We know that from the Riemann sum that $\Delta x = \mathrm{d}x$ where $n \to \infty$ and $\Delta x = \dfrac{b - a}{n}$. If, however, $x$ represents some length of interval, can we also say that $\displaystyle\lim_{n \to \infty}\dfrac{x}{n} = \mathrm{d}x$?
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No, $\lim_{n\rightarrow\infty} \frac{x}{n} = 0$. – Michael Jan 16 '21 at 01:52
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@Michael By that logic, couldn't one argue, then, that $\mathrm{d}x$ = 0 as well? – Kalcifer Jan 16 '21 at 01:53
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The differentials $dy$ and $dx$ are sort of nebulous. You might take a look at https://math.stackexchange.com/questions/200393/what-is-dx-in-integration – Austin Mohr Jan 16 '21 at 01:54
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2The $dx$ in $\int_a^b f, dx$ is just notation to indicate the variable of integration (and to show the end of the integrand, like a right parenthesis). That's really it. Like the $dx$ in a derivative $dy/dx$, it's not a number or function. (I'm ignoring matters like, e.g., differential forms here.) – anomaly Jan 16 '21 at 01:55
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@anomaly but geometrically, $\mathrm{d}x$ represents an infinitely small chunk of an interval $[a,b]$. So if $x$ were to represent some interval along a line, then taking an infinitely small chunk of $x$ would be equal to $\mathrm{d}x$, no? – Kalcifer Jan 16 '21 at 01:58
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2"Infinitely small" is meaningless. The idea is that you're summing across a partition of $[a, b]$ as the partition size goes to $0$, but making that rigorous, or even meaningful, is the whole point. The limit $\lim x/n$ in the post is definitely $0$, for example. I don't mean to dissuade you from this sort of train of thought, since it is the right intuition and you've got the right idea, but the sort of computation in the post doesn't make sense. It's like saying that because $\lim_{x \to 0} x$ and $\lim_{y\to 0} y$ are both $0$, then $\lim_{x\to 0} x = y$. The latter doesn't even make sense. – anomaly Jan 16 '21 at 02:10
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@Kalcifer The definitions of differentiation and integration can be rigorously defined without considering $dx$ as infinite small quantity. – Infinity_hunter Jan 16 '21 at 03:54
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In short: No. The limit of that sort of expression is, generally speaking, going to be a real expression (provided the limit exists and is finite). A "real expression" in this context means a mathematical expression in terms of $x$, where you can plug in real numbers as possible values of $x$ and get out real numbers as a result (like a function). On the other hand, $\mathrm{d}x$ is not a real expression; it is a different kind of object called a differential form, which is (to greatly oversimplify) "a thing that you can put under an integral sign."
The defining property of the real numbers is that you can take limits on them, so much so that a limit is usually assumed to be taken over the reals unless context implies some other set of numbers (e.g. the complex numbers, if an $i$ is present). But another important property is the archimedean property, which states that there are no infinities or infinitesimals. In other words, no real number is infinitely larger or smaller than any other (except for zero, which is infinitely smaller than all non-zero values). More formally, if you have two real numbers $x$ and $y$, you can add $x$ (or $-x$) to itself repeatedly, and the sum will eventually be greater than $y$ (or $-y$).
If you want to think of $\mathrm{d}x$ as an infinitesimal, then it immediately follows that $\mathrm{d}x$ cannot be a real number. However, in standard analysis, differentials are actually a much stranger sort of object, which arises out of abstract algebra (see link above). You don't need to know this algebra in order to do simple manipulations like the chain rule (cancelling differentials when multiplying two derivatives), but it is important to understand if you want to start talking about things like $\mathrm{d}x$ by itself (and not under an integral or part of a derivative).
Nonstandard analysis, which actually does use infinitesimals, most commonly uses the hyperreal numbers rather than the reals, and (usually) doesn't bother much with limits as such at all. In that context, it might make sense to write something like "Let $\mathrm{d}x = \frac{x}{\infty}$" (although we must take care not to confuse this $\mathrm{d}x$ with the previous paragraph's definition!). But this is a somewhat misleading thing to write, because the hyperreals do not just have one infinity - they have infinitely many. So you need to specify which infinity you want, or at least clarify that you're always using the same infinity. For simple cases, we might write $\omega$ to mean an arbitrary infinity, and thus have $\mathrm{d}x = \frac{x}{\omega}$. Then, if we write $\omega$ in another place, it will be clear that we mean the same infinity, and not $2\omega$ or something else.
Kevin
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I think this is quite a tricky topic to talk about because of how people interpret things vs how they are actually defined. I find it best to think of $dx$ as a separate value, rather than a function or scaling of the variable $x$. whilst we are talking about a small change in $x$ the key word here is change, so rather than $x/n\to0$ really we want the change in $x$, or the interval that we are looking at to be very small rather than the initial or end values.
One nice way to think of this is by looking at your average integral: $$I_{a,b}[f]=\int_a^bf(x)dx$$ now think of the integral symbol as a stretched out $S$, which is the first letter of sum. Now inside the integral (the part we are summing) is $f(x)\times dx$. In other words we can think of an integral as a sum over our domain $[a.b]$ where the height of each one is $f(x)$ and the width is $dx$. However for our error to tend to zero we require the rectangles to fit the function $f(x)$ well, and actually tend to it exactly. The best way for us to do this is to reduce the width $dx$ to a very small value close to zero, but it cannot actually equal zero or be "infinitely small". Whilst this may seem strange it does not actually matter as we do not need a numerical method for $dx$, unless using some methods for approximating and in that case the statement is not strictly true. Hope this helps
Henry Lee
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