What is the difference between $x\rightarrow 0$ and $x\ll 1$?

During the derivations of some physical equations, we see statements like $x\rightarrow 0$ and $x\ll1$.

1. Are these statements equivalent? If not, what is the difference between them?

2. Is there any reason for using $1$ in the second statement? Why doesn't anyone write $x\ll 0.25$ or $x\ll\pi$?

$\Delta x \rightarrow 0$ is the expression used in the context of limits, where we have precise mathematical definitions. As far as I know $\Delta x \ll 1$ does not have such a precise meaning in usual contexts.

In physics, when arguing something is "small" we've got to set a parameter by which we say something is small. If I use $\Delta x \rightarrow 0$ I imagine that the parameter is clear by the context of the situation/theory, but when $\Delta x \ll 1$ is used you are automatically setting your standards for "small", which is the unit.

A motivation for $\Delta x \ll 1$ instead of $\Delta x \ll 0.25$, for example, would be the usage of a power series, where it is necessary for the series convergence that we have powers of something smaller than the unit. By saying $\Delta x \ll 1$ in this context we are actually saying that "this converges quickly".

Of course in many cases being "much smaller than the unit" is also enough for characterizing "small" in a discussion, and both may used interchangeably.

Normally, by $\Delta x \to 0$ we are considering some quantity in the limit as $\Delta x \to 0$. For example, the Riemann integral is constructed from the Riemann sum by considering, schematically,

$$\int_a^b f(x) \, \mathrm dx = \lim_{\Delta x \to 0} \sum_{i}f(x_i) \Delta x.$$

The notation is also employed in asymptotic analysis; for example, we may be interested in the behaviour of a quantity in a certain limit. More precisely, one could say

$$A(x) \sim B(x), \quad x \to x_0$$

if it holds that,

$$\lim_{x \to x_0} \frac{A(x)}{B(x)} = 1.$$

The most common use of notation of the form $\epsilon \ll 1$, in physics, is probably encountered in Taylor series or perturbation theory, wherein they have arguably different precise meanings.

For example, you will find

$$f(x + \epsilon) \approx f(x) + \epsilon f'(x)$$

for $\epsilon \ll 1$ since the higher order terms are $\mathcal O(\epsilon^2).$ For the same reason, suppose we are dealing in quantum mechanics with perturbing a Hamiltonian,

$$H_0 \to H = H_0 + \lambda V$$

for some potential $V$ and coupling $\lambda$. We could then say perturbation theory may approximate the behaviour of the full system for $\lambda \ll 1$ (presuming it is dimensionless). It warrants ignoring higher terms and that truncating at a certain point will give us a decent approximation. (The error from truncating can be quantified.)

$\epsilon \to 0$ would not be appropriate as this case is of no interest; we know the system in this limit is the presumably soluble $H_0$. Rather, we are interested in $\epsilon \ll 1$ and we hope the full system is approximated by this, if the coupling is in the vicinity of zero.

The first is as $\Delta x$ approaches 0. This refers to behavior that you would see as you get arbitrarily close to 0 (often times being actually 0 leads to many other problems with the math and reality, so we approach it to see how it behaves).

The second one is just referencing that $\Delta x$ is very small. It does not necessarily imply that we are concerned with the behaviour near 0, but that it's value is small compared to 1.

You could choose $\Delta x \ll 0.25$, and that might be appropriate in some situations. Especially in situations where the behaviour might be effected by a value with the magnitude 0.25; but something of a far smaller magnitude will have minimal effect. As the other answers suggested, it's more about parameter comparison. Often times 1 is a good number because you aren't dealing with values orders of magnitude lower than 1. If everything you dealt with was on the scale of nanometres, saying $\Delta x \ll 1$ where $\Delta x$ is in metres doesn't really provide enough constraint using 1 as your value.

1. Depends on the context. Usually $x \rightarrow 0$ means that you are trying to set some parameter to be very small, whereas $x \ll 1$ usually means that $x$ is in fact very small compared to some other parameters in your system. In other words, the first is meant to be a statement about the parameter, and the second is parameter comparison. However, functionally speaking there isn't a great difference so they would be fairly interchangeable.
2. The reason people write $1$ is because one has in mind a Taylor series or geometric series, so $x < x^2$. It's meant to be schematic, so there is no need to worry about what particular number goes there.
• Almost. I think $x\rightarrow 0$ implies you want to explore the behavior at $x=0$ but that is not a value you can reach. On the other hand, $x\ll 1$ tends to imply you want to explore the behavior "in the vicinity of" zero, rather than "at" zero - things like small amplitude oscillations, where certain linear assumptions are valid. – Floris Feb 27 '17 at 20:58
• If you want to say that one parameter is much smaller than the other, you'd write $x\ll y$. – Norbert Schuch Feb 28 '17 at 10:18
• @norbertschuch I agree, but it is conventional to write $x\ll 1$ even when $x$ is dimensionful, in which case it is shorthand for comparison to other parameters of the system – Aaron Feb 28 '17 at 14:39
• @Aaron I don't think this is conventional. – Norbert Schuch Feb 28 '17 at 14:53

protected by Qmechanic♦Feb 27 '17 at 21:08

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