# Is Weyl's second form of the algebraic definition of the derivative of a polynomial complete?

On page 4 of Weyl's The Classical Groups: Their invariants and Representations, we are given two algebraic definitions of the derivative of a polynomial, which he presents as equivalent:

The derivative $$f^{\prime}\left(x\right)$$ of a polynomial $$f\left(x\right)$$ is introduced as the coefficient of $$t$$ in the expansion of $$f\left(x+t\right)$$ as a polynomial in $$t$$:

$$f\left(x+t\right)=f\left(x\right)+tf^{\prime}\left(x\right)+\dots.$$

$$\dots$$

One might restate the definition [above] as follows: there is a polynomial $$g\left(x,y\right)$$ satisfying the identity

$$f\left(y\right)-f\left(x\right)=\left(y-x\right)g\left(x,y\right);$$ $$f^{\prime}\left(x\right)$$ is $$=g\left(x,y\right).$$ *

• As indicated in the comments, I transcribed the final expression incorrectly. It should read $$f^{\prime}\left(x\right)$$ is $$=g\left(x,x\right).$$ So now it appears we have something similar to Horner's rule.

If we substitute $$y-x=t$$ in the first form of the definition and transpose $$f\left(x\right),$$ we get

$$f\left(y\right)-f\left(x\right)=tf^{\prime}\left(x\right)+\dots.$$

So I'm left wondering if either the ellipsis $$(\dots)$$ is superfluous in the first form of the definition, or else the second form should have some expression for higher order terms, which would make it identical to the standard calculus definition.

Consider the expansion

$$f\left(x+t\right)=\left(x+t\right)^{n}=\sum_{k=0}^{n}\begin{pmatrix}n\\ k \end{pmatrix}x^{n-k}t^{k}$$

$$f\left(x+t\right)-f\left(x\right)=\sum_{k=0}^{n}\begin{pmatrix}n\\ k \end{pmatrix}x^{n-k}t^{k}-x^{n}$$

$$=x^{n-0}t^{0}+\sum_{k=1}^{n}\begin{pmatrix}n\\ k \end{pmatrix}x^{n-k}t^{k}-x^{n}$$

$$=\sum_{k=1}^{n}\begin{pmatrix}n\\ k \end{pmatrix}x^{n-k}t^{k}$$

$$=\begin{pmatrix}n\\ 1 \end{pmatrix}x^{n-1}t^{1}+\sum_{k=2}^{n}\begin{pmatrix}n\\ k \end{pmatrix}x^{n-k}t^{k}$$

$$=nx^{n-1}t+\sum_{k=2}^{n}\begin{pmatrix}n\\ k \end{pmatrix}x^{n-k}t^{k}$$

$$=tf^{\prime}\left(x\right)+t^{2}\Delta$$

$$f\left(y\right)-f\left(x\right)=\left(y-x\right)f^{\prime}\left(x\right)+\left(y-x\right)^{2}\Delta$$

where $$\Delta$$ is an obvious abbreviation for a term which is, in general, non-zero. This sort of a term seems to be missing from Weyl's second form of the definition.

So, I ask, should the second form of the definition given by Weyl include an "error" term in order to be rigorously correct? The answer seems to be an obvious "yes". But this is unfamiliar territory for me.

• (i) there should be an ellipsis in the first definition. (ii) if $f(x)-f(y)=(x-y)g(x,y)$ then $f'(x)=g(x,x)$ (not $g(x,y)$). Oct 10 '19 at 19:17
• I see. I hadn't noticed that $x$ appears twice in $g\left(x,x\right)$ of the the last sentence. Oct 10 '19 at 19:28

No. The point is that the error terms are present in $$g(x,y)$$ but disappear when you compute $$g(x,x)$$. Writing $$y=x+t$$, the second equation becomes $$f(x+t)-f(x)=tg(x,x+t).$$ In other words, $$tg(x,x+t)$$ is what you call $$tf'(x)+t^2\Delta$$, so $$g(x,x+t)=f'(x)+t\Delta.$$ To evaluate $$g(x,x)$$, you plug in $$t=0$$ and get $$g(x,x)=f'(x).$$