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I'm curious to know if the definition given below of the second differential is actually used in mathematics. It is based on the following definition of the differential of a function:

Let $dx\in\mathbb{R}$ be an independent variable. Let $y:\mathbb{R}\to\mathbb{R}$ be a differentiable function of $x$. The differential $dy$ of $y=y\left[x\right]$ is defined as

$$ dy\equiv\frac{dy}{dx}dx, $$

where $\frac{dy}{dx}$ is the derivative of $y$ with respect to $x$ by the standard definition.

The second differential $d^{2}y$ of $y$ is defined as

$$ d^{2}y\equiv\frac{d^{2}y}{dx^{2}}dx^{2}. $$

I don't recall any case where this construct has been useful to me. Nonetheless, I have it in my notes, and am confident that it reflects a reputable, though dated, source.


Edit to provide source: Link: Calculus and Analytic Geometry: With Supplementary Problems, Classic Edition, by George B. Thomas $\S$ 2-7, Problem 22.


Is the above definition of the second differential used today in mathematics? This is the question for which I will accept an answer.

Can the above definition be brought into consonance with the definition of the differential of a differential form? Which, as I understand it goes as follows:

Let $\mathfrak{r}\equiv\left\{ x,y\right\} $ be a position variable in $\mathbb{R}^{2},$ the functions $P,Q:\mathbb{\mathbb{R}}^{2}\to\mathbb{R}$ be continuously differentiable over the neighborhood of interest, and $dx,dy:\mathbb{\mathbb{R}}^{2}\to\mathbb{R}$ be the coordinate projection mappings. That is, given a vector $\mathfrak{v}=\left\{ v^{x},v^{y}\right\} ,$ $dx\left[\mathfrak{v}\right]=v^{x}$ and $dy\left[\mathfrak{v}\right]=v^{y}.$ The the following product is defined $$ dxdy\equiv-dydx, $$

$$ dxdx\equiv dydy\equiv0. $$ A mapping which associates with every point $\mathfrak{r}$ a linear mapping of the form

$$ \omega_{\mathfrak{r}}=P\left[\mathfrak{r}\right]dx+Q\left[\mathfrak{r}\right]dy $$

is a differential 1-form. The differential of this differential form is defined to be $$ d\omega=\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy. $$

Given a continuously differentiable function $f:\mathbb{\mathbb{R}}^{2}\to\mathbb{R},$ its differential (form) is defined to be $$ df_{\mathfrak{r}}\equiv\left(\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy\right)_{\mathfrak{r}}. $$

The differential (exterior derivative) of $df$ is therefore

$$ d^{2}f\equiv ddf=\left(\frac{\partial^{2}f}{\partial x\partial y}-\frac{\partial^{2}f}{\partial y\partial x}\right)dxdy=0. $$

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  • $\begingroup$ What is the point of "the second differential"? $\endgroup$ Jul 7, 2018 at 22:19
  • $\begingroup$ I have no idea what it might be useful for. Perhaps it was just an problem intended to apply the lessons of the chapter. It is not referenced elsewhere in the index. $\endgroup$ Jul 8, 2018 at 1:46

3 Answers 3

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I will answer both questions. As @md2perpe said in the other answer, one use of the second-order differential is in quadratic approximation. More generally, all of the higher-order differentials together make up a Taylor series, which (for analytic functions, at least locally) is not just an approximation but exact. Yet another use of differentials is to take care of the Chain Rule when performing a change of variables, although you have to be careful here, because the second differential that you have doesn't do that (because it's missing a term). So that's two purposes. A third purpose, in theory, is local optimization, although I don't think that it's used in practice.

Before I can explain these applications, I'll need the total second differential of $ y $ when $ y = f ( x ) $. The first differential is $ \mathrm d y = f ' ( x ) \, \mathrm d x $, which depends on both $ x $ and $ \mathrm d x $, so its differential has two terms, which we can find with the help of the Product Rule: $$ \eqalign { \mathrm d ^ 2 y = \mathrm d ( \mathrm d y ) & = \mathrm d \big ( f ' ( x ) \, \mathrm d x \big ) = \mathrm d \big ( f ' ( x ) \big ) \, \mathrm d x + f ' ( x ) \, \mathrm d ( \mathrm d x ) \\ & = \big ( f ' ' ( x ) \, \mathrm d x \big ) \, \mathrm d x + f ' ( x ) \, \mathrm d ^ 2 x = f ' ' ( x ) \, \mathrm d x ^ 2 + f ' ( x ) \, \mathrm d ^ 2 x \text . } $$ This is the correct rule if you want to do change of variables by substitution; that is, if $ x = g ( t ) $, so that $ y = ( f \circ g ) ( t ) $, then using $ \mathrm d x = g ' ( t ) \, \mathrm d t $ and $ \mathrm d ^ 2 x = g ' ' ( t ) \, \mathrm d t ^ 2 + g ' ( t ) \, \mathrm d ^ 2 t $, we get: $$ \mathrm d y = f ' ( x ) \, \mathrm d x = f ' \big ( g ( t ) \big ) \big ( g ' ( t ) \, \mathrm d t \big ) = f ' \big ( g ( t ) \big ) g ' ( t ) \, \mathrm d t \text , $$ so $ ( f \circ g) ' ( t ) = f ' \big ( g ( t ) \big ) g ' ( t ) $, which you know is correct; and also $$ \eqalign { \mathrm d ^ 2 y & = f ' ' ( x ) \, \mathrm d x ^ 2 + f ' ( x ) \, \mathrm d ^ 2 x = f ' ' \big ( g ( t ) \big ) \big ( g ' ( t ) \, \mathrm d t \big ) ^ 2 + f ' \big ( g ( t ) \big ) \big ( g ' ' ( t ) \, \mathrm d t ^ 2 + g ' ( t ) \, \mathrm d ^ 2 t \big ) \\ & = \Big ( f ' ' \big ( g ( t ) \big ) g ' ( t ) ^ 2 + f ' \big ( g ( t ) \big ) g ' ' ( t ) \Big ) \, \mathrm d t ^ 2 + f ' \big ( g ( t ) \big ) g ' ( t ) \, \mathrm d ^ 2 t \text , } $$ so $ ( f \circ g ) ' ' ( t ) = f ' ' \big ( g ( t ) \big ) g ' ( t ) ^ 2 + f ' \big ( g ( t ) \big ) g ' ' ( t ) $, which is less famous but also correct. This is probably the most common application.

Now if you want to, you can partially evaluate the second differential $ \mathrm d ^ 2 y $ when $ \mathrm d ^ 2 x = 0 $, getting a partial second differential showing only the dependance on $ x $ and not on $ \mathrm d x $: $$ ( \partial ^ 2 y ) _ { \mathrm d x } = \mathrm d ^ 2 y \rvert _ { \mathrm d ^ 2 x = 0 } = f ' ' ( x ) \, \mathrm d x ^ 2 \text . $$ Then if you divide by $ \mathrm d x $, you could call that the partial derivative of $ \mathrm d y $ with respect to $ x $; but since $ \mathrm d y $ is itself a differential, we usually divide by $ \mathrm d x $ again to get the second derivative of $ y $ with respect to $ x $. Since $ y $ depends only on $ x $, this is really a total second derivative, which is why people usually write it as $ \mathrm d ^ 2 y / \mathrm d x ^ 2 $, even though it's not literally the quotient of $ \mathrm d ^ 2 y $ and $ \mathrm d x ^ 2 $, in contrast to the first derivative. (You could fairly write it as $ \partial ^ 2 y / \partial x ^ 2 $, or even $ ( \partial ^ 2 y / \partial x ^ 2 ) _ { \mathrm d x } $ to indicate what is held fixed, but this is unlikely to catch on; or if you want to be both pedantic and understood, you can still write $ ( \mathrm d / \mathrm d x ) ^ 2 y $.) Partial though it is, this second differential does have its uses, as in the quadratic approximation to $ f $ at $ a $: $$ Q ( x ) = f ( a ) + f ' ( a ) ( x - a ) + \frac 1 2 f ' ' ( a ) ( x - a ) ^ 2 = \Big ( y + \mathrm d y + \frac 1 2 \mathrm d ^ 2 y \Big ) \Big \rvert _ { x = a , \, \mathrm d x = x - a , \, \mathrm d ^ 2 x = 0 } \text . $$ More generally, we have the Taylor series of $ f $ at $ a $: $$ T ( x ) = \sum _ { n = 0 } ^ \infty \frac 1 { n ! } f ^ { ( n ) } ( a ) ( x - a ) ^ n = \sum _ { n = 0 } ^ \infty \frac 1 { n ! } \mathrm d ^ n y \Big \rvert _ { x = a , \, \mathrm d x = x - a , \, \mathrm d ^ n x = 0 \, \text {for} \, n \geq 2 } \text . $$ And if $ f $ is analytic at $ a $, then $ T ( x ) $ converges to $ f ( x ) $, at least on some neighbourhood of $ a $. Many authors (including George B. Thomas, apparently) will treat $ \mathrm d x $ as a constant, so that $ \mathrm d ^ n x = 0 $ for $ n \geq 2 $ automatically, which makes the development of this application a bit simpler. However, this isn't appropriate for the other applications.

Another potential application is local optimization; so assume that $ f $ is twice-differentiable at $ a $ and defined on a neighbourhood of $ a $. Normally we say that $ f $ has a (local) minimum at $ a $ only if $ f ' ( a ) = 0 $ and $ f ' ' ( a ) \geq 0 $, and that $ f $ has a mimimum at $ a $ if $ f ' ( a ) = 0 $ and $ f ' ' ( a ) > 0 $. In higher dimensions, the parts about $ f ' ' $ are generalized to saying that the Hessian matrix is positive (semi)-definite, but the parts about $ f ' $ are also still there (referring to the gradient vector). But you can combine each of these into a single statement: $ y $ has a minimum at $ x = a $ only if $ \mathrm d ^ 2 y \rvert _ { x = a } \geq 0 $ for all nonzero values (hence all values) of $ \mathrm d x $ and $ \mathrm d ^ 2 x $ (a kind of positive semidefiniteness); and $ y $ has a minimum at $ x = a $ if $ \mathrm d ^ 2 y \rvert _ { x = a } > 0 $ for all nonzero values of $ \mathrm d x $ and $ \mathrm d ^ 2 x $ (a kind of positive definiteness). This works unchanged in higher dimensions (taking $ x $ and $ a $ from $ \mathbb R ^ n $ instead of just $ \mathbb R $), and it can even handle points on the boundary of the domain if you're careful. I'm not sure how useful this is, because you still have to pull out the gradient vector and the Hessian matrix to analyse it with the tools of linear algebra, so this is only a potential application rather than anything that I've seen people use; but it's a nice way to think about it in my opinion.


Now to show the connection to differential forms, I want to say something about what $ \mathrm d ^ 2 x $, $ \mathrm d x ^ 2 $, and so forth really mean. As you probably know, one way to think of an exterior differential form is as a multilinear alternating (or antisymmetric) operation on tangent vectors. While $ \mathrm d ^ 2 x $ and $ \mathrm d x ^ 2 $ are not exterior differential forms, we can still think of them as generalized differential forms, giving operations on tangent vectors that are not necessarily multilinear, alternating, or antisymmetric.

So if you're working in $ \mathbb R ^ 2 $ (where a tangent vector at a given point is essentially just another point in $ \mathbb R ^ 2 $), with $ x $ and $ y $ as the standard coordinate functions, then the differential form $ 2 x \, \mathrm d x + 3 y ^ 2 \, \mathrm d y $ at a point $ ( x _ 0 , y _ 0 ) $ takes a vector $ ( v _ x , v _ y ) $ and returns $ 2 v _ x + 3 y _ 0 ^ 2 v _ y $. And the differential form $ x ^ 2 \, \mathrm d x \wedge \mathrm d y $ at a point $ ( x _ 0 , y _ 0 ) $ takes two vectors, $ ( v _ x , v _ y ) $ and $ ( w _ x , w _ y ) $, and returns $ x _ 0 ^ 2 ( v _ x w _ y - v _ y w _ x ) $ (or half that, depending on your convention). Similarly, the generalized differential form $ \sqrt { d x ^ 2 + d y ^ 2 } $ at a point $ ( x _ 0 , y _ 0 ) $ takes a vector $ ( v _ x , v _ y ) $ and returns $ \sqrt { v _ x ^ 2 + v _ y ^ 2 } $. This is not linear, but it still makes sense. And you can even define what it means to integrate this form along a curve and prove that the value of the integral is the arclength of the curve. So there is no reason that you cannot perform arbitrary operations on differentials.

As for $ \mathrm d ^ 2 x $ and $ \mathrm d ^ 2 y $, these also simply return the $ x $- or $ y $-component of a vector, only the interpretation of this vector is different. That is, while $ x $ and $ y $ return the $ x $- and $ y $-coordinates of a point thought of as representing position, and $ \mathrm d x $ and $ \mathrm d y $ return the $ x $- and $ y $-components of a vector thought of as representing velocity, so $ \mathrm d ^ 2 x $ and $ \mathrm d ^ 2 y $ return the $ x $- and $ y $-components of a vector thought of as representing acceleration, and so on. This is a little more subtle on a more general manifold, but if you work in local coordinates, then you don't really have to pay attention to the subtleties as long as your higher differentials respect the Chain Rule. So if $ y = f ( x ) $, then the second differential $ \mathrm d ^ 2 y = f ' ' ( x ) \, \mathrm d x ^ 2 + f ' ( x ) \, \mathrm d ^ 2 x $ at a point $ x = x _ 0 $ takes a velocity $ v $ and an acceleration $ a $ and returns $ f ' ' ( x _ 0 ) v ^ 2 + f ' ( x _ 0 ) a $, and similarly in more dimensions.

Now, there is another possible version of the second differential, which to avoid ambiguity I will write as $ \mathrm d \otimes \mathrm d x $ or $ \mathrm d ^ { \otimes 2 } x $ for short. But first I should say what $ \mathrm d x \otimes \mathrm d x $ or $ \mathrm d x \otimes \mathrm d y $ means. This is, like the exterior form $ \mathrm d x \wedge \mathrm d y $, an operation that acts on two tangent vectors (at a given point); $ \mathrm d x \otimes \mathrm d x $ multiples their $ x $-components together, and $ \mathrm d x \otimes \mathrm d y $ multiplies the $ x $-component of the first vector by the $ y $-component of the second vector. (Then $ \mathrm d x \wedge \mathrm d y $ itself is $ \mathrm d x \otimes \mathrm d y - \mathrm d y \otimes \mathrm d x $, or half that, depending on your convention.) This is multilinear, but it's not antisymmetric, so it's not an exterior differential form, but it's still a generalized differential form. Note that now both vectors represent a velocity, but they represent velocities along two different curves, or along two edges of a parallelogram (or triangle). Then $ \mathrm d ^ { \otimes 2 } x $ is another vector, still a kind of acceleration, but it indicates how the first velocity vector changes when moving in the direction of the second velocity vector (or how the second changes when moving in the direction of the first, which on an infinitesimal level is the same, essentially because of Schwarz's Theorem). Now if $ y = f ( x ) $, we have $$ \mathrm d \otimes \mathrm d y = f ' ' ( x ) \, \mathrm d x \otimes \mathrm d x + f ' ( x ) \, \mathrm d \otimes \mathrm d x \text , $$ which at a point $ x = x _ 0 $ takes two velocities $ v _ 1 $ and $ v _ 2 $ and an acceleration $ a $ and returns $ f ' ' ( x _ 0 ) v _ 1 v _ 2 + f ' ( x _ 0 ) a $.

Now antisymmetrize $ \otimes $ to $ \wedge $: $ \mathrm d \wedge \mathrm d y = f ' ' ( x ) \, \mathrm d x \wedge \mathrm d x + f ' ( x ) \, \mathrm d \wedge \mathrm d x $, which features the exterior product (aka wedge product) and exterior differential (aka exterior derivative) that you know from exterior differential forms, and so this all comes to zero. In more detail, if $ y = f ( x ) $, then $ \mathrm d \wedge \mathrm d y $ at a point $ x = x _ 0 $ takes two velocities $ v _ 1 $ and $ v _ 2 $ and an acceleration $ a $ and returns $ \big ( f ' ' ( x _ 0 ) v _ 1 v _ 2 + f ' ( x _ 0 ) a \big ) - \big ( f ' ' ( x _ 0 ) v _ 2 v _ 1 + f ' ( x _ 0 ) a \big ) = 0 $ (or half that, which is still $ 0 $). Of course, in more dimensions, there are more interesting exterior forms, but $ \mathrm d \wedge \mathrm d $ will still be zero. When working exclusively with exterior forms, one may leave out all of the wedges; this is often done with the exterior product and essentially always done with the exterior differential. But I have included all of the wedges here to contrast with the kind of multiplication and differentiation that appears in the second differential.

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    $\begingroup$ I just want to congratulate you for this answer, which in my opinion is much better, for detailed and explicative, than the one marked as correct! $\endgroup$
    – Hvjurthuk
    Aug 29, 2021 at 6:29
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    $\begingroup$ Thank you! But you can't blame people for not knowing this, because while the full second differential is often rediscovered, and the corresponding chain rule for the second derivative is known, it's not taught much, and not usually (or ever?) seen as part of a big picture that also includes exterior differential forms. (I really want to publish a paper about this, but I don't have time to work on it. Until then, there's ncatlab.org/nlab/show/cogerm+differential+form on the nLab and even more in the forum discussions linked at the bottom of it.) $\endgroup$ Aug 29, 2021 at 17:09
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    $\begingroup$ I agree. The topic is sometimes not well explained even in texts that somehow use these objects, which is quite annoying. I aim you to write that paper! $\endgroup$
    – Hvjurthuk
    Aug 29, 2021 at 22:12
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The ordinary "first order" differential can be seen as a linearization. The second order differential can then be seen as a "quadricization". This might best be seen in multidimensional analysis.

First order differential: $$df(v) = \sum_i \frac{\partial f}{\partial x^i} v^i,$$ where $v^i$ is the component of the vector $v$ in the $x^i$ direction.

Second order differential: $$d^2f(v_1, v_2) = \sum_{i,j} \frac{\partial^2 f}{\partial x^i \, \partial x^j} v_1^i v_2^j$$

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  • $\begingroup$ Could you please provide a reference to a formal (informal) discussion of this topic? $\endgroup$ Aug 7, 2018 at 15:18
  • $\begingroup$ There is a lot to be found on the web about first order differentials, but I don't know any about second order differentials. My answer was just the for me obvious. $\endgroup$
    – md2perpe
    Aug 7, 2018 at 16:46
  • $\begingroup$ So $$d^{2}f\left(\Delta\mathfrak{x}_{1},\Delta\mathfrak{x}_{2}\right)=\frac{\partial^{2}f}{\partial x^{i}\partial x^{j}}\Delta x_{1}^{i}\Delta x_{2}^{j}.$$ Or even $\Delta\mathfrak{x}\equiv d\mathfrak{x},$ $$d^{2}f\left(d\mathfrak{x}_{1},d\mathfrak{x}_{2}\right)=\frac{\partial^{2}f}{\partial x^{i}\partial x^{j}}dx_{1}^{i}dx_{2}^{j}.$$ If $$\frac{\partial^{2}f}{\partial x^{i}\partial x^{j}}dx_{1}^{i}dx_{2}^{j}\equiv\frac{\partial^{2}f}{\partial x^{i}\partial x^{j}}dx_{1}^{i}\wedge dx_{2}^{j},$$ we have a 2-form. Perhaps related: math.stackexchange.com/q/2848695/342834 $\endgroup$ Aug 8, 2018 at 12:56
  • $\begingroup$ Actually, $dx_1^i \, dx_2^j = dx_1^i \otimes dx_2^j.$ The 2-form that you have plotted down vanishes because of symmetry in the partial derivatives and anti-symmetry in the wedge product. $\endgroup$
    – md2perpe
    Aug 8, 2018 at 13:30
  • $\begingroup$ I didn't notice that the 2-form vanished. But I didn't intend it to be the only possible interpretation. I was just offering a comparison. In my first two expressions I intend $\Delta\mathfrak{x}\equiv d\mathfrak{x}\in\mathbb{R^{n}}$, and traditional multiplication implied. $\endgroup$ Aug 8, 2018 at 13:40
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I think that it's worth distinguishing second derivatives in the context of calculus and the second differential in the context of differential forms. In the latter case, the differential is defined with an antisymmetric wedge product that, as you noted, means that $d^2 y = 0$ for any $y$. Although this identity seems useless and trivial, it actually has plenty of applications.

For example, physicists like myself learn all sorts of multivariable calculus identities for vector fields in $\mathbb{R}^3$ as undergraduates. Two of these are $\nabla \times \nabla f = 0$ and $\nabla \cdot (\nabla \times \vec{f}) = 0$. These are really useful in solving problems in electrostatics and magnetostatics. They are also literally identical to the second differential equation above (applied to 0-forms and 1-forms, respectively). There are more such identities in higher dimensions, and it's much easier to remember/understand them when you realize that they are all "the same thing."

Differential forms unify many different identities of multivariable calculus, and this is only one example. For a physicist, the next most obvious example is that Green's Theorem and the various forms of Stokes' Theorem are all the "same thing." Another example from pure mathematics is that the equation $d^2 = 0$ is completely analogous to the equation $\partial^2 = 0$ where $\partial$ is the boundary operator in simplicial homology. This has a lot of deep implications. One simple implication is that this can be used to prove Euler's theorem on vertices, edges, and faces of polygons.

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