Is There a Difference Between $d^2x$ and $(dx)^2$?

Question

I've just started reading through Calculus Made Easy by Silvanus Thompson and am trying to solidify the concept of differentials in my mind before progressing too far through the text.

In Chapter 1 Thompson describes a differential $dx$ as "a little bit of $x$" or "an element of $x$" and then proceeds to describe $\int dx$ as "the sum of all the little bits of $x$." I'm okay upto this point.

In Chapter 2 he begins discussing the various degrees of "smallness" and this is where I begin to lose track. At times his explanation seems to make sense, and then again it does not.

Thompson starts his discussion of degrees of smallness with an appeal to time.

Obviously 1 minute is a very small quantity of time compared with a whole week. Indeed, our forefathers considered it small as compared with an hour, and called it "one minute," meaning a minute fraction - namely one sixtieth - of an hour. When they came to require still smaller subdivisions of time, they divided each minute into 60 still smaller parts, which, in Queen Elizabeth's days, they called "second minutes" (i.e. small quantities of the second order of minuteness).

Now, at this time it seems as though Thompson is making the point that $(dx)^2$ may be considered to be "a little bit of $dx$" or "a little bit of a little bit of $x$," which seems intuitive, but perhaps not consistent with the language.

However, a little later he describes $(dx)^2$ as "a little bit of a little bit of $x^2$. And in doing so, he backs up his statement with a figure of a square with sides of length $x + dx$ and notes that any one of the corners of the square represents the magnitude $(dx)^2$. While not necessarily appealing to my intuition, this description does seem to be more in line with his description of $dx$.

a little bit $\cdot$ $x$ $\cdot$ a little bit $\cdot$ $x$ $=$ a little bit $\cdot$ a little bit $\cdot$ $x^2$

Yet, I don't see how this second description is making the case for $(dx)^2$ being any less significant than $dx$. From the diagram, it actually seems more significant than $dx$, being $dx$ times greater than $dx$.

So I'm wondering, is there a difference between $d^2x$ and $(dx)^2$? Should I read $d^2x$ as "a little bit of a little bit of $x$" and $(dx)^2$ as "a little bit of a little bit of $x^2$?" Am I missing the point entirely? Have I gone completely mad?

In hopes of further clarifying the question, I'm including a little more of what Thompson said:

If $dx$ be a small bit of $x$, and relatively small of itself, it does not follow that such quantities as $x \cdot dx$, or $x^2 \cdot dx$, or $a^x \cdot dx$ are negligible. But $dx \cdot dx$ would be negligible, being a small quantity of the second order.

Furthermore, he states:

[I]n all cases we are justified in neglecting the small quantities of the second - or third (or higher) - orders, if only we take the small quantity of the first order small enough in itself.

Answer 1

Neither of these, standing alone, have a real meaning. The symbol $$ \frac{\mathrm{d}^2}{\mathrm{d}x^2}\tag{1} $$ means two derivatives with respect to $x$. That is, two applications of $\dfrac{\mathrm{d}}{\mathrm{d}x}$: $$ \left(\frac{\mathrm{d}}{\mathrm{d}x}\right)^2\tag{2} $$

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However, what Thompson is trying to explain, might be more easily understood using finite differences. After all, derivatives are limits of finite differences.

Define the finite difference operator $\Delta_h$ as $$ \Delta_hf(x)=f(x+h)-f(x)\tag{3} $$ Then, the derivative is given by $$ \lim_{h\to0}\frac{\Delta_hf(x)}{\Delta_hx}\tag{4} $$ Note that $\Delta_hx=h$.

The second derivative, when it exists, is given by $$ \lim_{h\to0}\frac{\Delta^2_hf(x)}{(\Delta_hx)^2}\tag{5} $$ Now we can start to see the similarity between $(5)$ and $(1)$. In this setting, $\mathrm{d}^2$ is the limit of the square of the forward difference operator (the second forward difference operator) and $\mathrm{d}x^2$ is the limit of the square of the first difference of $x$.

Answer 2

The notation $d^2x$ is a very unfortunate one in my opinion. In differential calculus, the exterior derivative $d$ satisfies $d^2=0$: if $dx$ is "a little bit of $x$", then "a little bit of a little bit of $x$" is $d(dx)=0$. An infinitesimal amount of an infinitesimal amount is $0$.

On the other hand we also have $(dx)^2=0$ because $(dx)^2$ is an infinitesimal area, but it is "flat" (collapsed to a segment). On the other hand $dx dy$ is also an infinitesimal area, but it is not zero if $dx$ and $dy$ are not colinear, because then it is not "flat". Here, $(dx)^2$ means $dx \wedge dx$, and the fact that it vanishes comes from the fact that the exterior algebra is anti-commutative.

In other words, formally we have $d^2x=0$ and $(dx)^2=0$ but for two different reasons.

When you see something like

$$\frac{d^2x}{dx^2},$$

this is really an abuse of notation for

$$\left(\frac{d}{dx}\otimes\frac{d}{dx}\right)x,$$

and the expression $\frac{d}{dx}\otimes \frac{d}{dx}$ lives in the tensor algebra, rather than in the exterior algebra. The tensor algebra is not anti-commutative, and in fact we do not have $\frac{d}{dx}\otimes \frac{d}{dx} = 0 $ unless $\frac{d}{dx}=0$.

Differential calculus has many subtleties which require quite a bit of machinery to elucidate in a rigorous manner (the basic language of differential calculus: tangent spaces, cotangent spaces, differential forms, vector fields, etc.). Until you are ready to learn about these things, it is probably best not to worry too much about the meaning of individual symbols like $d^2x$ and $dx^2$, and to simply concentrate on those expressions which obviously have a clear and accepted meaning (like "$d^2/dx^2$", which means "take the second derivative with respect to $x$")

Answer 3

I think this is a key point:

From the diagram, it actually seems more significant than dx, being dx times greater than dx.

Keep in mind that we are thinking of $dx$ as being an extremely tiny number. So if you multiply something by $dx$, what you get is much smaller than what you started with.

Answer 4

Parts of this is essentially just nonstandard analysis. The real numbers does not admit nonzero infinitesimals and they are not necessary for calculus but there is an alternate construction called nonstandard analysis which allows these infinitesimal numbers without contradiction.

Ostensibly, you are not being taught nonstandard analysis, it is simply being used to give an intuitive justification for complicated matters. If you have a number that represents a tiny quantity, the square of that tiny quantity will be "essentially zero". This is not mathematically precise but it serves as a good intuition.

Answer 5

Yes there obviously is, because $$ \frac{d^2x}{dx^2}=0 $$ rather than $1$ ;)

Answer 6

For those who may care, in reading a bit further I happened to stumble upon the author's (Thompson's) response to my very question.

What does $(dx)^2$ mean? Remember that $dx$ meant a bit - a little bit - of $x$. Then $(dx)^2$ will mean a little bit of a little bit of $x$... it is a small quantity of the second order of smallness.

So, rather than $(dx)^2$ representing "a little bit of a little bit of $x^2$" (which may or may not imply a value less than $dx$), it appears as though Thompson does in fact mean for $(dx)^2$ to represent a smaller portion of $x$ than $dx$.

Answer 7

Oh ya they are different.

You can freely multiply and divide by $(dx)^2$, but not by $d^2x$

Example : $$\left(\dfrac{dy}{dx}\right)^2 = x^5-x \Rightarrow \left(\dfrac{dx}{dy}\right)^2 = \dfrac{1}{x^5-x}$$

$ $

But $$\dfrac{d^2y}{dx^2} = x^4 \nRightarrow \dfrac{d^2x}{dy^2} = \dfrac{1}{x^4}$$