Understanding the difference between $\Delta x$ and $dx$

Question

I want to make sure I understand the difference between $\Delta x$ as the change in x and $dx$, an infinitesimally small change in $x$.

Instead of $x$ let's use $V$ like we're doing a thermodynamics problem involving a volume change. Now, if my understanding is correct, I technically can't assign a "size" to $dV$ because it's an infinitesimal. So it would be incorrect to say $dV$ = $\Delta V$

But, if I were using $dV$ in problem solving and needed to assign a size to this infinitesimal interval, all that I'd have to do is integrate it using a definite integral, correct?

$$\int_{1}^{2} dV = V(2) - V(1) = \Delta V$$

Thanks!

Answer 1

Note that $dV$ in the integral is just a symbol to identify the variable respect to we are integrating and $\Delta V$ in this case just indicate the difference $V(2)-V(1)$ according to the foundamental theorem of calculus

Answer 2

An integral is defined by the limit of Riemann sum as $$\int_{V_1}^{V_2}f(V) dV = \lim_{N\to\infty}\sum_{\substack{V=V_1\\\text{with increments }\frac{\Delta V}{N}}}^{V_2} f(V)\; \frac{\Delta V}{N},$$ where $\Delta V=V_2-V_1$. The founders of calculus picked the notation of integrals based on this definition:

• The symbol $\int$ is a short for $$\lim_{N\to\infty}\sum_{\substack{V=V_1\\\text{with increments }\frac{\Delta V}{N}}}^{V_2}$$
• and the symbol $dV$ is short for $$\frac{\Delta V}{N}$$ with the understanding that the limit of $N\to\infty$ will be taken after it is summed over in the above expression.

In your case $f(V)=1$, and you have $$\int_{V_1}^{V_2}dV = \lim_{N\to\infty}\sum_{\substack{V=V_1\\\text{with increments }\frac{\Delta V}{N}}}^{V_2} \frac{\Delta V}{N} = \lim_{N\to\infty}N \frac{\Delta V}{N} =\Delta V.$$

So back to your questions:

if my understanding is correct, I technically can't assign a "size" to $dV$ because it's an infinitesimal

That is true. From the above notation, you can see that $dV$ is, in fact, $\Delta V/N$ with the understanding that the limit of $N\to \infty$ will be taken after it is summed over. You cannot assign a value to it because the limit cannot be evaluated until the sum is computed. That is $dV$ does not have a meaning when taken out of the integral.

So it would be incorrect to say $dV = \Delta V$.

Yes, that would be incorrect.

I think there is an unfortunate confusion of notations here: if we had $\int_a^b dx$, with the above notation, $dx$ would be a symbol for $(b-a)/N$ with the understanding that $N\to \infty$ limit will be taken later. Sometimes the term $(b-a)/N$ is written as $\Delta x$ and the limit of $N\to \infty$ is expressed as limit of $\Delta x\to0.$ In this case, $dx$ is in fact a notation that represents $\Delta x$ with the understanding that the limit of $\Delta x\to 0$ will be taken after the sum. Now, in your problem, we already have something called $\Delta V$ and it is not an infinitesimal change, it is what we called $b-a$ in this example. I hope I didn't make it more confusing. If I did, just ignore this example.

if I were using $dV$ in problem-solving and needed to assign a size to this infinitesimal interval, all that I'd have to do is integrate it using a definite integral, correct?

Often in physics and engineering, when we need to calculate the total change in something, we would break it down to infinitesimals and then integrate over it. If that is what you mean, then you are correct. Basically, you are writing the total change as a sum of smaller changes of size $\Delta V/N$, adding them up, and then taking the limit of $N\to\infty$.

Answer 3

Contrary to what you might find in many contemporary textbooks (especially if they're elementary), you can well think of an infinitesimal as an entity and manipulate it according to its own rules (this has been rigorously set down in the last century in the so-called nonstandard calculus, which in fact I think is the more intuitive way to do calculus as manifest in their use by earlier mathematicians and many today).

I shall address each of your paragraphs in turn:

Yes, you're right. The symbol $\Delta$ is used in this context to symbolise a difference, or change, in a certain quantity, while the symbol $\rm d$ is used to mean that this difference is now to be thought of as infinitely small, or infinitesimal. Well, what could this mean? Infinitesimals capture our idea of things like an instant of time, a point of a line, a plane section of a volume, etc. Thus, it's often more easy to think of a varying infinitesimal -- that is, a continuous stream of infinitesimals, as it were. And that's often how we use them since we often take the differentials of varying quantities, the differentials of constants being zero. A semiformal way of talking about these things is to think of them as objects that combine the idea of a quantity with the fact that this quantity is ever approaching zero -- note, not the quantity itself, or the limit (those are functions and numbers respectively), but a combination of the idea of approaching zero and quantity, just like the vectors of physics were new objects to talk of the idea of directed quantities.

So, yes, the notion of size in the usual sense is inapt for an infinitesimal since it's not a number or numerical function, but an instantaneous part of a quantity, as it were; an evanescent quantity. So you can't have a change in volume and a differential change in volume to be ever equal, since $\Delta V$ is some number when evaluated, and $\rm d v$ is infinitely small. One may only "approximate" the latter by the former.

Integrating a differential (over a continuum) does not mean you've assigned a size to this differential. When you transform an object, it's no longer that object. So, whenever you integrate a continuum of differentials, for example, the result is no longer a differential, but a definite number. You've effected a transformation which may be intuitively thought of as accumulating all the infinitesimal changes of a certain quantity over a period. For example, think of water flowing from a tap; then the water emerging from the mouth can be thought of as being made up of infinitely many infinitely thin slices of water continuously merged -- these are the differentials, which between any two points in time accumulate to form a definite mass of water. So, a differential and the continuous sum (integral) of differentials are not the same thing, but rather the former have been changed into the latter.