Taylor formula for p-differentiable function

Posted December 18th, 2008 by Structure

$f(x)=f(x_0)+\frac{1}{1!}df(x_0)(x-x_0)+\frac{1}{2!}d^2f(x_0)(x-x_0,x-x_0)+...\frac{1}{(p-1)!}d^{p-1}f(x_0)(x-x_0,.....,x-x_0)+$

$+\frac{1}{p!}d^pf(\xi)(x-x_0,x-x_0,...x-x_0)$

$f(x_0)+\sum_{i_1=1}^{i_1=n}\frac{\partial f}{\partial x_{i_1}}(x_0)(x_{i_1}-x_{0,i_1})+\frac{1}{2!}\sum_{i_1=1}^{i_1=n}\sum_{i_2=1}^{i_2=n}\frac{\partial^2 f}{\partial x_{i_1}\partial x_{i_2}}(x_0)(x_{i_1}-x_{0,i_1})(x_{i_2}-x_{0,i_2})+$

$+\frac{1}{(p-1)!}\sum_{i_1=1}^{i_1=n}\sum_{i_2=1}^{i_2=n}...\sum_{i_{p-1}=1}^{i_{p-1}=n}\frac{\partial^{p-1} f}{\partial x_{i_1}\partial x_{i_2}...\partial x_{i_{p-1}}}(x_0)(x_{i_1}-x_{0,i_1})(x_{i_2}-x_{0,i_2}) ...(x_{i_{p-1}}-x_{0,i_{p-1}})+$

$+\frac{1}{p!}\sum_{i_1=1}^{i_1=n}\sum_{i_2=1}^{i_2=n}...\sum_{i_{p}=1}^{i_{p}=n}\frac{\partial^{p} f}{\partial x_{i_1}\partial x_{i_2}...\partial x_{i_{p}}}(\xi)(x_{i_1}-x_{0,i_1})(x_{i_2}-x_{0,i_2}) ...(x_{i_{p}}-x_{0,i_{p}})$

where

$\xi=x_0+t(x-x_0)$

$0<t<1$

Taylor formula for p-differentiable function

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