What's the criterion for neglecting terms "much smaller than" other terms?

Question

In a physics exercise relating some electric forces. I found the following equation for the restoring force:

$ F = \frac{kQq}{(y_0 + \Delta y)^2} - \frac{kQq}{y_0^2}$

Where $ \Delta y << y_0 $

Now, the resolution goes as:

$$ F \approx \frac{kQq}{(y_0^2 + 2y_0\Delta y)} - \frac{kQq}{y_0^2} \\ \approx \frac{-2y_0\Delta y kQq}{(y_0^4 + 2y_0^3\Delta y)} \\ \approx \frac{-2y_0\Delta y kQq}{y_0^4(1 + 2\frac{\Delta y}{y_0})} \\ \approx \frac{-2\Delta y kQq}{y_0^3} $$

However, I have some issues understanding the criteria to decide whether neglecting a term is okay or not.

In the first step for example, after resolving the expansion $(y_0 + \Delta y)^2 = (y_0^2 + 2y_0\Delta y + \Delta y^2) $, we are neglecting the term $ \Delta y^2 $ under the condition of $ \Delta y << y_0 $. But why couldn't we neglect it earlier $ (y_0 + \Delta y)^2 $ Is just because $ \Delta y^2 $ is a very small second-order term?

I know this would give 0 in the general force equation, but I think is still a valid question why with some sums we can neglect it and not with others.

Later, we again approximate with $ y_0^4(1 + 2\frac{\Delta y}{y_0}) \approx y_0^4 $ because $ \frac{\Delta y}{y_0} $ tends to 0 but in this case the order of $ \Delta y $ seems irrelevant.

Is there some sort of rules to decide when these "approximating" operations are allowed? Or is there some hidden formality that allows physicist to do it consistently?

Answer 1

I think things are clearer if you clear away the verbiage first, and then concentrate on the approximation. So write $$ F = \frac{kQq}{(y_0 + \Delta y)^2} - \frac{kQq}{y_0^2} =\frac{kQq}{y_0^2}\left((1+\frac{\Delta y}{y_0})^{-2}-1\right). $$

Now by Taylor's Theorem or the Binomial Theorem we know that $$ (1+\frac{\Delta y}{y_0})^{-2}=1-2\frac{\Delta y}{y_0}+o\left(\frac{\Delta y}{y_0}\right) $$ and we are done.

Answer 2

I hope you realize that if you define the electric force as a function of $y$ \begin{align} F(y)&:=\frac{kQq}{y^2} - \frac{kQq}{y_0^2}, \end{align} then all that computation is doing is calculating the first derivative at the point $y_0$: \begin{align} F'(y_0)&=\frac{-2kQq}{y_0^3} \end{align} (by essentially repeat (part of) the proof of the quotient rule for differentiation.)

By definition of what a derivative means, this just gives you the linear approximation to the force near the equilibrium position $y_0$: \begin{align} F(y_0+\Delta y)&=F(y_0)+F'(y_0)\cdot \Delta y + o(|\Delta y|)\\ &= F'(y_0)\cdot \Delta y+o(|\Delta y|) \end{align} (you can view this as a "first-order Taylor approximation" but really this is just the definition of $F$ being differentiable at $y_0$).

Now, the reason we "neglect" $(\Delta y)^2$ is because the first derivative (again by definition) only cares about approximating to linear order. The reason we're interested in the first derivative is because this is the first non-zero derivative of $F$, i.e the smallest $n\geq 0$ such that $F^{(n)}(y_0)\neq 0$.

In general, the number of terms you keep in the middle of the calculation depends on how good you want your final approximation to be. In connection to this, you may find this answer of mine helpful.

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Anyway, if it was me, I don't like to use $\approx$ symbols ever, unless I'm being super informal and writing notes only for myself (even then, rarely). I always like to explicitly write down the order of the "error in the expansion". So, I would write the computation down as follows: \begin{align} F(y_0+\Delta y)&=\frac{kQq}{(y_0+\Delta y)^2}-\frac{kQq}{y_0^2}\\ &=\frac{kQq}{y_0^2}\left[\frac{1}{\left(1+\epsilon\right)^2}-1\right]\tag{$\epsilon:=\frac{\Delta y}{y_0}$}\\ &=\frac{kQq}{y_0^2}\left[\left(1-\epsilon+\epsilon^2+\mathcal{O}(\epsilon^3)\right)^2-1\right]\\ &=\frac{kQq}{y_0^2}\left[(1-2\epsilon+ 3\epsilon^2 +\mathcal{O}(\epsilon^3))-1\right] \\ &=\frac{kQq}{y_0^2}\left(-2\epsilon+3\epsilon^2+\mathcal{O}(\epsilon^3)\right)\\ &:=\frac{-2kQq}{y_0^3}\Delta y+\frac{3kQq}{y_0^4}(\Delta y)^2 + \mathcal{O}\left(\left(\frac{\Delta y}{y_0}\right)^3\right) \end{align} In the middle, I used the formula for the geometric series (which is valid if $|\epsilon|<1$): \begin{align} \frac{1}{1+\epsilon}=\sum_{n=0}^{\infty}(-1)^n\epsilon^n=1-\epsilon+\epsilon^2+O(\epsilon^3). \end{align}

In the computation above, I intentionally wrote out the expansion valid up to second order (i.e so we have corrections of the third order), just to show how the computation might proceed in general (where the the idea is to always find the first non-vanishing term in the Taylor expansion of the function (if it exists)).