I'm considering the differential of a function of one variable $f$ to be defined as a function of two variables $df$ for which the following holds:

$$df(x, h) = f'(x)h$$

All good so far. The extension to higher-order differentials, however, is what puzzles me. This is a passage from Courant:

(...) it may be pointed out for the sake of completeness that we may also form second and higher differentials. For if we think of $h$ as chosen in any manner, but always the same for every value of $x$, then $dy=hf'(x)$ is a function of $x$, of which we can again form the differential. The result will be called the second differential of y, and will be denoted by the symbol $d^2y=d^2f(x)$. The increment of $hf'(x)$ being $h\{f'(x+h)-f'(x)\}$, the second differential is obtained by replacing the quantity in brackets by its linear part $hf''(x)$, so that $d^2y=h^2f''(x)$. We may naturally proceed further along the same lines, obtaining third, fourth, ... differentials of y, etc., which can be defined by the expressions $h^3f'''(x), h^4f^{iv}(x)$, and so on.

Two questions arise from this for me:

  1. The author writes of the second differential of a function. To arrive at it they had to arbitrarily fix some value of $h$ in order to define a new single-variable function $dy$ such that $dy(x) = hf'(x)$ for the given $h$, and form the differential of this new function. Doesn't this procedure give rise to an infinite number of different functions, each associated with a different value of $h$, which would, in turn, lead to an infinite number of different functions, all of which satisfy the definition of the second differential? Given that the first differential is a function of two variables, unlike the single-variable function from which it was formed, speaking of the second (and higher-order) differential as the differential of the differential doesn't seem accurate. Is there a procedure that circumvents the "fix a value of $h$" argument and still leads to a rigorous definition in this context?

  2. Why is it simply assumed that the $h$'s that multiply the higher-order derivatives in the definition of higher-order differentials are all the same ("third, fourth, ... differentials of y, etc., which can be defined by the expressions $h^3f'''(x), h^4f^{iv}(x)$, and so on")? Shouldn't each $h$ be independent of the other, such that a more accurate description of the higher-order differentials should be $h_1h_2h_3f'''(x)$, $h_1h_2h_3h_4f^{(4)}(x)$?


  • $\begingroup$ Look up the higher order Frechet derivatives. The definitions here (fixing $h$) seems ad-hoc due to the avoidance of linear algebra and higher-dimensional spaces. For the $k^{th}$ Frechet derivative of a function $f:V\to W$ at a point $a\in V$ ($V,W$ real/complex vector spaces) we have that $D^kf_a:V^k\to W$ is a $k$-multilinear map, so indeed we evaluate it as $D^kf_a(h_1,\dots, h_k)$. Now a nice fact is that these multilinear maps are symmetric, so given the 'monomial' $h\mapsto D^kf_a(h,\dots, h)$, we can recover the full multilinear map $D^kf_a$. $\endgroup$
    – peek-a-boo
    Apr 10 at 19:51
  • $\begingroup$ In the familiar 1-dimensional case $f:\Bbb{R}\to\Bbb{R}$, we have $Df_a(h)=f'(a)\cdot h$, $D^2f_a(h_1,h_2)=f''(a)\cdot h_1h_2$, and in general, $D^kf_a(h_1,\dots, h_k)=f^{(k)}(a)\cdot h_1\cdots h_k$. $\endgroup$
    – peek-a-boo
    Apr 10 at 19:52


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