Editing Taylor's theorem (section)

=== Derivation for the remainder of multivariate Taylor polynomials ===

We prove the special case, where <math>f:\mathbb R^n\to\mathbb R</math> has continuous partial derivatives up to the order <math>k+1</math> in some closed ball <math>B</math> with center <math>\boldsymbol{a}</math>.  The strategy of the proof is to apply the one-variable case of Taylor's theorem to the restriction of <math>f</math> to the line segment adjoining <math>\boldsymbol{x}</math> and <math>\boldsymbol{a}</math>.<ref>{{harvnb|Hörmander|1976|pp=12–13}}</ref>  Parametrize the line segment between <math>\boldsymbol{a}</math> and <math>\boldsymbol{x}</math> by <math>\boldsymbol{u}(t)=\boldsymbol{a}+t(\boldsymbol{x}-\boldsymbol{a})</math> We apply the one-variable version of Taylor's theorem to the function <math>g(t) = f(\boldsymbol{u}(t))</math>:

<math display="block"> f(\boldsymbol{x})=g(1)=g(0)+\sum_{j=1}^k\frac{1}{j!}g^{(j)}(0)\ +\ \int_0^1 \frac{(1-t)^k }{k!} g^{(k+1)}(t)\, dt.</math>

Applying the [[chain rule]] for several variables gives

<math display="block">\begin{align}
g^{(j)}(t)&=\frac{d^j}{dt^j}f(\boldsymbol{u}(t))\\
&= \frac{d^j}{dt^j} f(\boldsymbol{a}+t(\boldsymbol{x}-\boldsymbol{a}))\\
&= \sum_{|\alpha| =j} \left(\begin{matrix} j\\ \alpha\end{matrix} \right) (D^\alpha f) (\boldsymbol{a}+t(\boldsymbol{x}-\boldsymbol{a})) (\boldsymbol{x}-\boldsymbol{a})^\alpha
\end{align}</math>

where <math>\tbinom j \alpha</math> is the [[multinomial coefficient]]. Since <math>\tfrac{1}{j!}\tbinom j \alpha=\tfrac{1}{\alpha!}</math>, we get:

<math display="block"> f(\boldsymbol{x})= f(\boldsymbol{a}) + \sum_{1 \leq |\alpha| \leq k}\frac{1}{\alpha!} (D^\alpha f) (\boldsymbol{a})(\boldsymbol{x}-\boldsymbol{a})^\alpha+\sum_{|\alpha|=k+1}\frac{k+1}{\alpha!}
(\boldsymbol{x}-\boldsymbol{a})^\alpha \int_0^1 (1-t)^k (D^\alpha f)(\boldsymbol{a}+t(\boldsymbol{x}-\boldsymbol{a}))\,dt.</math>