Prove the following generalization of Taylor’s formula. Let \(f\), \(g\), and \(h\) be polynomials over a subfield of the complex numbers, with \(\deg f \leq n\). Then \[ f(g) = \sum_{k=0}^{n} \frac{1}{k!} f^{(k)}(h)(g - h)^k. \] (Here \(f(g)\) denotes ‘f of g.’)
Solution
Let \(R\) be the given subfield of \(\CC\). Introduce an indeterminate \(t\) and consider the polynomial \[ q(t)=f(h+t)\in R[x][t]. \] Since \(\deg f\le n\), Taylor’s formula for the polynomial \(q\) at \(t=0\) is an exact identity: \[ q(t)=\sum_{k=0}^n \frac{q^{(k)}(0)}{k!}t^k. \] Differentiating with respect to \(t\) gives \[ q^{(k)}(0)=f^{(k)}(h). \] Therefore \[ f(h+t)=\sum_{k=0}^n \frac{1}{k!}f^{(k)}(h)t^k. \] Now substitute \(t=g-h\). Since polynomials commute under multiplication, \[ f(g)=\sum_{k=0}^n \frac{1}{k!}f^{(k)}(h)(g-h)^k. \]