# Obtaining the higher-order differentials of a function of one variable

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 $$h_1h_2h_3f'''(x)$$, $$h_1h_2h_3h_4f^{(4)}(x)$$?

Thanks.

• 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 $D^kf_a$. Apr 10 at 19:51
• 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, $D^kf_a(h_1,\dots, h_k)=f^{(k)}(a)\cdot h_1\cdots h_k$. Apr 10 at 19:52