integral form of Taylor theorem remainder multivariable

Question

Let a function f be in $C^{k+1}(B(x_0,r)),r>0$, then $$f(x) = P_k(x;x_0)+ (k+1)\sum_{\vert \alpha \vert= k+1}\left(\int_0^1 (1-t)^kD^\alpha f(x_0+t(x-x_0))dt\right)\frac{(x-x_0)^\alpha}{\alpha !}$$ Where $P_k(x;x_0)$ is the Taylor polynomial centered at $x_0$.and $\alpha$ here is multivariable index $\alpha = (\alpha_1, \cdot \cdot \cdot, \alpha_n )$ and usual multivariable notation is used here. I know how to get the Taylor polynomial (just use integration by part repeatedly) but how can we get the remainder in that form?

Answer 1

Apply the usual Lagrange remainder form to the function $$ g(t)=f(x_0+t(x-x_0)) $$ (Of course $g$ is $C^{k+1}$ since $f$ is).

Spelling it out, we have $$ g'(t)=(x-x_0)\cdot\nabla f(x_0+t(x-x_0)) $$ and hence inductively $$ g^{(j)}(t)=\left(\frac{\mathrm{d}}{\mathrm{d}t}\right)^j (g(t)) =[(x-x_0)\cdot\nabla]^jf(x_0+t(x-x_0)) $$ Plugging that into the Lagrange remainder form $$ R_k(g;0,t_1)=\int_0^{t_1}g^{(k+1)}(t)\frac{(t_1-t)^k}{k!}\,\mathrm{d}t $$ and recall $$ (h\cdot\nabla)^j=\sum_{|\alpha|=j}\frac{j!}{\alpha!}h^\alpha\partial^\alpha $$ you get the answer with $t_1=1$.