# Taylor remainder of $f(x,y)=\sin x\cdot \cos y$

Given $f\colon \mathbb R^2\rightarrow\mathbb R,(x,y)\mapsto\sin x\cdot\cos y$ I want to show that there exists $M>0$ such that $$|f(x,y)-T_2(x,y)|\leq M(|x|+|y|)$$ for all $(x,y)\in\mathbb R^2$. $T_2$ is the taylor-polynomial of order $2$.

I know I have to consider the remainder of $T_2$. So I calculated all the third partial derivatives:

\begin{align*} \frac{\partial f^3}{\partial x^3}(x,y)&=-\cos(x)\cos(y)\\ \frac{\partial f^3}{\partial y^3}(x,y)&=\sin(x)\sin(y)\\ \frac{\partial f^3}{\partial y^2\partial x}(x,y)&=\frac{\partial f^3}{\partial x\partial y^2}(x,y)=\frac{\partial f^3}{\partial y\partial x\partial y}(x,y)=-\cos(x)\cos(y)\\ \frac{\partial f^3}{\partial x^2\partial y}(x,y)&=\frac{\partial f^3}{\partial x\partial y\partial x}(x,y)=\frac{\partial f^3}{\partial y\partial x^2}(x,y)=\sin(x)\sin(y) \end{align*}

But now I am really stuck. I looked up the formula for the remainder on wikipedia but I never get the term $|x|+|y|$. So how can you show that $|R_3(x,y)|\leq M(|x|+|y|)$?

• You can't. $f$ is bounded, and $T_2$ is a quadratic polynomial. Hence $\lvert R_3(x,\,y)\rvert$ has quadratic growth, not linear, as such an estimate would imply. – Daniel Fischer Jul 17 '13 at 7:00
• @DanielFischer Is there any chance to get an extimate like above if you change some conditions? – user31035 Jul 17 '13 at 7:02
• You get an estimate by $\lvert x\rvert + \lvert y\rvert$ if you take the first-order Taylor polynomial. However, Taylor approximations are really only useful near the centre point, the error far away is practically always too large to do anything with the approximation there. And near the centre point, a quadratic estimate is better than a linear one, cubic yet better ... Then you can find an $M > 0$ such that $\lvert R_3(x,y)\rvert < M(\lvert x\rvert + \lvert y\rvert)^3$, and near $(0,0)$, that gives a much much much better error bound. – Daniel Fischer Jul 17 '13 at 7:09
• @DanielFischer: Anyway, the statement in the question is correct, because $T_2(x,y)$ is simply $x$. – 23rd Jul 17 '13 at 7:15
• @Landscape Ah, duh. The $\sin x$ factor annihilates the second derivative of $\cos y$ at $(0,0)$. Didn't take that into account. – Daniel Fischer Jul 17 '13 at 7:20

First compute $T_2(x,y)$ which is given by

$$T_2(x,y) = x.$$

Now, we have

$$\Big| f(x,y)-T_2(x,y)\Big|= \Big| \sin(x)\cos(y)-x\Big|= \Big| \sin(x)\cos(y)-\sin(x)+\sin(x)-x\Big|$$

$$\leq |\sin(x)(\cos(y)-1)|+|\sin(x)|+|x|$$

$$=|(\cos(y)-1)|+2|x|$$

$$\leq |y|+2|x|$$

$$\leq 2(|x|+|y|).$$

Note: The following inequalities

$$|\sin(x)|\leq |x|,\quad |\cos(y)-1|\leq |y|$$

can be proved using the mean value theorem.