required and $v_0\sim\sqrt{\frac{k}{m}}\Delta\neq0$.