3

If $ K $ is a field of characteristic $ 0 $ then for all polynomials $ f[X] $ over $ K $ and elements $ x, h \in K $, we have $$ f(x + h) = f(x) + f'(x) h + \frac{1}{2} f''(x)h^2 + \cdots $$ This doesn't seem so hard (but tedious) to prove using induction: just write $ f = \sum a_n x^n $, calculate the first term, then suppose we have the $ n $th term, we can find the $ (n+1) $th term. But I'm wondering if there is an easier way to see this?

  • 1
    Yes: substitute $x+h$ for $h$, and read off the coefficients of the powers of $h$. And it works in arbitrary characteristic! – Lubin May 14 '18 at 01:03
  • @Lubin But in other characteristics how do you deal with the quotients $ \frac{1}{n} $? –  May 14 '18 at 02:28
  • 1
    The way to do it is to consider $(d/dx)^k f(x)/k!$ and apply it to $x^n$ resulting in ${n \choose k}x^{n-k}$ and binomial coefficients are integers. In other words, regard $(d/dx)^k f(x)/k!$ as an operator unit instead of the quotient of $(d/dx)^k f(x)$ and $k!$. – Somos May 14 '18 at 02:39
  • 1
    For a moment, @mathstudent_101, I worried about that too. But just follow my instructions with an example, and you’ll see that there are no quotients. This is an operation that uses only the ring axioms. The derivatives do not appear, nor the factorials. Maybe the moral of the story is that in characteristic $p$, Taylor’s Theorem does not apply without a heavy rewrite. Try an example! – Lubin May 14 '18 at 13:10
  • @Lubin thank you, I see it now that there is no worry with the factorials. –  May 15 '18 at 18:39

1 Answers1

2

The binomial theorem is $\;(x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k.\;$ If applied to $\;f(x) := x^n\;$ the result is $\;f(x+h) = x^n + nx^{n-1}h + n(n-1)x^{n-2}h^2/2 +\cdots = f(x) + f'(x)h + f''(x)h^2/2 + \cdots\;$ and since derivatives are linear, the equation hold for any sum of powers of $x$ multiples, or polynomials.

Somos
  • 35,251
  • 3
  • 30
  • 76