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In Linear Algrebra form Hoffman and Kunze, the Taylor's Formula is stated as follows:

Theorem 5. (Taylor's Formula) (page 129) Let $\mathbb{F}$ be a field of characteristic zero, $c\in \mathbb{F}$, and $n$ a positive integer. If $f$ is a polynomial over $\mathbb{F}[X]$ with $\deg f \leq n$, then $$f=\sum_{k=0}^{n}\frac{(D^{k}f)}{k!}(c)(x-c)^{k}.$$

After proving the Theorem, they make some comments."Although we shall not give any details, it is possible worth mentioning at this point that with the proper interpretation Taylor's Formula is also valid for polynomials over fields of finite characteristics. If the field $\mathbb{F}$ has finite characteristics then we may $k!=0$ in $\mathbb{F}$, in which case the division of $(D^{k})f(c)$ by $k!$ is meaningless. Nevertheless, sense can made out of the division of $D^{k}f$ by $k!$, because every coefficient of $D^{k}f$ is an element of $\mathbb{F}$ multiplied by an integer divisible by $k!$."

Is this the "proper interpretation"? I was hoping an interpretation over a field like $GF(2)$.

Thanks for your help.

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Writing "\deg f" instead of "deg f" has at least two effects: it sets "deg" in non-italicized letters, and it results in proper spacing between "deg" and "f". With some other things like \max or \sup, it also effects positioning of subscripts with the TeX is in a "displayed" as opposed to "inline" setting. So you get $\deg f$ instead of $deg f$. –  Michael Hardy Jan 26 '12 at 21:16
    
Thanks! I will do that next time. –  spohreis Jan 26 '12 at 21:23
    
.....and similar comments apply to \det, \log, \ln, \sin, \cos, \gcd (but not \lcm ; you have to use \operatorname{lcm}), and a variety of other things. –  Michael Hardy Jan 26 '12 at 21:35

1 Answer 1

up vote 8 down vote accepted

For instance, take $f(x) = (x^3+1)^3 = x^9+3x^6+3x^3+1$ over $\mathbb{Z}_5$, which has characteristic $5$. Then we have no problems with $D^kf$ for $k \le 4$, but we run into problems with $k=5$ since $5!=0$ in $\mathbb{Z}_5$. However, if we ignore this fact temporarily, we have that

$$\begin{align} \frac{D^5f(x)}{5!} &= \frac{9.8.7.6.5x^4 + 6.5.4.3.2.3x}{5!} \\ &= \frac{4.3.2.1.5x^4 + 1.5.4.3.2.3x}{5!} \\ &= \frac{5!x^4 + 5!.3x}{5!} \\ &= x^4+3x \end{align}$$

So we're fine.

So the "proper interpretation" is, for finite fields of prime order $p$, to work over $\mathbb{Z}[X]$ and then take the natural map $\mathbb{Z}[X] \rightarrow \mathbb{F}[X]$. You might run into issues with finite fields of order $p^n$ for $n>1$, but I hope this illustrates what the author of the book was trying to get at.

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To get around the "issues" with other fields of characteristic $p$ (where there might not be a natural map $\mathbb{Z}[X] \to F[X]$), we can define, on $F[X]$, the operator $D^k/k!$ for any $k$ by mapping the polynomial $\sum_{i \ge 0} a_i x^i$ to $\sum_{i \ge k} \binom{i}{k} a_i x^{i-k}$. Note that we have hidden the $k!$ into the binomial coefficient, which is always an integer, so the operator is valid for $F$ of any characteristic. Now we can interpret the $k$th term of the Taylor expansion as applying this operator to $f$ and then evaluating at $c$, then multiplying by $(x-c)^k$. –  Ted Jan 27 '12 at 4:06
    
The operator $D^k/k!$ is known as the Hasse derivative. There's a great elementary introduction to it here. –  Hamish Sep 23 '12 at 18:01

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