Taylor's theorem in field of characteristic $0$

Question

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?

Answer 1

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.