Formal Derivatives, Taylor Expansion

Taylor Expansion

In calculus, the taylor expansion of f, at x-a, is a sum of terms, where the j^th term is the j^th derivative of f, evaluated at x-a, times (x-a)^j over j factorial. A similar formula holds here, though I am going to change the symbols just a bit. Instead of x sliding away from a continuously, let x be the fixed point, and let b be an offset from x. We want to prove the following.

f(x+b) = f(x) + b*f′(x) + b²/2!*f′′(x) + b³/3!*f′′′(x) + …

You can run the series for as long as you like, but if f has degree n, the derivatives beyond n are all 0.

If this relationship holds for f and g, it holds for f+g. This is because derivative and sum commute. Thus it is enough to prove the theorem for a single term a_ixⁱ.

The constant carries along, so it is enough to prove the theorem for xⁱ.

Replace x with x+b and expand via the binomial theorem. Look at the j^th term. This is the same as the j^th derivative of xⁱ, times b^j, divided by j factorial. The taylor formula agrees with the binomial theorem, and that completes the proof.

If the ring has characteristic p, you might be troubled by p! in the denominator. However, the denominator is there to cancel factors in the numerator, thereby reconstructing the binomial coefficient. We're not really dividing by 0; just building a particular integer. If you interpret it properly, the formula is valid in every ring.