Neglecting higher order terms in expansion

Question

Suppose we have a function $v$ of $x$ with a minimum at $x=0$. We have, for $x$ close to zero, $$v'(x) = v'(0) +xv''(0) +\frac{x^2}{2}v'''(0)+\cdots$$ Then as $v'(0)=0$ $$v'(x)\approx xv''(0)$$ if $$|xv'''(0)|\ll v''(0)$$

Which is fine. I am unable to understand this statement:

Typically each extra derivative will bring with it a factor of $1/L $ where $L$ is the distance over which the function changes by a large fraction. So $$x\ll L$$

This is extracted from a physics derivation, and I cannot get how they tacked on a factor of $1/L$

Answer 1

This is a rule of thumb rather than a rigorous statement (which is indicated by the word "typically"). Just look at $v(x)=\sin\frac xL$. The length on which this function changes noticeably is of order $L$. Now try to differentiate a few times. The next exercise is to take an arbitrary smooth function $V$ and stretch it $L$ times horizontally to get $v(x)=V(x/L)$. Now the change in $V$ that you felt at distance $1$ is felt at distance $L$ in $v$. Try to differentiate here. I doubt anything much more profound than this simple scaling observation was meant.

Answer 2

If each derivative contributes $\frac{1}{L}$, then $|xv'''| << v'' \implies x(\frac{1}{L})^3 << (\frac{1}{L})^2$. Divide both sides by $(\frac{1}{L})^3$ and this becomes $x << L$.

That $\frac{1}{L}$ term is refering to the change in the function according to the difference method of derivatives (Definition via difference quotients) given in Wikipedia. If you calculate out the quotient between the second and third derivatives (or first and second), it should approximate to the result above given the context.