I know that for a function of one variable $f(x)$, the Taylor expansion is $$f(x)=f(x_0)+f'(x_0)(x-x_0)+...+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+R_n(x,x_0) $$
where \begin{align*} R_n(x,x_0)&=\int^x_{x_0}\frac{(x-t)^n}{n!} f^{(n+1)}(t)dt \\ &=f^{(n+1)}(\xi)\int^x_{x_0} \frac{(x-t)^n}{n!}dt \\ &=f^{(n+1)}(\xi) \frac{(x-x_0)^{n+1}}{(n+1)!}. \end{align*}

My question is for $f(x,y)$ with a Taylor expansion $$f(x)=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)+...+R_n(x,x_0,y,y_0) $$ What is $R_n(x,x_0,y,y_0)$?

  • $\begingroup$ Not sure, but it's suppose to be the $n+1$ term evaluated at some point $(t,k)$ such that $x_0 < t < x, y_0 < k < y$ $\endgroup$ – Rab Sep 21 '17 at 18:39
  • $\begingroup$ I do not know if you have seen this but it may be of some help. sites.math.washington.edu/~folland/Math425/taylor2.pdf $\endgroup$ – Derek Sep 23 '17 at 23:07
  • $\begingroup$ Yes I have seen it but I am confused. I have no idea on how to apply what it says to something like $x^y$. $\endgroup$ – AzJ Sep 23 '17 at 23:09

We consider $x,x_0,y,y_0\in\mathbb{R}$ and define intervals $I,J$ with \begin{align*} &I=[\mathrm{min}(x,x_0),\mathrm{max}(x,x_0)]\qquad\text{and}\qquad J=[\mathrm{min}(y,y_0),\mathrm{max}(y,y_0)] \end{align*}

Let $f:I\times J\rightarrow\mathbb{R}$ be an $(n+1)$-times continuously differentiable function in $x$ and $y$. There exist values $\xi\in I^\circ,\eta\in J^\circ$, so that following is valid \begin{align*} f(x,y)&=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\\ &\qquad+\frac{1}{2}\left(f_{xx}(x_0,y_0)(x-x_0)^2+2f_{xy}(x_0,y_0)(x-x_0)(y-y_0)\right.\\ &\qquad\qquad\quad\left.+f_{yy}(x_0,y_0)(y-y_0)^2\right)\\ &\qquad\,\,\,\vdots\\ &\qquad+\frac{1}{n!}\left(f_{x^n}(x_0,y_0)(x-x_0)^n+\binom{n}{1}f_{x^{n-1}y}(x_0,y_0)(x-x_0)^{n-1}(y-y_0)\right.\\ &\qquad\qquad\quad+\cdots+f_{y^n}(x_0,y_0)(y-y_0)^n\Big)\\ &\qquad+R_n(x,y) \end{align*} The Lagrange remainder term $R_n(x,y)$ is given as \begin{align*} \color{blue}{R_n(x,y)}&\color{blue}{=\frac{1}{(n+1)!}\left(f_{x^{n+1}}(\xi,\eta)(x-x_0)^{n+1}+\binom{n+1}{1}f_{x^{n-1}y}(\xi,\eta)(x-x_0)^n(y-y_0)\right.}\\ &\qquad\qquad\qquad\quad\color{blue}{+\cdots+f_{y^{n+1}}(\xi,\eta)(y-y_0)^{n+1}\Big)} \end{align*}

A more compact notation can be given with the total derivative. We obtain \begin{align*} df&=(x-x_0)f_x+(y-y_0)f_y,\\ d^2f&=((x-x_0)f_x+(y-y_0)f_y)^{(2)}\\ &=((x-x_0)^2f_{xx}+2 (x-x_0)(y-y_0) f_{xy}+(y-y_0)^2f_{yy}\\ &\,\,\,\vdots\\ d^nf&=((x-x_0)f_x+(y-y_0)f_y)^{(n)}\\ &=(x-x_0)^nf_{x^n}+\binom{n}{1}(x-x_0)^{n-1}(y-y_0)f_{x^{n-1}y}\\ &\qquad+\cdots+(y-y_0)^nf_{y^n} \end{align*}

and we can write
\begin{align*} f(x,y)=f(x_0,y_0)+df(x_0,y_0)+\frac{1}{2}d^2f(x_0,y_0)+\cdots+\frac{1}{n!}d^nf(x_0,y_0)+R_n(x,y) \end{align*} with \begin{align*} \color{blue}{R_n(x,y)=\frac{1}{(n+1)!}d^{n+1}f(\xi,\eta)} \end{align*}

Note: A derivation similar to this answer can be found in section 6.3 of Introduction to Calculus and Analysis II by Richard Courant.

  • $\begingroup$ Just so that I fully understand, the remainder for a Taylor series up to 2nd order, $R_2(x,y)$ is \begin{align} R_2(x,y)&=\frac{1}{3!}(f_{x^3}(\xi,\eta)(x-x_0)^3 \\ &+3f_{x^2y}(\xi,\eta)(x-x_0)^2(y-y_0)+3f_{xy^2}(\xi,\eta)(x-x_0)(y-y_0)^2\\ &+f_{y^3}(\xi,\eta)(y-y_0)^3) \end{align} $\endgroup$ – AzJ Sep 24 '17 at 18:07
  • 1
    $\begingroup$ @AzJ: Sorry, I've misread your comment. Yes, you're right. The Taylor series has terms up to the 2nd order, implying the remainder term shows derivations of 3rd order. $\endgroup$ – Markus Scheuer Sep 24 '17 at 18:23
  • $\begingroup$ @AzJ: Thanks a lot for accepting my answer and granting the bounty! :-) $\endgroup$ – Markus Scheuer Oct 2 '17 at 6:42

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